Contents / Real Analysis / Advanced Topics

Chapter 8

Advanced Topics

Functions of bounded variation and the Jordan decomposition; absolute continuity, the Cantor function, and the link to the integral; the Lebesgue differentiation theorem via the Hardy-Littlewood maximal function; L^p spaces with Young, Holder, Minkowski and Riesz-Fischer completeness; Fourier series with orthogonality, Bessel, Parseval, Fejer and the Gibbs phenomenon; and rigorous multivariable calculus through Fubini-Tonelli, change of variables and Stokes.

Introduction

This chapter collects the four subjects that a first course in analysis reaches only at its end, and it is long because they are genuinely four subjects. What ties them together is a single question left open by the earlier chapters: which functions does calculus actually work for?

The Riemann theory answered "continuous ones, more or less", and Riemann Integration sharpened that to Lebesgue's criterion — integrability is continuity almost everywhere. The Lebesgue theory answered "measurable ones", and the convergence theorems of Lebesgue Integration made limits and integrals commute under hypotheses that are easy to check. Neither theory says anything about differentiation. The fundamental theorem of calculus, in the form proved in Riemann Integration, assumes at the outset that $f^{'}$ exists and is integrable, which quietly rules out every interesting example.

The first two parts of this chapter repair that. Bounded variation is the condition that makes a function differentiable almost everywhere — a weak conclusion, but true for a huge class, monotone functions included. Absolute continuity is the strictly stronger condition that makes the fundamental theorem true in both directions: $f$ is an indefinite Lebesgue integral of its own derivative exactly when it is absolutely continuous. Between the two classes sits the Cantor function, continuous, of bounded variation, with derivative zero almost everywhere and yet climbing from $0$ to $1$ ; it is the counterexample that makes the distinction necessary rather than pedantic. The third part, on differentiation theory, proves that the repair is not an accident: the Lebesgue differentiation theorem says averages of any locally integrable function converge back to the function at almost every point, and the engine that drives it — the Hardy–Littlewood maximal function — is the first genuinely modern tool in the book.

The fourth and fifth parts change register. $L^{p}$ spaces are the setting in which the convergence theorems become geometry: a set of functions with a norm, complete, and for $p = 2$ carrying an inner product. Hölder's and Minkowski's inequalities are what make the norm a norm, and the Riesz–Fischer theorem — completeness — is the single result that makes Lebesgue's integral worth its cost, since the same construction over the Riemann integral produces an incomplete space. Fourier series are then the $L^{2}$ theory made concrete, and the place where the difference between pointwise, uniform and mean-square convergence stops being a technicality: the same series can converge in $L^{2}$ , fail to converge at a point, and overshoot a jump by a fixed $8.9%$ no matter how many terms are taken.

The last part, rigorous multivariable calculus, is independent of the rest and could be read first. Its content is that the derivative in several variables is a linear map, not a list of partials, and that every major theorem — chain rule, inverse function theorem, change of variables, Stokes — is a statement about that linear map.

Throughout, $m$ denotes Lebesgue measure, $∣ E ∣$ the measure of a set $E$ , "a.e." means outside a set of measure zero, and $χ_{E}$ is the indicator function of $E$ . Measurability, the integral, and the convergence theorems are taken from Lebesgue Integration without reproof.

8.1Definition via Total Variation

Monotone functions are easy to handle: they have one-sided limits everywhere, countably many discontinuities, and — as we will prove — a derivative almost everywhere. Most functions of interest are not monotone, but many are built out of monotone pieces, and the right way to say "built out of monotone pieces" is to measure how much total up-and-down motion a function performs.

Definition 8.1 (Variation over a partition and total variation). Let $f : [a, b] \to R$ and let $P : a = x_{0} < x_{1} < \dots < x_{n} = b$ be a partition. The variation of $f$ over $P$ is

V (f, P) = i = 1 \sum n f (x_{i}) - f (x_{i - 1}) .

The total variation of $f$ on $[a, b]$ is

V_{a}^{b} (f) = P sup V (f, P),

the supremum taken over all partitions of $[a, b]$ . If $V_{a}^{b} (f) < \infty$ we say $f$ is of bounded variation on $[a, b]$ and write $f \in B V [a, b]$ .

The supremum is the right operation because refining a partition can only increase the sum: this is the first thing to prove, and everything else rests on it.

Proof. It suffices to treat $Q = P \cup {y}$ with $x_{k - 1} < y < x_{k}$ , since a general refinement is obtained by adding one point at a time. The two sums differ only in the $k$ -th block, and there

∣ f (x_{k}) - f (x_{k - 1}) ∣ = ∣ (f (x_{k}) - f (y)) + (f (y) - f (x_{k - 1})) ∣ \leq ∣ f (x_{k}) - f (y) ∣ + ∣ f (y) - f (x_{k - 1}) ∣

by the triangle inequality proved in The Real Number System. So the refined sum is at least as large, and induction on the number of added points finishes the argument.∎

Example 8.3 (Variation of a monotone function and of a sawtooth). Compute $V_{0}^{1} (f)$ for (a) $f (x) = x^{3}$ , and (b) the function $g$ on $[0, 3]$ that is linear on each of $[0, 1], [1, 2], [2, 3]$ with $g (0) = 0$ , $g (1) = 2$ , $g (2) = - 1$ , $g (3) = 4$ .

Solution.

(a) $f$ is increasing, so every term $∣ f (x_{i}) - f (x_{i - 1}) ∣ = f (x_{i}) - f (x_{i - 1})$ and the sum telescopes to $f (1) - f (0) = 1$ for every partition. The supremum of a constant family is that constant: $V_{0}^{1} (x^{3}) = 1$ .
This is general: for increasing $f$ , $V_{a}^{b} (f) = f (b) - f (a)$ , and for decreasing $f$ it is $f (a) - f (b)$ .
(b) On each monotone piece the variation is the size of the jump in value, and the partition ${0, 1, 2, 3}$ already separates the pieces. Adding points inside a linear piece changes nothing, by the same telescoping.
Hence $V_{0}^{3} (g) = ∣2 - 0∣ + ∣ - 1 - 2∣ + ∣4 - (- 1) ∣ = 2 + 3 + 5 = 10$ .
Sanity check: $∣ g (3) - g (0) ∣ = 4$ , and $4 \leq 10$ as it must be — the net change is never more than the total motion.

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Two facts about $B V$ deserve to be recorded immediately, because they are used constantly and both follow from the definition in a line.

Proposition 8.4 (Boundedness and additivity). If $f \in B V [a, b]$ then $f$ is bounded, with $∣ f (x) ∣ \leq ∣ f (a) ∣ + V_{a}^{b} (f)$ for all $x$ . Moreover, for any $c \in (a, b)$ ,

V_{a}^{b} (f) = V_{a}^{c} (f) + V_{c}^{b} (f) .

Proof. For the bound, the two-point partition ${a, x, b}$ gives $∣ f (x) - f (a) ∣ \leq V_{a}^{b} (f)$ , hence $∣ f (x) ∣ \leq ∣ f (a) ∣ + V_{a}^{b} (f)$ .

For additivity, take partitions $P_{1}$ of $[a, c]$ and $P_{2}$ of $[c, b]$ ; their union is a partition of $[a, b]$ with $V (f, P_{1}) + V (f, P_{2}) = V (f, P_{1} \cup P_{2}) \leq V_{a}^{b} (f)$ . Taking suprema separately gives $V_{a}^{c} (f) + V_{c}^{b} (f) \leq V_{a}^{b} (f)$ . Conversely, given a partition $P$ of $[a, b]$ , the refinement $P \cup {c}$ splits into partitions of the two halves, so by Lemma Refinement increases variation, $V (f, P) \leq V (f, P \cup {c}) \leq V_{a}^{c} (f) + V_{c}^{b} (f)$ ; take the supremum over $P$ .∎

Pitfall. Continuity does not imply bounded variation. The function $f (x) = x sin (1/ x)$ for $x \in (0, 1]$ , $f (0) = 0$ , is continuous on $[0, 1]$ yet has infinite total variation, because the peaks at $x = 2/ ((2 k + 1) π)$ have heights $\sim 2/ ((2 k + 1) π)$ and $\sum 1/ k$ diverges. Raising the damping to $x^{2} sin (1/ x)$ makes the heights summable and restores bounded variation. Continuity controls local oscillation; bounded variation controls its total.

Exercises for §8.1 →

8.2Jordan Decomposition: Bounded Variation as a Difference of Increasing Functions

The pitfall above shows that bounded variation is not a smoothness condition. What is it, then? Jordan's theorem answers exactly: it is the condition of being a difference of two increasing functions, and nothing more. Everything true of $B V$ functions is therefore a consequence of something true of monotone functions.

The key object is the variation accumulated so far.

Definition 8.5 (The variation function). For $f \in B V [a, b]$ define $v : [a, b] \to [0, \infty)$ by $v (x) = V_{a}^{x} (f)$ , with $v (a) = 0$ . Call $v$ the variation function of $f$ .

Theorem 8.6 (Jordan decomposition). A function $f : [a, b] \to R$ is of bounded variation if and only if $f = g - h$ with $g, h$ increasing on $[a, b]$ . One such decomposition is

g = \frac{1}{2} (v + f), h = \frac{1}{2} (v - f),

where $v$ is the variation function; in it, $g$ and $h$ may be taken with $g (a) = \frac{1}{2} f (a)$ , $h (a) = - \frac{1}{2} f (a)$ .

Proof. ( $\Leftarrow$ ) Suppose $f = g - h$ with both increasing. For any partition,

i \sum ∣ f (x_{i}) - f (x_{i - 1}) ∣ \leq i \sum (g (x_{i}) - g (x_{i - 1})) + i \sum (h (x_{i}) - h (x_{i - 1})) = (g (b) - g (a)) + (h (b) - h (a)),

using that each increment of an increasing function is non-negative so the sums telescope. The right-hand side does not depend on the partition, so $V_{a}^{b} (f) < \infty$ .

( $\Rightarrow$ ) Suppose $f \in B V [a, b]$ . First, $v$ is increasing: for $x < y$ , additivity (Proposition Boundedness and additivity) gives $v (y) - v (x) = V_{x}^{y} (f) \geq 0$ . Second, $v - f$ is increasing: for $x < y$ , the two-point partition of $[x, y]$ shows

f (y) - f (x) \leq ∣ f (y) - f (x) ∣ \leq V_{x}^{y} (f) = v (y) - v (x),

so $(v - f) (y) \geq (v - f) (x)$ . Third, $v + f$ is increasing by the same computation applied to $- f$ , since $V_{x}^{y} (- f) = V_{x}^{y} (f)$ and hence $f (x) - f (y) \leq v (y) - v (x)$ .

Now set $g = \frac{1}{2} (v + f)$ and $h = \frac{1}{2} (v - f)$ . Both are increasing by the previous paragraph, and $g - h = f$ identically.∎

The decomposition is not unique — adding the same increasing function to $g$ and to $h$ changes neither difference — but the one built from $v$ is minimal in a precise sense, and this is worth stating because it identifies the variation function as canonical.

Proposition 8.7 (Minimality of the Jordan decomposition). If $f = G - H$ with $G, H$ increasing, then for every $x \in [a, b]$

V_{a}^{x} (f) \leq (G (x) - G (a)) + (H (x) - H (a)) .

So no decomposition accumulates less total increase than the one built from the variation function, and equality for all $x$ forces $G - g$ and $H - h$ to be the same constant.

Proof. The inequality is the computation in the ( $\Leftarrow$ ) half of Theorem Jordan decomposition, applied on $[a, x]$ rather than on $[a, b]$ . Suppose equality holds for all $x$ . Since $G - H = g - h = f$ , the function $φ = G - g = H - h$ is well defined, and the equality reads $(G (x) + H (x)) - (G (a) + H (a)) = v (x) = (g (x) + h (x)) - (g (a) + h (a))$ , i.e. $2 φ (x) = 2 φ (a)$ . So $φ$ is constant.∎

Intuition. Think of $f$ as the position of a walker on a line. The variation function $v (x)$ is the odometer: total distance walked, which only ever goes up. Position is $g - h$ , where $g$ is the total distance walked forwards and $h$ the total distance walked backwards, and both of those are odometers too. "Bounded variation" just says the odometer does not run away to infinity in finite time. That is why the class is closed under so much: any statement about odometers transfers.

Example 8.8 (Jordan decomposition of an explicit function). Let $f (x) = x^{2} - x$ on $[0, 1]$ . Compute $V_{0}^{1} (f)$ and exhibit the Jordan decomposition.

Solution.

$f^{'} (x) = 2 x - 1$ , so $f$ decreases on $[0, \frac{1}{2}]$ and increases on $[\frac{1}{2}, 1]$ , with $f (0) = 0$ , $f (\frac{1}{2}) = - \frac{1}{4}$ , $f (1) = 0$ .
Variation is additive over the monotone pieces, and on each piece it is the net change in absolute value: $V_{0}^{1} (f) = ∣ - \frac{1}{4} - 0∣ + ∣0 - (- \frac{1}{4}) ∣ = \frac{1}{2}$ .
The variation function accumulates the motion: for $x \leq \frac{1}{2}$ the function is decreasing, so $v (x) = - (x^{2} - x) = x - x^{2}$ ; for $x \geq \frac{1}{2}$ we add the rise since $\frac{1}{2}$ , giving $v (x) = \frac{1}{4} + ((x^{2} - x) + \frac{1}{4}) = x^{2} - x + \frac{1}{2}$ .
Then $g = \frac{1}{2} (v + f)$ and $h = \frac{1}{2} (v - f)$ . On $[0, \frac{1}{2}]$ : $g (x) = \frac{1}{2} ((x - x^{2}) + (x^{2} - x)) = 0$ and $h (x) = x - x^{2}$ . On $[\frac{1}{2}, 1]$ : $g (x) = \frac{1}{2} ((x^{2} - x + \frac{1}{2}) + (x^{2} - x)) = x^{2} - x + \frac{1}{4}$ and $h (x) = \frac{1}{4}$ .
Check: $g$ is constant then increasing, $h$ increasing then constant, and $g - h$ equals $- (x - x^{2}) = x^{2} - x$ on the left piece and $x^{2} - x + \frac{1}{4} - \frac{1}{4}$ on the right. Both give $f$ . And $v (1) = \frac{1}{2}$ matches step 2.

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Corollary 8.9 (The vector space $B V [a, b]$ ). $B V [a, b]$ is a vector space, and it is closed under products: if $f_{1}, f_{2} \in B V [a, b]$ then $f_{1} f_{2} \in B V [a, b]$ . If in addition $∣ f_{1} ∣ \geq c > 0$ then $1/ f_{1} \in B V [a, b]$ .

Proof. For sums, $V_{a}^{b} (f_{1} + f_{2}) \leq V_{a}^{b} (f_{1}) + V_{a}^{b} (f_{2})$ term by term from the triangle inequality inside each partition sum, and $V_{a}^{b} (λ f) = ∣ λ ∣ V_{a}^{b} (f)$ is immediate. For products, each $f_{j}$ is bounded by Proposition Boundedness and additivity, say by $M$ , and

∣ f_{1} f_{2} (x_{i}) - f_{1} f_{2} (x_{i - 1}) ∣ \leq ∣ f_{1} (x_{i}) ∣ ∣ f_{2} (x_{i}) - f_{2} (x_{i - 1}) ∣ + ∣ f_{2} (x_{i - 1}) ∣ ∣ f_{1} (x_{i}) - f_{1} (x_{i - 1}) ∣,

so summing gives $V_{a}^{b} (f_{1} f_{2}) \leq M (V_{a}^{b} (f_{1}) + V_{a}^{b} (f_{2}))$ . For the reciprocal, $1/ f_{1} (x_{i}) - 1/ f_{1} (x_{i - 1}) = ∣ f_{1} (x_{i}) - f_{1} (x_{i - 1}) ∣/∣ f_{1} (x_{i}) f_{1} (x_{i - 1}) ∣ \leq c^{- 2} ∣ f_{1} (x_{i}) - f_{1} (x_{i - 1}) ∣$ , so $V_{a}^{b} (1/ f_{1}) \leq c^{- 2} V_{a}^{b} (f_{1})$ .∎

Exercises for §8.2 →

8.3Differentiability and Discontinuities of BV Functions

Jordan's theorem converts every question about $B V$ functions into a question about monotone functions. Two such questions have classical answers, and together they say that a $B V$ function is well behaved in exactly the two ways one could hope for: its discontinuities are few, and it has a derivative nearly everywhere.

Theorem 8.10 (Discontinuities of a monotone function). Let $g : [a, b] \to R$ be increasing. Then $g (x^{-})$ and $g (x^{+})$ exist at every interior $x$ , every discontinuity is a jump, and the set of discontinuities is at most countable.

Proof. Fix $x \in (a, b)$ . The set ${g (t) : t < x}$ is bounded above by $g (x)$ , so it has a supremum by the completeness axiom of The Real Number System; a routine $ε$ -argument shows this supremum is $lim_{t \to x^{-}} g (t) = g (x^{-})$ , and symmetrically $g (x^{+}) = in f {g (t) : t > x}$ . Monotonicity gives $g (x^{-}) \leq g (x) \leq g (x^{+})$ , so the only possible failure of continuity is $g (x^{+}) - g (x^{-}) > 0$ : a jump.

To count them, let $J_{x} = g (x^{+}) - g (x^{-}) > 0$ be the jump at a discontinuity $x$ . For distinct discontinuities $x_{1} < \dots < x_{N}$ in $(a, b)$ , the intervals $(g (x_{j}^{-}), g (x_{j}^{+}))$ are pairwise disjoint subintervals of $(g (a), g (b))$ , so $\sum_{j = 1}^{N} J_{x_{j}} \leq g (b) - g (a)$ . Hence for each $n$ , the set $D_{n} = {x : J_{x} > 1/ n}$ has at most $n (g (b) - g (a))$ elements, and the discontinuity set $⋃_{n} D_{n}$ is a countable union of finite sets.∎

Corollary 8.11 (Discontinuities of a BV function). A function of bounded variation has one-sided limits at every point and at most countably many discontinuities, all of them jumps. In particular a $B V$ function has no oscillatory discontinuity.

Proof. Write $f = g - h$ by Theorem Jordan decomposition. One-sided limits of $f$ exist because they exist for $g$ and $h$ by Theorem Discontinuities of a monotone function, and limits subtract. $f$ is continuous wherever both $g$ and $h$ are, so its discontinuity set is contained in the union of two countable sets. Since $f (x^{\pm})$ both exist, no discontinuity can be of oscillatory type in the classification of Limits & Continuity.∎

The differentiability statement is deeper, and its proof is postponed to the section on Vitali covering arguments where the right tool is built. We state it now because the rest of the part uses it.

Theorem 8.12 (Lebesgue's differentiation theorem for monotone functions). An increasing function $g : [a, b] \to R$ is differentiable at almost every point of $[a, b]$ , $g^{'}$ is measurable and non-negative a.e., and

\int_{a}^{b} g^{'} (x) d x \leq g (b) - g (a) .

The inequality can be strict.

Proof. Sketch, with the covering lemma assumed. The almost-everywhere differentiability is proved in the section Proof Ideas (Vitali Covering or Rising Sun in 1D) below; granting it, the integral inequality is a short application of Fatou's lemma from Lebesgue Integration. Extend $g$ by $g (x) = g (b)$ for $x > b$ and set

g_{n} (x) = n (g (x + \frac{1}{n}) - g (x)) \geq 0.

Then $g_{n} \to g^{'}$ a.e., and each $g_{n}$ is measurable as a difference of monotone functions. Fatou gives

\int_{a}^{b} g^{'} \leq n lim inf \int_{a}^{b} g_{n} = n lim inf n (\int_{a + 1/ n}^{b + 1/ n} g - \int_{a}^{b} g) = n lim inf (n \int_{b}^{b + 1/ n} g - n \int_{a}^{a + 1/ n} g) .

The first term equals $g (b)$ exactly, by the extension; the second is at least $g (a)$ since $g \geq g (a)$ there. Hence $\int_{a}^{b} g^{'} \leq g (b) - g (a)$ .∎

Corollary 8.13 (BV implies differentiable a.e.). If $f \in B V [a, b]$ then $f^{'} (x)$ exists and is finite for almost every $x \in [a, b]$ , and $f^{'} \in L^{1} [a, b]$ with $\int_{a}^{b} ∣ f^{'} ∣ \leq V_{a}^{b} (f)$ .

Proof. Write $f = g - h$ with $g, h$ increasing (Theorem Jordan decomposition). Each is differentiable a.e. by Theorem Lebesgue's differentiation theorem for monotone functions, and the union of two null sets is null, so $f^{'} = g^{'} - h^{'}$ exists a.e. For the integral bound, $∣ f^{'} ∣ \leq g^{'} + h^{'}$ a.e., so $\int ∣ f^{'} ∣ \leq (g (b) - g (a)) + (h (b) - h (a)) = v (b) = V_{a}^{b} (f)$ for the minimal decomposition.∎

Example 8.14 (Where strictness comes from). Let $f$ be the jump function $f = χ_{[1/2, 1]}$ on $[0, 1]$ . Compute $V_{0}^{1} (f)$ , $f^{'}$ a.e., and compare $\int_{0}^{1} f^{'}$ with $f (1) - f (0)$ .

Solution.

$f$ is increasing with a single jump of size $1$ , so $V_{0}^{1} (f) = 1$ .
$f$ is locally constant off the single point $x = \frac{1}{2}$ , so $f^{'} (x) = 0$ for every $x \neq = \frac{1}{2}$ — that is, almost everywhere.
Therefore $\int_{0}^{1} f^{'} = 0$ , while $f (1) - f (0) = 1$ . The inequality in Theorem Lebesgue's differentiation theorem for monotone functions is strict, by exactly the size of the jump.
The moral: the derivative cannot see a jump, because a jump lives on a null set. Any theorem recovering $f$ from $f^{'}$ must therefore exclude jumps — and in fact must exclude more, as the Cantor function will show.

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Exercises for §8.3 →

8.4Absolute Continuity: a Preview of the Stronger Condition

Example Where strictness comes from shows that $B V$ is too weak for the fundamental theorem: a jump destroys it. One might guess that continuity plus bounded variation is enough. It is not, and the counterexample is important enough to construct carefully.

