---
title: "The Lp Spaces"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 12
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces
---

# Chapter 12 — The Lp Spaces

The Lebesgue integral was built for analysis; the $L^p$ spaces are where that analysis lives. They are Banach spaces (Riesz–Fischer) — the [completions](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#thm-b3-complete-completion) that [Exercise 7.1](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#exo-b3-complete-1) showed the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functions lacked — and they support a smoothing technology, *convolution against [mollifiers](#def-b3-lp-mollifier)*, that approximates any $L^p$ function by $\mathcal C^\infty$ ones. This chapter proves the integral versions of Hölder and Minkowski, [completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete), the [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) theorems, and the regularization machine, ending with the inclusion and interpolation geography of the $L^p$ scale. Throughout, $(X, \mathcal A, \mu)$ is a [measure space](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) and functions are complex-valued; on $\R^d$, the [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) is $\lambda_d$.

## 12.1 Definition; Hölder and Minkowski

**Definition 12.1.**

For $1 \leq p < \infty$, $\mathcal L^p(\mu)$ is the set of [measurable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-measurable) $f$ with $\norm f_p = \bigl(\int\abs
f^p\dd\mu\bigr)^{1/p} < \infty$, and $\mathcal L^\infty(\mu)$ the set of $f$ bounded outside a null set, with $\norm
f_\infty$ the *essential sup* — the least $M$ with $\abs f \leq M$ a.e. (the inf is attained: intersect the null sets for $M + \frac1n$). Since $\norm f_p = 0$ only forces $f
= 0$ *a.e.* ([Exercise 10.5](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#exo-b3-lebesgue-5)), we define

$$
L^p(\mu) = \mathcal L^p(\mu)/\{f = 0 \text{ a.e.}\} :
$$

elements are classes of functions modulo null sets, and $\norm\cdot_p$ is a genuine norm on $L^p$.

**Theorem 12.2 (Hölder’s inequality).**

Let $1 \leq p, q \leq \infty$ with $\frac1p + \frac1q = 1$ (conjugate exponents). For [measurable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-measurable) $f, g$:

$$
\norm{fg}_1 \leq \norm f_p\,\norm g_q ,
$$

with equality (for $1 < p < \infty$, finite norms, $f,g\ne0$) iff $\abs f^p$ and $\abs g^q$ are proportional a.e.

**Proof.** The cases $\{p, q\} = \{1, \infty\}$ are direct ($\abs{fg} \leq
\norm g_\infty\abs f$ a.e.). Let $1 < p < \infty$; normalize $\norm f_p = \norm g_q = 1$ (homogeneity; zero or infinite norms are trivial). Young’s inequality $ab \leq \frac{a^p}p +
\frac{b^q}q$ ($a, b \geq 0$; concavity of $\ln$, as in [Problem 8.1](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#pb-b3-banach-1)) gives pointwise $\abs{f g} \leq
\frac{\abs f^p}p + \frac{\abs g^q}q$; integrate: $\norm{fg}_1
\leq \frac1p + \frac1q = 1$. Equality forces a.e. equality in Young, i.e. $\abs f^p = \abs g^q$ a.e. (after the normalization; undoing it, proportionality). ∎

**Theorem 12.3 (Minkowski’s inequality).**

For $1 \leq p \leq \infty$: $\norm{f + g}_p \leq \norm f_p +
\norm g_p$.

**Proof.** $p = 1, \infty$: pointwise/a.e. triangle inequality. For $1 <
p < \infty$, assume $\norm{f+g}_p < \infty$ (else use $\abs{f+g}^p \leq 2^{p-1}(\abs f^p + \abs g^p)$, from convexity of $t^p$, to see the left side is finite when the right is). Then

$$
\norm{f{+}g}_p^p \leq \int\abs f\,\abs{f{+}g}^{p-1} +
\int\abs g\,\abs{f{+}g}^{p-1}
\leq \bigl(\norm f_p + \norm g_p\bigr)\,
\bigl\|\abs{f{+}g}^{p-1}\bigr\|_q
$$

by Hölder, and $\norm{\abs{f+g}^{p-1}}_q =
\norm{f+g}_p^{p/q}$ since $(p-1)q = p$; divide by $\norm{f+g}_p^{p/q}$ (if nonzero; else trivial) and use $p -
\frac pq = 1$. ∎

## 12.2 Completeness and its companions

**Theorem 12.4 (Riesz–Fischer).**

For $1 \leq p \leq \infty$, $L^p(\mu)$ is a Banach space. Moreover, every sequence converging in $L^p$ has a subsequence converging *[almost everywhere](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1)* (with an $L^p$ dominator in the case $p < \infty$).

**Proof.** $p = \infty$: a $\norm\cdot_\infty$-Cauchy sequence is, off a single null set (union of countably many), uniformly Cauchy: it converges uniformly off it; done. Let $p < \infty$. By [Exercise 7.1](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#exo-b3-complete-1)(b) it suffices to sum absolutely convergent series: let $\sum\norm{f_k}_p = M < \infty$. Set $G_n = \sum_{k \leq n}\abs{f_k}$ and $G = \sum_k\abs{f_k}$ (pointwise in $[0,\infty]$): by Minkowski $\norm{G_n}_p \leq
M$, and MCT ($G_n^p \nearrow G^p$) gives $\int G^p \leq M^p$: $G < \infty$ a.e., so the series $\sum f_k(x)$ converges absolutely for a.e. $x$; call the sum $S(x)$ (any value on the null set). Then $\abs{S - \sum_{k\leq n}f_k}^p \leq (2G)^p
\in L^1$, and DCT gives $\norm{S - \sum_{k\leq n}f_k}_p \to 0$: the series converges in $L^p$.

The subsequence statement: if $f_n \to f$ in $L^p$, pick $n_k$ with $\norm{f_{n_{k+1}} - f_{n_k}}_p \leq 2^{-k}$; the series $\sum(f_{n_{k+1}} - f_{n_k})$ falls under the previous argument: it converges absolutely a.e., dominated by a $G \in
L^p$, so $f_{n_k} \to f_{n_1} + \sum(\cdots)$ a.e., and this a.e.-limit must be (a representative of) $f$ (both are $L^p$ limits). The dominator: $\abs{f_{n_k}} \leq \abs{f_{n_1}} + G$. ∎

**Remark 12.5.**

$L^p$ convergence does not imply a.e. convergence (the *typewriter* sequence, [Exercise 12.3](#exo-b3-lp-3)), nor conversely (escaping bumps): the two modes are linked only through subsequences and domination. Keeping the counterexamples of [Exercise 12.3](#exo-b3-lp-3) in mind is the best vaccine.

## 12.3 Density theorems

**Theorem 12.6.**

Let $1 \leq p < \infty$.

1. [Simple functions](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-simple) (with [finite-measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) supports) are dense in $L^p(\mu)$ .
2. In $L^p(\R^d)$ , the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) compactly supported functions $\mathcal C_c(\R^d)$ are dense.
3. Translation is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $L^p(\R^d)$ : writing $\tau_hf = f(\cdot - h)$ , $\norm{\tau_hf - f}_p \to 0$ as $h \to 0$ .

None of the three holds for $p = \infty$.

**Proof.** (1) For $f \geq 0$: the dyadic $s_n \nearrow f$ of [Theorem 10.4](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#thm-b3-lebesgue-approximation) satisfy $\abs{f - s_n}^p
\leq f^p \in L^1$: DCT. (Each $s_n \leq f$ lies in $L^p$, and its level sets have finite [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) where the value is positive: $\mu(s_n \geq c) \leq c^{-p}\int f^p$.) Split general $f$ into four nonnegative parts.

(2) By (1) it suffices to approximate $\mathbf 1_A$, $A$ Borel with $\lambda_d(A) < \infty$. Regularity (proof as in [Theorem 9.13](https://one-course.com/books/math/5/en/chapter/9-measure-theory#thm-b3-measure-regularity)) gives [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K \subseteq A
\subseteq U$ open with $\lambda_d(U\setminus K) <
\varepsilon$; Urysohn’s function

$$
\varphi(x) = \frac{d(x, \R^d\setminus U)}{d(x, \R^d\setminus U)
+ d(x, K)}
$$

is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), $1$ on $K$, $0$ outside $U$, and can be taken compactly supported (shrink $U$ to a bounded open first). Then $\norm{\mathbf 1_A - \varphi}_p^p \leq \lambda_d(U\setminus K)
< \varepsilon$.

(3) For $g \in \mathcal C_c$: uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) gives $\norm{\tau_hg - g}_\infty \to 0$, with supports in a fixed [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) for $\abs h \leq 1$: $\norm{\tau_hg - g}_p \to 0$. For general $f$: pick $g \in \mathcal C_c$ with $\norm{f - g}_p <
\varepsilon$; then $\norm{\tau_hf - f}_p \leq 2\norm{f - g}_p +
\norm{\tau_hg - g}_p$ (translation invariance of the norm).

