Hardy's Inequality: Problems $3.14$ and $3.15$ in Rudin's RCA

Question

In Problem $3.14$, we prove (a) Hardy's inequality, (b) the condition for equality, and I shall talk about (c), (d) below. Problem $3.15$ is the discrete case of Hardy's inequality. I have asked three related questions in a single post itself, since all of them are related to Hardy's inequality, and none should be too involved.

There are some existing posts on MSE related to these topics, so I shall link them right away and point out that my question is not a duplicate: Post 1, Post 2, Post 3, Post 4.

---

For the sake of mentioning it, Hardy's inequality is: For $p\in (1,\infty)$, $f\in L^p((0,\infty))$ relative to the Lebesgue measure, and $$F(x) = \frac{1}{x}\int_0^x f(t)\ dt\quad (0 < x < \infty)$$ we have $$\|F\|_p \le \frac{p}{p-1} \|f\|_p$$

---

Question 1: This is Problem $3.14(c)$ in Rudin's book.

Prove that the constant $p/(p-1)$ cannot be replaced by a smaller one.

In one of the linked posts, there is some discussion on how this is the best constant, but I was unable to follow it. My sense is that it suffices to find a counterexample, i.e. for every constant $\beta$ smaller than $p/(p-1)$, we need a function $f_\beta\in L^p((0,\infty))$ which does not satisfy the required inequality. Why are we complicating things? If I'm thinking right, could someone help me find a counterexample?

Question 2: This appears as Problem $3.14(d)$ of Rudin's book. The author is trying to emphasize that the inequality is not for $p = 1$.

If $f > 0$ and $f\in L^1$, prove that $F\notin L^1$.

I found an example, $f(x) = e^{-x}$. Then $F(x) = \frac{1-e^{-x}}{x}$. $F$'s integral diverges, since the integral of $1/x$ diverges (use limit comparison test for integrals). However, as @David C. Ullrich pointed out, this is not enough.

Question 3: This is Problem $3.15$ in the same book and is the discrete case of Hardy's inequality.

Suppose $\{a_n\}$ is a sequence of positive numbers. Prove that $$\sum_{N=1}^\infty \left(\frac{1}{N} \sum_{n=1}^N a_n \right)^p \le \left(\frac{p}{p-1} \right)^p \sum_{n=1}^\infty a_n^p$$ if $1 < p < \infty$. If $a_n\ge a_{n+1}$, the result can be made to follow from Hardy's inequality. This special case implies the general one.

I took $f = \sum_{n=1}^\infty a_n \mathbf{1}_{[n,n+1]}$. Then $f\in L^p$ only if $\sum_{n=1}^\infty a^p_n < \infty$. If $f\notin L^p$, the inequality is trivial. So let's take $f\in L^p$. Now using Hardy's inequality, we have $\|F\|_p \le \frac{p}{p-1} \|f\|_p$. What is $F$? $$F(x) = \frac{1}{x}\int_0^x \sum_{n=1}^\infty a_n\mathbf{1}_{[n,n+1]}(t)\ dt = \frac{1}{x}\left(\sum_{n=1}^{\lfloor x\rfloor} a_n + (x - \lfloor x\rfloor)a_{\lfloor x\rfloor + 1} \right)$$ How do I proceed?

---

P.S. I have already solved Problems $3.14(a)$ and $3.14(b)$, i.e. proving Hardy's inequality and showing that equality holds iff $f = 0$ a.e.

Answer 1

May details are left to you. The important part is that you review Post 4 where they show why $p/(p-1)$ is the best contact in Hardy's inequality. I give a brief explanation (but do not re do any of the constructions there) of why this is the case.

---

Question 1: The answers to the Post 4 (one of which follow the Hint of your textbook) gives the optimal bound in Hardy's inequality. To see this, notice that each solution there explicitly constructs a sequence of functions $\{f_n:n\in\mathbb{N}\}\subset L_p$, such that

1. $\|f_n\|_p=1$,
2. and $\lim_n\|Hf_n\|_p=\frac{p}{p-1}$.