Definition 8.15 (The Cantor function). Let $C \subseteq [0, 1]$ be the middle-thirds Cantor set. The Cantor function (or devil's staircase) $Φ : [0, 1] \to [0, 1]$ is defined by writing $x$ in base $3$ as $x = \sum_{k \geq 1} a_{k} 3^{- k}$ with $a_{k} \in {0, 1, 2}$ : if some $a_{k} = 1$ , truncate at the first such $k$ ; then

Φ (x) = k = 1 \sum K \frac{a _{k} /2}{2 ^{k}} with the convention above,

equivalently, $Φ$ is constant on each interval removed at stage $n$ of the construction, taking there the dyadic value $(2 j - 1) / 2^{n}$ .

Theorem 8.16 (Properties of the Cantor function). $Φ$ is continuous and increasing on $[0, 1]$ with $Φ (0) = 0$ , $Φ (1) = 1$ ; it is of bounded variation with $V_{0}^{1} (Φ) = 1$ ; and $Φ^{'} (x) = 0$ for almost every $x$ , so that

\int_{0}^{1} Φ^{'} (x) d x = 0 \neq = 1 = Φ (1) - Φ (0) .

Proof. Monotone and surjective. By construction $Φ$ maps the $2^{n}$ closed intervals surviving at stage $n$ onto the $2^{n}$ dyadic intervals $[(j - 1) 2^{- n}, j 2^{- n}]$ in order, so $Φ$ is increasing and its range contains all dyadic rationals in $[0, 1]$ .

Continuous. An increasing function fails continuity only by a jump (Theorem Discontinuities of a monotone function), and a jump would leave an interval of values unattained. But the range contains a dense set, and the range of an increasing function omitting an interval cannot be dense there. So $Φ$ has no jump and is continuous.

Bounded variation. $Φ$ is increasing, so $V_{0}^{1} (Φ) = Φ (1) - Φ (0) = 1$ by Example Variation of a monotone function and of a sawtooth.

Derivative zero a.e. The complement $[0, 1] ∖ C$ is a countable union of open intervals on each of which $Φ$ is constant, so $Φ^{'} = 0$ there. The total length removed is $\sum_{n \geq 1} 2^{n - 1} 3^{- n} = 1$ , so $∣ C ∣ = 0$ and the set where $Φ^{'} \neq = 0$ is contained in a null set.∎

So a continuous, increasing, bounded-variation function can rise by $1$ while its derivative integrates to $0$ . The whole increase happens on the null set $C$ . The condition that forbids this is a uniform strengthening of continuity — instead of controlling $∣ f (y) - f (x) ∣$ for one short interval, it controls the sum over finitely many disjoint short intervals.

Definition 8.17 (Absolute continuity, first statement). $f : [a, b] \to R$ is absolutely continuous if for every $ε > 0$ there is $δ > 0$ such that for every finite collection of pairwise disjoint intervals $(a_{1}, b_{1}), \dots, (a_{N}, b_{N})$ in $[a, b]$ ,

k = 1 \sum N (b_{k} - a_{k}) < δ ⟹ k = 1 \sum N ∣ f (b_{k}) - f (a_{k}) ∣ < ε .

Proposition 8.18 (The Cantor function is not absolutely continuous). $Φ$ fails the condition of Definition Absolute continuity, first statement.

Proof. Take $ε = \frac{1}{2}$ and let $δ > 0$ be arbitrary. At stage $n$ of the Cantor construction the surviving set is a union of $2^{n}$ closed intervals of total length $(2/3)^{n}$ , and $Φ$ increases by exactly $2^{- n}$ across each, hence by $2^{n} \cdot 2^{- n} = 1$ in total across them. Choose $n$ with $(2/3)^{n} < δ$ . These $2^{n}$ disjoint intervals have total length below $δ$ yet total increment $1 > \frac{1}{2} = ε$ .∎

Figure 8.19. The Cantor function climbs from

0

1

while its derivative is

0

almost everywhere. Shown is the stage-5 approximation; the plateaus are the removed middle thirds.

Intuition. Ordinary continuity says: a small interval maps to a small set of values. Absolute continuity says: a small total length, however it is scattered, maps to a small total change. The Cantor function passes the first test and fails the second, because it does all its climbing on a set of length zero that can be covered by many tiny intervals at once. Absolute continuity is precisely the ban on doing work on a null set.

Exercises for §8.4 →

8.5Riemann–Stieltjes Integration against BV Integrators

Bounded variation is not only a differentiability hypothesis; it is the natural hypothesis on an integrator. The Riemann–Stieltjes integral $\int_{a}^{b} f d α$ was defined in Riemann Integration as a limit of sums $\sum f (t_{i}) (α (x_{i}) - α (x_{i - 1}))$ . The following theorem explains why $α \in B V$ is the right condition: it is exactly what makes those sums bounded uniformly in the partition.

Theorem 8.20 (Existence against a BV integrator). If $f$ is continuous on $[a, b]$ and $α \in B V [a, b]$ , then $\int_{a}^{b} f d α$ exists, and

\int_{a}^{b} f d α \leq ([a, b] sup ∣ f ∣) V_{a}^{b} (α) .

Proof. By Theorem Jordan decomposition write $α = g - h$ with $g, h$ increasing; by linearity of the Stieltjes sums in $α$ it suffices to treat an increasing integrator, which is the case proved in Riemann Integration under the title A sufficient condition for existence (continuous integrand, monotone integrator). Existence follows.

For the bound, estimate any Stieltjes sum directly:

i \sum f (t_{i}) (α (x_{i}) - α (x_{i - 1})) \leq sup ∣ f ∣ i \sum α (x_{i}) - α (x_{i - 1}) \leq sup ∣ f ∣ \cdot V_{a}^{b} (α),

by Definition Variation over a partition and total variation. The bound is uniform over partitions and tags, so it passes to the limit.∎

Remark. The inequality says that $f \mapsto \int f d α$ is a bounded linear functional on $C [a, b]$ with norm at most $V_{a}^{b} (α)$ . The Riesz representation theorem — beyond this chapter — says every bounded linear functional on $C [a, b]$ arises this way, from a unique normalised $α \in B V$ . Bounded variation is thus not one possible hypothesis among many; it is the exact dual of continuity.

Proposition 8.21 (Stieltjes integration against a jump). Let $α = c χ_{[s, b]}$ on $[a, b]$ with $a < s < b$ and $c \in R$ . Then for every $f$ continuous at $s$ ,

\int_{a}^{b} f d α = c f (s) .

Proof. Let $P$ be a partition of mesh $< δ$ and let $[x_{k - 1}, x_{k}]$ be a subinterval containing $s$ . Every increment $α (x_{i}) - α (x_{i - 1})$ vanishes except for the ones straddling $s$ , where the increments total $c$ . Hence the Stieltjes sum equals $c f (t_{k})$ for a tag $t_{k}$ within $δ$ of $s$ (or splits as $c$ between two adjacent tags, both within $δ$ of $s$ ). By continuity at $s$ , $f (t_{k}) \to f (s)$ as $δ \to 0$ , so the sums converge to $c f (s)$ .∎

Example 8.22 (A mixed integrator). On $[0, 2]$ let $α (x) = x$ for $x \in [0, 1)$ and $α (x) = x + 3$ for $x \in [1, 2]$ . Compute $V_{0}^{2} (α)$ and $\int_{0}^{2} x^{2} d α$ .

Solution.

$α$ is increasing, so $V_{0}^{2} (α) = α (2) - α (0) = 5 - 0 = 5$ .
Split $α = α_{ac} + α_{jump}$ with $α_{ac} (x) = x$ and $α_{jump} = 3 χ_{[1, 2]}$ . Stieltjes integration is linear in the integrator.
Against $α_{ac}$ the integral is ordinary: $\int_{0}^{2} x^{2} d x = \frac{8}{3}$ .
Against $α_{jump}$ , Proposition Stieltjes integration against a jump gives $3 \cdot 1^{2} = 3$ .
Total: $\int_{0}^{2} x^{2} d α = \frac{8}{3} + 3 = \frac{17}{3}$ .
Sanity check against Theorem Existence against a BV integrator: $sup ∣ x^{2} ∣ \cdot V_{0}^{2} (α) = 4 \cdot 5 = 20$ , and $\frac{17}{3} \approx 5.67 \leq 20$ . $✓$

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Corollary 8.23 (Integration by parts for BV integrators). If $f$ is continuous and $α \in B V [a, b]$ , and one of $\int f d α$ , $\int α df$ exists, then so does the other and

\int_{a}^{b} f d α + \int_{a}^{b} α df = f (b) α (b) - f (a) α (a) .

Proof. This is the theorem Stieltjes integration by parts of Riemann Integration; the content added here is only that the hypothesis " $α$ monotone" used there may be relaxed to " $α \in B V$ ", which follows by writing $α = g - h$ and adding the two identities.∎

Exercises for §8.5 →

8.6The ε-δ Definition: Uniform Control on Total Increments

Definition Absolute continuity, first statement was given in passing to expose the Cantor function. It is now the central object, so we restate it and place it correctly among the conditions already known.

Definition 8.24 (Absolutely continuous function). $f : [a, b] \to R$ is absolutely continuous, written $f \in A C [a, b]$ , if for every $ε > 0$ there exists $δ > 0$ such that for every finite family of pairwise disjoint open intervals $(a_{k}, b_{k}) \subseteq [a, b]$ , $1 \leq k \leq N$ ,

k = 1 \sum N (b_{k} - a_{k}) < δ ⟹ k = 1 \sum N f (b_{k}) - f (a_{k}) < ε .

Taking $N = 1$ recovers uniform continuity verbatim, so $A C \subseteq$ uniformly continuous. The content of the definition is that $δ$ must work for all $N$ at once — the same budget of total length, however finely it is chopped up and scattered.

Proposition 8.25 (Lipschitz implies absolutely continuous). If $∣ f (x) - f (y) ∣ \leq L ∣ x - y ∣$ for all $x, y \in [a, b]$ , then $f \in A C [a, b]$ . In particular every continuously differentiable function on a closed bounded interval is absolutely continuous.

Proof. Given $ε > 0$ take $δ = ε / L$ . For disjoint intervals of total length $< δ$ ,

k \sum ∣ f (b_{k}) - f (a_{k}) ∣ \leq L k \sum (b_{k} - a_{k}) < L δ = ε .

For the second claim, if $f \in C^{1} [a, b]$ then $f^{'}$ is continuous on a compact set, hence bounded by some $L$ by the Extreme Value Theorem of Limits & Continuity, and the Mean Value Theorem of Differentiation gives the Lipschitz bound with that $L$ .∎

Theorem 8.26 (The chain of inclusions). On $[a, b]$ ,

C^{1} ⊊ Lipschitz ⊊ A C ⊊ (C \cap B V) ⊊ B V,

and every inclusion is strict.

Proof. The inclusions themselves: the first two are Proposition Lipschitz implies absolutely continuous; $A C \subseteq C$ because $N = 1$ gives uniform continuity; $A C \subseteq B V$ is Theorem Absolutely continuous implies bounded variation below; and $C \cap B V \subseteq B V$ is trivial.

Strictness, one example each. $f (x) = ∣ x ∣$ on $[- 1, 1]$ is Lipschitz but not $C^{1}$ . $f (x) = x$ on $[0, 1]$ is absolutely continuous (it is increasing with $\int_{0}^{1} f^{'} = 1 = f (1) - f (0)$ , and Theorem Absolutely continuous means indefinite integral applies) but not Lipschitz, since $f^{'} (x) = 1/ (2 x)$ is unbounded. The Cantor function $Φ$ lies in $C \cap B V$ but not in $A C$ , by Proposition The Cantor function is not absolutely continuous. And $χ_{[1/2, 1]}$ is in $B V$ but not continuous.∎

Example 8.27 (Testing the definition on a non-example). Show directly that $f (x) = x$ on $[0, 1]$ is not Lipschitz but does satisfy the $ε$ - $δ$ condition when the intervals are forced to avoid $0$ — and explain what happens at $0$ .

Solution.

Not Lipschitz: $(f (h) - f (0)) / h = h / h = h^{- 1/2} \to \infty$ as $h \to 0^{+}$ , so no single constant $L$ works.
Away from $0$ : on $[c, 1]$ with $c > 0$ the derivative is bounded by $1/ (2 c)$ , so $f$ is Lipschitz there, hence absolutely continuous on $[c, 1]$ by Proposition Lipschitz implies absolutely continuous.
At $0$ the single interval $(0, η)$ contributes $η$ , and $η < ε$ as soon as $η < ε^{2}$ . So the bad point costs a smaller $δ$ , not an impossible one.
Combining: given $ε$ , put $c = (ε /2)^{2}$ ; intervals meeting $[0, c]$ contribute at most $c = ε /2$ in total by monotonicity, and the rest are handled by the Lipschitz bound on $[c, 1]$ with $δ = c ε$ . So $f \in A C [0, 1]$ after all.
The contrast with $Φ$ is instructive: $x$ has one point of unbounded derivative, and one point is negligible; $Φ$ has an uncountable set of them, of measure zero but of positive "variation mass".

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Pitfall. Uniform continuity is not absolute continuity. $Φ$ is continuous on the compact $[0, 1]$ , hence uniformly continuous by Heine-Cantor, and is still not absolutely continuous. The distinction is invisible for one interval and only appears when arbitrarily many disjoint intervals are allowed simultaneously.

Exercises for §8.6 →

8.7Equivalence with Being an Indefinite Integral

This is the central theorem of the part, and it is the fundamental theorem of calculus in its final form. Both directions matter: one says indefinite integrals are absolutely continuous, the other says nothing else is.

Theorem 8.28 (Absolute continuity of the integral). Let $g \in L^{1} [a, b]$ . For every $ε > 0$ there is $δ > 0$ such that $∣ E ∣ < δ$ implies $\int_{E} ∣ g ∣ < ε$ , for every measurable $E \subseteq [a, b]$ .

Proof. Let $g_{n} = min (∣ g ∣, n)$ . Then $g_{n} ↑ ∣ g ∣$ pointwise, so by the Monotone Convergence Theorem of Lebesgue Integration, $\int_{a}^{b} g_{n} \to \int_{a}^{b} ∣ g ∣$ , which is finite. Choose $n$ with $\int_{a}^{b} (∣ g ∣ - g_{n}) < ε /2$ , and put $δ = ε / (2 n)$ . Then for $∣ E ∣ < δ$ ,

\int_{E} ∣ g ∣ = \int_{E} (∣ g ∣ - g_{n}) + \int_{E} g_{n} \leq \frac{ε}{2} + n ∣ E ∣ < \frac{ε}{2} + \frac{ε}{2} = ε .

∎

Corollary 8.29 (Indefinite integrals are absolutely continuous). If $g \in L^{1} [a, b]$ and $F (x) = F (a) + \int_{a}^{x} g (t) d t$ , then $F \in A C [a, b]$ .

Proof. Given $ε$ , take $δ$ from Theorem Absolute continuity of the integral. For disjoint intervals with $\sum (b_{k} - a_{k}) < δ$ , put $E = ⋃_{k} (a_{k}, b_{k})$ , so $∣ E ∣ < δ$ and

k \sum ∣ F (b_{k}) - F (a_{k}) ∣ = k \sum \int_{a_{k}}^{b_{k}} g \leq \int_{E} ∣ g ∣ < ε .

∎

Theorem 8.30 (Fundamental Theorem of Calculus for the Lebesgue integral). For $f : [a, b] \to R$ the following are equivalent.

$f \in A C [a, b]$ .
$f^{'}$ exists a.e., $f^{'} \in L^{1} [a, b]$ , and $f (x) = f (a) + \int_{a}^{x} f^{'} (t) d t$ for every $x \in [a, b]$ .
$f$ is an indefinite integral: $f (x) = f (a) + \int_{a}^{x} g$ for some $g \in L^{1} [a, b]$ .

When these hold, $g = f^{'}$ a.e.

Proof. $(3) \Rightarrow (1)$ is Corollary Indefinite integrals are absolutely continuous. $(2) \Rightarrow (3)$ is immediate with $g = f^{'}$ .

$(1) \Rightarrow (2)$ , the substantial direction, in three steps. Step A. $f \in A C \Rightarrow f \in B V$ (Theorem Absolutely continuous implies bounded variation below), so by Corollary BV implies differentiable a.e. the derivative $f^{'}$ exists a.e. and lies in $L^{1}$ .

Step B. Set $F (x) = f (a) + \int_{a}^{x} f^{'}$ . By Corollary Indefinite integrals are absolutely continuous, $F \in A C$ , and by the Lebesgue differentiation theorem (stated in the next part, Lebesgue differentiation theorem) $F^{'} = f^{'}$ a.e. Hence $h = f - F$ is absolutely continuous with $h^{'} = 0$ a.e. and $h (a) = 0$ .

Step C. An absolutely continuous function with vanishing derivative a.e. is constant. Fix $c \in (a, b]$ and $ε > 0$ , and let $δ$ be as in Definition Absolutely continuous function for this $ε$ . The set $E = {x \in (a, c) : h^{'} (x) = 0}$ has full measure in $(a, c)$ . For each $x \in E$ and each $η > 0$ there are arbitrarily short intervals $[x, x + ℓ]$ with $∣ h (x + ℓ) - h (x) ∣ \leq η ℓ$ ; these form a Vitali cover of $E$ , so by the Vitali covering lemma (used again in Proof Ideas (Vitali Covering or Rising Sun in 1D)) finitely many disjoint such intervals cover all but $δ$ of $E$ . Across those intervals the total increment of $h$ is at most $η (c - a)$ ; across the leftover set, of measure $< δ$ , it is at most $ε$ by absolute continuity. Hence $∣ h (c) - h (a) ∣ \leq η (c - a) + ε$ , and letting $η, ε \to 0$ gives $h (c) = h (a) = 0$ . So $f = F$ .∎

Example 8.31 (Recovering a function from its derivative, and failing to). For each of $f_{1} (x) = x^{2/3}$ on $[0, 1]$ and $f_{2} = Φ$ , decide whether $f (x) = f (0) + \int_{0}^{x} f^{'}$ .

Solution.

$f_{1}^{'} (x) = \frac{2}{3} x^{- 1/3}$ for $x > 0$ , unbounded at $0$ but integrable: $\int_{0}^{1} \frac{2}{3} x^{- 1/3} d x = \frac{2}{3} \cdot \frac{3}{2} x^{2/3}_{0}^{1} = 1$ .
And $f_{1} (1) - f_{1} (0) = 1 - 0 = 1$ . The identity holds, so $f_{1} \in A C$ by Theorem Fundamental Theorem of Calculus for the Lebesgue integral. Unboundedness of $f^{'}$ is no obstruction; only non-integrability would be.
$Φ^{'} = 0$ a.e. by Theorem Properties of the Cantor function, so $\int_{0}^{x} Φ^{'} = 0$ for every $x$ , while $Φ (1) = 1$ . The identity fails at every $x$ where $Φ (x) \neq = 0$ .
By the theorem, this is equivalent to $Φ \in / A C$ — we now have two independent proofs of that fact, the direct one and this one.

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Exercises for §8.7 →

8.8Properties Inherited from Integrals

Because $A C$ functions are exactly indefinite integrals, every stability property of the integral becomes a stability property of $A C$ , usually with a one-line proof. This section collects the ones that get used.

Theorem 8.32 (Algebra of absolutely continuous functions). Let $f, g \in A C [a, b]$ and $λ \in R$ . Then $f + g$ , $λ f$ , $f g$ and $∣ f ∣$ are in $A C [a, b]$ ; and if $∣ g ∣ \geq c > 0$ then $f / g \in A C [a, b]$ .

Proof. Sums and scalars: add the $ε /2$ -versions of the defining condition. For $∣ f ∣$ , the reverse triangle inequality gives $∣ f (b_{k}) ∣ - ∣ f (a_{k}) ∣ \leq ∣ f (b_{k}) - f (a_{k}) ∣$ , so the same $δ$ works.

For products, $f$ and $g$ are continuous on a compact interval hence bounded by some $M$ , and

∣ f g (b_{k}) - f g (a_{k}) ∣ \leq M ∣ g (b_{k}) - g (a_{k}) ∣ + M ∣ f (b_{k}) - f (a_{k}) ∣;

given $ε$ , use the $δ$ that serves $ε / (2 M)$ for each factor. For quotients, $∣1/ g (b_{k}) - 1/ g (a_{k}) ∣ \leq c^{- 2} ∣ g (b_{k}) - g (a_{k}) ∣$ and then apply the product case.∎

Theorem 8.33 (Integration by parts and substitution). If $f, g \in A C [a, b]$ then $f^{'} g + f g^{'} \in L^{1} [a, b]$ and

\int_{a}^{b} (f^{'} g + f g^{'}) = f (b) g (b) - f (a) g (a) .

Proof. By Theorem Algebra of absolutely continuous functions, $f g \in A C$ , so by Theorem Fundamental Theorem of Calculus for the Lebesgue integral applied to $f g$ ,

\int_{a}^{b} (f g)^{'} = f (b) g (b) - f (a) g (a),

and the integrand is in $L^{1}$ . It remains to see $(f g)^{'} = f^{'} g + f g^{'}$ a.e., which is the product rule of Differentiation applied at each of the almost-every points where both $f^{'}$ and $g^{'}$ exist — the intersection of two full-measure sets.∎

Proposition 8.34 (Composition). If $φ$ is Lipschitz on an interval containing the range of $f$ , and $f \in A C [a, b]$ , then $φ \circ f \in A C [a, b]$ . The composition $f \circ φ$ of two absolutely continuous functions need not be absolutely continuous.

Proof. For the positive statement, with Lipschitz constant $L$ for $φ$ , $∣ φ (f (b_{k})) - φ (f (a_{k})) ∣ \leq L ∣ f (b_{k}) - f (a_{k}) ∣$ , so the $δ$ serving $ε / L$ for $f$ serves $ε$ for $φ \circ f$ .