For $p = \infty$: uniform approximation of $\mathbf
1_{\intoo0\infty}$ by [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functions is impossible (jump), and $\norm{\tau_h\mathbf 1_{\intoo0\infty} - \mathbf
1_{\intoo0\infty}}_\infty = 1$ for $h \neq 0$. ∎

## 12.4 Convolution and regularization

**Theorem 12.7 (Young’s inequality).**

Let $1 \leq p \leq \infty$, $f \in L^1(\R^d)$, $g \in
L^p(\R^d)$. Then $f * g$ is defined a.e., belongs to $L^p$, and

$$
\norm{f * g}_p \leq \norm f_1\,\norm g_p .
$$

**Proof.** $p = \infty$: direct bound. $p = 1$: [Theorem 11.9](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-convolution). Let $1 < p < \infty$, $q$ conjugate. Split $\abs{f(y)} = \abs{f(y)}^{1/q}\cdot
\abs{f(y)}^{1/p}$ and apply Hölder:

$$
\int\abs{f(y)}\,\abs{g(x{-}y)}\,\dd y
\leq \Bigl(\int\abs f\Bigr)^{1/q}
\Bigl(\int\abs{f(y)}\,\abs{g(x - y)}^p\,\dd y\Bigr)^{1/p} .
$$

Raise to the $p$-th power and integrate in $x$; Tonelli on the second factor gives $\norm f_1^{p/q}\cdot\norm f_1\norm g_p^p$, i.e. $\norm{f*g}_p^p \leq \norm f_1^{1 + p/q}\norm g_p^p =
(\norm f_1\norm g_p)^p$ — and the finiteness of the Tonelli integral justifies a.e. absolute convergence as in [Theorem 11.9](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-convolution). ∎

**Definition 12.8 (Mollifiers).**

The function

$$
\rho(x) = \begin{cases}
c\,\exp\Bigl(-\dfrac{1}{1 - \norm x^2}\Bigr) & \norm x < 1,\\
0 & \norm x \geq 1,
\end{cases}
$$

with $c$ normalizing $\int\rho = 1$, is $\mathcal C^\infty$ on $\R^d$: the point is that $t \mapsto \eu^{-1/t}\mathbf
1_{t>0}$ is $\mathcal C^\infty$ on $\R$, all its derivatives at $0^+$ being $0$ (each derivative is $P(1/t)\eu^{-1/t}$ for a polynomial $P$, which tends to $0$; induction). For $\varepsilon > 0$ set $\rho_\varepsilon(x) =
\varepsilon^{-d}\rho(x/\varepsilon)$: supported in $\bar B(0,
\varepsilon)$, still of integral $1$.

**Theorem 12.9 (Regularization).**

Let $1 \leq p < \infty$ and $f \in L^p(\R^d)$. Then:

1. $f * \rho_\varepsilon \in \mathcal C^\infty(\R^d)$ , with $\partial^\alpha(f * \rho_\varepsilon) = f *  \partial^\alpha\rho_\varepsilon$ ;
2. $\norm{f * \rho_\varepsilon - f}_p \to 0$ as $\varepsilon \to 0$ ;
3. consequently $\mathcal C^\infty_c(\R^d)$ is dense in $L^p(\R^d)$ .

**Proof.** (1) Differentiation under the integral ([Theorem 10.15](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#thm-b3-lebesgue-paramdiff)) in $x$: for $x$ in a ball $B$, $\abs{\partial_{x_i}\rho_\varepsilon(x - y)} \leq
C_\varepsilon\,\mathbf 1_{K}(y)$ with $K$ [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) ($y$ within $\varepsilon$ of $B$), and $\abs f\,\mathbf 1_K \in L^1$ (Hölder against $\mathbf 1_K$): the theorem applies; iterate for higher derivatives.

(2) Since $\int\rho_\varepsilon = 1$:

$$
(f * \rho_\varepsilon)(x) - f(x)
= \int \bigl(f(x - y) - f(x)\bigr)\rho_\varepsilon(y)\,\dd y ,
$$

and Minkowski’s *integral* inequality — or directly: Hölder/Jensen with the probability [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) $\rho_\varepsilon\dd y$ and Tonelli —

$$
\norm{f*\rho_\varepsilon - f}_p^p
\leq \int\Bigl(\int\abs{f(x-y) -
f(x)}^p\dd x\Bigr)\rho_\varepsilon(y)\,\dd y
= \int \norm{\tau_yf - f}_p^p\;\rho_\varepsilon(y)\,\dd y
$$

(the middle step: apply Jensen’s inequality, [Exercise 12.10](#exo-b3-lp-10), to the inner $y$-integral, then Tonelli). The integrand is supported in $\norm y \leq \varepsilon$ and tends to $0$ there uniformly as $\varepsilon \to 0$ ([Theorem 12.6](#thm-b3-lp-density)(3)): the whole expression tends to $0$.

(3) Approximate $f$ by $g \in \mathcal C_c$ ([Theorem 12.6](#thm-b3-lp-density)(2)), then $g$ by $g *
\rho_\varepsilon \in \mathcal C_c^\infty$ ([compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) support: sum of supports). ∎

**Example 12.10 (Mollifying ∣x∣\abs x∣x∣, with rates).**

Take $f(x) = \abs x$ on $\R$ (locally $L^1$; the theorem applies on every bounded window) and a symmetric [mollifier](#def-b3-lp-mollifier) $\rho_\varepsilon$. Then

$$
f_\varepsilon(x) = (f * \rho_\varepsilon)(x)
= \int\abs{x - y}\,\rho_\varepsilon(y)\,\dd y
$$

is $\mathcal C^\infty$; away from the kink, nothing happens: for $\abs x \geq \varepsilon$, $\abs{x - y}$ is linear in $x$ on the support of $\rho_\varepsilon$, so $f_\varepsilon(x) =
\abs x$ *exactly* (symmetry kills the correction). Near $0$, smoothing costs precisely

$$
0 \leq f_\varepsilon(0) = \int\abs
y\,\rho_\varepsilon(y)\,\dd y \leq \varepsilon,
\qquad
\norm{f_\varepsilon - f}_\infty \leq \varepsilon :
$$

the approximation error is confined to the $\varepsilon$-neighborhood of the singularity and is of its size. Meanwhile $f_\varepsilon'' \geq 0$ everywhere ($f$ is convex, and convolution against $\rho_\varepsilon
\geq 0$ preserves convexity), with $\int f_\varepsilon'' =
f_\varepsilon'(\infty) - f_\varepsilon'(-\infty) = 2$: the second derivative is a bump of mass $2$ squeezed into width $O(\varepsilon)$, so $\norm{f_\varepsilon''}_\infty \gtrsim
\varepsilon^{-1}$. Smoothing is a trade: uniform error $O(\varepsilon)$ against derivative blow-up $O(\varepsilon^{-1})$ — the exact exchange rate that quantitative analysis (interpolation inequalities, [Problem 12.1](#pb-b3-lp-1)’s circle of ideas) formalizes.

**Corollary 12.11 (Fundamental lemma of the calculus of variations).**

Let $f \in L^1_{\mathrm{loc}}(\R^d)$ ([integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact)) with $\int f\varphi = 0$ for every $\varphi \in \mathcal
C^\infty_c(\R^d)$. Then $f = 0$ a.e.

**Proof.** Fix a ball $B = B(0, R)$ and let $g = f\mathbf 1_{B(0, R+1)}
\in L^1$. For $x \in B$ and $\varepsilon < 1$: $(g *
\rho_\varepsilon)(x) = \int f(y)\rho_\varepsilon(x - y)\dd y =
0$, the test function being $y \mapsto \rho_\varepsilon(x-y)
\in \mathcal C_c^\infty$. But $g * \rho_\varepsilon \to g$ in $L^1$ ([Theorem 12.9](#thm-b3-lp-regularization)): $g = 0$ a.e. on $B$; exhaust $\R^d$. ∎

## 12.5 The $L^p$ geography

**Proposition 12.12.**

(a) If $\mu(X) < \infty$ and $1 \leq p \leq q \leq \infty$, then $L^q \subseteq L^p$ with $\norm f_p \leq
\mu(X)^{\frac1p - \frac1q}\,\norm f_q$. (b) On $\R^d$ (infinite [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure)) there are no inclusions: for $p \neq q$ there are functions in $L^p\setminus L^q$. (c) (Interpolation) If $p < r < q$ and $\alpha \in \intoo01$ is defined by $\frac1r = \frac\alpha p + \frac{1 - \alpha}q$, then

$$
\norm f_r \leq \norm f_p^{\alpha}\,\norm f_q^{1 - \alpha} ;
$$

in particular $L^p \cap L^q \subseteq L^r$.

**Proof.** (a) Hölder with exponents $\frac qp$ and its conjugate: $\int\abs f^p\cdot 1 \leq \norm{\abs f^p}_{q/p}\,\norm
1_{(q/p)'} = \norm f_q^p\,\mu(X)^{1 - p/q}$ ($q = \infty$ directly). (b) Near $0$ and near $\infty$, powers $x^{-\alpha}$ calibrate: [Exercise 12.2](#exo-b3-lp-2). (c) Write $\abs f^r = \abs
f^{r\alpha}\abs f^{r(1-\alpha)}$ and apply Hölder with the conjugate pair $\frac p{r\alpha}$, $\frac q{r(1-\alpha)}$ (conjugates precisely by the definition of $\alpha$): $\int\abs f^r \leq \norm f_p^{r\alpha}\norm f_q^{r(1 -
\alpha)}$. ∎

**Method 12.13.**

The $L^p$ toolkit, as used everywhere below: to prove an identity or inequality for all $f \in L^p$ — prove it on a dense class ($\mathcal C_c^\infty$ via [Theorem 12.9](#thm-b3-lp-regularization)) and extend by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ([Theorem 7.2](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#thm-b3-complete-extension), both sides being $L^p$-continuous); to prove $f = 0$, test against $\mathcal
C_c^\infty$ ([Corollary 12.11](#cor-b3-lp-fundlemma)); to gain smoothness, convolve; to trade exponents, Hölder and interpolation. The Fourier theory of [Chapter 14](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#ch-b3-fouriertransform) is one long application of this method.

