University Mathematics — Year 3 · Bachelor Year 3

12The 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 that Exercise 7.1 showed the continuous functions lacked — and they support a smoothing technology, convolution against mollifiers, that approximates any $L^p$ function by $\mathcal C^\infty$ ones. This chapter proves the integral versions of Hölder and Minkowski, completeness, the density 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 and functions are complex-valued; on $\R^d$ , the 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 $f$ 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), 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 $f, g$ :

\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) 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 (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(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), nor conversely (escaping bumps): the two modes are linked only through subsequences and domination. Keeping the counterexamples of Exercise 12.3 in mind is the best vaccine.

12.3 Density theorems

Theorem 12.6

Let $1 \leq p < \infty$ .

Simple functions (with finite-measure supports) are dense in $L^p(\mu)$ .
In $L^p(\R^d)$ , the continuous compactly supported functions $\mathcal C_c(\R^d)$ are dense.
Translation is continuous 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 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 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) gives 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, $1$ 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 gives $\norm{\tau_hg - g}_\infty \to 0$ , with supports in a fixed 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 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. 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. ∎

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:

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

Proof. (1) Differentiation under the integral (Theorem 10.15) 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 ( $y$ 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 $\rho_\varepsilon\dd y$ 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, 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(3)): the whole expression tends to $0$ .

(3) Approximate $f$ by $g \in \mathcal C_c$ (Theorem 12.6(2)), then $g$ by $g * \rho_\varepsilon \in \mathcal C_c^\infty$ (compact support: sum of supports). ∎

Example 12.10 (Mollifying $\abs x$ , with rates)

Take $f(x) = \abs x$ on $\R$ (locally $L^1$ ; the theorem applies on every bounded window) and a symmetric 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’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 on compacts) 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): $g = 0$ 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) 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. (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) and extend by continuity (Theorem 7.2, both sides being $L^p$ -continuous); to prove $f = 0$ , test against $\mathcal C_c^\infty$ (Corollary 12.11); to gain smoothness, convolve; to trade exponents, Hölder and interpolation. The Fourier theory of Chapter 14 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

Solution of Exercise 12.1.

(a) $p = q = 2$ in Theorem 12.2: $\abs{\int f\bar g\,\dd\mu} \leq \int\abs{fg} \leq \norm f_2\norm g_2$ — 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) 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

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; 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 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. (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

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; lower bound by integrating over $\{\abs f > \norm f_\infty - \varepsilon\}$ , of positive measure.)

Solution

Solution of Exercise 12.4.

Upper: $\norm f_p \leq \mu(X)^{1/p}\norm f_\infty$ (Proposition 12.12(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(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 representative.

Solution

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 support) degenerates for $p = \infty$ only in that the density of $\mathcal C_c$ fails there — that is the real gap: step (2) of Theorem 12.6 has no $L^\infty$ analogue.

(b) If $f$ has a uniformly continuous 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, 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 ( $\norm{\tau_hf_\varepsilon - f_\varepsilon}_\infty \leq \norm{\tau_hf - f}_\infty$ , by averaging), and a uniform limit of uniformly continuous functions is one: $f$ agrees a.e. with a uniformly continuous 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. (Continuity of translation in $L^p$ ; treat $p \in \{1, \infty\}$ separately — for $p = \infty$ use translation continuity on the $L^1$ factor.)

Solution

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 ( $p < \infty$ ):

\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(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 to $f - c$ with $c = \int f\chi$ .)

Solution

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 (localized on $\intoo ab$ ), $f = c$ a.e.

Exercise 12.8 ★★★

(Smooth Urysohn) Let $K \subseteq U \subseteq \R^d$ , $K$ 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 covered by finitely many open sets.

Solution

Solution of Exercise 12.8.

Let $3\delta < d(K, \R^d\setminus U)$ (positive: Exercise 6.6(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(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 $K_i \subseteq U_i$ 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(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

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(c) gives $\norm f_p \leq \norm f_1^{1/p}\norm f_\infty^{1 - 1/p}$ .

(b) Taking logarithms in Proposition 12.12(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, $f \in L^1(\mu)$ 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(b).

Solution

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: $\Phi$ is continuous; integrability 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(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); $q = 1$ ( $L^p$ -stability of convolution by an integrable kernel); $p = q = 1$ ( $L^1$ is a convolution algebra, Theorem 11.9). (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

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’s engine). $p = q = 1$ gives $r = 1$ : the convolution algebra (Theorem 11.9).

(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

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 with compact support in $\intoo0{+\infty}$ , and let $F(x) = \int_0^xf$ .

Show that $Hf \in L^p$ : near $0$ , $F$ vanishes on a neighborhood of $0$ ; near $\infty$ , $F$ is bounded, so $(Hf)(x) = O(1/x)$ , and $x \mapsto \frac1x$ belongs to $L^p(\intoo1{+\infty})$ for $p > 1$ .
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.)
Apply Hölder to the right side and deduce $\norm{Hf}_p \leq \frac p{p-1}\norm f_p$ for such $f$ .
Extend to all of $L^p$ : for $f \geq 0$ , construct $f_n$ continuous with 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.

For $A > 1$ let $f_A(t) = t^{-1/p}\,\mathbf 1_{\intcc1A}(t)$ . Compute $\norm{f_A}_p^p = \ln A$ and, for $1 \leq x \leq A$ ,

$(Hf_A)(x) = \frac p{p-1}\;x^{-1/p}\, \bigl(1 - x^{-(1 - 1/p)}\bigr).$
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.
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.

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).
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.

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 .

(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 and triple length (greedy: repeatedly pick the longest interval disjoint from those already picked).
(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 $K \subseteq \{Mf > t\}$ (inner regularity), cover it by finitely many $B_x$ , apply question 11, and exhaust.
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.
(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) 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.)
(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 $f$ . In general write $f = g + h$ , $g$ continuous with compact support, $\norm h_1 < \varepsilon$ (Theorem 12.6); the set where $\limsup_{r\to0}$ of the averaged oscillation exceeds $\delta$ sits inside $\{Mh > \delta/2\} \cup \{\abs h > \delta/2\}$ , of measure $O(\varepsilon/\delta)$ ; let $\varepsilon \to 0$ , then $\delta \to 0$ along a sequence.)
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), which showed the converse half can fail.
(Density points) For 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 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).

Part VI — Variations on the theme of averaging.

(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).
(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).
(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).
(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).
(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.

(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 $\eu$ .
(The constant $\eu$ is optimal) Test $a_n = \frac1n\,\mathbf 1_{n\leq N}$ : using the Stirling bracketing of Problem 11.1, 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.)
(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

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: $Hf \in L^p$ .

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(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, 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 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 (dominated convergence in $x$ ), and a supremum of continuous 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 $K \subseteq \{Mf > t\}$ : 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) 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 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) 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 $g$ : $A_rg(x) \to g(x)$ everywhere. Given $\varepsilon > 0$ , split $f = g + h$ with $g$ continuous of compact support and $\norm h_1 < \varepsilon$ (Theorem 12.6); 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 $\leq \frac{6\varepsilon}\delta + \frac{2\varepsilon}\delta$ (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 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) 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 $\frac{\lambda(A\cap\intcc{x-r}{x+r})}{2r} \to 1$ . Steinhaus: around a density point take $r$ with density $> \frac34$ ; 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 $x^\alpha\dd x$ (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.

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 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.

12.1 Definition; Hölder and Minkowski

12.2 Completeness and its companions

12.3 Density theorems

12.4 Convolution and regularization

12.5 The LpL^pLp geography

12.6 Exercises

12.7 Problem: Hardy’s inequality

12.5 The $L^p$ geography