For the negative statement, take $f (x) = x$ on $[0, 1]$ , absolutely continuous by Example Testing the definition on a non-example, and $ψ (t) = t^{2} sin (t^{- 2})$ with $ψ (0) = 0$ . One checks that $ψ$ is of bounded variation and absolutely continuous on $[0, 1]$ , since its oscillations are damped by $t^{2}$ . But

(ψ \circ f) (x) = x sin (1/ x),

which is not even of bounded variation, by the pitfall in Definition via Total Variation. Absolute continuity is therefore not preserved by composition on the inside.∎

Example 8.35 (Variation of an AC function as an integral). Let $f (x) = sin x$ on $[0, 2 π]$ . Compute $V_{0}^{2 π} (f)$ two ways.

Solution.

By monotone pieces. $sin$ increases on $[0, π /2]$ from $0$ to $1$ , decreases on $[π /2, 3 π /2]$ from $1$ to $- 1$ , increases on $[3 π /2, 2 π]$ from $- 1$ to $0$ . Total: $1 + 2 + 1 = 4$ .
By the integral. For $f \in A C$ one has $V_{a}^{b} (f) = \int_{a}^{b} ∣ f^{'} ∣$ — the variation function is itself the indefinite integral of $∣ f^{'} ∣$ . Here $\int_{0}^{2 π} ∣ cos x ∣ d x = 4 \int_{0}^{π /2} cos x d x = 4$ .
The two agree, as they must.

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Theorem 8.36 (Variation of an AC function). If $f \in A C [a, b]$ then $V_{a}^{x} (f) = \int_{a}^{x} ∣ f^{'} (t) ∣ d t$ for all $x$ .

Proof. ( $\leq$ ) For any partition of $[a, x]$ , $\sum_{i} ∣ f (x_{i}) - f (x_{i - 1}) ∣ = \sum_{i} \int_{x_{i - 1}}^{x_{i}} f^{'} \leq \int_{a}^{x} ∣ f^{'} ∣$ ; take the supremum.

( $\geq$ ) Let $s = sign (f^{'})$ , measurable, and approximate: by Lusin's theorem (from Lebesgue Integration) and density of step functions in $L^{1}$ , choose a step function $σ$ with values in $[- 1, 1]$ and $\int_{a}^{x} ∣ s - σ ∣∣ f^{'} ∣ < ε$ . With $σ$ constant on the blocks of a partition $P$ ,

\int_{a}^{x} ∣ f^{'} ∣ = \int_{a}^{x} s f^{'} \leq \int_{a}^{x} σ f^{'} + ε = i \sum σ_{i} (f (x_{i}) - f (x_{i - 1})) + ε \leq V (f, P) + ε \leq V_{a}^{x} (f) + ε,

and $ε$ was arbitrary.∎

Exercises for §8.8 →

8.9Relation to Bounded Variation and Singular Functions

We owe a proof of the inclusion $A C \subseteq B V$ used twice already, and we can then state the decomposition theorem that puts the Cantor function in its place inside the general theory.

Theorem 8.37 (Absolutely continuous implies bounded variation). If $f \in A C [a, b]$ then $f \in B V [a, b]$ . The converse is false: the Cantor function is continuous and of bounded variation but not absolutely continuous.

Proof. Apply Definition Absolutely continuous function with $ε = 1$ to get a $δ > 0$ . Choose an integer $N > (b - a) / δ$ and split $[a, b]$ into $N$ consecutive closed intervals $I_{1}, \dots, I_{N}$ each of length $(b - a) / N < δ$ .

Fix $j$ and let $P_{j}$ be any partition of $I_{j}$ . Its subintervals are pairwise disjoint with total length $∣ I_{j} ∣ < δ$ , so the defining condition applies and gives $V (f, P_{j}) < 1$ . Taking the supremum over $P_{j}$ yields $V_{I_{j}} (f) \leq 1$ .

By additivity of variation (Proposition Boundedness and additivity, applied $N - 1$ times),

V_{a}^{b} (f) = j = 1 \sum N V_{I_{j}} (f) \leq N < \frac{b - a}{δ} + 1 < \infty.

The converse fails by Theorem Properties of the Cantor function together with Proposition The Cantor function is not absolutely continuous: $Φ \in C \cap B V$ and $Φ \in / A C$ .∎

Definition 8.38 (Singular function). A function $f : [a, b] \to R$ is singular if it is continuous, of bounded variation, non-constant, and $f^{'} (x) = 0$ for almost every $x$ .

The Cantor function is the standard example, and Theorem Properties of the Cantor function verifies each clause. The next theorem says that singular functions and absolutely continuous ones account for every $B V$ function once jumps are separated off.

Theorem 8.39 (Lebesgue decomposition of a BV function). Every $f \in B V [a, b]$ can be written uniquely, up to additive constants, as

f = f_{ac} + f_{sc} + f_{jump},

where $f_{ac} \in A C [a, b]$ , $f_{sc}$ is continuous with $f_{sc}^{'} = 0$ a.e., and $f_{jump}$ is a jump function — a sum $\sum_{n} c_{n} χ_{[s_{n}, b]}$ over the (countably many) discontinuities with $\sum_{n} ∣ c_{n} ∣ < \infty$ .

Proof. Sketch; the measure-theoretic ingredient is quoted. Write $f = g - h$ with $g, h$ increasing (Theorem Jordan decomposition), so it suffices to treat increasing $f$ . By Corollary Discontinuities of a BV function the discontinuities ${s_{n}}$ are countable with jumps $c_{n} = f (s_{n}^{+}) - f (s_{n}^{-}) \geq 0$ summing to at most $f (b) - f (a)$ ; define $f_{jump}$ as that sum and check that $f_{c} = f - f_{jump}$ is continuous and increasing. Next set $f_{ac} (x) = \int_{a}^{x} f^{'}$ , which is $A C$ by Corollary Indefinite integrals are absolutely continuous and has the same derivative a.e. as $f_{c}$ by the Lebesgue differentiation theorem. Then $f_{sc} = f_{c} - f_{ac}$ is continuous with derivative $0$ a.e.

Uniqueness: a difference of two such decompositions is an $A C$ function with derivative $0$ a.e. and no jumps, hence constant by Step C of Theorem Fundamental Theorem of Calculus for the Lebesgue integral. What is being assumed is the correspondence between increasing right-continuous functions and Lebesgue–Stieltjes measures, together with the Lebesgue decomposition of a measure, which is the subject of the next section.∎

Example 8.40 (Decomposing an explicit function). Decompose $f (x) = x + Φ (x) + 2 χ_{[1/2, 1]} (x)$ on $[0, 1]$ and compute $V_{0}^{1} (f)$ .

Solution.

The three summands are already of the three types: $f_{ac} (x) = x$ (Lipschitz, hence $A C$ ), $f_{sc} = Φ$ (singular), $f_{jump} = 2 χ_{[1/2, 1]}$ .
$f$ is a sum of increasing functions, hence increasing, so $V_{0}^{1} (f) = f (1) - f (0)$ .
$f (1) = 1 + 1 + 2 = 4$ and $f (0) = 0 + 0 + 0 = 0$ , so $V_{0}^{1} (f) = 4$ .
Note $\int_{0}^{1} f^{'} = \int_{0}^{1} 1 = 1$ , strictly less than $4$ : the deficit $3 = 1 + 2$ is exactly the singular mass plus the jump.

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Exercises for §8.9 →

8.10Absolute Continuity of Measures and Radon–Nikodym

Every statement of this part has a measure-theoretic shadow, and in the shadow the theorems are cleaner. A $B V$ function corresponds to a signed measure; absolute continuity of the function corresponds to absolute continuity of the measure; and the fundamental theorem becomes the Radon–Nikodym theorem.

Definition 8.41 (Absolute continuity and mutual singularity of measures). Let $μ, ν$ be ( $σ$ -finite, signed or positive) measures on a measurable space $(X, A)$ . Say $ν$ is absolutely continuous with respect to $μ$ , written $ν ≪ μ$ , if $μ (E) = 0 \Rightarrow ν (E) = 0$ . Say they are mutually singular, $ν ⊥ μ$ , if $X = A ⊔ B$ with $∣ ν ∣ (A) = 0$ and $μ (B) = 0$ .

Proposition 8.42 (The ε-δ form for finite measures). If $ν$ is a finite measure, then $ν ≪ μ$ if and only if for every $ε > 0$ there is $δ > 0$ with $μ (E) < δ \Rightarrow ν (E) < ε$ .

Proof. ( $\Leftarrow$ ) If $μ (E) = 0$ then $μ (E) < δ$ for every $δ$ , so $ν (E) < ε$ for every $ε$ , so $ν (E) = 0$ .

( $\Rightarrow$ ) Suppose the $ε$ - $δ$ statement fails for some $ε_{0}$ . Then for each $n$ there is $E_{n}$ with $μ (E_{n}) < 2^{- n}$ and $ν (E_{n}) \geq ε_{0}$ . Put $E = lim sup_{n} E_{n} = ⋂_{N} ⋃_{n \geq N} E_{n}$ . Then $μ (E) \leq \sum_{n \geq N} 2^{- n} = 2^{1 - N}$ for every $N$ , so $μ (E) = 0$ ; while $ν (⋃_{n \geq N} E_{n}) \geq ε_{0}$ for every $N$ and $ν$ is finite, so continuity from above gives $ν (E) \geq ε_{0} > 0$ . This contradicts $ν ≪ μ$ .∎

Remark. Finiteness of $ν$ is essential in the ( $\Rightarrow$ ) direction. On $(0, 1)$ with $μ$ Lebesgue measure and $d ν = x^{- 1} d x$ we have $ν ≪ μ$ , yet $ν ((0, δ)) = \infty$ for every $δ$ .

Theorem 8.43 (Radon–Nikodym). Let $μ$ be $σ$ -finite and $ν$ a $σ$ -finite measure with $ν ≪ μ$ . Then there is a $μ$ -a.e. unique measurable $g \geq 0$ with

ν (E) = \int_{E} g d μ for every measurable E .

One writes $g = \frac{d ν}{d μ}$ , the Radon–Nikodym derivative.

Proof. Sketch (von Neumann's argument), for finite $μ, ν$ ; the $σ$ -finite case follows by exhaustion. What is assumed: the Riesz representation theorem for bounded linear functionals on a Hilbert space, which is not proved in this book.

Set $λ = μ + ν$ and consider $T (f) = \int_{X} f d ν$ on $L^{2} (λ)$ . By Cauchy–Schwarz, $∣ T (f) ∣ \leq ∥ f ∥_{L^{2} (ν)} ν (X)^{1/2} \leq ∥ f ∥_{L^{2} (λ)} ν (X)^{1/2}$ , so $T$ is bounded, and Riesz representation supplies $h \in L^{2} (λ)$ with $\int f d ν = \int f h d λ$ for all $f \in L^{2} (λ)$ .

Testing on indicators shows $0 \leq h \leq 1$ $λ$ -a.e. Rearranging, $\int f (1 - h) d ν = \int f h d μ$ . The set ${h = 1}$ then has $μ$ -measure $0$ , hence $ν$ -measure $0$ by $ν ≪ μ$ , so $h < 1$ a.e. Putting $g = h / (1 - h)$ and taking $f = χ_{E} / (1 - h)$ (justified by monotone convergence on truncations) gives $ν (E) = \int_{E} g d μ$ .

Uniqueness: if $\int_{E} g_{1} d μ = \int_{E} g_{2} d μ$ for all $E$ , test on $E = {g_{1} > g_{2}}$ and on its complement to force $g_{1} = g_{2}$ $μ$ -a.e.∎

Corollary 8.44 (The function form). A finite Borel measure $ν$ on $[a, b]$ satisfies $ν ≪ m$ (Lebesgue) if and only if its distribution function $F (x) = ν ([a, x])$ is absolutely continuous, and then $F^{'} = d ν / d m$ a.e.

Proof. If $ν ≪ m$ , Radon–Nikodym gives $g \in L^{1}$ with $F (x) = \int_{a}^{x} g d m$ , so $F \in A C$ by Corollary Indefinite integrals are absolutely continuous, and $F^{'} = g$ a.e. by the Lebesgue differentiation theorem. Conversely if $F \in A C$ then $F (x) = \int_{a}^{x} F^{'}$ by Theorem Fundamental Theorem of Calculus for the Lebesgue integral; a set of Lebesgue measure zero then has $ν$ -measure zero, since the integral of an $L^{1}$ function over a null set vanishes.∎

Example 8.45 (Reading off a Radon–Nikodym derivative). On $[0, 1]$ let $ν (E) = \int_{E} 3 x^{2} d x$ . Compute $ν ([0, 1/2])$ , identify $d ν / d m$ , and decide whether $m ≪ ν$ .

Solution.

$ν ([0, 1/2]) = \int_{0}^{1/2} 3 x^{2} d x = x^{3}_{0}^{1/2} = 1/8$ .
By definition $d ν / d m = 3 x^{2}$ , and $ν ≪ m$ because an integral over a null set is $0$ .
For the reverse: $ν ({0}) = 0$ trivially, but consider $E = {0}$ — it is $m$ -null too, so that is no obstruction. Since $3 x^{2} > 0$ for $x \neq = 0$ , any set with $ν (E) = 0$ has $m (E ∖ {0}) = 0$ , hence $m (E) = 0$ . So in fact $m ≪ ν$ as well, and $d m / d ν = 1/ (3 x^{2})$ .
The measures are therefore equivalent: each is absolutely continuous with respect to the other. That fails as soon as the density vanishes on a set of positive measure.

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Exercises for §8.10 →

8.11The Lebesgue Differentiation Theorem

The fundamental theorem just proved has a local counterpart that holds for every integrable function, absolutely continuous or not: averages over shrinking intervals converge back to the function almost everywhere. This is the result the previous part kept quoting.

Definition 8.46 (Averaging operator). For $f \in L_{loc}^{1} (R^{n})$ and a ball $B (x, r)$ write

A_{r} f (x) = \frac{1}{∣ B ( x , r ) ∣} \int_{B (x, r)} f (y) d y .

In dimension one this is $\frac{1}{2 r} \int_{x - r}^{x + r} f$ .

Theorem 8.47 (Lebesgue differentiation theorem). Let $f \in L_{loc}^{1} (R^{n})$ . Then

r \to 0^{+} lim A_{r} f (x) = f (x) for almost every x .

In dimension one, consequently, $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$ a.e.

Proof. Granting the maximal inequality proved in the next section. Fix $N$ and work on the ball $B (0, N)$ ; it suffices to prove the statement there. Define the oscillation

Ω f (x) = r \to 0^{+} lim sup A_{r} f (x) - f (x) .

The claim is $Ω f = 0$ a.e. Let $ε > 0$ . By density of continuous functions in $L^{1}$ (proved in Lebesgue Integration), choose $g$ continuous with $∥ f - g ∥_{L^{1}} < ε$ . For continuous $g$ , $A_{r} g (x) \to g (x)$ at every $x$ by continuity, so $Ω g \equiv 0$ . Writing $h = f - g$ and using $A_{r}$ linear,

Ω f \leq Ω g + Ω h = Ω h \leq M h + ∣ h ∣,

where $M$ is the Hardy–Littlewood maximal operator of the next section. Hence for $α > 0$ ,

{Ω f > α} \leq {M h > α /2} + {∣ h ∣ > α /2} \leq \frac{2 C _{n}}{α} ∥ h ∥_{1} + \frac{2}{α} ∥ h ∥_{1} \leq \frac{2 ( C _{n} + 1 ) ε}{α},

using the weak-type inequality for the first term and Chebyshev's inequality for the second. Since $ε > 0$ was arbitrary, $∣ {Ω f > α} ∣ = 0$ for every $α > 0$ , and taking $α = 1/ k$ and a countable union gives $Ω f = 0$ a.e.∎

The structure of that argument is worth naming, because it recurs everywhere in analysis: a dense class on which the statement is trivial, plus a quantitative bound that controls the error uniformly. Density alone proves nothing — convergence a.e. is not preserved under $L^{1}$ limits — and the maximal inequality is precisely the missing uniformity.

Example 8.48 (Averages of a function with a jump). Let $f = χ_{[0, \infty)}$ on $R$ . Compute $A_{r} f (x)$ for $x = 0$ and for $x > 0$ with $r < x$ , and identify the exceptional set.

Solution.

For $x > 0$ and $r < x$ , the whole interval $(x - r, x + r)$ lies in $[0, \infty)$ , so $A_{r} f (x) = 1 = f (x)$ . Likewise $A_{r} f (x) = 0 = f (x)$ for $x < 0$ , $r < ∣ x ∣$ .
At $x = 0$ : $A_{r} f (0) = \frac{1}{2 r} \int_{- r}^{r} χ_{[0, \infty)} = \frac{1}{2 r} \cdot r = \frac{1}{2}$ for every $r > 0$ , so the limit is $\frac{1}{2} \neq = 1 = f (0)$ .
The exceptional set is ${0}$ , of measure zero, exactly as the theorem permits. Note the limit exists at $0$ — it simply equals the average of the two one-sided values.

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Corollary 8.49 (Density points). Let $E \subseteq R^{n}$ be measurable. Then for almost every $x \in E$ ,

r \to 0^{+} lim \frac{∣ E \cap B ( x , r ) ∣}{∣ B ( x , r ) ∣} = 1,

and the same limit is $0$ for almost every $x \in / E$ .

Proof. Apply Theorem Lebesgue differentiation theorem to $f = χ_{E \cap B (0, N)}$ , noting $A_{r} χ_{E} (x) = ∣ E \cap B (x, r) ∣/∣ B (x, r) ∣$ , and let $N \to \infty$ .∎

Intuition. Density points say that a measurable set has no genuinely "fuzzy" interior: almost every point of $E$ is a point where $E$ occupies $99.99%$ of every sufficiently small ball. A set like the rationals, which feels dense everywhere, has measure zero and therefore no density points — its apparent ubiquity is topological, not measure-theoretic.

Exercises for §8.11 →

8.12The Hardy–Littlewood Maximal Function and the Weak-Type Inequality

The maximal function replaces a limit that may not exist by a supremum that always does, at the cost of an inequality instead of an identity. Controlling it is the whole game.

Definition 8.50 (Hardy–Littlewood maximal function). For $f \in L_{loc}^{1} (R^{n})$ ,

M f (x) = r > 0 sup \frac{1}{∣ B ( x , r ) ∣} \int_{B (x, r)} ∣ f (y) ∣ d y = r > 0 sup A_{r} ∣ f ∣ (x) .

Note at once that $M$ does not map $L^{1}$ into $L^{1}$ : if $f \neq = 0$ then $M f (x) ≳ ∥ f ∥_{1} /∣ x ∣^{n}$ for large $∣ x ∣$ , which is not integrable on $R^{n}$ . The correct substitute is a bound on the distribution function.

Lemma 8.51 (Vitali covering lemma, finite form). Let $B_{1}, \dots, B_{N}$ be balls in $R^{n}$ . There is a pairwise disjoint subcollection $B_{i_{1}}, \dots, B_{i_{k}}$ with

j = 1 ⋃ N B_{j} \leq 3^{n} l = 1 \sum k ∣ B_{i_{l}} ∣.

Proof. Greedy selection. Pick $B_{i_{1}}$ of largest radius. Having chosen $B_{i_{1}}, \dots, B_{i_{l}}$ , pick $B_{i_{l + 1}}$ of largest radius among those disjoint from all previously chosen; stop when none remain.

Every $B_{j}$ meets some chosen $B_{i_{l}}$ — otherwise it would have been selectable — and the first chosen ball it meets has radius at least that of $B_{j}$ , by the greedy order. If two balls meet and $rad (B_{i_{l}}) \geq rad (B_{j})$ , then $B_{j} \subseteq 3 B_{i_{l}}$ , the concentric triple. Hence $⋃_{j} B_{j} \subseteq ⋃_{l} 3 B_{i_{l}}$ and

j ⋃ B_{j} \leq l \sum ∣3 B_{i_{l}} ∣ = 3^{n} l \sum ∣ B_{i_{l}} ∣,

since dilating a ball by $3$ multiplies its measure by $3^{n}$ .∎

Theorem 8.52 (Hardy–Littlewood maximal inequality (weak type (1,1))). For every $f \in L^{1} (R^{n})$ and every $α > 0$ ,

{x : M f (x) > α} \leq \frac{3 ^{n}}{α} ∥ f ∥_{L^{1}} .

Proof. Let $E_{α} = {M f > α}$ and let $K \subseteq E_{α}$ be compact; by inner regularity it suffices to bound $∣ K ∣$ .

For each $x \in K$ there is, by Definition Hardy–Littlewood maximal function, a radius $r_{x}$ with

\int_{B (x, r_{x})} ∣ f ∣ > α ∣ B (x, r_{x}) ∣.

These balls cover $K$ ; by compactness finitely many $B_{1}, \dots, B_{N}$ suffice. Apply Lemma Vitali covering lemma, finite form to get a disjoint subfamily $B_{i_{1}}, \dots, B_{i_{k}}$ with $∣ K ∣ \leq ∣ ⋃ B_{j} ∣ \leq 3^{n} \sum_{l} ∣ B_{i_{l}} ∣$ . Then

∣ K ∣ \leq 3^{n} l = 1 \sum k ∣ B_{i_{l}} ∣ < \frac{3 ^{n}}{α} l = 1 \sum k \int_{B_{i_{l}}} ∣ f ∣ \leq \frac{3 ^{n}}{α} \int_{R^{n}} ∣ f ∣,

the last step because the $B_{i_{l}}$ are pairwise disjoint. Taking the supremum over compact $K \subseteq E_{α}$ gives the claim.∎

Corollary 8.53 (Strong type for $p > 1$ ). For $1 < p \leq \infty$ there is $C_{n, p}$ with $∥ M f ∥_{L^{p}} \leq C_{n, p} ∥ f ∥_{L^{p}}$ .