## 12.6 Exercises

**Exercise 12.1 ★.**

(a) State and prove the Cauchy–Schwarz inequality in $L^2(\mu)$ as the case $p = q = 2$ of Hölder. (b) On a *probability* space, show $p \mapsto \norm f_p$ is nondecreasing. (c) When is Hölder an equality for $p = 1$, $q = \infty$?

**Solution of Exercise 12.1.**

(a) $p = q = 2$ in [Theorem 12.2](#thm-b3-lp-holder): $\abs{\int f\bar
g\,\dd\mu} \leq \int\abs{fg} \leq \norm f_2\norm g_2$ — Cauchy–Schwarz, with equality iff $\abs f, \abs g$ proportional and the phases aligned.

(b) On a probability space, for $p \leq q$: apply Jensen ([Exercise 12.10](#exo-b3-lp-10)) with the convex $\Phi(t) =
\abs t^{q/p}$ to the function $\abs f^p$: $\bigl(\int\abs f^p\bigr)^{q/p} \leq \int\abs f^q$, i.e. $\norm f_p \leq \norm f_q$.

(c) $\int\abs{fg} = \norm f_1\norm g_\infty$ iff $\abs{g} =
\norm g_\infty$ a.e. on $\{f \neq 0\}$ (the inequality $\abs{fg} \leq \abs f\norm g_\infty$ must be an a.e. equality).

**Exercise 12.2 ★.**

For which $p \in \intco1\infty$ do the following belong to $L^p$?

$$
x^{-1/2}\mathbf 1_{\intoo01},\qquad
x^{-1/2}\mathbf 1_{\intoo1\infty},\qquad
\frac{1}{x^{1/2}(1 + \abs{\ln x})}\ \text{on } \intoo01,
\qquad
\frac1{1 + \abs x}\ \text{on } \R .
$$

Conclude: on $\intoo01$ small $p$ is easier, on $\intoo1\infty$ large $p$ is easier, and no $L^p$ contains another on $\R$.

**Solution of Exercise 12.2.**

$\int_0^1 x^{-p/2}\dd x < \infty$ iff $p < 2$: the first is in $L^p$ for $p \in \intco12$. $\int_1^\infty x^{-p/2}\dd x <
\infty$ iff $p > 2$: the second for $p \in \intoo2\infty$ (and $p = \infty$: it is bounded — include it). Third: for $p <
2$, dominated by $x^{-p/2}$: [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1); for $p = 2$, substitute $u = -\ln x$: $\int_0^1\frac{\dd x}{x(1 + \abs{\ln
x})^2} = \int_0^\infty\frac{\dd u}{(1 + u)^2} < \infty$; for $p > 2$ the power dominates: divergent. So $p \in \intcc12$. Fourth: $\int_\R\frac{\dd x}{(1 + \abs x)^p} < \infty$ iff $p >
1$; bounded, so also $L^\infty$: $p \in
\intoc1\infty$. Moral: [integrability](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) at $0$ likes small $p$, at $\infty$ large $p$; combining both obstructions, no inclusion between $L^p(\R)$ spaces.

**Exercise 12.3 ★★.**

(The typewriter) Enumerate the dyadic intervals $I_1 =
\intcc01$, $I_2 = \intcc0{\frac12}$, $I_3 = \intcc{\frac12}1$, $I_4 = \intcc0{\frac14}$, … and let $f_n = \mathbf
1_{I_n}$. (a) Show $f_n \to 0$ in every $L^p(\intcc01)$, $p < \infty$, but $(f_n(x))$ diverges for *every* $x \in \intcc01$. (b) Exhibit the a.e.-convergent subsequence promised by [Theorem 12.4](#thm-b3-lp-complete). (c) Conversely give a sequence converging a.e. but not in $L^1$, and one converging in $L^1$ but in no $L^p$, $p > 1$.

**Solution of Exercise 12.3.**

(a) $\norm{f_n}_p^p = \lambda(I_n) \to 0$ (at dyadic level $k$ the length is $2^{-k}$). But every $x$ lies in one interval of each dyadic level: $f_n(x) = 1$ infinitely often and $= 0$ infinitely often (intervals of the same level not containing $x$): no convergence at any point.

(b) $f_{n_k} = \mathbf 1_{\intcc0{2^{-k}}}$ (the first interval of each level) converges to $0$ at every $x > 0$: a.e.

(c) a.e. but not $L^1$: $n\mathbf 1_{\intoo0{1/n}} \to 0$ a.e., integral $1$. In $L^1$ but in no $L^p$ ($p > 1$): $g_n =
\eu^n\,\mathbf 1_{(0,\ \eu^{-n}/n)}$: $\norm{g_n}_1 = \frac1n
\to 0$, while $\norm{g_n}_p^p = \eu^{(p-1)n}/n \to \infty$ for every $p > 1$.

**Exercise 12.4 ★★.**

Let $\mu(X) < \infty$ and $f \in L^\infty(\mu)$, $f \neq 0$. Show that $\norm f_p \to \norm f_\infty$ as $p \to \infty$. *(Upper bound by (a) of [Proposition 12.12](#prop-b3-lp-inclusions); lower bound by integrating over $\{\abs f > \norm f_\infty -
\varepsilon\}$, of positive [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure).)*

**Solution of Exercise 12.4.**

Upper: $\norm f_p \leq \mu(X)^{1/p}\norm f_\infty$ ([Proposition 12.12](#prop-b3-lp-inclusions)(a) with $q = \infty$), and $\mu(X)^{1/p} \to 1$. Lower: for $\varepsilon > 0$, $A =
\{\abs f > \norm f_\infty - \varepsilon\}$ has $\mu(A) > 0$ (definition of the essential sup), and

$$
\norm f_p \geq \Bigl(\int_A \abs f^p\Bigr)^{1/p}
\geq (\norm f_\infty - \varepsilon)\,\mu(A)^{1/p}
\xrightarrow[p\to\infty]{} \norm f_\infty - \varepsilon .
$$

**Exercise 12.5 ★★.**

(a) Where exactly does the proof of [Theorem 12.6](#thm-b3-lp-density)(3) use $p < \infty$? (b) Show that $f \in L^\infty(\R)$ satisfies $\norm{\tau_hf -
f}_\infty \to 0$ iff $f$ has a uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) representative.

**Solution of Exercise 12.5.**

(a) Twice: the conversion $\norm{\tau_hg - g}_p \leq
C^{1/p}\norm{\tau_hg - g}_\infty$ ([finite-measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) support) degenerates for $p = \infty$ only in that the *[density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) of $\mathcal C_c$* fails there — that is the real gap: step (2) of [Theorem 12.6](#thm-b3-lp-density) has no $L^\infty$ analogue.

(b) If $f$ has a uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) representative $g$: $\norm{\tau_hf - f}_\infty = \sup_x\abs{g(x - h) - g(x)} \to
0$. Conversely, suppose $\norm{\tau_hf - f}_\infty \to 0$. The mollifications $f_\varepsilon = f * \rho_\varepsilon$ are [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), and

$$
\norm{f_\varepsilon - f}_\infty
\leq \sup_{\norm y \leq \varepsilon}\norm{\tau_yf -
f}_\infty \longrightarrow 0
$$

(the convolution is an average of translates). Each $f_\varepsilon$ is uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ($\norm{\tau_hf_\varepsilon - f_\varepsilon}_\infty \leq
\norm{\tau_hf - f}_\infty$, by averaging), and a uniform limit of uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functions is one: $f$ agrees a.e. with a uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) function.

**Exercise 12.6 ★★.**

Let $p, q$ be conjugate, $f \in L^p(\R^d)$, $g \in
L^q(\R^d)$. Show that $f * g$ is defined *everywhere*, bounded, with $\norm{f*g}_\infty \leq \norm f_p\norm g_q$, and *uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)*. *([Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of translation in $L^p$; treat $p \in \{1, \infty\}$ separately — for $p =
\infty$ use translation [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on the $L^1$ factor.)*

**Solution of Exercise 12.6.**

By Hölder, for every $x$ the integrand $y \mapsto f(x -
y)g(y)$ is in $L^1$ with $\abs{(f*g)(x)} \leq \norm f_p\norm
g_q$: everywhere defined and bounded. Uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ($p <
\infty$):

$$
\abs{(f*g)(x + h) - (f*g)(x)}
= \Bigl|\int\bigl(\tau_{-h}f - f\bigr)(x - y)\,g(y)\dd y\Bigr|
\leq \norm{\tau_{-h}f - f}_p\,\norm g_q
\xrightarrow[h\to0]{} 0,
$$

uniformly in $x$ ([Theorem 12.6](#thm-b3-lp-density)(3)). If $p =
\infty$, then $q = 1$: write $f * g = g * f$ and run the same bound with the translation acting on $g \in L^1$.