If the bound $p/(p-1)$ were not optimal, and say $\|Hf\|_p\leq c\|f\|_p$ for all $f\in L_p$, for some constant $c<\frac{p}{p-1}$, then for any of the sequences built in the aforementioned posting, you would have that $$c\|f_n\|=c\geq \|Hf_n\|_p\xrightarrow{n\rightarrow\infty}\frac{p}{p-1}$$ which is a contradiction to $c<p/(p-1)$.

---

Question 3: suppose $\{a_n:n\in\mathbb{N}\}$ is a monotone non increasing sequence of positive numbers such that $\sum_na^p_n<\infty$, and define $f$ as $$ f(x):=\sum^\infty_{n=1}a_n\mathbb{1}_{(n-1,n]}(x)$$ and consider the Hardy transform $Hf$ of $f$, i.e., $Hf(x):=\frac1x\int^x_0f(t)\,dt$ Then $$Hf(x)=\left\{ \begin{matrix} a_1 &\text{if}&0<x\leq1\\ \tfrac{a_1+\ldots + a_n+(x-n)a_{n+1}}{x} & \text{if} &1\leq n<x\leq n+1 \end{matrix} \right. $$ The assumption on $a_n$ implies that for $1\leq n<x\leq n+1$ $$\begin{align} \frac{a_1+\ldots + a_n+(x-n)a_{n+1}}{x}&=\frac{(a_1-a_{n+1})+\ldots +(a_n-a_{n+1})}{x}+a_{n+1}\\ &\geq \frac{a_1+\ldots+ a_n-na_{n+1}}{n+1} +a_{n+1}=\frac{a_1+\ldots + a_{n+1}}{n+1} \end{align}$$ Hence $$\begin{align} \int^\infty_0(Hf)^p&=\sum^\infty_{n=0}\int^{n+1}_n(Hf)^p\\ &\geq\sum^\infty_{n=0}\Big(\frac1{n+1}\sum^{n+1}_{k=1}a_k\Big)^p =\sum^\infty_{n=1}\Big(\frac1{n}\sum^{n}_{k=1}a_k\Big)^p \end{align}$$ Provided you have proved Hardy's inequality for $L_p((0,\infty),\mathscr{B}((0,\infty),\lambda)$, where $\lambda$ is Lebesgue's measure, then The conclusion of Question 2 follows for the special case $0\leq a_{n+1}\leq a_n$ for all $n\in\mathbb{N}$.

See if you can get the general case ($a_n\geq0$ and $\sum^\infty_{n=1}a^p_n<\infty$) from this (consider for example finite segments of $a_n$, that is $b^{(m)}_n=a_n\mathbb{1}_{(0,m]}(n)$ and apply the result to a "reordering" of $b^{(m)}$).

---

Question 2: As David C. Ulrich mentioned in his comment and answer, the problem asks to show that $Hf\notin L_1(0,\infty)$ for all $f\in L_1(0,\infty)$ with $f>0$. Here is a proof using Fubini's theorem. Since $g(t, x)=\frac{1}{x}f(t)\mathbb{1}_{(0,x]}(t)\geq0\,$ and measurable in $(0,\infty)\times(0,\infty)$, one can iterate integrals to get \begin{align} \int^\infty_0\int^x_0\frac{1}{x}f(t)\,dt\,dx&=\int^\infty_0\int^\infty_t\frac{1}{x}f(t)\,dx\,dt=\int^\infty_0f(t)\int^\infty_t\frac{1}{x}\,dx\,dt=\infty \end{align} A simpler proof is given by David C. Ulrich here

Although not asked in the OP, one can easily show that the Hardy operator is of weak type (1,1), that is, there is a constant $C>0$ such that $\displaystyle \lambda(|Hf|> t)\leq \frac{C\|f\|_1}{t}$ for all $f\in L_1(0,\infty)$ where $\lambda$ denotes Lebesgue measure. Indeed, since $$|Hf|\leq\frac1x\int^x_0|f|\,d\lambda\leq\frac1x\|f\|_1,$$ $\left\{x>0:|Hf(x)|>t\right\}\subset\left\{x>0: \frac{\|f\|_1}{x}>t\right\}=\left(0,\frac{\|f\|_1}{t}\right)$. Consequently $$\lambda(|Hf|> t)\leq \frac{\|f\|_1}{t}\quad\forall t>0$$

Answer 2

People seem to be misinterpreting Q2. It says

If $f > 0$ and $f\in L^1$, prove that $F\notin L^1$.