Proof. Sketch; the interpolation step is quoted. For $p = \infty$ the bound is immediate with $C = 1$ , since every average of $∣ f ∣$ is at most $∥ f ∥_{\infty}$ . For $1 < p < \infty$ , the Marcinkiewicz interpolation theorem applied between the weak- $(1, 1)$ bound of Theorem Hardy–Littlewood maximal inequality (weak type (1,1)) and the strong $(\infty, \infty)$ bound yields the strong $(p, p)$ bound. The interpolation theorem itself is beyond this chapter; the elementary route is to split $f = f χ_{{∣ f ∣ > α /2}} + f χ_{{∣ f ∣ \leq α /2}}$ , apply the weak bound to the first piece, and integrate the distribution function via $∥ M f ∥_{p}^{p} = p \int_{0}^{\infty} α^{p - 1} ∣ {M f > α} ∣ d α$ .∎

Example 8.54 (The maximal function of an indicator). Compute $M f$ for $f = χ_{[- 1, 1]}$ on $R$ , and verify the weak-type bound at $α = 1/4$ .

Solution.

For $∣ x ∣ \leq 1$ the ball $B (x, r)$ with small $r$ lies inside $[- 1, 1]$ , giving average $1$ ; no average exceeds $1$ , so $M f (x) = 1$ there.
For $∣ x ∣ > 1$ , the best interval is the smallest one containing all of $[- 1, 1]$ , namely $r = ∣ x ∣ + 1$ : then $M f (x) = \frac{2}{2 ( ∣ x ∣ + 1 )} = \frac{1}{∣ x ∣ + 1}$ .
So ${M f > 1/4} = {∣ x ∣ < 3}$ , of measure $6$ .
The bound gives $3^{1} ∥ f ∥_{1} / α = 3 \cdot 2/ (1/4) = 24 \geq 6$ . $✓$
Note $M f \sim 1/∣ x ∣$ at infinity, confirming $M f \in / L^{1}$ even though $f \in L^{1}$ : the weak-type form is not a weakness of the proof but of the operator.

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Exercises for §8.12 →

8.13Proof Ideas: Vitali Covering and the Rising Sun Lemma

Two arguments carry the differentiation theory. The first is the covering idea already used; the second is a one-dimensional trick that gives monotone differentiability with bare hands.

Lemma 8.55 (Vitali covering lemma). Let $E \subseteq R^{n}$ with $∣ E ∣ < \infty$ and let $V$ be a family of closed balls that covers $E$ in the Vitali sense: every $x \in E$ lies in balls of $V$ of arbitrarily small radius. Then for every $ε > 0$ there are finitely many pairwise disjoint $B_{1}, \dots, B_{N} \in V$ with

E ∖ j = 1 ⋃ N B_{j} < ε .

Proof. Sketch. Enclose $E$ in an open $U$ with $∣ U ∣ < ∣ E ∣ + ε$ and discard balls not contained in $U$ ; the family still covers $E$ in the Vitali sense. Select greedily: having chosen disjoint $B_{1}, \dots, B_{k}$ , if they do not already cover $E$ up to measure $ε$ , choose $B_{k + 1}$ disjoint from them with radius at least half the supremum of admissible radii. Disjointness inside $U$ forces $\sum_{j} ∣ B_{j} ∣ \leq ∣ U ∣ < \infty$ , so radii tend to $0$ . A point of $E$ not covered by $B_{1}, \dots, B_{N}$ lies in some admissible small ball, which must meet some $B_{j}$ with $j > N$ and hence lies in $5 B_{j}$ ; therefore the uncovered part has measure at most $5^{n} \sum_{j > N} ∣ B_{j} ∣ \to 0$ . Choose $N$ making this $< ε$ .∎

Lemma 8.56 (Rising sun lemma). Let $g$ be continuous on $[a, b]$ and let

E = {x \in (a, b) : g (y) > g (x) for some y \in (x, b]}

be the set of points "in shadow". Then $E$ is open, $E = ⨆_{k} (a_{k}, b_{k})$ , and $g (a_{k}) \leq g (b_{k})$ for each $k$ .

Proof. $E$ is open because $g$ is continuous: if $g (y) > g (x)$ then the same $y$ works for all $x^{'}$ near $x$ . Write $E$ as a countable disjoint union of open intervals $(a_{k}, b_{k})$ , its connected components.

Fix $k$ and $x \in (a_{k}, b_{k})$ . We claim $g (x) \leq g (b_{k})$ . Let $s = sup {y \in [x, b] : g (y) \geq g (x)}$ ; the set is non-empty ( $y = x$ ) and closed by continuity, so the supremum is attained and $g (s) \geq g (x)$ . If $s < b_{k}$ then $s \in E$ , so some $y > s$ has $g (y) > g (s) \geq g (x)$ , contradicting maximality of $s$ . Hence $s \geq b_{k}$ ; if $s > b_{k}$ then $b_{k} \in E$ , impossible since $b_{k}$ is an endpoint of a component and $E$ is open — so $s = b_{k}$ and $g (b_{k}) \geq g (x)$ . Letting $x ↓ a_{k}$ and using continuity gives $g (a_{k}) \leq g (b_{k})$ .∎

Theorem 8.57 (Monotone functions are differentiable a.e.). An increasing $g : [a, b] \to R$ is differentiable at almost every point.

Proof. Sketch, using the covering lemma. For $0 < u < v$ let

E_{u, v} = {x : h \to 0 lim inf \frac{g ( x + h ) - g ( x )}{h} < u < v < h \to 0 lim sup \frac{g ( x + h ) - g ( x )}{h}},

the set where the derivative fails to exist with the two Dini limits separated by the rationals $u < v$ . The non-differentiability set is $⋃_{u < v rational} E_{u, v}$ , a countable union, so it suffices to show each $E_{u, v}$ is null.

Put $s = ∣ E_{u, v} ∣^{*}$ (outer measure) and fix $ε > 0$ . Intervals on which the increment ratio is $< u$ form a Vitali cover of $E_{u, v}$ ; by Lemma Vitali covering lemma finitely many disjoint ones $I_{1}, \dots, I_{N}$ cover all but $ε$ , and summing gives $\sum_{j} (g (right) - g (left)) < u (s + ε)$ . Inside those intervals, intervals with ratio $> v$ form a Vitali cover of $E_{u, v} \cap ⋃ I_{j}$ , whose outer measure is at least $s - ε$ ; selecting disjoint ones $J_{1}, \dots, J_{K}$ covering all but $ε$ gives $\sum_{l} (g (right) - g (left)) > v (s - 2 ε)$ . Since the $J_{l}$ sit inside the $I_{j}$ and $g$ is increasing, the second sum is at most the first, so

v (s - 2 ε) < u (s + ε) .

Letting $ε \to 0$ gives $v s \leq u s$ , and since $v > u$ this forces $s = 0$ . Finiteness of the derivative a.e. follows from the integral bound in Theorem Lebesgue's differentiation theorem for monotone functions.∎

Remark. The rising sun lemma gives a second, wholly one-dimensional route: applied to $g (x) - λ x$ it produces the estimate $∣ {x : \overline{D} g (x) > λ} ∣ \leq λ^{- 1} (g (b) - g (a))$ , a weak-type bound for the derivative rather than the average, from which the same $u, v$ argument runs. The name is Riesz's: the shadowed set $E$ is what a horizontal sun rising from the right leaves in darkness.

Exercises for §8.13 →

8.14Differentiation of Measures and the Radon–Nikodym Derivative

Averaging a function over small balls generalises to comparing two measures over small balls, and the limit is again a Radon–Nikodym derivative — with the Cantor function's pathology appearing as the singular part.

Definition 8.58 (Symmetric derivative of a measure). Let $ν$ be a signed Borel measure on $R^{n}$ , finite on compacts. Its derivative with respect to Lebesgue measure at $x$ is

D ν (x) = r \to 0^{+} lim \frac{ν ( B ( x , r ) )}{∣ B ( x , r ) ∣},

when the limit exists.

Theorem 8.59 (Lebesgue decomposition). Let $μ$ be $σ$ -finite and $ν$ $σ$ -finite on $(X, A)$ . Then $ν$ decomposes uniquely as

ν = ν_{ac} + ν_{sing}, ν_{ac} ≪ μ, ν_{sing} ⊥ μ .

Proof. Sketch, assuming Radon–Nikodym for the pair $(ν, μ + ν)$ . Since $ν ≪ μ + ν$ and $μ ≪ μ + ν$ , Theorem Radon–Nikodym supplies densities; write $d ν = h d (μ + ν)$ with $0 \leq h \leq 1$ . Let $A = {h = 1}$ and $B = {h < 1}$ . On $A$ one gets $μ (A) = 0$ , so $ν_{sing} = ν ∣_{A} ⊥ μ$ . On $B$ , setting $g = h / (1 - h)$ gives $ν ∣_{B} (E) = \int_{E} g d μ$ , so $ν_{ac} = ν ∣_{B} ≪ μ$ .

Uniqueness: if $ν = ν_{1} + σ_{1} = ν_{2} + σ_{2}$ are two such decompositions, then $ν_{1} - ν_{2} = σ_{2} - σ_{1}$ is simultaneously $≪ μ$ and $⊥ μ$ , and a measure both absolutely continuous and singular with respect to $μ$ is zero.∎

Theorem 8.60 (Differentiating a measure). Let $ν$ be a finite signed Borel measure on $R^{n}$ with Lebesgue decomposition $ν = f d m + ν_{sing}$ . Then $D ν$ exists a.e. and

D ν (x) = f (x) for a.e. x, D ν_{sing} (x) = 0 for a.e. x .

Proof. Sketch. The absolutely continuous part is Theorem Lebesgue differentiation theorem verbatim, since $ν_{ac} (B (x, r)) /∣ B (x, r) ∣ = A_{r} f (x)$ .

For the singular part, let $S$ carry $ν_{sing}$ with $∣ S ∣ = 0$ . Replacing $M$ by the maximal operator $M ν (x) = sup_{r} ∣ ν ∣ (B (x, r)) /∣ B (x, r) ∣$ , the same Vitali argument as in Theorem Hardy–Littlewood maximal inequality (weak type (1,1)) gives $∣ {M ν_{sing} > α} ∣ \leq 3^{n} ∣ ν_{sing} ∣ (R^{n} ∖ K) / α$ for compacts $K$ exhausting $S$ ; regularity of $∣ ν_{sing} ∣$ makes the right side arbitrarily small off $S$ , so $M ν_{sing} = 0$ a.e. off $S$ , hence a.e.∎

Example 8.61 (The measure induced by the Cantor function). Let $ν_{Φ}$ be the Lebesgue–Stieltjes measure with $ν_{Φ} ([0, x]) = Φ (x)$ . Find its Lebesgue decomposition and $D ν_{Φ}$ .

Solution.

$ν_{Φ}$ is a probability measure on $[0, 1]$ (total mass $Φ (1) - Φ (0) = 1$ ) supported on the Cantor set $C$ : every removed interval is a plateau of $Φ$ and so has $ν_{Φ}$ -measure $0$ .
$∣ C ∣ = 0$ by Theorem Properties of the Cantor function, so $ν_{Φ} ⊥ m$ with the splitting $[0, 1] = C ⊔ C^{c}$ .
Hence $ν_{ac} = 0$ and $ν_{sing} = ν_{Φ}$ : the decomposition is entirely singular.
By Theorem Differentiating a measure, $D ν_{Φ} = 0$ a.e., which is the statement $Φ^{'} = 0$ a.e. read as a statement about measures.
Note $ν_{Φ}$ has no atoms either ( $Φ$ is continuous), so it is an example of a continuous singular measure — neither a density nor a sum of point masses.

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Exercises for §8.14 →

8.15Lebesgue Points and Lusin's Condition (N)

Two refinements close this part. The first strengthens the differentiation theorem from "averages converge" to "averages of the deviation converge". The second gives the missing metric characterisation of absolute continuity.

Definition 8.62 (Lebesgue point). A point $x$ is a Lebesgue point of $f \in L_{loc}^{1}$ if

r \to 0^{+} lim \frac{1}{∣ B ( x , r ) ∣} \int_{B (x, r)} f (y) - f (x) d y = 0.

Theorem 8.63 (Almost every point is a Lebesgue point). If $f \in L_{loc}^{1} (R^{n})$ then almost every $x$ is a Lebesgue point of $f$ .

Proof. For $q \in Q$ apply Theorem Lebesgue differentiation theorem to the function $y \mapsto ∣ f (y) - q ∣$ , which is locally integrable: there is a null set $N_{q}$ off which

r \to 0 lim A_{r} (∣ f - q ∣) (x) = ∣ f (x) - q ∣.

Let $N = ⋃_{q \in Q} N_{q}$ , still null. Fix $x \in / N$ and $ε > 0$ , and choose $q$ rational with $∣ f (x) - q ∣ < ε /2$ . Then

A_{r} (∣ f - f (x) ∣) (x) \leq A_{r} (∣ f - q ∣) (x) + ∣ q - f (x) ∣ ⟶ ∣ f (x) - q ∣ + ∣ q - f (x) ∣ < ε .

Since $ε$ was arbitrary, the limsup is $0$ .∎

Remark. The difference from the plain differentiation theorem matters: $A_{r} f (x) \to f (x)$ allows cancellation between places where $f$ is too big and places where it is too small, as the jump in Example Averages of a function with a jump illustrates. A Lebesgue point forbids the cancellation, which is what makes Lebesgue points the right notion of "the value of $f$ at $x$ " for an equivalence class of functions agreeing a.e.

Definition 8.64 (Lusin's condition (N)). $f : [a, b] \to R$ satisfies condition (N) if $∣ E ∣ = 0 \Rightarrow ∣ f (E) ∣ = 0$ for every $E \subseteq [a, b]$ : $f$ maps null sets to null sets.

Theorem 8.65 (Banach–Zarecki). $f \in A C [a, b]$ if and only if $f$ is continuous, of bounded variation, and satisfies condition (N).

Proof. ( $\Rightarrow$ ) Continuity and bounded variation are Theorem The chain of inclusions and Theorem Absolutely continuous implies bounded variation. For (N), let $∣ E ∣ = 0$ and $ε > 0$ , and take $δ$ from Definition Absolutely continuous function. Cover $E$ by disjoint intervals of total length $< δ$ ; on each, $f$ attains its max and min at points $α_{k} \leq β_{k}$ inside, and the image of that interval is an interval of length $∣ f (β_{k}) - f (α_{k}) ∣$ . The intervals $(α_{k}, β_{k})$ are disjoint of total length $< δ$ , so $\sum_{k} ∣ f (β_{k}) - f (α_{k}) ∣ < ε$ , giving $∣ f (E) ∣^{*} < ε$ .

( $\Leftarrow$ ) Sketch. Given continuity, $B V$ and (N), one shows $V_{a}^{x} (f) = \int_{a}^{x} ∣ f^{'} ∣$ , which by Theorem Variation of an AC function characterises $A C$ . The step that uses (N) is that the image of the set where $f^{'}$ fails to exist — null by Corollary BV implies differentiable a.e. — contributes no length, so no variation is hidden there; the remaining variation is accounted for by the derivative via a Vitali covering argument.∎

Example 8.66 (The Cantor function violates (N)). Show that $Φ$ maps a null set onto a set of measure $1$ , and reconcile this with Theorem Banach–Zarecki.

Solution.

$Φ$ maps $[0, 1]$ onto $[0, 1]$ and is constant on each removed interval, so $Φ$ of the removed set is a countable set of dyadic values, of measure $0$ .
Since $Φ ([0, 1]) = [0, 1]$ has measure $1$ , and the removed part contributes measure $0$ , we get $∣Φ (C) ∣ = 1$ .
But $∣ C ∣ = 0$ . So $Φ$ carries a null set to a set of full measure: condition (N) fails spectacularly.
This is consistent: $Φ$ is continuous and $B V$ , and Theorem Banach–Zarecki then says exactly one thing can be wrong, namely (N) — which is a third independent proof that $Φ \in / A C$ .
The picture to keep: absolute continuity is the analytic name for "does not manufacture measure out of nothing".

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Exercises for §8.15 →

8.16 $L^{p}$ Spaces: Definition and Basic Properties

We now change subject. The convergence theorems of Lebesgue Integration say when limits may be taken inside an integral; $L^{p}$ theory turns those statements into the geometry of a normed space, where "limit" means limit in a norm and the theorems become statements about completeness.

Definition 8.67 (The space $L^{p}$ ). Let $(X, A, μ)$ be a measure space and $1 \leq p < \infty$ . For measurable $f : X \to R$ put

∥ f ∥_{p} = (\int_{X} ∣ f ∣^{p} d μ)^{1/ p},

and let $L^{p} (μ)$ be the set of measurable $f$ with $∥ f ∥_{p} < \infty$ , with two functions identified when they agree $μ$ -a.e. For $p = \infty$ set

∥ f ∥_{\infty} = ess sup ∣ f ∣ = in f {M : ∣ f ∣ \leq M a.e.},

the essential supremum, and $L^{\infty} (μ) = {f : ∥ f ∥_{\infty} < \infty}$ .

The identification of a.e.-equal functions is not a technicality to be waved through: without it $∥ \cdot ∥_{p}$ is only a seminorm, since $∥ f ∥_{p} = 0$ says exactly $f = 0$ a.e., not $f = 0$ . An element of $L^{p}$ is an equivalence class, and Theorem Almost every point is a Lebesgue point is what lets us speak of "the value at $x$ " for a canonical representative.

Proposition 8.68 ( $∥ \cdot ∥_{p}$ is a norm, given Minkowski). For $1 \leq p \leq \infty$ , $L^{p} (μ)$ is a vector space and $∥ \cdot ∥_{p}$ is a norm on it.

Proof. Homogeneity, $∥ λ f ∥_{p} = ∣ λ ∣ ∥ f ∥_{p}$ , is immediate from pulling $∣ λ ∣^{p}$ out of the integral. Definiteness holds because of the a.e. identification, as noted. Closure under addition and the triangle inequality are both Theorem Minkowski's inequality, proved two sections below; note that $∣ f + g ∣^{p} \leq 2^{p} (∣ f ∣^{p} + ∣ g ∣^{p})$ already gives closure under addition without it. For $p = \infty$ , the triangle inequality follows because $∣ f ∣ \leq ∥ f ∥_{\infty}$ and $∣ g ∣ \leq ∥ g ∥_{\infty}$ outside a null set, whose union is null.∎

Proposition 8.69 ( $∥ f ∥_{p} \to ∥ f ∥_{\infty}$ ). If $μ (X) < \infty$ and $f \in L^{\infty} (μ)$ , then $∥ f ∥_{p} \to ∥ f ∥_{\infty}$ as $p \to \infty$ .

Proof. Write $M = ∥ f ∥_{\infty}$ . Upper bound: $∥ f ∥_{p} \leq (M^{p} μ (X))^{1/ p} = M μ (X)^{1/ p} \to M$ . Lower bound: for $ε > 0$ the set $E = {∣ f ∣ > M - ε}$ has $μ (E) > 0$ by definition of the essential supremum, and

∥ f ∥_{p} \geq (\int_{E} ∣ f ∣^{p})^{1/ p} \geq (M - ε) μ (E)^{1/ p} \to M - ε .

Since $ε$ was arbitrary, $lim inf ∥ f ∥_{p} \geq M$ .∎

That proposition is the reason for the notation: $L^{\infty}$ is the limiting case of $L^{p}$ , and the essential supremum rather than the supremum is what makes the limit work — a function altered on a null set has the same $∥ \cdot ∥_{p}$ for every finite $p$ , so the limit cannot see an ordinary supremum.

Example 8.70 (Which $L^{p}$ does $x^{- a}$ belong to?). On $(0, 1)$ with Lebesgue measure, for which $p \in [1, \infty)$ is $f (x) = x^{- 1/2}$ in $L^{p}$ ? Compute $∥ f ∥_{1}$ .

Solution.

$\int_{0}^{1} x^{- p /2} d x$ converges exactly when $p /2 < 1$ , i.e. $p < 2$ .
So $f \in L^{p} (0, 1)$ for $1 \leq p < 2$ and $f \in / L^{2} (0, 1)$ .
For $p = 1$ : $\int_{0}^{1} x^{- 1/2} d x = 2 x^{1/2}_{0}^{1} = 2$ , so $∥ f ∥_{1} = 2$ .
Sanity check on the boundary: at $p = 2$ the integral is $\int_{0}^{1} x^{- 1} d x = \infty$ , confirming exclusion. The example shows the $L^{p}$ scale is genuinely a scale: membership is a sharp threshold, not a yes/no for all $p$ at once.

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Example 8.71 (The same question at infinity). On $(1, \infty)$ , for which $p$ is $g (x) = x^{- 1}$ in $L^{p}$ ? Compute $∥ g ∥_{2}$ .

Solution.

$\int_{1}^{\infty} x^{- p} d x$ converges exactly when $p > 1$ .
So $g \in L^{p} (1, \infty)$ for $p > 1$ but $g \in / L^{1} (1, \infty)$ — the opposite threshold to the previous example.
$∥ g ∥_{2}^{2} = \int_{1}^{\infty} x^{- 2} d x = 1$ , so $∥ g ∥_{2} = 1$ .
Comparing the two examples: near a singularity, small $p$ helps; near infinity, large $p$ helps. This is why no inclusion between $L^{p}$ spaces holds on $R$ , as the section on relations will make precise.

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Exercises for §8.16 →

8.17Hölder's and Minkowski's Inequalities

Two inequalities make $L^{p}$ work, and both come from a single convexity fact about the logarithm. We prove that fact first, since deriving Hölder from it and Minkowski from Hölder is the cleanest sequence in this chapter.

Definition 8.72 (Conjugate exponents). $p, q \in [1, \infty]$ are conjugate exponents if

\frac{1}{p} + \frac{1}{q} = 1,

with the conventions $1/\infty = 0$ , so that $1$ and $\infty$ are conjugate, and $2$ is self-conjugate.

Lemma 8.73 (Young's inequality). Let $p, q \in (1, \infty)$ be conjugate. Then for all $a, b \geq 0$ ,

ab \leq \frac{a ^{p}}{p} + \frac{b ^{q}}{q},

with equality if and only if $a^{p} = b^{q}$ .

Proof. If $a = 0$ or $b = 0$ both sides behave trivially, so assume $a, b > 0$ . The logarithm is concave — its second derivative is $- 1/ t^{2} < 0$ — so for weights $1/ p + 1/ q = 1$ ,

lo g (\frac{a ^{p}}{p} + \frac{b ^{q}}{q}) \geq \frac{1}{p} lo g (a^{p}) + \frac{1}{q} lo g (b^{q}) = lo g a + lo g b = lo g (ab) .