**Exercise 12.7 ★★.**

Let $f \in L^1_{\mathrm{loc}}(\intoo ab)$ with $\int f\varphi' = 0$ for every $\varphi \in \mathcal
C^\infty_c(\intoo ab)$. Show that $f$ is a.e. equal to a constant. *(Fix $\chi \in \mathcal C_c^\infty$ with $\int\chi = 1$; any $\psi \in \mathcal C_c^\infty$ with $\int\psi = 0$ is a $\varphi'$; write a general test function as $\psi + (\int\psi)\chi$ and apply [Corollary 12.11](#cor-b3-lp-fundlemma) to $f - c$ with $c = \int
f\chi$.)*

**Solution of Exercise 12.7.**

Fix $\chi \in \mathcal C_c^\infty(\intoo ab)$ with $\int\chi =
1$, and set $c = \int f\chi$. Let $\varphi \in \mathcal
C_c^\infty$ be arbitrary and $\psi = \varphi -
\bigl(\int\varphi\bigr)\chi$: then $\int\psi = 0$, so $\Phi(x)
= \int_a^x\psi$ defines $\Phi \in \mathcal C_c^\infty(\intoo
ab)$ (it vanishes near both ends: near $a$ trivially, near $b$ because the total integral is $0$) with $\Phi' = \psi$. The hypothesis gives $\int f\psi = \int f\Phi' = 0$, hence

$$
\int f\varphi = \Bigl(\int\varphi\Bigr)\int f\chi = \int
c\,\varphi
\quad\text{for every } \varphi:
\qquad \int(f - c)\varphi = 0 .
$$

By [Corollary 12.11](#cor-b3-lp-fundlemma) (localized on $\intoo ab$), $f =
c$ a.e.

**Exercise 12.8 ★★★.**

(Smooth Urysohn) Let $K \subseteq U \subseteq \R^d$, $K$ [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact), $U$ open. Construct $\varphi \in \mathcal
C^\infty_c(\R^d)$ with $0 \leq \varphi \leq 1$, $\varphi = 1$ on $K$, $\operatorname{supp}\varphi \subseteq U$. *(Mollify the indicator of the $\delta$-neighborhood $K_\delta$ of $K$ with $\rho_{\delta/2}$, for $\delta$ small.)* Deduce a $\mathcal C^\infty$ partition-of-unity statement for a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) covered by finitely many [open sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology).

**Solution of Exercise 12.8.**

Let $3\delta < d(K, \R^d\setminus U)$ (positive: [Exercise 6.6](https://one-course.com/books/math/5/en/chapter/6-general-topology#exo-b3-topology-6)(b); if $U = \R^d$ any $\delta$ works), $K_\delta = \{x : d(x, K) \leq \delta\}$, and

$$
\varphi = \mathbf 1_{K_\delta} * \rho_{\delta/2} .
$$

Then $\varphi \in \mathcal C^\infty$ ([Theorem 12.9](#thm-b3-lp-regularization)(1); the indicator is $L^1$), $0 \leq \varphi \leq 1$ ($\int\rho = 1$), $\varphi = 1$ on $K$ (for $x \in K$, $\bar B(x, \delta/2) \subseteq K_\delta$, so the convolution integrates $\rho$ fully), and $\operatorname{supp}\varphi \subseteq K_{3\delta/2} \subseteq
U$: compactly supported ($K_\delta$ is bounded). Partition of unity: given $K \subseteq U_1\cup\dots\cup U_m$, choose (by compactness) [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K_i \subseteq U_i$ with $K \subseteq
\bigcup \mathring K_i$, take $\varphi_i$ as above for $(K_i,
U_i)$, and set $\psi_i = \varphi_i\prod_{j <i}(1 -
\varphi_j)$: each $\psi_i \in \mathcal C_c^\infty(U_i)$, and $\sum_i\psi_i = 1 - \prod_i(1 - \varphi_i) = 1$ on $K$.

**Exercise 12.9 ★★.**

Using interpolation ([Proposition 12.12](#prop-b3-lp-inclusions)(c)): (a) show that $L^1(\R)\cap L^\infty(\R) \subseteq L^p(\R)$ for all $p$, with $\norm f_p \leq \norm f_1^{1/p}\norm
f_\infty^{1 - 1/p}$; (b) show that $f \mapsto \norm f_p$ is, for fixed $f$, log-convex in $\frac1p$, and give an example where $f \in L^p$ exactly for $p$ in a given interval $(p_0, p_1)$.

**Solution of Exercise 12.9.**

(a) The interpolation exponent for $(p_0, q_0) = (1, \infty)$ at $r = p$ is $\alpha = \frac1p$: [Proposition 12.12](#prop-b3-lp-inclusions)(c) gives $\norm f_p \leq \norm
f_1^{1/p}\norm f_\infty^{1 - 1/p}$.

(b) Taking logarithms in [Proposition 12.12](#prop-b3-lp-inclusions)(c): $\ln\norm f_r \leq
\alpha\ln\norm f_p + (1-\alpha)\ln\norm f_q$ where $\frac1r$ is the same convex combination of $\frac1p, \frac1q$: $\frac1p
\mapsto \ln\norm f_p$ is convex. Example with $L^p$-membership exactly on $\intoo{p_0}{p_1}$:

$$
f(x) = x^{-1/p_1}\,\mathbf 1_{\intoo01}(x) +
x^{-1/p_0}\,\mathbf 1_{\intco1\infty}(x):
$$

the first term is $L^p$ iff $p < p_1$, the second iff $p >
p_0$.

**Exercise 12.10 ★★.**

(Jensen) Let $\mu$ be a *probability* [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure), $f \in
L^1(\mu)$ real, and $\Phi \colon \R \to \R$ convex. Show

$$
\Phi\Bigl(\int f\,\dd\mu\Bigr) \leq \int \Phi\circ
f\,\dd\mu
$$

*(support line of $\Phi$ at the point $m = \int f$)*. Deduce the arithmetic–geometric inequality and the monotonicity of $p \mapsto \norm f_p$ of [Exercise 12.1](#exo-b3-lp-1)(b).

**Solution of Exercise 12.10.**

Let $m = \int f\,\dd\mu \in \R$. Convexity provides a support line at $m$: there is $s$ with $\Phi(t) \geq \Phi(m) + s(t -
m)$ for all $t$ (take $s$ between the one-sided derivatives, which exist for convex functions). Substitute $t = f(x)$ and integrate against the probability $\mu$:

$$
\int\Phi\circ f\,\dd\mu \geq \Phi(m) + s\Bigl(\int f - m\Bigr)
= \Phi\Bigl(\int f\,\dd\mu\Bigr)
$$

([measurability](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-measurable): $\Phi$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity); [integrability](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) of the negative part of $\Phi\circ f$ is guaranteed by the support line). AM–GM: on a finite set with weights $w_i$, take $\Phi = \exp$ and $f = \sum(\ln a_i)\mathbf 1_i$: $\exp\bigl(\sum w_i\ln a_i\bigr) \leq \sum w_ia_i$, i.e. $\prod a_i^{w_i} \leq \sum w_ia_i$. The norm monotonicity is [Exercise 12.1](#exo-b3-lp-1)(b).

**Exercise 12.11 ★★★.**

(Young’s convolution inequality) Let $1 \leq p, q, r \leq
\infty$ with $\frac1p + \frac1q = 1 + \frac1r$, and $f \in
L^p(\R^d)$, $g \in L^q(\R^d)$. (a) Prove $\norm{f * g}_r \leq \norm f_p\,\norm g_q$. *(Write, for conjugate exponents worked out from $p, q,
r$, $$\abs{f(y)g(x-y)} = \bigl(\abs f^p\abs g^q\bigr)^{1/r}
\cdot\abs f^{\,p(1/p - 1/r)}\cdot\abs g^{\,q(1/q - 1/r)},$$ and apply the three-factor Hölder inequality with exponents $r$, $\frac{pr}{r - p}$, $\frac{qr}{r - q}$; then integrate in $x$ by Tonelli.)* (b) Check the three special cases already known: $r =
\infty$ (Hölder, [Exercise 12.6](#exo-b3-lp-6)); $q = 1$ ($L^p$-stability of convolution by an [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) kernel); $p = q = 1$ ($L^1$ is a convolution algebra, [Theorem 11.9](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-convolution)). (c) Why is there no inequality with $\frac1p + \frac1q < 1 +
\frac1r$? *(Test on dilations $f_\lambda(x) =
f(\lambda x)$ and compare the scalings of both sides.)*

**Solution of Exercise 12.11.**

(a) Assume first $p, q, r < \infty$ and $f, g \geq 0$ (replace by absolute values). The three exponents $r$, $\alpha = \frac{pr}{r-p}$, $\beta = \frac{qr}{r-q}$ satisfy $\frac1r + \frac1\alpha + \frac1\beta = \frac1r +
\frac1p - \frac1r + \frac1q - \frac1r = 1$ (the scaling relation). Split, for fixed $x$,

$$
f(y)g(x-y) = \bigl[f(y)^pg(x-y)^q\bigr]^{1/r}\cdot
f(y)^{1 - p/r}\cdot g(x-y)^{1 - q/r},
$$

and Hölder with the three exponents gives

$$
(f*g)(x) \leq \Bigl(\int f^pg(x-\cdot)^q\Bigr)^{1/r}
\norm f_p^{\,p/\alpha\cdot\alpha/p}\cdots
$$

more precisely: the second factor is $\bigl(\int f^{(1-p/r)\alpha}\bigr)^{1/\alpha} =
\norm f_p^{p(1/p - 1/r)}$ since $(1 - \frac pr)\alpha = p$, and likewise the third is $\norm g_q^{q(1/q - 1/r)}$. Raise to the $r$-th power and integrate in $x$ (Tonelli on the first factor):

$$
\norm{f*g}_r^r \leq \norm f_p^p\,\norm g_q^q\cdot
\norm f_p^{\,rp(1/p - 1/r)}\,\norm g_q^{\,rq(1/q - 1/r)}
= \norm f_p^r\,\norm g_q^r .
$$

The endpoint cases ($r = \infty$ or an exponent equal to its bound) are plain Hölder or direct estimates.