You can't do that by giving an example. Luckily it's trivial:

Since $f>0$ some elementary measure theory(see below) shows that there exists a bounded set $E$ with $m(E)>0$ and a $\delta>0$ so $f\ge\delta$ on $E$. Hence if $x>\sup E$ we have $$F(x)=\frac1x\int_0^x f \ge\frac1x\int_E\delta=\frac{\delta m(E)}{x},$$so $F\notin L^1$.

Below

(edit) If $f(x)>0$ there exists $n\in\Bbb N$ with $f(x)>1/n$; hence $$(0,\infty)=\bigcup_{n=1}^\infty I_n,$$ where $$I_n=\{x:f(x)>1/n\}.$$So countable additivity shows there exists $n_0$ with $$m(I_{n_0})>0.$$ Again, $$I_{n_0}=\bigcup_{k=1}^\infty(I_{n_0}\cap(0,k)),$$so we can let $E=I_{n_0}\cap(0,k)$ for a suitable $k$; any choice of $k$ gives a bounded set on which $f>\delta>0$, and if $k$ is large enough then $m(E)>0$.

Answer 3

Here another approach to question 1, finding simple counter-examples.

Given $a>p$, let's define this function : $$f_a(x)=x^{-1/a}\times\chi_{(0,1]}(x)$$ where $\chi_{(0,1]}(x)=1$ if $x\in(0,1]$ and $0$ otherwise.

Then $$(\lVert f_a\rVert_p)^p=\int_0^1x^{-p/a}dx=\dfrac{a}{a-p} $$ Let's compute $F_a$.

if $x\ge 1$ : $$\begin{equation}\begin{aligned} F_a(x)&=\dfrac{1}{x}\int_0^xt^{-1/a}dt\\\\ &=\dfrac{a}{a-1}x^{-1/a} \end{aligned}\end{equation}$$

if $x>1$ : $$\begin{equation}\begin{aligned} F_a(x)&=\dfrac{1}{x}\int_0^1t^{-1/a}dt\\\\ &=\dfrac{a}{a-1}x^{-1} \end{aligned}\end{equation}$$

Thus

$$\begin{equation}\begin{aligned} (\lVert F_a\lVert_p)^p&=\left(\dfrac{a}{a-1}\right)^p \left(\int_0^1x^{-p/a}dx+\int_1^{+\infty}x^{-p}dx\right)\\\\ &=\left(\dfrac{a}{a-1}\right)^p\left(\dfrac{a}{a-p}+\dfrac{1}{p-1}\right)\\\\ =&\left(\dfrac{a}{a-1}\right)^p\left(\dfrac{p(a-1)}{(a-p)(p-1)}\right) \end{aligned}\end{equation}$$

And now we can compute $$\begin{equation}\begin{aligned} \left(\dfrac{\lVert F_a\rVert_p}{\lVert f_a\rVert_p}\right)^p&= \left(\dfrac{a}{a-1}\right)^p\left(\dfrac{p(a-1)}{(a-p)(p-1)}\right)\times\dfrac{a-p}{a}\\\\ &=\left(\dfrac{a}{a-1}\right)^{p-1}\times \dfrac{p}{p-1} \end{aligned}\end{equation}$$

The function $\varphi : x\mapsto \dfrac{x}{x-1}$ is decreasing. Thus the last equality allows, as $a>p$ : $$ \dfrac{\lVert F_a\rVert_p}{\lVert f_a\rVert_p}\ge \dfrac{a}{a-1} $$

For any $\varepsilon>0$, as $\varphi$ is continuous, there exists $a>p$ such as $$ \dfrac{a}{a-1}>\dfrac{p}{p-1}-\varepsilon $$

So finally, for any $\varepsilon>0$, there exists a $f_a$ function so that $\lVert F_a\rVert_p>\left(\dfrac{p}{p-1}-\varepsilon\right)\lVert f_a\rVert_p$

Thus $\dfrac{p}{p-1}$ is the smallest constant.