Exponentiating, using that $exp$ is increasing, gives the inequality. Strict concavity makes the middle step an equality exactly when $a^{p} = b^{q}$ .∎

Theorem 8.74 (Hölder's inequality). Let $p, q$ be conjugate exponents in $[1, \infty]$ . For measurable $f, g$ ,

∥ f g ∥_{1} = \int_{X} ∣ f g ∣ d μ \leq ∥ f ∥_{p} ∥ g ∥_{q} .

In particular $f g \in L^{1}$ whenever $f \in L^{p}$ and $g \in L^{q}$ .

Proof. The cases $p = 1, q = \infty$ are immediate: $∣ f g ∣ \leq ∣ f ∣ ∥ g ∥_{\infty}$ a.e., so integrate.

Let $1 < p < \infty$ . If $∥ f ∥_{p} = 0$ then $f = 0$ a.e. and both sides vanish; likewise for $g$ . If either norm is infinite the right side is $\infty$ . So assume $0 < ∥ f ∥_{p}, ∥ g ∥_{q} < \infty$ and normalise:

F = \frac{∣ f ∣}{∥ f ∥ _{p}}, G = \frac{∣ g ∣}{∥ g ∥ _{q}}, so \int F^{p} = \int G^{q} = 1.

Apply Lemma Young's inequality pointwise with $a = F (x)$ , $b = G (x)$ :

F (x) G (x) \leq \frac{F ( x ) ^{p}}{p} + \frac{G ( x ) ^{q}}{q} .

Integrate over $X$ , using monotonicity and linearity of the integral:

\int_{X} FG d μ \leq \frac{1}{p} \int F^{p} + \frac{1}{q} \int G^{q} = \frac{1}{p} + \frac{1}{q} = 1.

Multiplying through by $∥ f ∥_{p} ∥ g ∥_{q}$ gives $\int ∣ f g ∣ \leq ∥ f ∥_{p} ∥ g ∥_{q}$ .∎

Corollary 8.75 (Cauchy–Schwarz). $\int_{X} ∣ f g ∣ d μ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ , with equality iff $∣ f ∣$ and $∣ g ∣$ are proportional a.e.

Proof. Take $p = q = 2$ in Theorem Hölder's inequality. The equality case comes from the equality case of Lemma Young's inequality: $F^{2} = G^{2}$ a.e., i.e. $∣ f ∣/∥ f ∥_{2} = ∣ g ∣/∥ g ∥_{2}$ a.e.∎

Theorem 8.76 (Minkowski's inequality). For $1 \leq p \leq \infty$ and $f, g \in L^{p} (μ)$ ,

∥ f + g ∥_{p} \leq ∥ f ∥_{p} + ∥ g ∥_{p} .

Proof. For $p = 1$ integrate the pointwise triangle inequality; for $p = \infty$ see Proposition $∥ \cdot ∥_{p}$ is a norm, given Minkowski. Let $1 < p < \infty$ and assume $∥ f + g ∥_{p} \neq = 0$ , which is finite because $∣ f + g ∣^{p} \leq 2^{p} (∣ f ∣^{p} + ∣ g ∣^{p})$ .

Split the integrand by writing $∣ f + g ∣^{p} = ∣ f + g ∣ ∣ f + g ∣^{p - 1}$ and using $∣ f + g ∣ \leq ∣ f ∣ + ∣ g ∣$ :

\int ∣ f + g ∣^{p} \leq \int ∣ f ∣ ∣ f + g ∣^{p - 1} + \int ∣ g ∣ ∣ f + g ∣^{p - 1} .

Apply Theorem Hölder's inequality to each term with exponents $p$ and $q = p / (p - 1)$ :

\int ∣ f + g ∣^{p} \leq (∥ f ∥_{p} + ∥ g ∥_{p}) (\int ∣ f + g ∣^{(p - 1) q})^{1/ q} = (∥ f ∥_{p} + ∥ g ∥_{p}) ∥ f + g ∥_{p}^{p / q},

since $(p - 1) q = p$ and $p / q = p - 1$ . Divide both sides by the finite non-zero quantity $∥ f + g ∥_{p}^{p - 1}$ and use $p - (p - 1) = 1$ :

∥ f + g ∥_{p} = ∥ f + g ∥_{p}^{p - (p - 1)} \leq ∥ f ∥_{p} + ∥ g ∥_{p} .

∎

Example 8.77 (Hölder in a concrete estimate). On $(0, 1)$ , bound $\int_{0}^{1} x^{- 1/3} (1 - x)^{- 1/3} d x$ using Hölder with $p = q = 2$ , and compare with the exact value $B (\frac{2}{3}, \frac{2}{3}) \approx 2.0534$ .

Solution.

Take $f = x^{- 1/3}$ , $g = (1 - x)^{- 1/3}$ and apply Corollary Cauchy–Schwarz.
$∥ f ∥_{2}^{2} = \int_{0}^{1} x^{- 2/3} d x = 3 x^{1/3}_{0}^{1} = 3$ , so $∥ f ∥_{2} = 3$ .
By the symmetry $x \mapsto 1 - x$ , $∥ g ∥_{2} = 3$ too.
Hence the integral is at most $3 \cdot 3 = 3$ .
The true value is about $2.053$ , so the bound is valid but lossy by a factor near $1.46$ — expected, since equality in Cauchy–Schwarz needs $∣ f ∣ \propto ∣ g ∣$ , and here the two factors are singular at opposite endpoints.

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Pitfall. Hölder's inequality requires conjugate exponents, and the commonest error is to use $p$ and $p$ instead of $p$ and $p / (p - 1)$ . Check by homogeneity: replacing $f$ by $λ f$ must scale both sides by $∣ λ ∣$ , and replacing $g$ by $λ g$ likewise, which pins down $1/ p + 1/ q = 1$ uniquely.

Figure 8.78. Unit balls

{∣ x ∣^{p} + ∣ y ∣^{p} \leq 1}

for

p = 1

p = 2

and

p = \infty

R^{2}

. The

ℓ^{1}

ball is inscribed in the

ℓ^{2}

ball, which is inscribed in the

ℓ^{\infty}

ball, matching

∥ v ∥_{\infty} \leq ∥ v ∥_{2} \leq ∥ v ∥_{1}

Exercises for §8.17 →

8.18Completeness: the Riesz–Fischer Theorem

Completeness is the property for which the Lebesgue integral was built. The same definition of $∥ \cdot ∥_{p}$ applied to Riemann-integrable functions produces a space that is not complete — Cauchy sequences of continuous functions can converge to discontinuous limits — and every limiting argument in analysis then has to carry its own existence proof.

Theorem 8.79 (Riesz–Fischer: $L^{p}$ is complete). For $1 \leq p \leq \infty$ and any measure space, $L^{p} (μ)$ is a Banach space: every Cauchy sequence in $∥ \cdot ∥_{p}$ converges in $∥ \cdot ∥_{p}$ to an element of $L^{p} (μ)$ .

Proof. Take $1 \leq p < \infty$ (the case $p = \infty$ is at the end). Let $(f_{n})$ be Cauchy. It suffices to produce a subsequence converging in norm to some $f$ , since a Cauchy sequence with a convergent subsequence converges — a fact from Sequences & Series that holds in any metric space.

Step 1: extract a rapidly Cauchy subsequence. Choose $n_{1} < n_{2} < \dots$ with

∥ f_{n_{k + 1}} - f_{n_{k}} ∥_{p} \leq 2^{- k} .

Step 2: a dominating function. Put

g_{K} = k = 1 \sum K f_{n_{k + 1}} - f_{n_{k}}, g = K \to \infty lim g_{K} \in [0, \infty] .

By Theorem Minkowski's inequality applied $K$ times, $∥ g_{K} ∥_{p} \leq \sum_{k = 1}^{K} 2^{- k} < 1$ . The $g_{K}$ increase to $g$ , so the Monotone Convergence Theorem of Lebesgue Integration gives $\int g^{p} = lim_{K} \int g_{K}^{p} \leq 1$ . Hence $g \in L^{p}$ and in particular $g < \infty$ a.e.

Step 3: pointwise limit. Where $g (x) < \infty$ the telescoping series

f_{n_{1}} (x) + k \geq 1 \sum (f_{n_{k + 1}} (x) - f_{n_{k}} (x))

converges absolutely, hence converges; call its sum $f (x)$ , and set $f = 0$ on the null set where $g = \infty$ . Then $f_{n_{k}} \to f$ a.e., and $∣ f ∣ \leq ∣ f_{n_{1}} ∣ + g \in L^{p}$ , so $f \in L^{p}$ .

Step 4: convergence in norm. $∣ f - f_{n_{k}} ∣^{p} \leq (∣ f ∣ + ∣ f_{n_{k}} ∣)^{p} \leq (2∣ f_{n_{1}} ∣ + 2 g)^{p} \in L^{1}$ , a fixed integrable dominating function, and $∣ f - f_{n_{k}} ∣^{p} \to 0$ a.e. The Dominated Convergence Theorem of Lebesgue Integration gives $\int ∣ f - f_{n_{k}} ∣^{p} \to 0$ , i.e. $∥ f - f_{n_{k}} ∥_{p} \to 0$ .

The case $p = \infty$ . If $(f_{n})$ is Cauchy in $L^{\infty}$ , then for each pair $m, n$ the inequality $∣ f_{n} - f_{m} ∣ \leq ∥ f_{n} - f_{m} ∥_{\infty}$ holds off a null set $N_{m, n}$ ; off the countable union $N = ⋃ N_{m, n}$ , still null, $(f_{n})$ is uniformly Cauchy, hence converges uniformly to some bounded $f$ , and $∥ f_{n} - f ∥_{\infty} \to 0$ .∎

Corollary 8.80 (Norm convergence gives an a.e. convergent subsequence). If $f_{n} \to f$ in $L^{p} (μ)$ , $1 \leq p < \infty$ , then some subsequence converges to $f$ $μ$ -a.e.

Proof. $(f_{n})$ is Cauchy in norm; Step 1 to Step 3 of Theorem Riesz–Fischer: $L^{p}$ is complete produce a subsequence converging a.e. to some $h \in L^{p}$ and in norm to $h$ . Norm limits are unique, so $h = f$ in $L^{p}$ , i.e. $h = f$ a.e.∎

Example 8.81 (Norm convergence without pointwise convergence). On $[0, 1]$ let $f_{n}$ run through the indicators of the dyadic intervals $[j 2^{- k}, (j + 1) 2^{- k}]$ , $0 \leq j < 2^{k}$ , taken in order $k = 0, 1, 2, \dots$ — the "typewriter sequence". Show $f_{n} \to 0$ in $L^{1}$ but $f_{n} (x) \neq \to 0$ for any $x$ .

Solution.

The $n$ -th function is the indicator of an interval of length $2^{- k}$ where $k \to \infty$ , so $∥ f_{n} ∥_{1} = 2^{- k} \to 0$ .
Fix $x \in [0, 1]$ . For each $k$ exactly one of the $2^{k}$ intervals of that generation contains $x$ , so $f_{n} (x) = 1$ for infinitely many $n$ and $f_{n} (x) = 0$ for infinitely many $n$ .
Hence $f_{n} (x)$ diverges at every $x$ : $L^{1}$ convergence does not imply a.e. convergence.
Corollary Norm convergence gives an a.e. convergent subsequence is not violated. Take the subsequence $f_{n_{k}} = χ_{[0, 2^{- k}]}$ , the first function of each generation: it tends to $0$ at every $x \neq = 0$ . A subsequence converges a.e.; the full sequence need not.

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Corollary 8.82 ( $L^{2}$ is a Hilbert space). $L^{2} (μ)$ with $⟨ f, g ⟩ = \int_{X} f \overset{g}{ˉ} d μ$ is a Hilbert space, and $∥ f ∥_{2}^{2} = ⟨ f, f ⟩$ .

Proof. The integral defining $⟨ \cdot, \cdot ⟩$ is finite by Corollary Cauchy–Schwarz, and it is sesquilinear and positive definite modulo the a.e. identification. That it induces $∥ \cdot ∥_{2}$ is immediate. Completeness is Theorem Riesz–Fischer: $L^{p}$ is complete at $p = 2$ .∎

Remark. $p = 2$ is the only exponent for which $∥ \cdot ∥_{p}$ comes from an inner product. The test is the parallelogram law $∥ f + g ∥^{2} + ∥ f - g ∥^{2} = 2∥ f ∥^{2} + 2∥ g ∥^{2}$ ; taking $f, g$ indicators of disjoint sets of measure $1$ gives $2 \cdot 2^{2/ p}$ on the left and $4$ on the right, which agree only when $p = 2$ . Everything specific to Fourier series in the next part is a consequence of this one accident.

Exercises for §8.18 →

8.19Duality and Reflexivity

Hölder's inequality says that each $g \in L^{q}$ defines a bounded linear functional on $L^{p}$ . The duality theorem says these are all of them.

Definition 8.83 (The dual pairing). For conjugate $p, q$ and $g \in L^{q} (μ)$ , define $Λ_{g} : L^{p} (μ) \to R$ by

Λ_{g} (f) = \int_{X} f g d μ .

Proposition 8.84 (The pairing is isometric). For $1 \leq p < \infty$ with conjugate $q$ and $g \in L^{q}$ , the functional $Λ_{g}$ is bounded with operator norm exactly $∥ g ∥_{q}$ .

Proof. $∣ Λ_{g} (f) ∣ \leq ∥ f ∥_{p} ∥ g ∥_{q}$ is Theorem Hölder's inequality, so $∥ Λ_{g} ∥ \leq ∥ g ∥_{q}$ .

For the reverse, assume $0 < ∥ g ∥_{q} < \infty$ and $1 < q < \infty$ , and take the extremal $f = sign (g) ∣ g ∣^{q - 1}$ . Then $∣ f ∣^{p} = ∣ g ∣^{(q - 1) p} = ∣ g ∣^{q}$ , so $∥ f ∥_{p} = ∥ g ∥_{q}^{q / p}$ , while

Λ_{g} (f) = \int ∣ g ∣^{q} = ∥ g ∥_{q}^{q} .

Hence $∥ Λ_{g} ∥ \geq ∥ g ∥_{q}^{q} /∥ g ∥_{q}^{q / p} = ∥ g ∥_{q}^{q - q / p} = ∥ g ∥_{q}$ , using $q - q / p = q (1 - 1/ p) = q / q = 1$ . The cases $q = 1$ and $q = \infty$ are handled by testing on indicators of sets where $∣ g ∣$ is nearly maximal.∎

Theorem 8.85 (Duality of $L^{p}$ ). Let $μ$ be $σ$ -finite and $1 \leq p < \infty$ with conjugate $q$ . Then $g \mapsto Λ_{g}$ is an isometric isomorphism $L^{q} (μ) \to (L^{p} (μ))^{*}$ . In particular $(L^{p})^{*} ≅ L^{q}$ and, for $1 < p < \infty$ , $L^{p}$ is reflexive.

Proof. Sketch; the Radon–Nikodym theorem is the substantial input. Injectivity and isometry are Proposition The pairing is isometric. For surjectivity, let $Λ \in (L^{p})^{*}$ and first suppose $μ (X) < \infty$ . Define a set function $ν (E) = Λ (χ_{E})$ . Countable additivity of $ν$ follows from continuity of $Λ$ and $∥ χ_{⋃_{n \leq N} E_{n}} - χ_{⋃_{n} E_{n}} ∥_{p} \to 0$ . If $μ (E) = 0$ then $χ_{E} = 0$ in $L^{p}$ , so $ν (E) = 0$ : thus $ν ≪ μ$ , and Theorem Radon–Nikodym gives $g$ with $ν (E) = \int_{E} g d μ$ . Linearity extends $Λ (f) = \int f g$ from indicators to simple functions and then, by density of simple functions in $L^{p}$ and continuity, to all of $L^{p}$ . A truncation argument shows $∥ g ∥_{q} = ∥Λ∥ < \infty$ . The $σ$ -finite case follows by exhausting $X$ with sets of finite measure.

Reflexivity for $1 < p < \infty$ : applying the isomorphism twice, $(L^{p})^{**} ≅ (L^{q})^{*} ≅ L^{p}$ , and one checks the composite is the canonical embedding.∎

Remark. The theorem fails at $p = \infty$ : $(L^{\infty})^{*}$ is strictly larger than $L^{1}$ , containing functionals built from finitely additive measures (Banach limits), and $L^{1}$ and $L^{\infty}$ are therefore not reflexive. The hypothesis $p < \infty$ in Theorem Duality of $L^{p}$ is essential, not technical.

Example 8.86 (Computing an operator norm by duality). Let $Λ (f) = \int_{0}^{1} f (x) x d x$ on $L^{3} (0, 1)$ . Find $∥Λ∥$ .

Solution.

The conjugate of $p = 3$ is $q = 3/2$ , since $1/3 + 2/3 = 1$ .
By Proposition The pairing is isometric, $∥Λ∥ = ∥ g ∥_{3/2}$ with $g (x) = x$ .
$∥ g ∥_{3/2}^{3/2} = \int_{0}^{1} x^{3/2} d x = \frac{2}{5}$ .
So $∥Λ∥ = (2/5)^{2/3} \approx 0.5429$ .
Sanity check: the crude bound $∣Λ (f) ∣ \leq ∥ f ∥_{3} \cdot ∥ x ∥_{3/2}$ is exactly what we computed, and it is attained at $f = x^{1/2}$ up to scaling, since the extremal is $f = ∣ g ∣^{q - 1} = x^{1/2}$ .

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Exercises for §8.19 →

8.20Relations between $L^{p}$ Spaces

On a general measure space no $L^{p}$ contains another. Two hypotheses restore order in opposite directions: finite measure, and a measure bounded below on its atoms.

Theorem 8.87 (Inclusions on a finite measure space). If $μ (X) < \infty$ and $1 \leq r < s \leq \infty$ , then $L^{s} (μ) \subseteq L^{r} (μ)$ and

∥ f ∥_{r} \leq μ (X)^{\frac{1}{r} - \frac{1}{s}} ∥ f ∥_{s} .

Proof. For $s = \infty$ the claim is $∥ f ∥_{r} \leq μ (X)^{1/ r} ∥ f ∥_{\infty}$ , immediate by integrating $∣ f ∣^{r} \leq ∥ f ∥_{\infty}^{r}$ .

For $s < \infty$ apply Theorem Hölder's inequality to the product $∣ f ∣^{r} \cdot 1$ with exponents $s / r > 1$ and its conjugate $t = \frac{s / r}{s / r - 1} = \frac{s}{s - r}$ :

\int_{X} ∣ f ∣^{r} d μ \leq (\int_{X} ∣ f ∣^{r \cdot s / r})^{r / s} (\int_{X} 1^{t})^{1/ t} = ∥ f ∥_{s}^{r} μ (X)^{\frac{s - r}{s}} .

Take $r$ -th roots: $∥ f ∥_{r} \leq ∥ f ∥_{s} μ (X)^{\frac{s - r}{rs}} = μ (X)^{\frac{1}{r} - \frac{1}{s}} ∥ f ∥_{s}$ .∎

Proposition 8.88 (Reverse inclusions for counting measure). For $ℓ^{p} = L^{p} (N, counting measure)$ and $1 \leq r < s \leq \infty$ , we have $ℓ^{r} \subseteq ℓ^{s}$ with $∥ x ∥_{s} \leq ∥ x ∥_{r}$ — the inclusion runs the other way.

Proof. Normalise $∥ x ∥_{r} = 1$ , so $∣ x_{n} ∣ \leq 1$ for every $n$ . Then $∣ x_{n} ∣^{s} \leq ∣ x_{n} ∣^{r}$ because $s > r$ and the base is at most $1$ , hence $\sum ∣ x_{n} ∣^{s} \leq \sum ∣ x_{n} ∣^{r} = 1$ and $∥ x ∥_{s} \leq 1 = ∥ x ∥_{r}$ . Homogeneity extends this to all $x$ .∎

Theorem 8.89 (Interpolation of $L^{p}$ norms). Let $1 \leq r < p < s \leq \infty$ and write $\frac{1}{p} = \frac{θ}{r} + \frac{1 - θ}{s}$ for a unique $θ \in (0, 1)$ . Then

∥ f ∥_{p} \leq ∥ f ∥_{r}^{θ} ∥ f ∥_{s}^{1 - θ},

so $L^{r} \cap L^{s} \subseteq L^{p}$ for every intermediate $p$ .

Proof. Take $s < \infty$ ; split $∣ f ∣^{p} = ∣ f ∣^{θp} \cdot ∣ f ∣^{(1 - θ) p}$ and apply Theorem Hölder's inequality with exponents $a = \frac{r}{θp}$ and $b = \frac{s}{( 1 - θ ) p}$ . These are conjugate because

\frac{1}{a} + \frac{1}{b} = \frac{θp}{r} + \frac{( 1 - θ ) p}{s} = p (\frac{θ}{r} + \frac{1 - θ}{s}) = p \cdot \frac{1}{p} = 1.

Hölder then gives $\int ∣ f ∣^{p} \leq (\int ∣ f ∣^{r})^{θp / r} (\int ∣ f ∣^{s})^{(1 - θ) p / s} = ∥ f ∥_{r}^{θp} ∥ f ∥_{s}^{(1 - θ) p}$ ; take $p$ -th roots.∎

Example 8.90 (No inclusion on the whole line). Exhibit $f \in L^{1} (R) ∖ L^{2} (R)$ and $g \in L^{2} (R) ∖ L^{1} (R)$ .

Solution.

Take $f (x) = x^{- 1/2} χ_{(0, 1)}$ . By Example Which $L^{p}$ does $x^{- a}$ belong to?, $∥ f ∥_{1} = 2$ and $f \in / L^{2}$ .
Take $g (x) = x^{- 1} χ_{(1, \infty)}$ . By Example The same question at infinity, $∥ g ∥_{2} = 1$ and $g \in / L^{1}$ .
Neither space contains the other, so Theorem Inclusions on a finite measure space really does need $μ (X) < \infty$ : on $R$ the measure is infinite and both failures are available at once.
Their sum $f + g$ lies in neither $L^{1}$ nor $L^{2}$ . It does lie in $L^{p}$ for $1 < p < 2$ and only there: the singular part at $0$ requires $p < 2$ and the tail at infinity requires $p > 1$ .