(b) $r = \infty$ forces $q = p'$: $\abs{f*g(x)} \leq \norm
f_p\norm g_{p'}$ — Hölder after translation-reflection. $q = 1$ gives $r = p$: $\norm{f*g}_p \leq \norm
g_1\norm f_p$, the mollification workhorse ([Theorem 12.9](#thm-b3-lp-regularization)’s engine). $p = q = 1$ gives $r = 1$: the convolution algebra ([Theorem 11.9](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-convolution)).

(c) Replace $f, g$ by $f_\lambda = f(\lambda\cdot)$, $g_\lambda = g(\lambda\cdot)$: then $f_\lambda*g_\lambda =
\lambda^{-d}(f*g)(\lambda\cdot)$, and comparing norms,

$$
\text{LHS} \sim \lambda^{-d - d/r}, \qquad
\text{RHS} \sim \lambda^{-d/p - d/q} :
$$

an inequality valid for all $f, g$ forces the two scaling exponents to match, i.e. $1 + \frac1r = \frac1p + \frac1q$ exactly. Any other combination dies at $\lambda \to 0$ or $\infty$: Young’s relation is not a convenience but a scaling law.

**Exercise 12.12 ★★.**

(Equality cases) (a) In Hölder’s inequality $\int\abs{fg} \leq \norm f_p\norm g_q$ ($1 < p < \infty$), show that equality holds iff $\abs f^p$ and $\abs g^q$ are proportional a.e. *(Track the equality case of Young’s inequality $ab \leq \frac{a^p}p + \frac{b^q}q$, which is $a^p
= b^q$.)* (b) In Minkowski’s inequality $\norm{f + g}_p \leq \norm f_p
+ \norm g_p$ ($1 < p < \infty$), show that equality with $f,
g \neq 0$ forces $g = cf$ a.e. with $c > 0$. (c) Contrast with $p = 1$ and $p = \infty$: describe the (much larger) equality cases there, on examples.

**Solution of Exercise 12.12.**

(a) Normalize $\norm f_p = \norm g_q = 1$. The proof of Hölder integrates Young’s inequality $\abs{fg} \leq
\frac{\abs f^p}p + \frac{\abs g^q}q$; equality of the integrals forces equality a.e. in Young, which (strict convexity of $\exp$; equality iff $a^p = b^q$) means $\abs
f^p = \abs g^q$ a.e. Undoing the normalization: $\abs f^p\norm g_q^q = \abs g^q\norm f_p^p$ a.e. — proportionality.

(b) Minkowski is two Hölders applied to $\abs{f + g}^{p-1}
\abs f$ and $\abs{f+g}^{p-1}\abs g$; equality forces (a)’s proportionality in both: $\abs f^p$ and $\abs g^p$ each proportional to $\abs{f+g}^{(p-1)q} = \abs{f+g}^p$, so $\abs g = t\abs f$ a.e. for a constant $t \geq 0$; and the initial pointwise triangle inequality $\abs{f + g} \leq
\abs f + \abs g$ must also be an a.e. equality, which for complex values means $f$ and $g$ have a.e. the same argument where both are nonzero. Combining: $g = tf$ a.e., $t > 0$ (both nonzero).

(c) $p = 1$: equality in $\int\abs{f + g} = \int\abs f +
\int\abs g$ holds whenever $f, g$ have the same sign pattern (same argument a.e.) — no proportionality needed: $f = \mathbf 1_{\intcc01}$ and $g = \mathbf 1_{\intcc02}$ work. $p = \infty$: $\norm{f+g}_\infty = \norm f_\infty +
\norm g_\infty$ as soon as the two functions peak compatibly at a common point (or along a common sequence): $f = g\,$ near one point suffices regardless of behavior elsewhere. The strict convexity of the $L^p$ balls for $1 <
p < \infty$ — and its failure at the endpoints — is exactly what these equality cases witness.

## 12.7 Problem: Hardy’s inequality

**Problem 12.1.**

Weekend problem — Hardy’s inequality and its sharp constant

For $f \in L^p(\intoo0{+\infty})$, $1 < p < \infty$, define the *Hardy operator*

$$
(Hf)(x) = \frac1x\int_0^x f(t)\,\dd t .
$$

Hardy’s inequality (1920) asserts

$$
\norm{Hf}_p \;\leq\; \frac{p}{p-1}\,\norm f_p ,
$$

and the constant $\frac p{p-1}$ is optimal and not attained. This problem proves everything, then extends to series.

**Part I — The inequality.** Assume first $f
\geq 0$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) support in $\intoo0{+\infty}$, and let $F(x) = \int_0^xf$.

1. Show that $Hf \in L^p$ : near $0$ , $F$ vanishes on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $0$ ; near $\infty$ , $F$ is bounded, so $(Hf)(x) = O(1/x)$ , and $x \mapsto \frac1x$ belongs to $L^p(\intoo1{+\infty})$ for $p > 1$ .
2. Integrate by parts to show $$\int_0^\infty \Bigl(\frac Fx\Bigr)^p\dd x  = \frac{p}{p-1}\int_0^\infty\Bigl(\frac  Fx\Bigr)^{p-1}f(x)\,\dd x .$$ *(Differentiate $x^{1-p}F^p$; boundary terms vanish — justify both ends.)*
3. Apply Hölder to the right side and deduce $\norm{Hf}_p \leq \frac p{p-1}\norm f_p$ for such $f$ .
4. Extend to all of $L^p$ : for $f \geq 0$ , construct $f_n$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) support in $\intoo0{+\infty}$ , $0 \leq f_n \nearrow f$ a.e. *(truncate, then approximate monotonically — justify the construction)* ; then $Hf_n \nearrow Hf$ pointwise (MCT inside the average) and MCT passes the inequality to the limit. For signed or complex $f$ , conclude with $\abs{Hf} \leq H\abs f$ .

**Part II — Optimality.**

5. For $A > 1$ let $f_A(t) = t^{-1/p}\,\mathbf  1_{\intcc1A}(t)$. Compute $\norm{f_A}_p^p = \ln A$ and, for $1 \leq x \leq A$, $$(Hf_A)(x) = \frac p{p-1}\;x^{-1/p}\,  \bigl(1 - x^{-(1 - 1/p)}\bigr).$$
6. Deduce $\liminf_{A\to\infty} \norm{Hf_A}_p/  \norm{f_A}_p \geq \frac{p}{p-1}$ , and conclude that the constant is optimal.
7. Show that equality $\norm{Hf}_p = \frac  p{p-1}\norm f_p$ with $f \neq 0$ is impossible. *(Track the equality case of Hölder in question 3: it would force $f = cx^{-1/p}$-type behavior, which is not in $L^p$.)*

**Part III — The discrete inequality.**

8. For a nonincreasing $g \geq 0$ on $(0,\infty)$ and $a_n = g(n)$, compare $\sum a_n^p$ and $\int g^p$, and $H$-averages accordingly, to deduce from Part I *Hardy’s discrete inequality*: for $a_n \geq 0$, $$\sum_{n\geq1}\Bigl(\frac{a_1 + \dots +  a_n}{n}\Bigr)^{p}  \;\leq\;  \Bigl(\frac{p}{p-1}\Bigr)^{p}\,\sum_{n\geq1}a_n^p$$ — prove it first for nonincreasing $(a_n)$ via the comparison above, then reduce the general case to the nonincreasing one by rearrangement *(admit, with a one-line justification, that sorting $(a_n)$ in decreasing order can only increase the left side while fixing the right)*.
9. Deduce: if $\sum a_n^p < \infty$ then the Cesàro means of $(a_n)$ are again $\ell^p$ — and give an example ( $p = 2$ ) where $(a_n) \in \ell^2$ but $a_n$ is not summable, yet Hardy still controls the means.

**Part IV — Epilogue.**

10. Show that Hardy’s inequality fails for $p = 1$ : with $f = \mathbf 1_{\intcc01}$ , compute $Hf$ and observe $Hf \notin L^1$ . Where does the proof break?