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Exercises for §8.20 →

8.21Applications of $L^{p}$ Spaces

Three settings where the $L^{p}$ framework does real work, each stated with the result that makes it work.

Theorem 8.91 (Density of nice functions). For $1 \leq p < \infty$ , the continuous compactly supported functions $C_{c} (R^{n})$ are dense in $L^{p} (R^{n})$ . The same fails for $p = \infty$ .

Proof. Sketch. Simple functions are dense in $L^{p}$ (approximate $f^{+}$ and $f^{-}$ from below by simple functions and dominate). A simple function is a finite combination of indicators of finite-measure sets, so it suffices to approximate $χ_{E}$ with $∣ E ∣ < \infty$ . By regularity of Lebesgue measure there are $K$ compact and $U$ open with $K \subseteq E \subseteq U$ and $∣ U ∖ K ∣ < ε^{p}$ ; Urysohn's lemma supplies continuous $φ$ with $χ_{K} \leq φ \leq χ_{U}$ , and $∥ χ_{E} - φ ∥_{p}^{p} \leq ∣ U ∖ K ∣ < ε^{p}$ .

For $p = \infty$ the closure of $C_{c}$ is the space of continuous functions vanishing at infinity, which does not contain $χ_{[0, 1]}$ : the uniform distance from any continuous function to an indicator is at least $\frac{1}{2}$ .∎

Theorem 8.92 (Convolution and Young's convolution inequality). For $f \in L^{1} (R^{n})$ and $g \in L^{p} (R^{n})$ , $1 \leq p \leq \infty$ , the convolution $(f * g) (x) = \int f (y) g (x - y) d y$ is defined for a.e. $x$ and

∥ f * g ∥_{p} \leq ∥ f ∥_{1} ∥ g ∥_{p} .

Proof. For $p = 1$ this is Tonelli's theorem: $\int \int ∣ f (y) g (x - y) ∣ d y d x = ∥ f ∥_{1} ∥ g ∥_{1}$ after the substitution $x \mapsto x + y$ in the inner integral. For $1 < p < \infty$ , write $∣ f (y) g (x - y) ∣ = (∣ f (y) ∣^{1/ q}) (∣ f (y) ∣^{1/ p} ∣ g (x - y) ∣)$ and apply Theorem Hölder's inequality in $y$ with exponents $q, p$ :

∣ (f * g) (x) ∣ \leq ∥ f ∥_{1}^{1/ q} (\int ∣ f (y) ∣∣ g (x - y) ∣^{p} d y)^{1/ p} .

Raise to the $p$ -th power, integrate in $x$ , and use Tonelli again on the inner double integral to get $∥ f * g ∥_{p}^{p} \leq ∥ f ∥_{1}^{p / q} \cdot ∥ f ∥_{1} ∥ g ∥_{p}^{p} = ∥ f ∥_{1}^{p} ∥ g ∥_{p}^{p}$ . The $p = \infty$ case is Hölder with $(1, \infty)$ .∎

Remark. The three applications worth naming. Probability: $L^{2} (Ω, P)$ is the space of finite-variance random variables, the inner product is covariance about the mean, and conditional expectation is orthogonal projection onto a sub- $σ$ -algebra — so the whole theory of least-squares prediction is Hilbert-space geometry. Partial differential equations: Sobolev spaces $W^{k, p}$ are built from $L^{p}$ by requiring weak derivatives up to order $k$ to lie in $L^{p}$ , and existence proofs work by finding a weak solution in such a space and then upgrading its regularity. Fourier analysis: Theorem Convolution and Young's convolution inequality with an approximate identity is how one proves that smoothing converges, and it is the engine behind Fejér's theorem in the next part.

Example 8.93 (An approximate identity). Let $φ \geq 0$ be continuous with $\int_{R} φ = 1$ and set $φ_{t} (x) = t^{- 1} φ (x / t)$ . Show $∥ φ_{t} ∥_{1} = 1$ for every $t > 0$ and bound $∥ φ_{t} * g ∥_{2}$ for $g \in L^{2}$ .

Solution.

Substituting $u = x / t$ , $d x = t d u$ : $\int φ_{t} = \int t^{- 1} φ (u) t d u = \int φ = 1$ . Since $φ \geq 0$ , $∥ φ_{t} ∥_{1} = 1$ .
By Theorem Convolution and Young's convolution inequality with $p = 2$ , $∥ φ_{t} * g ∥_{2} \leq ∥ φ_{t} ∥_{1} ∥ g ∥_{2} = ∥ g ∥_{2}$ .
So smoothing never increases the $L^{2}$ norm, uniformly in $t$ — exactly the uniform bound needed to run the density argument of Theorem Lebesgue differentiation theorem again, this time proving $φ_{t} * g \to g$ in $L^{2}$ .

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Exercises for §8.21 →

8.22Trigonometric Polynomials and Orthogonality

Fourier series are the concrete realisation of the $L^{2}$ theory: an orthonormal basis for $L^{2}$ of the circle, written down explicitly. Everything begins with one computation.

Definition 8.94 (Fourier coefficients and partial sums). For $f \in L^{1} (- π, π)$ and $n \in Z$ , the Fourier coefficients and partial sums are

\hat{f} (n) = \frac{1}{2 π} \int_{- π}^{π} f (x) e^{- in x} d x, S_{N} f (x) = ∣ n ∣ \leq N \sum \hat{f} (n) e^{in x} .

In real form, $a_{n} = \frac{1}{π} \int_{- π}^{π} f cos n x d x$ , $b_{n} = \frac{1}{π} \int_{- π}^{π} f sin n x d x$ , and $S_{N} f = \frac{a _{0}}{2} + \sum_{n = 1}^{N} (a_{n} cos n x + b_{n} sin n x)$ .

Theorem 8.95 (Orthogonality of the trigonometric system). With $e_{n} (x) = e^{in x}$ and the inner product $⟨ f, g ⟩ = \frac{1}{2 π} \int_{- π}^{π} f \overset{g}{ˉ}$ , the system ${e_{n}}_{n \in Z}$ is orthonormal:

⟨ e_{n}, e_{m} ⟩ = δ_{nm} .

Equivalently, for $n, m \geq 1$ : $\int_{- π}^{π} cos n x cos m x d x = π δ_{nm}$ , $\int_{- π}^{π} sin n x sin m x d x = π δ_{nm}$ , and $\int_{- π}^{π} sin n x cos m x d x = 0$ for all $n, m \geq 0$ .

Proof. If $n = m$ then $e_{n} \overset{e}{ˉ}_{m} = 1$ and $\frac{1}{2 π} \int_{- π}^{π} 1 = 1$ .

If $n \neq = m$ , put $k = n - m \neq = 0$ . Then

⟨ e_{n}, e_{m} ⟩ = \frac{1}{2 π} \int_{- π}^{π} e^{ik x} d x = \frac{1}{2 π} [\frac{e ^{ik x}}{ik}]_{- π}^{π} = \frac{e ^{ikπ} - e ^{- ikπ}}{2 πik} = \frac{2 i sin ( kπ )}{2 πik} = 0,

since $sin (kπ) = 0$ for integer $k$ .

For the real form, use the product-to-sum identities: $cos n x cos m x = \frac{1}{2} (cos (n - m) x + cos (n + m) x)$ . Each cosine with non-zero frequency integrates to $0$ over a full period, leaving $\frac{1}{2} \int_{- π}^{π} 1 = π$ exactly when $n = m$ . Similarly $sin n x sin m x = \frac{1}{2} (cos (n - m) x - cos (n + m) x)$ and $sin n x cos m x = \frac{1}{2} (sin (n + m) x + sin (n - m) x)$ , the latter an odd function, whose integral over a symmetric interval is $0$ .∎

Theorem 8.96 (Best approximation in $L^{2}$ ). For $f \in L^{2} (- π, π)$ , among all trigonometric polynomials $T = \sum_{∣ n ∣ \leq N} c_{n} e_{n}$ of degree $\leq N$ , the one minimising $∥ f - T ∥_{2}$ is $T = S_{N} f$ , i.e. $c_{n} = \hat{f} (n)$ .

Proof. Expand, using orthonormality (Theorem Orthogonality of the trigonometric system):

∥ f - T ∥^{2} = ∥ f ∥^{2} - ∣ n ∣ \leq N \sum (c_{n} \overline{\hat{f} (n)} + \overset{c}{ˉ}_{n} \hat{f} (n)) + ∣ n ∣ \leq N \sum ∣ c_{n} ∣^{2} = ∥ f ∥^{2} - ∣ n ∣ \leq N \sum ∣ \hat{f} (n) ∣^{2} + ∣ n ∣ \leq N \sum c_{n} - \hat{f} (n)^{2} .

Only the last sum depends on the $c_{n}$ , and it is minimised, at value $0$ , exactly when $c_{n} = \hat{f} (n)$ .∎

Corollary 8.97 (Bessel's inequality). For $f \in L^{2} (- π, π)$ , $n \in Z \sum \hat{f} (n)^{2} \leq ∥ f ∥^{2} = \frac{1}{2 π} \int_{- π}^{π} ∣ f ∣^{2}$ . In particular $\hat{f} (n) \to 0$ as $∣ n ∣ \to \infty$ .

Proof. Put $c_{n} = \hat{f} (n)$ in the expansion of the proof above: $0 \leq ∥ f - S_{N} f ∥^{2} = ∥ f ∥^{2} - \sum_{∣ n ∣ \leq N} ∣ \hat{f} (n) ∣^{2}$ , so the partial sums of $\sum ∣ \hat{f} (n) ∣^{2}$ are bounded by $∥ f ∥^{2}$ for every $N$ . A non-decreasing bounded sequence converges, by the monotone convergence theorem for sequences in Sequences & Series, and the limit is at most $∥ f ∥^{2}$ . Convergence of the series forces its terms to $0$ .∎

Example 8.98 (Fourier coefficients of a square wave). Let $f (x) = - 1$ on $(- π, 0)$ and $f (x) = 1$ on $(0, π)$ . Compute $a_{n}$ and $b_{n}$ .

Solution.

$f$ is odd, and $cos n x$ is even, so $f cos n x$ is odd and $a_{n} = 0$ for all $n$ .
For $b_{n}$ , use oddness to fold the integral: $b_{n} = \frac{2}{π} \int_{0}^{π} sin n x d x = \frac{2}{π} \cdot \frac{1 - c o s nπ}{n}$ .
$cos nπ = (- 1)^{n}$ , so $b_{n} = \frac{2 ( 1 - ( - 1 ) ^{n} )}{πn}$ , which is $0$ for even $n$ and $\frac{4}{πn}$ for odd $n$ .
Hence

f (x) \sim \frac{4}{π} (sin x + \frac{sin 3 x}{3} + \frac{sin 5 x}{5} + \dots) .

Sanity check via Bessel: $\sum_{n} ∣ \hat{f} (n) ∣^{2} \leq ∥ f ∥^{2} = 1$ , and in real form $\frac{1}{2} \sum b_{n}^{2} = \frac{1}{2} \cdot \frac{16}{π ^{2}} \sum_{k odd} k^{- 2} = \frac{1}{2} \cdot \frac{16}{π ^{2}} \cdot \frac{π ^{2}}{8} = 1$ . Equality — as Parseval's identity will explain.

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Exercises for §8.22 →

8.23Pointwise Convergence: Dirichlet and Jordan

Bessel's inequality controls the coefficients but says nothing about whether $S_{N} f (x) \to f (x)$ at a given $x$ . The classical answer is local: convergence at $x$ depends only on $f$ near $x$ , and bounded variation near $x$ suffices.

Definition 8.99 (Dirichlet kernel).

D_{N} (t) = ∣ n ∣ \leq N \sum e^{in t} = \frac{sin ( ( N + \frac{1}{2} ) t )}{sin ( t /2 )}, S_{N} f (x) = \frac{1}{2 π} \int_{- π}^{π} f (x - t) D_{N} (t) d t .

Proposition 8.100 (Properties of the Dirichlet kernel). $\frac{1}{2 π} \int_{- π}^{π} D_{N} = 1$ for every $N$ , $D_{N}$ is even, and $\int_{- π}^{π} ∣ D_{N} (t) ∣ d t \sim \frac{4}{π} lo g N \to \infty$ .

Proof. The integral of $D_{N}$ is the sum of the integrals of $e^{\int}$ , all zero except $n = 0$ which gives $2 π$ ; divide by $2 π$ . Evenness is clear from the sum, since the $\pm n$ terms combine to $2 cos n t$ . For the closed form, multiply $\sum_{∣ n ∣ \leq N} e^{\int}$ by $e^{i t /2} - e^{- i t /2} = 2 i sin (t /2)$ and telescope. The growth of $\int ∣ D_{N} ∣$ follows from $∣ sin (t /2) ∣ \leq ∣ t ∣/2$ giving $∣ D_{N} (t) ∣ \geq \frac{2∣ s i n (( N + 1/2 ) t ) ∣}{∣ t ∣}$ and summing the resulting harmonic-type series over the humps $∣ t ∣ \in [\frac{kπ}{N}, \frac{( k + 1 ) π}{N}]$ .∎

The divergence of $\int ∣ D_{N} ∣$ — the Lebesgue constants — is the source of every pathology in this part. A kernel whose $L^{1}$ norm blows up cannot converge for all continuous functions.

Theorem 8.101 (Riemann–Lebesgue lemma). If $g \in L^{1} (- π, π)$ then $\int_{- π}^{π} g (t) e^{- iλ t} d t \to 0$ as $∣ λ ∣ \to \infty$ .

Proof. For $g = χ_{(a, b)}$ the integral is $(e^{- iλa} - e^{- iλb}) / (iλ)$ , of modulus at most $2/∣ λ ∣ \to 0$ . By linearity the result holds for step functions, which are dense in $L^{1}$ by Theorem Density of nice functions. Given $g \in L^{1}$ and $ε > 0$ , pick a step function $s$ with $∥ g - s ∥_{1} < ε /2$ ; then

\int g e^{- iλ t} \leq ∥ g - s ∥_{1} + \int s e^{- iλ t} < \frac{ε}{2} + \frac{ε}{2}

for $∣ λ ∣$ large.∎

Theorem 8.102 (Dini's criterion and the localisation principle). Let $f \in L^{1} (- π, π)$ and $x$ be such that for some $δ > 0$

\int_{0}^{δ} \frac{∣ f ( x + t ) + f ( x - t ) - 2 L ∣}{t} d t < \infty.

Then $S_{N} f (x) \to L$ . Consequently, if two integrable functions agree on a neighbourhood of $x$ , their Fourier series converge or diverge together at $x$ .

Proof. Using $\frac{1}{2 π} \int D_{N} = 1$ and evenness,

S_{N} f (x) - L = \frac{1}{2 π} \int_{0}^{π} (f (x + t) + f (x - t) - 2 L) D_{N} (t) d t .

Write $g (t) = \frac{f ( x + t ) + f ( x - t ) - 2 L}{s i n ( t /2 )}$ . Since $sin (t /2) \sim t /2$ near $0$ , the Dini hypothesis makes $g$ integrable on $(0, δ)$ ; on $(δ, π)$ the denominator is bounded below, so $g \in L^{1} (0, π)$ . The displayed integral is $\frac{1}{2 π} \int_{0}^{π} g (t) sin ((N + \frac{1}{2}) t) d t$ , which tends to $0$ by Theorem Riemann–Lebesgue lemma.

Localisation: if $f_{1} = f_{2}$ near $x$ then $f_{1} - f_{2}$ satisfies Dini's condition at $x$ with $L = 0$ .∎

Theorem 8.103 (Dirichlet–Jordan). If $f \in L^{1} (- π, π)$ is of bounded variation on a neighbourhood of $x$ , then

S_{N} f (x) ⟶ \frac{f ( x ^{+} ) + f ( x ^{-} )}{2} .

If $f$ is of bounded variation and continuous on $[- π, π]$ with $f (- π) = f (π)$ , the convergence is uniform.

Proof. Sketch; the second mean value theorem for integrals is assumed. By Theorem Jordan decomposition it suffices to treat $f$ increasing near $x$ , and by localisation (Theorem Dini's criterion and the localisation principle) only the behaviour on $(x - δ, x + δ)$ matters. Write $ψ (t) = f (x + t) + f (x - t) - f (x^{+}) - f (x^{-})$ , which is of bounded variation on $[0, δ]$ with $ψ (0^{+}) = 0$ . Applying the second mean value theorem to $\int_{0}^{δ} ψ (t) \frac{s i n (( N + 1/2 ) t )}{t} d t$ splits it into terms controlled by $sup_{0 < t < η} ∣ ψ (t) ∣$ times the uniformly bounded sine integrals $\int_{0}^{u} \frac{s i n s}{s} d s$ . Choosing $η$ small makes the first factor small, and the remaining range is handled by Riemann–Lebesgue.∎

Example 8.104 (Evaluating a series by pointwise convergence). Use the square wave of Example Fourier coefficients of a square wave at $x = π /2$ to evaluate $1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots$ .

Solution.

$f$ is of bounded variation and continuous at $π /2$ , with $f (π /2) = 1$ , so Theorem Dirichlet–Jordan gives $S_{N} f (π /2) \to 1$ .
$sin (kπ /2)$ is $0$ for even $k$ and $(- 1)^{(k - 1) /2}$ for odd $k$ .
Hence $1 = \frac{4}{π} (1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots)$ .
Therefore $1 - \frac{1}{3} + \frac{1}{5} - \dots = π /4 \approx 0.7854$ : the Leibniz series, obtained here as a corollary of Fourier convergence.

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Exercises for §8.23 →

8.24 $L^{2}$ Convergence and Parseval's Identity

Pointwise convergence is delicate. Mean-square convergence is not: it always holds, for every $f \in L^{2}$ , with no regularity hypothesis whatever. This is the pay-off of completeness.

Theorem 8.105 (Completeness of the trigonometric system). Trigonometric polynomials are dense in $L^{2} (- π, π)$ . Equivalently, if $\hat{f} (n) = 0$ for all $n$ then $f = 0$ a.e.

Proof. Let $f \in L^{2}$ and $ε > 0$ . By Theorem Density of nice functions choose $g$ continuous and $2 π$ -periodic with $∥ f - g ∥_{2} < ε /2$ . By Theorem Fejér below, the Cesàro means $σ_{N} g$ converge to $g$ uniformly, hence in $L^{2}$ on a finite interval; choose $N$ with $∥ g - σ_{N} g ∥_{2} < ε /2$ . Since $σ_{N} g$ is a trigonometric polynomial, $∥ f - σ_{N} g ∥_{2} < ε$ .

For the equivalent form: if all coefficients vanish then $f ⊥ T$ for every trigonometric polynomial, hence $f ⊥ L^{2}$ by density, hence $f ⊥ f$ , so $∥ f ∥_{2} = 0$ .∎

Theorem 8.106 (Mean-square convergence and Parseval's identity). For every $f \in L^{2} (- π, π)$ , $∥ f - S_{N} f ∥_{2} \to 0$ , and

n \in Z \sum \hat{f} (n)^{2} = \frac{1}{2 π} \int_{- π}^{π} ∣ f (x) ∣^{2} d x .

In real form, $\frac{a _{0}^{2}}{2} + \sum_{n \geq 1} (a_{n}^{2} + b_{n}^{2}) = \frac{1}{π} \int_{- π}^{π} ∣ f ∣^{2}$ . More generally $\sum_{n} \hat{f} (n) \overline{\overset{g}{^} (n)} = ⟨ f, g ⟩$ .

Proof. By Theorem Best approximation in $L^{2}$ , $∥ f - S_{N} f ∥_{2} \leq ∥ f - T ∥_{2}$ for every trigonometric polynomial $T$ of degree $\leq N$ . Given $ε > 0$ , Theorem Completeness of the trigonometric system supplies a trigonometric polynomial $T$ , of some degree $N_{0}$ , with $∥ f - T ∥_{2} < ε$ ; then $∥ f - S_{N} f ∥_{2} < ε$ for all $N \geq N_{0}$ . So $S_{N} f \to f$ in $L^{2}$ .

For the identity, the expansion in the proof of Theorem Best approximation in $L^{2}$ at $c_{n} = \hat{f} (n)$ reads

∥ f - S_{N} f ∥^{2} = ∥ f ∥^{2} - ∣ n ∣ \leq N \sum ∣ \hat{f} (n) ∣^{2} .

Letting $N \to \infty$ , the left side tends to $0$ , so the sum tends to $∥ f ∥^{2}$ . That is Parseval. The polarised version follows by applying it to $f + g$ and $f - g$ and subtracting.∎

Example 8.107 (Parseval and the Basel problem). Apply Parseval to $f (x) = x$ on $(- π, π)$ to evaluate $\sum_{n \geq 1} 1/ n^{2}$ .

Solution.

$f$ is odd so $a_{n} = 0$ , and $b_{n} = \frac{2}{π} \int_{0}^{π} x sin n x d x$ . Integrating by parts, $\int_{0}^{π} x sin n x d x = [- \frac{x c o s n x}{n}]_{0}^{π} + \frac{1}{n} \int_{0}^{π} cos n x d x = - \frac{π ( - 1 ) ^{n}}{n}$ .
So $b_{n} = \frac{2 ( - 1 ) ^{n + 1}}{n}$ .
Parseval in real form: $\sum_{n \geq 1} b_{n}^{2} = \frac{1}{π} \int_{- π}^{π} x^{2} d x = \frac{1}{π} \cdot \frac{2 π ^{3}}{3} = \frac{2 π ^{2}}{3}$ .
The left side is $\sum_{n \geq 1} \frac{4}{n ^{2}}$ , so $\sum_{n \geq 1} \frac{1}{n ^{2}} = \frac{1}{4} \cdot \frac{2 π ^{2}}{3} = \frac{π ^{2}}{6} \approx 1.6449$ .
Sanity check: the first four terms already give $1 + 0.25 + 0.111 + 0.0625 = 1.4236$ , climbing toward $1.6449$ . $✓$

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Example 8.108 (Parseval for the square wave). Apply Parseval to the square wave to evaluate $\sum_{k \geq 0} (2 k + 1)^{- 2}$ .

Solution.

From Example Fourier coefficients of a square wave, $a_{n} = 0$ and $b_{2 k + 1} = \frac{4}{π ( 2 k + 1 )}$ .
Parseval: $\sum_{n} b_{n}^{2} = \frac{1}{π} \int_{- π}^{π} 1 d x = 2$ .
So $\frac{16}{π ^{2}} \sum_{k \geq 0} \frac{1}{( 2 k + 1 ) ^{2}} = 2$ , giving $\sum_{k \geq 0} \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π ^{2}}{8} \approx 1.2337$ .
Consistency with the previous example: $\sum 1/ n^{2} = \sum_{odd} + \sum_{even} = \frac{π ^{2}}{8} + \frac{1}{4} \sum 1/ n^{2}$ , so $\frac{3}{4} \sum 1/ n^{2} = \frac{π ^{2}}{8}$ and $\sum 1/ n^{2} = \frac{π ^{2}}{6}$ . $✓$

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Exercises for §8.24 →

8.25Cesàro Means, Fejér's Theorem, and Uniform Convergence

The Dirichlet kernel is not positive and its $L^{1}$ norms diverge, which is why $S_{N}$ misbehaves. Averaging the partial sums repairs both defects at once.

Definition 8.109 (Fejér means and kernel).

σ_{N} f = \frac{1}{N + 1} k = 0 \sum N S_{k} f, K_{N} (t) = \frac{1}{N + 1} k = 0 \sum N D_{k} (t) = \frac{1}{N + 1} (\frac{sin ( \frac{( N + 1 ) t}{2} )}{sin ( t /2 )})^{2} .

Lemma 8.110 (The Fejér kernel is an approximate identity). $K_{N} \geq 0$ ; $\frac{1}{2 π} \int_{- π}^{π} K_{N} = 1$ for every $N$ ; and for each $δ > 0$ , $\int_{δ \leq ∣ t ∣ \leq π} K_{N} (t) d t \to 0$ as $N \to \infty$ .

Proof. Non-negativity is immediate from the closed form, a square divided by $N + 1 > 0$ . The closed form itself follows by summing $\sum_{k = 0}^{N} sin ((k + \frac{1}{2}) t)$ , which telescopes after multiplying by $2 sin (t /2)$ to give $1 - cos ((N + 1) t) = 2 sin^{2} (\frac{( N + 1 ) t}{2})$ . The mass is $1$ because each $D_{k}$ has mass $1$ (Proposition Properties of the Dirichlet kernel) and $K_{N}$ is their average.

For concentration, on $δ \leq ∣ t ∣ \leq π$ we have $sin^{2} (t /2) \geq sin^{2} (δ /2) > 0$ , and the numerator is at most $1$ , so $K_{N} (t) \leq \frac{1}{( N + 1 ) s i n ^{2} ( δ /2 )}$ there; integrating over a set of measure at most $2 π$ gives a bound $O (1/ N) \to 0$ .∎

Theorem 8.111 (Fejér). If $f$ is continuous and $2 π$ -periodic, then $σ_{N} f \to f$ uniformly on $R$ . If $f \in L^{1}$ and both one-sided limits exist at $x$ , then $σ_{N} f (x) \to \frac{f ( x ^{+} ) + f ( x ^{-} )}{2}$ .

Proof. Since $\frac{1}{2 π} \int K_{N} = 1$ ,

σ_{N} f (x) - f (x) = \frac{1}{2 π} \int_{- π}^{π} (f (x - t) - f (x)) K_{N} (t) d t .

$f$ is continuous and periodic, hence uniformly continuous by Heine-Cantor applied on $[- 2 π, 2 π]$ , and bounded, say $∣ f ∣ \leq M$ . Given $ε > 0$ pick $δ > 0$ with $∣ f (x - t) - f (x) ∣ < ε /2$ whenever $∣ t ∣ < δ$ , uniformly in $x$ . Split the integral:

∣ σ_{N} f (x) - f (x) ∣ \leq \frac{1}{2 π} \int_{∣ t ∣ < δ} \frac{ε}{2} K_{N} (t) d t + \frac{1}{2 π} \int_{δ \leq ∣ t ∣ \leq π} 2 M K_{N} (t) d t .

The first term is at most $ε /2$ , using $K_{N} \geq 0$ and total mass $1$ . The second tends to $0$ as $N \to \infty$ by Lemma The Fejér kernel is an approximate identity, uniformly in $x$ because the bound does not involve $x$ ; choose $N_{0}$ making it $< ε /2$ . Then $∥ σ_{N} f - f ∥_{\infty} < ε$ for $N \geq N_{0}$ .

For the second statement, replace $f (x)$ by the average of the one-sided limits and use evenness of $K_{N}$ to pair $t$ with $- t$ ; the same split works.∎

Corollary 8.112 (Weierstrass approximation, trigonometric form). Trigonometric polynomials are dense in $(C (T), ∥ \cdot ∥_{\infty})$ , and a continuous periodic function is determined by its Fourier coefficients.

Proof. $σ_{N} f$ is a trigonometric polynomial — it is a finite combination of the $e_{n}$ , $∣ n ∣ \leq N$ — and it converges uniformly to $f$ by Theorem Fejér. If all coefficients of $f$ vanish then every $S_{k} f = 0$ , so every $σ_{N} f = 0$ , so $f = 0$ uniformly.∎

Intuition. $S_{N}$ chops the frequency spectrum off abruptly at $N$ ; a sharp cut in frequency produces ringing in space, which is the Dirichlet kernel's oscillating tail. $σ_{N}$ tapers the coefficients linearly to zero — $\hat{σ_{N} f} (n) = (1 - \frac{∣ n ∣}{N + 1}) \hat{f} (n)$ — and a smooth taper gives a positive, concentrated kernel with no ringing. Every practical signal-processing window is a version of this trade.

Exercises for §8.25 →

8.26Divergence, the Gibbs Phenomenon, and Carleson's Theorem

Fejér repairs convergence at the cost of changing the sums being taken. Left alone, $S_{N}$ genuinely misbehaves, in two distinct ways.

Theorem 8.113 (Existence of a continuous function with divergent Fourier series). There exists $f \in C (T)$ whose Fourier series diverges at $x = 0$ . The set of such $f$ is comeager in $C (T)$ .

Proof. Sketch, using the uniform boundedness principle. The maps $Λ_{N} (f) = S_{N} f (0)$ are bounded linear functionals on $C (T)$ with $∥ Λ_{N} ∥ = \frac{1}{2 π} \int_{- π}^{π} ∣ D_{N} ∣ \to \infty$ by Proposition Properties of the Dirichlet kernel — the norm is attained in the limit by testing against continuous approximations to $sign D_{N}$ . If $S_{N} f (0)$ converged for every $f$ , the sequence $(Λ_{N} (f))$ would be bounded for each $f$ , and the uniform boundedness principle (a consequence of the Baire category theorem of Metric Spaces) would force $sup_{N} ∥ Λ_{N} ∥ < \infty$ , a contradiction. Baire's theorem also gives the comeagerness.∎

Theorem 8.114 (Gibbs phenomenon). Let $f$ have a jump of size $J = f (x_{0}^{+}) - f (x_{0}^{-}) > 0$ at $x_{0}$ and be smooth on each side. Then

N \to \infty lim sup x sup S_{N} f (x) - f (x_{0}^{+}) = J (\frac{1}{π} \int_{0}^{π} \frac{sin s}{s} d s - \frac{1}{2}) \approx 0.0895 J,

an overshoot of about $8.95%$ of the jump that does not shrink as $N \to \infty$ ; it merely migrates toward $x_{0}$ .

Proof. Sketch, for the square wave; the general case follows by subtracting a smooth function and using localisation. For the square wave of Example Fourier coefficients of a square wave, differentiating $S_{N}$ term by term shows the first maximum to the right of $0$ occurs at $x_{N} = π / (N + 1)$ . Evaluating,

S_{N} f (x_{N}) = \frac{4}{π} k odd \leq N \sum \frac{sin ( kπ / ( N + 1 ))}{k} ⟶ \frac{2}{π} \int_{0}^{π} \frac{sin s}{s} d s,

the sum being a Riemann sum for that integral. Numerically $\int_{0}^{π} \frac{s i n s}{s} d s = Si (π) \approx 1.8519$ , so the limit is $\approx 1.1790$ against the true value $1$ : an overshoot of $17.9%$ of the half-jump, i.e. $8.95%$ of the full jump $J = 2$ .∎

Figure 8.115. Partial Fourier sums

S_{1}

S_{5}

and

S_{19}

of the square wave near the jump at

x = 0

. The overshoot does not decay: it stays at about

9%

of the jump and slides toward the discontinuity.

Theorem 8.116 (Carleson–Hunt). If $f \in L^{p} (T)$ with $1 < p < \infty$ — in particular if $f$ is continuous — then $S_{N} f (x) \to f (x)$ for almost every $x$ . For $p = 1$ the conclusion fails: Kolmogorov constructed $f \in L^{1}$ whose Fourier series diverges everywhere.

Proof. Proof sketch only; a complete proof is a research paper and is not attempted here. The statement is equivalent to a weak-type bound on the Carleson maximal operator $C f (x) = sup_{N} ∣ S_{N} f (x) ∣$ , namely $∥ C f ∥_{L^{p}} \leq C_{p} ∥ f ∥_{L^{p}}$ , by exactly the density-plus-maximal-inequality scheme used in Theorem Lebesgue differentiation theorem: trigonometric polynomials are dense and the statement is trivial for them. What is assumed and not proved is the maximal bound itself, whose proof requires a time-frequency decomposition of the operator into "tiles" and a delicate combinatorial selection argument. Kolmogorov's counterexample at $p = 1$ shows no such bound can hold there.∎

Pitfall. Three modes of convergence are in play and they are genuinely different. $S_{N} f \to f$ in $L^{2}$ for every $f \in L^{2}$ (Theorem Mean-square convergence and Parseval's identity); $S_{N} f \to f$ a.e. for $f \in L^{p}$ , $p > 1$ (Theorem Carleson–Hunt); and $S_{N} f \to f$ uniformly only under extra regularity, since a continuous function can have a divergent series. Gibbs shows that even where pointwise convergence holds, uniform convergence can fail near a jump — the sup-norm error stays bounded away from $0$ .

Exercises for §8.26 →

8.27Applications and Extensions

Definition 8.117 (Fourier transform on $R$ ). For $f \in L^{1} (R)$ ,

\hat{f} (ξ) = \int_{R} f (x) e^{- 2 πi x ξ} d x .

Theorem 8.118 (Plancherel). The Fourier transform, defined first on $L^{1} \cap L^{2}$ , extends uniquely to a unitary operator on $L^{2} (R)$ :

∥ \hat{f} ∥_{2} = ∥ f ∥_{2}, ⟨ \hat{f}, \overset{g}{^} ⟩ = ⟨ f, g ⟩ .

Proof. Sketch. On the Gaussian $g (x) = e^{- π x^{2}}$ one computes $\overset{g}{^} = g$ directly by completing the square and a contour shift. Using the multiplication formula $\int \hat{f} g = \int f \overset{g}{^}$ and dilates of the Gaussian as an approximate identity, one obtains $∥ \hat{f} ∥_{2} = ∥ f ∥_{2}$ for $f$ in the dense subspace of Schwartz functions. A densely defined isometry on a complete space extends uniquely to an isometry of the whole space — this is where Theorem Riesz–Fischer: $L^{p}$ is complete at $p = 2$ is used. Surjectivity follows from the inversion formula on the dense subspace.∎

Example 8.119 (Solving the heat equation by Fourier series). Solve $u_{t} = u_{xx}$ on $(0, π)$ with $u (0, t) = u (π, t) = 0$ and $u (x, 0) = sin 3 x$ . Evaluate $u (π /6, 1)$ to four decimal places.

Solution.

Separating variables, $u (x, t) = \sum_{n} c_{n} e^{- n^{2} t} sin n x$ satisfies the equation and the boundary conditions term by term.
Matching $t = 0$ : the initial datum is already a single mode, so $c_{3} = 1$ and all other $c_{n} = 0$ .
Hence $u (x, t) = e^{- 9 t} sin 3 x$ .
At $x = π /6$ , $sin (π /2) = 1$ , so $u = e^{- 9} \approx 0.0001234$ .
The moral: the Fourier basis diagonalises $\partial_{x}^{2}$ , turning a PDE into an ODE in each coordinate. That is the same statement as Theorem Orthogonality of the trigonometric system, used in earnest.

□

Theorem 8.120 (Weyl equidistribution). If $α$ is irrational then the sequence $(n α mod 1)_{n \geq 1}$ is equidistributed in $[0, 1)$ : for every interval $I \subseteq [0, 1)$ , $\frac{1}{N} # {n \leq N : n α \in I} \to ∣ I ∣$ .

Proof. By Corollary Weierstrass approximation, trigonometric form and a sandwich of $χ_{I}$ between continuous functions, it suffices to prove $\frac{1}{N} \sum_{n \leq N} e^{2 πikn α} \to 0$ for every non-zero integer $k$ . That is a geometric sum with ratio $z = e^{2 πik α} \neq = 1$ (here irrationality of $α$ is used), so

\frac{1}{N} n = 1 \sum N z^{n} = \frac{1}{N} \frac{z ( z ^{N} - 1 )}{z - 1} \leq \frac{2}{N ∣ z - 1∣} ⟶ 0.

∎

Remark. Three further directions, stated without proof. Wavelets replace the globally supported $e^{in x}$ by localised bases, giving unconditional bases of $L^{p}$ for $p \neq = 2$ where the trigonometric system fails to be one. The Poisson summation formula $\sum_{n} f (n) = \sum_{k} \hat{f} (k)$ links Fourier series to the Fourier transform and is the analytic heart of the functional equation of the Riemann zeta function. Uncertainty principles quantify the observation in the Gibbs analysis, that $f$ and $\hat{f}$ cannot both be concentrated: $∥ x f ∥_{2} ∥ ξ \hat{f} ∥_{2} \geq ∥ f ∥_{2}^{2} / (4 π)$ .

Exercises for §8.27 →

8.28Differentiability in Several Variables

The single idea of this last part is that the derivative of a map $R^{n} \to R^{m}$ is a linear map, and that partial derivatives are merely its matrix entries in one chosen basis. Confusing the two is the source of every standard error.

Definition 8.121 (Total derivative). Let $U \subseteq R^{n}$ be open and $F : U \to R^{m}$ . $F$ is differentiable at $a \in U$ if there is a linear map $D F (a) : R^{n} \to R^{m}$ with

h \to 0 lim \frac{F ( a + h ) - F ( a ) - D F ( a ) h}{∥ h ∥} = 0.

$D F (a)$ is then unique and is called the total derivative; its matrix in the standard bases is the Jacobian $(\partial F_{i} / \partial x_{j})_{i, j}$ .

Proposition 8.122 (Uniqueness, and differentiability implies continuity). The map $D F (a)$ in Definition Total derivative is unique, and differentiability at $a$ implies continuity at $a$ and the existence of every directional derivative, with $\partial_{v} F (a) = D F (a) v$ .

Proof. Uniqueness. If $L_{1}, L_{2}$ both work, then $∥ (L_{1} - L_{2}) h ∥/∥ h ∥ \to 0$ as $h \to 0$ . Fix a unit vector $v$ and take $h = t v$ , $t \to 0$ : by linearity $∥ (L_{1} - L_{2}) v ∥ = ∥ (L_{1} - L_{2}) (t v) ∥/ t \to 0$ , but the left side does not depend on $t$ , so it is $0$ . Hence $L_{1} = L_{2}$ .

Continuity. The defining limit gives $∥ F (a + h) - F (a) ∥ \leq ∥ D F (a) h ∥ + o (∥ h ∥) \leq (∥ D F (a) ∥ + 1) ∥ h ∥$ for small $h$ , which tends to $0$ .

Directional derivatives. Put $h = t v$ : $(F (a + t v) - F (a)) / t = D F (a) v + o (∥ t v ∥) / t \to D F (a) v$ .∎

Theorem 8.123 (Continuous partials imply differentiability). If all partial derivatives $\partial F_{i} / \partial x_{j}$ exist on a neighbourhood of $a$ and are continuous at $a$ , then $F$ is differentiable at $a$ .

Proof. It suffices to treat $m = 1$ , applying the result to each component. Write $h = (h_{1}, \dots, h_{n})$ and interpolate one coordinate at a time: with $a^{(0)} = a$ and $a^{(j)} = a^{(j - 1)} + h_{j} e_{j}$ ,

f (a + h) - f (a) = j = 1 \sum n (f (a^{(j)}) - f (a^{(j - 1)})) .

Each summand changes only the $j$ -th coordinate, so the Mean Value Theorem of Differentiation in one variable applies: it equals $\partial_{j} f (c_{j}) h_{j}$ for some $c_{j}$ on the segment, and $c_{j} \to a$ as $h \to 0$ . Therefore

\frac{f ( a + h ) - f ( a ) - \sum _{j} \partial _{j} f ( a ) h _{j}}{∥ h ∥} \leq j = 1 \sum n \partial_{j} f (c_{j}) - \partial_{j} f (a) \frac{∣ h _{j} ∣}{∥ h ∥} \leq j = 1 \sum n \partial_{j} f (c_{j}) - \partial_{j} f (a),

which tends to $0$ by continuity of the partials at $a$ .∎

Example 8.124 (Partials exist, differentiability fails). Let $f (x, y) = \frac{x y}{x ^{2} + y ^{2}}$ for $(x, y) \neq = (0, 0)$ and $f (0, 0) = 0$ . Compute $\partial_{x} f (0, 0)$ , $\partial_{y} f (0, 0)$ , and decide whether $f$ is differentiable at the origin.

Solution.

Along the axes $f \equiv 0$ , so $\partial_{x} f (0, 0) = \partial_{y} f (0, 0) = 0$ : both partials exist.
If $f$ were differentiable, $D f (0, 0)$ would be the zero map by Proposition Uniqueness, and differentiability implies continuity, and then $f (h) /∥ h ∥ \to 0$ .
Take $h = (t, t)$ : $f (t, t) = t^{2} / (2 t^{2}) = \frac{1}{2}$ , so $f (h) /∥ h ∥ = \frac{1}{2} / (t 2) \to \infty$ .
So $f$ is not differentiable at the origin — indeed it is not even continuous there, since $f \equiv \frac{1}{2}$ along the diagonal and $f \equiv 0$ along the axes.
The hypothesis of Theorem Continuous partials imply differentiability fails: the partials exist everywhere but are not continuous at $0$ . Existence of partials is a statement along $n$ lines; differentiability is a statement about all directions at once, uniformly.

□

Theorem 8.125 (Symmetry of second derivatives). If $\partial_{i} \partial_{j} f$ and $\partial_{j} \partial_{i} f$ exist near $a$ and are continuous at $a$ , then $\partial_{i} \partial_{j} f (a) = \partial_{j} \partial_{i} f (a)$ .

Proof. Work in the $(x_{i}, x_{j})$ plane and consider the second difference

Δ (s, t) = f (a + s e_{i} + t e_{j}) - f (a + s e_{i}) - f (a + t e_{j}) + f (a) .

Applying the Mean Value Theorem twice, first in $x_{i}$ then in $x_{j}$ , gives $Δ (s, t) = s t \partial_{j} \partial_{i} f (ξ)$ for some $ξ$ in the rectangle; applying it in the other order gives $Δ (s, t) = s t \partial_{i} \partial_{j} f (η)$ . Divide by $s t$ and let $s, t \to 0$ : both $ξ, η \to a$ , so continuity of the two mixed partials at $a$ forces the limits, hence the values at $a$ , to agree.∎

Exercises for §8.28 →

8.29The Chain Rule, Inverse and Implicit Function Theorems

Theorem 8.126 (Chain rule in several variables). Let $F : U \to R^{m}$ be differentiable at $a$ and $G : V \to R^{k}$ differentiable at $b = F (a)$ , with $F (U) \subseteq V$ . Then $G \circ F$ is differentiable at $a$ and

D (G \circ F) (a) = D G (b) \circ D F (a),

i.e. the Jacobian of a composite is the product of the Jacobians.

Proof. Write $A = D F (a)$ , $B = D G (b)$ , and define the error terms

F (a + h) = b + A h + φ (h), G (b + k) = G (b) + B k + ψ (k),

with $∥ φ (h) ∥ = o (∥ h ∥)$ and $∥ ψ (k) ∥ = o (∥ k ∥)$ . Put $k (h) = A h + φ (h)$ , so $∥ k (h) ∥ \leq (∥ A ∥ + 1) ∥ h ∥$ for small $h$ . Then

G (F (a + h)) - G (b) = B (A h + φ (h)) + ψ (k (h)) = B A h + Bφ (h) + ψ (k (h)) .

Now $∥ Bφ (h) ∥ \leq ∥ B ∥ o (∥ h ∥) = o (∥ h ∥)$ , and $∥ ψ (k (h)) ∥ = o (∥ k (h) ∥) = o (∥ h ∥)$ by the linear bound on $∥ k (h) ∥$ . So the total error is $o (∥ h ∥)$ , which is the definition of $D (G \circ F) (a) = B A$ .∎

Theorem 8.127 (Inverse function theorem). Let $F : U \to R^{n}$ be $C^{1}$ on an open $U \subseteq R^{n}$ and suppose $D F (a)$ is invertible. Then there are open $V ∋ a$ and $W ∋ F (a)$ with $F : V \to W$ a bijection, $F^{- 1} : W \to V$ of class $C^{1}$ , and

D (F^{- 1}) (F (x)) = (D F (x))^{- 1} for x \in V .

Proof. Sketch; the fixed-point step is complete, the $C^{1}$ regularity of the inverse is outlined. Normalise $a = 0$ , $F (0) = 0$ , $D F (0) = I$ by composing with the linear map $D F (a)^{- 1}$ and translating. Set $Φ_{y} (x) = x - F (x) + y$ ; a fixed point of $Φ_{y}$ is a solution of $F (x) = y$ . Since $D Φ_{y} (x) = I - D F (x)$ is continuous and vanishes at $0$ , choose $r$ with $∥ I - D F (x) ∥ \leq \frac{1}{2}$ on $\overline{B (0, r)}$ ; the mean value inequality then makes $Φ_{y}$ a contraction with constant $\frac{1}{2}$ , mapping $\overline{B (0, r)}$ into itself for $∥ y ∥ < r /2$ . The Banach fixed-point theorem of Metric Spaces supplies a unique fixed point, giving a bijection onto an open set $W$ .