**Part V — The maximal function, and Lebesgue’s differentiation theorem.** Hardy averages from the origin; Hardy–Littlewood average *around each point*. For $f \in L^1(\R)$ define

$$
Mf(x) = \sup_{r>0}\ \frac1{2r}\int_{x-r}^{x+r}\abs
f\,\dd\lambda .
$$

11. (Vitali, finite version) Let $B_1, \dots, B_N$ be open intervals. Show there is a disjoint subfamily $B_{i_1}, \dots, B_{i_k}$ with $\bigcup_jB_j  \subseteq \bigcup_l3B_{i_l}$ , where $3B$ denotes the interval with the same [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) and triple length *(greedy: repeatedly pick the longest interval disjoint from those already picked)* .
12. (Weak type $(1,1)$) Show that for every $t > 0$, $$\lambda\bigl(\{Mf > t\}\bigr) \;\leq\;  \frac3t\,\norm f_1 :$$ each $x$ with $Mf(x) > t$ owns a centered interval $B_x$ with $\int_{B_x}\abs f > t\,\lambda(B_x)$; take a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K \subseteq \{Mf > t\}$ (inner regularity), cover it by finitely many $B_x$, apply question 11, and exhaust.
13. Compute $M\mathbf 1_{\intcc01}(x)$ for $x > 1$ and deduce that $Mf \notin L^1$ for every $f \neq 0$ ( $Mf(x) \geq \frac c{\abs x}$ at infinity): at $p =  1$ , the weak inequality of question 12 is the best possible statement.
14. (Strong type for $p > 1$) For $f \in L^p$: split $f =  f\,\mathbf 1_{\abs f > t/2} + f\,\mathbf 1_{\abs f  \leq t/2}$, observe $Mf \leq M\bigl(f\mathbf 1_{\abs  f > t/2}\bigr) + \frac t2$, and combine question 12 with the layer-cake formula ([Proposition 11.8](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#prop-b3-product-layercake)) and Tonelli to prove $$\norm{Mf}_p^p \;\leq\;  \frac{6p\,2^{p-1}}{p-1}\,\norm f_p^p .$$ (The blow-up as $p \downarrow 1$ is question 13’s failure, quantified.)
15. (Lebesgue differentiation theorem) Prove: for $f \in  L^1(\R)$, $$\frac1{2r}\int_{x-r}^{x+r}f\,\dd\lambda  \;\longrightarrow\; f(x)  \qquad (r \to 0)\quad\text{for a.e.\ }x .$$ *(Clear for [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$. In general write $f  = g + h$, $g$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) support, $\norm h_1 < \varepsilon$ ([Theorem 12.6](#thm-b3-lp-density)); the set where $\limsup_{r\to0}$ of the averaged oscillation exceeds $\delta$ sits inside $\{Mh > \delta/2\} \cup \{\abs h  > \delta/2\}$, of [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) $O(\varepsilon/\delta)$; let $\varepsilon \to 0$, then $\delta \to 0$ along a sequence.)*
16. Deduce: (a) almost every point is a *Lebesgue point* of $f$ ; (b) for $f \in L^1$ , the primitive $F(x) = \int_0^xf$ is differentiable a.e. with $F' =  f$ a.e. — the integral half of the fundamental theorem of calculus in the Lebesgue world, closing the circle opened by the staircase ( [Problem 9.1](https://one-course.com/books/math/5/en/chapter/9-measure-theory#pb-b3-measure-1) ), which showed the converse half can fail.
17. ( [Density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) points) For [measurable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-measurable) $A \subseteq \R$ , show that almost every $x \in A$ satisfies $\frac{\lambda(A\cap\intcc{x-r}{x+r})}{2r} \to 1$ : [measurable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-measurable) sets are locally full at almost all of their points. Sketch, in two lines, how this yields yet another proof of Steinhaus’ theorem ( [Exercise 9.8](https://one-course.com/books/math/5/en/chapter/9-measure-theory#exo-b3-measure-8) ).

**Part VI — Variations on the theme of averaging.**

18. (Weighted Hardy) For $\alpha < p - 1$ and $f \geq 0$, show $$\int_0^\infty\Bigl(\frac{F(x)}x\Bigr)^{p}  x^{\alpha}\,\dd x \;\leq\;  \Bigl(\frac{p}{p - 1 - \alpha}\Bigr)^{p}  \int_0^\infty f(x)^p\,x^{\alpha}\,\dd x$$ by the same integration by parts, and check that the borderline $\alpha = p - 1$ is genuinely forbidden (adapt question 10’s counterexample).
19. (The adjoint) Let $H^*f(x) =  \int_x^{\infty}\frac{f(t)}t\,\dd t$ . Show $\langle  Hf, g\rangle = \langle f, H^*g\rangle$ for nonnegative $f, g$ (Tonelli), and prove $\norm{H^*f}_p \leq p\,\norm f_p$ *(directly by parts, or from Hardy on the conjugate exponent by duality — mind which exponent picks up which constant)* .
20. (A Hilbert-type inequality) Deduce that for nonnegative $f \in L^p$, $g \in L^q$: $$\int_0^\infty\!\!\int_0^\infty  \frac{f(x)\,g(y)}{\max(x,y)}\,\dd x\,\dd y  \;\leq\; (p + q)\,\norm f_p\,\norm g_q$$ *(split along $y < x$ / $y \geq x$: each half is a pairing of one function against a Hardy transform of the other)*.
21. (Optimality, discrete) Show that the constant $\bigl(\frac p{p-1}\bigr)^p$ of question 8 is also optimal: test on $a_n = n^{-1/p}\,\mathbf 1_{n \leq  N}$ , compare both sides with integrals, and let $N  \to \infty$ (the discrete mirror of Part II).
22. (Synthesis) Three averaging operators appeared in this problem: Hardy’s $H$ , the discrete Cesàro mean, and the maximal operator $M$ . State in one line each what its boundedness says, observe that all three fail exactly at $p = 1$ , and explain why it is the same failure three times (the harmonic tail $\frac1x$ ).

**Part VII — Carleman’s inequality, and how sharp is sharp.**

23. (Carleman’s inequality) Let $a_n \geq 0$ with $\sum  a_n < \infty$. Apply the discrete Hardy inequality of question 8 to $b_n = a_n^{1/p}$, use the arithmetic–geometric mean inequality, and let $p \to  \infty$ (show that $p \mapsto  \bigl(\frac p{p-1}\bigr)^p$ decreases to $\eu$) to obtain $$\sum_{n\geq1}\bigl(a_1a_2\cdots a_n\bigr)^{1/n}  \;\leq\; \eu\,\sum_{n\geq1}a_n :$$ the geometric means of a summable sequence are summable, at cost at most $\eu$.
24. (The constant $\eu$ is optimal) Test $a_n =  \frac1n\,\mathbf 1_{n\leq N}$ : using the Stirling bracketing of [Problem 11.1](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#pb-b3-product-1) , show $(n!)^{-1/n} = \frac\eu n\bigl(1 +  O\bigl(\frac{\ln n}n\bigr)\bigr)$ , deduce that both sides of Carleman grow like $\eu\ln N$ , and conclude that no constant smaller than $\eu$ can work. (Observe the pattern: the optimizers of Hardy and of Carleman are both the harmonic-type sequences that just fail to be in the space.)
25. (How slowly is “sharp” approached?) Take $p = 2$. For $f = \mathbf 1_{\intcc01}$, compute $\norm{Hf}_2/\norm f_2 = \sqrt2$, against the bound $2$. For the near-optimizers $f_A$ of question 5, prove the exact identity $$\norm{Hf_A}_2^2 = 4\ln A - 8 + \frac{8}{\sqrt A},  \qquad\text{so}\qquad  \frac{\norm{Hf_A}_2^2}{\norm{f_A}_2^2}  = 4 - \frac{8 - 8A^{-1/2}}{\ln A} .$$ Evaluate at $A = \eu^{10}$ (ratio $\approx 1.790$) and comment: the supremum $2$ is approached at speed $1/\ln A$ only — an optimal constant can be all but invisible numerically.

**Solution of Problem 12.1.**

**1.** $f$ has support in some $[\alpha, \beta]
\subseteq \intoo0\infty$, so $F = 0$ on $[0, \alpha]$ and $F
\equiv F(\beta)$ on $[\beta, \infty)$: $Hf$ vanishes near $0$ and is $O(1/x)$ at infinity; $\int_1^\infty x^{-p}\dd x <
\infty$ for $p > 1$, and $Hf$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): $Hf \in L^p$.

**2.** $\bigl(x^{1-p}F(x)^p\bigr)' = (1-p)x^{-p}F^p +
p\,x^{1-p}F^{p-1}f$. Both boundary values vanish: at $0$ because $F = 0$ near $0$; at $\infty$ because $x^{1-p}F^p \leq
F(\beta)^p x^{1-p} \to 0$ ($p > 1$). Integrating the identity over $\intoo0\infty$:

$$
0 = (1 - p)\int_0^\infty\Bigl(\frac Fx\Bigr)^p\dd x
+ p\int_0^\infty\Bigl(\frac Fx\Bigr)^{p-1}f(x)\,\dd x,
$$

which is the displayed relation.

**3.** Hölder with exponents $q = \frac p{p-1}$ and $p$:

$$
\int\Bigl(\frac Fx\Bigr)^{p-1}f
\leq \Bigl(\int\Bigl(\frac
Fx\Bigr)^{p}\Bigr)^{1 - 1/p}\,\norm f_p ,
$$

so $\norm{Hf}_p^p \leq \frac p{p-1}\norm{Hf}_p^{p-1}\norm
f_p$; divide (finite by question 1, and if $0$ there is nothing to prove).

**4.** Let $f \in L^p$, $f \geq 0$. Choose $\varphi_k \in
\mathcal C_c(\intoo0\infty)$ with $\varphi_k \to f$ in $L^p$ ([Theorem 12.6](#thm-b3-lp-density)(2), intersected with the open half-line — approximate $f\mathbf 1_{[1/k, k]}$ and diagonalize), and replace $\varphi_k$ by $\abs{\varphi_k}$ (still [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), and closer to $f \geq 0$). For every fixed $x > 0$, Hölder on $\intoo0x$ gives

$$
\abs{H\varphi_k(x) - Hf(x)} \leq
\frac1x\,x^{1 - 1/p}\,\norm{\varphi_k - f}_p \to 0 :
$$

$H\varphi_k \to Hf$ pointwise. Fatou and question 3:

$$
\int (Hf)^p \leq \liminf_k\int(H\varphi_k)^p
\leq \Bigl(\frac p{p-1}\Bigr)^p\liminf_k\norm{\varphi_k}_p^p
= \Bigl(\frac p{p-1}\Bigr)^p\norm f_p^p .
$$

For signed or complex $f$: $\abs{Hf} \leq H\abs f$ pointwise, and the nonnegative case applies to $\abs f$.