For regularity, the contraction estimate gives $∥ F (x_{1}) - F (x_{2}) ∥ \geq \frac{1}{2} ∥ x_{1} - x_{2} ∥$ , so $F^{- 1}$ is Lipschitz, hence continuous; feeding that into the difference quotient and using invertibility of $D F$ near $a$ shows $F^{- 1}$ is differentiable with the stated derivative, and the formula for the inverse of a matrix is continuous in its entries, so $F^{- 1}$ is $C^{1}$ .∎

Theorem 8.128 (Implicit function theorem). Let $G : R^{n} \times R^{m} \to R^{m}$ be $C^{1}$ near $(a, b)$ with $G (a, b) = 0$ , and suppose the $m \times m$ block $D_{y} G (a, b)$ is invertible. Then there are neighbourhoods $A ∋ a$ , $B ∋ b$ and a unique $C^{1}$ map $g : A \to B$ with $g (a) = b$ and

G (x, g (x)) = 0 for x \in A, D g (x) = - (D_{y} G)^{- 1} D_{x} G .

Proof. Apply Theorem Inverse function theorem to $F (x, y) = (x, G (x, y))$ , a $C^{1}$ map $R^{n + m} \to R^{n + m}$ . Its Jacobian at $(a, b)$ is block triangular,

D F (a, b) = (I D_{x} G 0 D_{y} G),

with determinant $det D_{y} G (a, b) \neq = 0$ , hence invertible. So $F$ has a local $C^{1}$ inverse of the form $F^{- 1} (x, z) = (x, h (x, z))$ , and $g (x) = h (x, 0)$ satisfies $G (x, g (x)) = 0$ . Differentiating that identity by Theorem Chain rule in several variables gives $D_{x} G + D_{y} G D g = 0$ , i.e. the stated formula.∎

Example 8.129 (Implicit differentiation on a circle). The curve $x^{2} + y^{2} = 1$ near $(3/5, 4/5)$ : verify the hypotheses and compute $d y / d x$ there.

Solution.

$G (x, y) = x^{2} + y^{2} - 1$ is $C^{1}$ with $G (3/5, 4/5) = 0$ .
$D_{y} G = 2 y = 8/5 \neq = 0$ , so Theorem Implicit function theorem applies and $y = g (x)$ locally.
$D_{x} G = 2 x = 6/5$ , so $g^{'} (x) = - D_{x} G / D_{y} G = - (6/5) / (8/5) = - 3/4$ .
Check against the explicit solution $y = 1 - x^{2}$ : $y^{'} = - x / 1 - x^{2} = - (3/5) / (4/5) = - 3/4$ . $✓$
At $(1, 0)$ the hypothesis fails, $D_{y} G = 0$ , and indeed $y$ is not a function of $x$ there — the circle has a vertical tangent.

□

Exercises for §8.29 →

8.30Multiple Integrals, Fubini and Tonelli

Theorem 8.130 (Tonelli). Let $(X, μ)$ , $(Y, ν)$ be $σ$ -finite and $f : X \times Y \to [0, \infty]$ measurable for the product $σ$ -algebra. Then $x \mapsto \int_{Y} f (x, y) d ν$ is measurable and

\int_{X \times Y} f d (μ \times ν) = \int_{X} (\int_{Y} f d ν) d μ = \int_{Y} (\int_{X} f d μ) d μ .

Proof. Sketch. For $f = χ_{E}$ with $E$ in the product $σ$ -algebra the statement is the definition of the product measure together with a monotone class argument: the collection of $E$ for which it holds contains measurable rectangles, is closed under increasing unions and proper differences by $σ$ -finiteness, hence contains the generated $σ$ -algebra. Linearity extends it to non-negative simple functions, and the Monotone Convergence Theorem of Lebesgue Integration extends it to all non-negative measurable $f$ by taking simple $s_{n} ↑ f$ .∎

Theorem 8.131 (Fubini). With the same hypotheses, if $f \in L^{1} (μ \times ν)$ then $y \mapsto f (x, y)$ is in $L^{1} (ν)$ for $μ$ -a.e. $x$ , the a.e.-defined function $x \mapsto \int_{Y} f (x, y) d ν$ is in $L^{1} (μ)$ , and the iterated integrals both equal $\int f d (μ \times ν)$ .

Proof. Apply Theorem Tonelli to $∣ f ∣$ : the iterated integral of $∣ f ∣$ is finite, so $\int_{Y} ∣ f (x, y) ∣ d ν < \infty$ for $μ$ -a.e. $x$ . Now split $f = f^{+} - f^{-}$ , both non-negative and dominated by $∣ f ∣$ , hence with finite iterated integrals. Tonelli applies to each, and subtracting the two identities gives the result; the subtraction is legitimate because both quantities are finite.∎

Pitfall. Tonelli needs $f \geq 0$ but no integrability; Fubini needs integrability but no sign. Using Fubini without checking $\int ∣ f ∣ < \infty$ is the standard error, and it is not harmless. On $[0, 1]^{2}$ with $f (x, y) = \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ the two iterated integrals are $- π /4$ and $+ π /4$ . Nothing is wrong with the theorems; $\int ∣ f ∣ = \infty$ , so neither applies. The safe order is always: Tonelli on $∣ f ∣$ first, Fubini on $f$ second.

Example 8.132 (An iterated integral over a triangle). Compute $\int_{0}^{1} \int_{x}^{1} e^{y^{2}} d y d x$ .

Solution.

The inner integral has no elementary antiderivative, so swap the order. The region is ${0 \leq x \leq y \leq 1}$ .
The integrand is positive and the region has finite measure, so Theorem Tonelli permits the swap with no further check.
$\int_{0}^{1} \int_{0}^{y} e^{y^{2}} d x d y = \int_{0}^{1} y e^{y^{2}} d y$ .
Substituting $u = y^{2}$ : $\frac{1}{2} \int_{0}^{1} e^{u} d u = \frac{e - 1}{2} \approx 0.8591$ .

□

Proposition 8.133 (Volume of a region as an iterated integral). If $E \subseteq R^{n} \times R^{m}$ is measurable, then $∣ E ∣ = \int_{R^{n}} E_{x} d x$ , where $E_{x} = {y : (x, y) \in E}$ is the slice.

Proof. Apply Theorem Tonelli to $f = χ_{E} \geq 0$ , noting $\int_{R^{m}} χ_{E} (x, y) d y = ∣ E_{x} ∣$ .∎

Exercises for §8.30 →

8.31Change of Variables in Multiple Integrals

Theorem 8.134 (Change of variables). Let $T : U \to V$ be a $C^{1}$ diffeomorphism between open subsets of $R^{n}$ . Then for every measurable $f \geq 0$ on $V$ (and for every $f \in L^{1} (V)$ ),

\int_{V} f (y) d y = \int_{U} f (T (u)) det D T (u) d u .

Proof. Sketch; the key measure-theoretic estimate is stated rather than proved in full. The content is the special case $f = χ_{T (E)}$ , i.e.

T (E) = \int_{E} ∣ det D T ∣.

For a linear $T$ this is the determinant's defining property: $T$ maps the unit cube to a parallelepiped of volume $∣ det T ∣$ , and the formula extends to all measurable sets by translation-invariance and countable additivity. For a general $C^{1}$ diffeomorphism, cover $U$ by small cubes $Q$ ; on each, $T (u) \approx T (u_{0}) + D T (u_{0}) (u - u_{0})$ with error $o (diam Q)$ , so $∣ T (Q) ∣$ and $∣ det D T (u_{0}) ∣ ∣ Q ∣$ agree to relative error tending to $0$ as the cubes shrink. Summing over the cubes and passing to the limit, using uniform continuity of $D T$ on compacts, gives the identity for $χ_{E}$ ; monotone convergence then gives it for non-negative $f$ and linearity for $f \in L^{1}$ . What is assumed is that the error terms can be controlled uniformly, which requires the Vitali covering machinery of the earlier part.∎

Corollary 8.135 (Polar and spherical coordinates). For $T (r, θ) = (r cos θ, r sin θ)$ on $(0, \infty) \times (0, 2 π)$ , $det D T = r$ . For $T (ρ, φ, θ) = (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)$ , $det D T = ρ^{2} sin φ$ .

Proof. In two dimensions,

D T = (cos θ sin θ - r sin θ r cos θ), det D T = r cos^{2} θ + r sin^{2} θ = r .

In three, expanding the $3 \times 3$ Jacobian along the last row and using $sin^{2} + cos^{2} = 1$ twice gives $det D T = ρ^{2} sin φ$ , which is non-negative for $φ \in (0, π)$ .∎

Figure 8.136. The polar map sends the rectangle

[1, 2] \times [0, π /2]

in the

(r, θ)

plane to a quarter annulus. A small rectangle of sides

d r

d θ

becomes a curved box of sides

d r

and

r d θ

— hence the factor

r

Example 8.137 (The Gaussian integral). Evaluate $\int_{R} e^{- x^{2}} d x$ by squaring and changing to polar coordinates.

Solution.

Let $I = \int_{R} e^{- x^{2}} d x$ . Then $I^{2} = \int_{R} \int_{R} e^{- (x^{2} + y^{2})} d x d y$ , the product being legitimate by Theorem Tonelli since the integrand is positive.
Change to polar by Corollary Polar and spherical coordinates: $x^{2} + y^{2} = r^{2}$ and $d x d y = r d r d θ$ .
$I^{2} = \int_{0}^{2 π} \int_{0}^{\infty} e^{- r^{2}} r d r d θ = 2 π \cdot [- \frac{1}{2} e^{- r^{2}}]_{0}^{\infty} = 2 π \cdot \frac{1}{2} = π$ .
Hence $I = π \approx 1.7725$ , taking the positive root since the integrand is positive.
The whole trick is the factor $r$ : without it the substitution $u = r^{2}$ would not close.

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Example 8.138 (Volume of a ball in spherical coordinates). Compute the volume of ${x^{2} + y^{2} + z^{2} \leq R^{2}}$ .

Solution.

By Corollary Polar and spherical coordinates the element is $ρ^{2} sin φ d ρ d φ d θ$ , with $ρ \in (0, R)$ , $φ \in (0, π)$ , $θ \in (0, 2 π)$ .
The integral factors: $\int_{0}^{R} ρ^{2} d ρ \cdot \int_{0}^{π} sin φ d φ \cdot \int_{0}^{2 π} d θ$ .
These are $R^{3} /3$ , $2$ , and $2 π$ .
Volume $= \frac{R ^{3}}{3} \cdot 2 \cdot 2 π = \frac{4 π R ^{3}}{3}$ . For $R = 1$ this is $\approx 4.1888$ .

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Exercises for §8.31 →

8.32Differential Forms, the Exterior Derivative, and Stokes' Theorem

The final result of the chapter unifies the fundamental theorem of calculus, Green's theorem, the divergence theorem and the classical Stokes theorem into a single line. The unification requires the right objects: differential forms, which are exactly the things that can be integrated over oriented $k$ -dimensional pieces.

Definition 8.139 (Differential $k$ -form and exterior derivative). A $k$ -form on an open $U \subseteq R^{n}$ is an expression

ω = i_{1} < \dots < i_{k} \sum f_{i_{1} \dots i_{k}} d x_{i_{1}} \land \dots \land d x_{i_{k}}

with smooth coefficients, where $\land$ is associative, bilinear and alternating: $d x_{i} \land d x_{j} = - d x_{j} \land d x_{i}$ , hence $d x_{i} \land d x_{i} = 0$ . The exterior derivative is

d ω = i_{1} < \dots < i_{k} \sum j = 1 \sum n \frac{\partial f _{i_{1} \dots i_{k}}}{\partial x _{j}} d x_{j} \land d x_{i_{1}} \land \dots \land d x_{i_{k}},

a $(k + 1)$ -form.

Proposition 8.140 ( $d^{2} = 0$ ). For every smooth form $ω$ , $d (d ω) = 0$ .

Proof. By linearity it suffices to take $ω = f d x_{I}$ . Then

dd ω = j, k \sum \frac{\partial ^{2} f}{\partial x _{k} \partial x _{j}} d x_{k} \land d x_{j} \land d x_{I} .

The coefficient $\partial^{2} f / \partial x_{k} \partial x_{j}$ is symmetric in $j, k$ by Theorem Symmetry of second derivatives, while $d x_{k} \land d x_{j} = - d x_{j} \land d x_{k}$ is antisymmetric. Pairing the $(j, k)$ and $(k, j)$ terms, each pair cancels, and the diagonal terms vanish because $d x_{j} \land d x_{j} = 0$ .∎

Theorem 8.141 (Stokes' theorem). Let $M$ be a compact oriented smooth $k$ -manifold with boundary $\partial M$ carrying the induced orientation, and let $ω$ be a smooth $(k - 1)$ -form on $M$ . Then

\int_{M} d ω = \int_{\partial M} ω .

Proof. Sketch; the machinery of partitions of unity and manifold charts is assumed. Using a partition of unity subordinate to a cover by charts, write $ω = \sum_{α} ω_{α}$ with each $ω_{α}$ supported in one chart; both sides are linear in $ω$ , so it suffices to treat a single chart.

Two cases. If the chart is interior, $ω$ has compact support in the open half-space's interior, and $\int d ω$ reduces to $\int_{R^{k}} \partial_{j} f d x$ , which is $0$ by the one-variable fundamental theorem of Riemann Integration applied in the $x_{j}$ direction — the compact support kills both endpoint terms. And $\int_{\partial M} ω = 0$ since $ω$ vanishes near the boundary. If the chart meets the boundary, model it on the half-space ${x_{k} \geq 0}$ ; the same one-variable computation now leaves exactly one surviving endpoint term, the value on ${x_{k} = 0}$ , which is precisely $\int_{\partial M} ω$ . Orientation conventions are fixed so that the sign comes out $+$ .

What is assumed and not proved here: the existence of smooth partitions of unity, the invariance of $\int ω$ under orientation-preserving changes of chart (which is Theorem Change of variables together with the transformation rule $T^{*} (f d x_{1} \land \dots \land d x_{k}) = (f \circ T) det D T d u_{1} \land \dots \land d u_{k}$ ), and the definition of the induced boundary orientation.∎

Corollary 8.142 (The classical theorems as special cases). Stokes' theorem specialises to: the Fundamental Theorem of Calculus ( $k = 1$ ), Green's theorem ( $k = 2$ in the plane), the classical Stokes/curl theorem ( $k = 2$ in $R^{3}$ ), and the divergence theorem ( $k = 3$ in $R^{3}$ ).

Proof. Each is a matter of writing out $d ω$ in coordinates. For $k = 1$ , $M = [a, b]$ , $ω = f$ a $0$ -form, $d ω = f^{'} d x$ , $\partial M = {b} - {a}$ , and the theorem reads $\int_{a}^{b} f^{'} = f (b) - f (a)$ .

For Green, $ω = P d x + Q d y$ gives $d ω = (\partial_{x} Q - \partial_{y} P) d x \land d y$ after cancelling the terms with a repeated $d x$ or $d y$ .

For the divergence theorem, $ω = P d y \land d z + Q d z \land d x + R d x \land d y$ gives $d ω = (\partial_{x} P + \partial_{y} Q + \partial_{z} R) d x \land d y \land d z = (\nabla \cdot F) d V$ , and $\int_{\partial M} ω$ is the flux $\int F \cdot n d S$ . The curl theorem is the intermediate case.∎

Example 8.143 (Green's theorem computing an area). Use $ω = \frac{1}{2} (x d y - y d x)$ to compute the area enclosed by the ellipse $x = 3 cos t$ , $y = 2 sin t$ , $t \in [0, 2 π]$ .

Solution.

$d ω = \frac{1}{2} (d x \land d y - d y \land d x) = d x \land d y$ , so $\int_{M} d ω$ is the area.
By Theorem Stokes' theorem, the area equals $\oint_{\partial M} \frac{1}{2} (x d y - y d x)$ .
On the curve, $x d y = 3 cos t \cdot 2 cos t d t = 6 cos^{2} t d t$ and $y d x = 2 sin t \cdot (- 3 sin t) d t = - 6 sin^{2} t d t$ .
So the integrand is $\frac{1}{2} (6 cos^{2} t + 6 sin^{2} t) d t = 3 d t$ , and the integral is $3 \cdot 2 π = 6 π \approx 18.8496$ .
Check against $πab = π \cdot 3 \cdot 2 = 6 π$ . $✓$

□

Intuition. The pattern in every case is the same: $d$ measures how much a form fails to be "already a boundary value", and Stokes says the total of that failure inside equals what is seen on the edge. $d^{2} = 0$ is the statement that a boundary has no boundary — the edge of a disc is a circle, and a circle has no endpoints. Every conservation law in physics is that identity in disguise.

Exercises for §8.32 →

Common Mistakes

Assuming continuity implies bounded variation. $x sin (1/ x)$ is continuous on $[0, 1]$ with infinite variation. Continuity bounds local oscillation; bounded variation bounds its total.
Assuming bounded variation implies absolute continuity. The Cantor function is continuous, increasing and $B V$ , yet climbs by $1$ with derivative $0$ a.e. Three separate proofs of its non-absolute-continuity appear above; all say the same thing — it manufactures measure on a null set.
Using the fundamental theorem without checking absolute continuity. $f (x) = f (a) + \int_{a}^{x} f^{'}$ is equivalent to $f \in A C$ , not a consequence of $f^{'}$ existing a.e.
Confusing uniform continuity with absolute continuity. Every continuous function on a compact interval is uniformly continuous; only some are absolutely continuous. The difference appears only when arbitrarily many disjoint intervals are allowed at once.
Expecting $M f \in L^{1}$ when $f \in L^{1}$ . The maximal function decays like $∣ x ∣^{- n}$ at infinity and is never integrable unless $f = 0$ . Weak type $(1, 1)$ is the correct — and sharp — statement.
Using non-conjugate exponents in Hölder. $1/ p + 1/ q = 1$ , so the conjugate of $3$ is $3/2$ , not $3$ . Homogeneity in $f$ and in $g$ pins the relation down uniquely.
Assuming $L^{p} \subseteq L^{r}$ in general. The inclusion $L^{s} \subseteq L^{r}$ for $r < s$ needs $μ (X) < \infty$ ; on $N$ with counting measure the inclusion reverses; on $R$ neither holds.
Believing $L^{1}$ convergence implies pointwise convergence. The typewriter sequence converges to $0$ in $L^{1}$ and at no point at all. Only a subsequence converges a.e.
Applying the parallelogram law or orthogonal projection in $L^{p}$ , $p \neq = 2$ . Only $L^{2}$ has an inner product; the entire Fourier-basis argument is unavailable elsewhere.
Forgetting that $S_{N} f$ need not converge pointwise for continuous $f$ . The Dirichlet kernel's $L^{1}$ norms diverge like $lo g N$ , and uniform boundedness turns that into a genuine counterexample. Use Fejér means if uniform convergence is needed.
Thinking the Gibbs overshoot disappears with more terms. It does not: it stays at about $8.95%$ of the jump and only moves closer to the discontinuity. Pointwise convergence and uniform convergence differ exactly here.
Reading Parseval as pointwise equality. It is an identity between norms; it says nothing about the value of the series at any particular $x$ .
Deducing differentiability from existence of partial derivatives. $x y / (x^{2} + y^{2})$ has both partials at the origin and is not even continuous there. Continuous partials are what suffice.
Applying Fubini without checking $\int ∣ f ∣ < \infty$ . Use Tonelli on $∣ f ∣$ first. Two iterated integrals of $\frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ over the unit square differ in sign.
Dropping the absolute value or the Jacobian in the change of variables formula. The factor is $∣ det D T ∣$ ; the area element in polar coordinates is $r d r d θ$ , and omitting $r$ is why $\int e^{- x^{2}}$ appears intractable.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter

Introduction

8.1Definition via Total Variation

8.2Jordan Decomposition: Bounded Variation as a Difference of Increasing Functions

8.3Differentiability and Discontinuities of BV Functions

8.4Absolute Continuity: a Preview of the Stronger Condition

8.5Riemann–Stieltjes Integration against BV Integrators

8.6The ε-δ Definition: Uniform Control on Total Increments

8.7Equivalence with Being an Indefinite Integral

8.8Properties Inherited from Integrals

8.9Relation to Bounded Variation and Singular Functions

8.10Absolute Continuity of Measures and Radon–Nikodym

8.11The Lebesgue Differentiation Theorem

8.12The Hardy–Littlewood Maximal Function and the Weak-Type Inequality

8.13Proof Ideas: Vitali Covering and the Rising Sun Lemma

8.14Differentiation of Measures and the Radon–Nikodym Derivative

8.15Lebesgue Points and Lusin's Condition (N)

8.16Lp Spaces: Definition and Basic Properties

8.17Hölder's and Minkowski's Inequalities

8.18Completeness: the Riesz–Fischer Theorem

8.19Duality and Reflexivity

8.20Relations between Lp Spaces

8.21Applications of Lp Spaces

8.22Trigonometric Polynomials and Orthogonality

8.23Pointwise Convergence: Dirichlet and Jordan

8.24L2 Convergence and Parseval's Identity

8.25Cesàro Means, Fejér's Theorem, and Uniform Convergence

8.26Divergence, the Gibbs Phenomenon, and Carleson's Theorem

8.27Applications and Extensions

8.28Differentiability in Several Variables

8.29The Chain Rule, Inverse and Implicit Function Theorems

8.30Multiple Integrals, Fubini and Tonelli

8.31Change of Variables in Multiple Integrals

8.32Differential Forms, the Exterior Derivative, and Stokes' Theorem

8.16 $L^{p}$ Spaces: Definition and Basic Properties

8.20Relations between $L^{p}$ Spaces

8.21Applications of $L^{p}$ Spaces

8.24 $L^{2}$ Convergence and Parseval's Identity