**5.** $\norm{f_A}_p^p = \int_1^A t^{-1}\dd t = \ln A$. For $1 \leq x \leq A$:

$$
(Hf_A)(x) = \frac1x\int_1^x t^{-1/p}\dd t
= \frac{x^{1 - 1/p} - 1}{(1 - \tfrac1p)\,x}
= \frac p{p-1}\,x^{-1/p}\bigl(1 - x^{-(1 - 1/p)}\bigr).
$$

**6.** Fix $\varepsilon > 0$ and $X_0$ with $(1 -
x^{-(1-1/p)})^p \geq 1 - \varepsilon$ for $x \geq X_0$. Then

$$
\norm{Hf_A}_p^p \geq \int_{X_0}^A\Bigl(\frac
p{p-1}\Bigr)^p\frac{1 - \varepsilon}{x}\,\dd x
= \Bigl(\frac p{p-1}\Bigr)^p(1 - \varepsilon)\,(\ln A - \ln
X_0),
$$

so $\dfrac{\norm{Hf_A}_p^p}{\norm{f_A}_p^p} \geq \bigl(\frac
p{p-1}\bigr)^p(1 - \varepsilon)\bigl(1 - \frac{\ln X_0}{\ln
A}\bigr) \to \bigl(\frac p{p-1}\bigr)^p(1 - \varepsilon)$ as $A \to \infty$: the constant cannot be improved.

**7.** Equality in question 3 forces equality in Hölder: $f^p$ proportional to $\bigl(\frac Fx\bigr)^{(p-1)q}
= \bigl(\frac Fx\bigr)^p$ a.e., i.e. $f = \gamma\,\frac Fx$ a.e. for some $\gamma \geq 0$. Since $F(x) = \int_0^xf$ is absolutely [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with $F' = f$ a.e., $F$ solves $F' =
\gamma F/x$: on any interval where $F > 0$, $(\ln F)' =
\gamma/x$, so $F = c\,x^{\gamma}$ and $f = c\gamma
x^{\gamma - 1}$ there. But no nonzero power $x^{\gamma-1}$ belongs to $L^p(\intoo0\infty)$ ($p(\gamma - 1) < -1$ needed at $\infty$ and $> -1$ at $0$: incompatible), and $F$ cannot vanish identically unless $f = 0$. So equality demands $f =
0$.

**8.** Given $(a_n)$ nonincreasing $\geq 0$, define the step function $g(t) = a_{\lceil t\rceil}$ on $\intoo0{+\infty}$: nonincreasing, with $\int_0^\infty g^p =
\sum_na_n^p$ and $\int_0^ng = a_1 + \dots + a_n$, so $(Hg)(n) = \frac{a_1 + \dots + a_n}n$. The average of a nonincreasing function is nonincreasing, so $Hg$ is, and

$$
\sum_{n\geq1}\Bigl(\frac{a_1{+}\dots{+}a_n}n\Bigr)^p
= \sum_{n\geq1}(Hg)(n)^p
\leq \sum_{n\geq1}\int_{n-1}^n (Hg)(t)^p\,\dd t
= \norm{Hg}_p^p
\leq \Bigl(\frac p{p-1}\Bigr)^p\sum_n a_n^p
$$

by Part I. For a general nonnegative sequence, let $(a_n^*)$ be its nonincreasing rearrangement (possible when $a_n \to 0$, which we may assume — otherwise both sides are infinite): the right side is unchanged, and each partial sum $a_1 + \dots
+ a_n$ is at most $a_1^* + \dots + a_n^*$ (the $n$ largest terms): the left side only grows. Hence the inequality for all $(a_n)$.

**9.** If $(a_n) \in \ell^p$, the sequence of Cesàro means is in $\ell^p$ with norm $\leq \frac p{p-1}\norm
a_p$. Example ($p = 2$): $a_n = \frac1{\sqrt n\,\ln n}$ ($n
\geq 2$): $\sum a_n^2 = \sum\frac1{n\ln^2n} < \infty$, yet $\sum a_n = \infty$ (integral test); Hardy still guarantees $\sum_n\bigl(\frac{a_1 + \dots + a_n}n\bigr)^2 < \infty$.

**10.** For $f = \mathbf 1_{\intcc01}$: $Hf(x) = 1$ on $\intoc01$ and $= \frac1x$ for $x \geq 1$: $\int_0^\infty Hf =
1 + \int_1^\infty\frac{\dd x}x = \infty$, while $\norm f_1 =
1$. The proof collapses at two points: the constant $\frac p{p-1}$ blows up as $p \to 1$, and the boundary term $x^{1-p}F^p$ no longer vanishes at infinity for $p = 1$. Hardy’s inequality is an honest $p > 1$ phenomenon.

**11.** Pick the longest interval $B_{i_1}$; discard every interval meeting it; pick the longest survivor $B_{i_2}$; iterate (finitely many intervals). The chosen ones are disjoint by construction, and every discarded $B$ met a chosen interval at least as long: an interval meeting a longer-or-equal one is contained in its triple, $B \subseteq
3B_{i_l}$.

**12.** $\{Mf > t\}$ is open: each average $x \mapsto
\frac1{2r}\int_{x-r}^{x+r}\abs f$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (dominated convergence in $x$), and a supremum of [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functions is lower semicontinuous. Each $x$ in it owns $B_x =
\intoo{x-r_x}{x+r_x}$ with $\int_{B_x}\abs f >
t\,\lambda(B_x)$. For [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K \subseteq \{Mf > t\}$: finitely many $B_x$ cover $K$, Vitali (question 11) extracts disjoint $B_1', \dots, B_k'$ with $K \subseteq
\bigcup_l3B_l'$, so

$$
\lambda(K) \leq 3\sum_l\lambda(B_l') <
\frac3t\sum_l\int_{B_l'}\abs f \leq \frac3t\,\norm f_1
$$

by disjointness; inner regularity ([Theorem 9.13](https://one-course.com/books/math/5/en/chapter/9-measure-theory#thm-b3-measure-regularity)) concludes.

**13.** For $x > 1$: with $r \in \intcc{x-1}x$ the average is $\frac{r - x + 1}{2r}$, increasing in $r$; for $r
\geq x$ it is $\frac1{2r}$, decreasing: the supremum is $\frac1{2x}$, attained at $r = x$. So $M\mathbf 1_{\intcc01}
\notin L^1$. In general, if $\int_I\abs f = c > 0$ on a bounded interval $I \subseteq \intcc{-C}C$, then $Mf(x) \geq
\frac{c}{2(\abs x + C)}$ for all $x$: never [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) unless $f = 0$ a.e.

**14.** With $f_t = f\,\mathbf 1_{\abs f > t/2}$: $M(f - f_t) \leq \frac t2$, so $\{Mf > t\} \subseteq \{Mf_t
> \frac t2\}$ and question 12 gives $\lambda(Mf > t) \leq
\frac6t\int_{\abs f > t/2}\abs f$. Layer cake ([Proposition 11.8](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#prop-b3-product-layercake)) and Tonelli:

$$
\norm{Mf}_p^p = p\int_0^\infty t^{p-1}\lambda(Mf > t)\dd t
\leq 6p\int\abs f\int_0^{2\abs f}t^{p-2}\,\dd t\,\dd\lambda
= \frac{6p\,2^{p-1}}{p-1}\,\norm f_p^p .
$$

**15.** Write $A_rf(x) = \frac1{2r}\int_{x-r}^{x+r}f$. For [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $g$: $A_rg(x) \to g(x)$ everywhere. Given $\varepsilon > 0$, split $f = g + h$ with $g$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) support and $\norm h_1 < \varepsilon$ ([Theorem 12.6](#thm-b3-lp-density)); then

$$
\limsup_{r\to0}\,\abs{A_rf(x) - f(x)} \leq Mh(x) +
\abs{h(x)},
$$

so $\Omega_\delta = \{\limsup_r\abs{A_rf - f} > \delta\}
\subseteq \{Mh > \tfrac\delta2\} \cup \{\abs h >
\tfrac\delta2\}$ has [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) $\leq \frac{6\varepsilon}\delta
+ \frac{2\varepsilon}\delta$ (question 12; Markov). $\varepsilon$ arbitrary: $\lambda(\Omega_\delta) = 0$; union over $\delta = \frac1k$: $A_rf \to f$ a.e.

**16.** (a) For each $q \in \Q$, question 15 applied to $\abs{f - q}$ gives $A_r\abs{f - q}(x) \to \abs{f(x) - q}$ a.e.; on the intersection of these [full-measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) sets, choose $q$ with $\abs{f(x) - q} < \eta$: $\limsup_rA_r\abs{f - f(x)}(x) \leq 2\eta$ for every $\eta$: almost every $x$ is a Lebesgue point. (b) At a Lebesgue point,

$$
\Bigl|\frac{F(x + h) - F(x)}h - f(x)\Bigr|
= \Bigl|\frac1h\int_x^{x+h}\bigl(f - f(x)\bigr)\Bigr|
\leq 2\,A_{\abs h}\abs{f - f(x)}(x) \to 0 :
$$

$F' = f$ a.e. — primitives of $L^1$ functions do differentiate back; the staircase ([Problem 9.1](https://one-course.com/books/math/5/en/chapter/9-measure-theory#pb-b3-measure-1)) is the counterexample to the *converse* direction only.

**17.** Apply question 15 to $\mathbf 1_{A\cap[-n,n]}$ and let $n$ grow: for a.e. $x \in A$ the [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) $\frac{\lambda(A\cap\intcc{x-r}{x+r})}{2r} \to 1$. Steinhaus: around a [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) point take $r$ with [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) $> \frac34$; for $\abs t < \frac r2$, $A$ and $A + t$ each fill more than $\frac32r$ of an interval of length $\leq \frac52r$, hence intersect: $\intoo{-r/2}{r/2} \subseteq A - A$.

**18.** Let $G(x) = x^{\alpha+1-p}F(x)^p$: $G$ vanishes at $0$ ($F$ vanishes near $0$) and at $\infty$ ($F$ bounded, $\alpha + 1 - p < 0$), so $\int_0^\infty G' = 0$ with

$$
G' = (\alpha + 1 - p)\,x^{\alpha-p}F^p +
p\,x^{\alpha+1-p}F^{p-1}f
= (\alpha + 1 - p)\Bigl(\frac Fx\Bigr)^px^\alpha +
p\Bigl(\frac Fx\Bigr)^{p-1}f\,x^\alpha .
$$

Hence $\int(F/x)^px^\alpha = \frac{p}{p-1-\alpha}
\int(F/x)^{p-1}f\,x^\alpha$; Hölder for the [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) $x^\alpha\dd x$ (exponents $\frac p{p-1}$ and $p$) finishes as in question 3. Borderline $\alpha = p - 1$: with $f(t) =
\frac1t\mathbf 1_{\intcc1A}$, the right side is $\ln A$ while the left contains $\int_1^A\frac{(\ln x)^p}x\dd x =
\frac{(\ln A)^{p+1}}{p+1}$: no constant survives $A \to
\infty$.

**19.** Tonelli on $\{0 < t < x\}$:

$$
\langle Hf, g\rangle =
\int_0^\infty\frac{g(x)}x\int_0^xf(t)\,\dd t\,\dd x
= \int_0^\infty f(t)\int_t^\infty\frac{g(x)}x\,\dd x\,\dd t
= \langle f, H^*g\rangle .
$$

Duality: $\norm{H^*f}_p = \sup\{\langle f, Hg\rangle : g
\geq 0, \norm g_q \leq 1\} \leq \norm f_p\cdot
\sup\norm{Hg}_q \leq \frac q{q-1}\norm f_p = p\,\norm f_p$, Hardy being invoked in $L^q$, whose constant $\frac q{q-1}$ equals $p$.

**20.** Split along the (null) diagonal. On $\{y \leq
x\}$:

$$
\iint_{y\leq x}\frac{f(x)g(y)}{x}\,\dd y\,\dd x
= \int f(x)\,(Hg)(x)\,\dd x \leq \norm f_p\,\norm{Hg}_q
\leq p\,\norm f_p\norm g_q,
$$

since Hardy in $L^q$ carries the constant $\frac q{q-1} =
p$. Symmetrically, $\iint_{y>x} = \int g\,(Hf) \leq
\norm g_q\norm{Hf}_p \leq q\,\norm f_p\norm g_q$ ($\frac
p{p-1} = q$). Total: $(p + q)\,\norm f_p\norm g_q$.

**21.** For $a_n = n^{-1/p}$, $n \leq N$: the right side is $\bigl(\frac p{p-1}\bigr)^p\sum_{n\leq N}\frac1n =
\bigl(\frac p{p-1}\bigr)^p\ln N + O(1)$. On the left, for $n
\leq N$: $a_1 + \dots + a_n \geq \int_1^{n+1}t^{-1/p}\dd t =
\frac{p}{p-1}\bigl((n+1)^{1-1/p} - 1\bigr)$, so the $n$-th Cesàro mean is $\geq \frac p{p-1}n^{-1/p}(1 -
o(1))$ uniformly for $n$ in any range $n \geq n_0(\eta)$; raising to the $p$ and summing, the left side is $\geq
\bigl(\frac p{p-1}\bigr)^p(1 - \eta)\ln N + O_\eta(1)$. Dividing and letting $N \to \infty$, then $\eta \to 0$: no constant smaller than $\bigl(\frac p{p-1}\bigr)^p$ can work.

**22.** $H$ bounded on $L^p$: cumulative averages do not inflate $p$-norms (constant $\frac p{p-1}$); Cesàro on $\ell^p$: the same, discretized; $M$ bounded on $L^p$: even the best local average stays under control (constant $O(\frac1{p-1})$). All three fail at $p = 1$, and for one reason: averaging a concentrated unit of mass produces a $\frac1x$ tail (questions 10 and 13), and $\frac1x$ belongs to every $L^p$ near infinity except $L^1$. Smoothing spreads mass exactly to the harmonic frontier of [integrability](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1).

**23.** Set $b_n = a_n^{1/p}$, so $\sum b_n^p = \sum
a_n$. Question 8 gives

$$
\sum_{n\geq1}\Bigl(\frac{b_1 + \dots + b_n}n\Bigr)^{p}
\leq \Bigl(\frac p{p-1}\Bigr)^{p}\sum_{n\geq1}a_n,
$$

and AM–GM bounds each summand from below:

$$
\Bigl(\frac{b_1 + \dots + b_n}n\Bigr)^{p}
\geq \bigl(b_1\cdots b_n\bigr)^{p/n}
= \bigl(a_1\cdots a_n\bigr)^{1/n}.
$$

Hence $\sum(a_1\cdots a_n)^{1/n} \leq \bigl(\frac
p{p-1}\bigr)^p\sum a_n$ for *every* $p > 1$. With $m =
p - 1$,

$$
\Bigl(\frac p{p-1}\Bigr)^{p} =
\Bigl(1 + \frac1m\Bigr)^{m+1},
$$

which decreases to $\eu$ as $m \to \infty$ (the classical monotone upper sequence for $\eu$). Taking the infimum over $p$ gives Carleman’s inequality with constant $\eu$.

**24.** For $a_n = \frac1n$, $(a_1\cdots a_n)^{1/n} =
(n!)^{-1/n}$. The bracketing of [Problem 11.1](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#pb-b3-product-1) gives $\sqrt{2\pi n}\,(n/\eu)^n \leq n! \leq \sqrt{2\pi
n}\,(n/\eu)^n\eu^{1/(12n)}$, so

$$
(n!)^{1/n} = \frac n\eu\,(2\pi n)^{1/(2n)}
\eu^{O(1/n^2)}
= \frac n\eu\Bigl(1 +
O\Bigl(\frac{\ln n}{n}\Bigr)\Bigr),
$$

since $(2\pi n)^{1/(2n)} = \exp\bigl(\frac{\ln(2\pi
n)}{2n}\bigr)$. Inverting, $(n!)^{-1/n} = \frac\eu n(1 +
O(\frac{\ln n}n))$, and summing over $n \leq N$:

$$
\sum_{n\leq N}(n!)^{-1/n}
= \eu\ln N + O(1),
\qquad
\eu\sum_{n\leq N}\frac1n = \eu\ln N + O(1)
$$

(the error series $\sum\frac{\ln n}{n^2}$ converges). A Carleman inequality with constant $c$ would force $\eu\ln N
+ O(1) \leq c\,(\ln N + O(1))$, hence $c \geq \eu$ upon dividing by $\ln N$. The optimizing sequences line up: Hardy’s constant is approached by $n^{-1/p}$ (question 21), Carleman’s by $n^{-1}$ — in each case the harmonic-type sequence sitting just outside the space being averaged.

**25.** For $f = \mathbf 1_{\intcc01}$: $Hf(x) = 1$ on $\intoc01$ and $Hf(x) = \frac1x$ for $x > 1$, so $\norm{Hf}_2^2 = 1 + \int_1^\infty x^{-2}\dd x = 2$ and the ratio is $\sqrt2 \approx 1.414$, about $71\%$ of the sharp bound. For $f_A$ ($p = 2$): $F(x) = 2(\sqrt x - 1)$ on $\intcc1A$, so on that range $Hf_A = 2x^{-1/2} - 2x^{-1}$ and, for $x > A$, $Hf_A(x) = 2(\sqrt A - 1)/x$. Squaring and integrating,

$$
\begin{align*}
\int_1^A\bigl(2x^{-1/2} - 2x^{-1}\bigr)^2\dd x
&= 4\ln A - 12 + \frac{16}{\sqrt A} - \frac4A,\\
\int_A^{\infty}\frac{4(\sqrt A - 1)^2}{x^2}\,\dd x
&= 4 - \frac{8}{\sqrt A} + \frac4A,
\end{align*}
$$

whence $\norm{Hf_A}_2^2 = 4\ln A - 8 + 8A^{-1/2}$; dividing by $\norm{f_A}_2^2 = \ln A$ gives the stated identity. At $A
= \eu^{10}$: $4 - \frac{8(1 - \eu^{-5})}{10} = 3.2054$, so the ratio is $\sqrt{3.2054} \approx 1.790 < 2$. The defect $4 - \norm{Hf_A}_2^2/\norm{f_A}_2^2 \sim 8/\ln A$ decays only logarithmically: to reach ratio $1.99$ one would need $\ln A \approx 200$, i.e. $A \approx 10^{87}$. Sharp constants are theorems, not experiments.