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CONVENTION. All functions in this post are nonnegative and defined on $[0, \infty)$.

The Hardy operator is $$ Hf(x)=\frac1x\int_0^x f(y)\, dy, $$ and, as it is well known, it satisfies the inequality $$\tag{H} \|Hf\|_{p}\le \frac{p}{p-1}\|f\|_{p},\quad \text{for }p>1, $$ where the constant $p/(p-1)$ is the best possible, in the sense that (H) fails if it is replaced by any strictly smaller one.

In a vague sense, the function $f(x)=x^{-1/p}$ is a maximizer for (H), if one "neglects a logarithmic divergence in both sides" (as said in this answer of Terry Tao). More precisely, as shown, for example, in the answers to this old question of mine, some sequences that approximate $f$ also saturate (H); it is the case of $$ f_n(x)=x^{-\frac1p -\frac1n}\mathbf 1_{(1, \infty)} $$ and of $$ f_n(x)=x^{-\frac1p}\mathbf 1_{(1, n)}.$$

My question is, roughly speaking: is $f(x)=x^{-1/p}$ the "only" maximizer to (H)? More precisely:

Question. Suppose that $$ \frac{\|Hf_n\|_p}{\|f_n\|_p}\to \frac{p}{p-1}.$$ Does there exist a sequence $\lambda_n>0$ such that $$f_n(\lambda_nx)\to x^{-\frac1 p}\mathbf 1_{(1, \infty)},$$ almost everywhere on $(0, \infty)$?

Remark. The presence of $\lambda_n$ is necessary to prevent trivial counterexamples; indeed, the ratio $\|Hf\|_p / \|f\|_p$ is invariant under the scaling transformation $$f\mapsto f_\lambda(x)=f(\lambda x).$$

FINAL NOTE. David C. Ullrich gave an exhaustive negative answer to the present question. His answer was edited heavily and so it may be a bit hard to read. Please see my "Summary to David C. Ullrich's answer".

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  • $\begingroup$ What is a definition of $\|f\|_p?$ $\endgroup$ – Yuri Negometyanov Jan 5 '19 at 4:14
  • $\begingroup$ @YuriNegometyanov: It is $\|f\|_p:=(\int_0^\infty f^p)^{1/p}$. $\endgroup$ – Giuseppe Negro Jan 5 '19 at 9:34
  • $\begingroup$ Thank you. I hope to finish my answer in time, $\endgroup$ – Yuri Negometyanov Jan 5 '19 at 20:50
  • $\begingroup$ Ready, but..... $\endgroup$ – Yuri Negometyanov Jan 6 '19 at 21:06
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Edit: No, it turns out that the maximizing sequence is far from unique. See Final Edit at the bottom...

This is not an actual answer to your question, just a suggestion regarding how to look at it, that it seems to me may be useful.

Observe that $$Hf(x)=\int_0^1 f(tx)\,dt.$$

So if we define $$f_t(x)=f(tx)$$ we can regard $$Hf=\int_0^1 f_t\,dt$$ as a vector-valued integral. This makes (H) more or less obvious: $$||Hf||_p\le\int_0^1||f_t||_p\,dt=||f||_p\int_0^1t^{-1/p}\,dt=\frac p{p-1}||f||_p.$$This makes it clear why there is no $f\ne0$ with $||Hf||_p=\frac p{p-1}||f||_p$; that would require that the norm of the integral equal the integral of the norm, implying that for almost every $t$ we have $f_t=c_tf$, certainly impossible for $f\in L^p$, $f\ne0$.

So regarding your question, perhaps you can get somewhere by investigating (maybe by Hahn-Banach) what follows if the norm of the integral is almost the integral of the norm...

Edit in answer to the question of what makes $x^{-1/p}$ special: Note first that there's no actual rigorous math from this point on, just fuzzy heuristics.

We've seen that if $f$ were a maximizer for $H$ then $f_t=c_tf$ for every $t>0$, which is impossible for a (non-zero) $L^p$ function. The obvious way to get $f_t=c_tf$ is to set $f(x)=x^\alpha$ (I think it's clear that this is essentially the only way, at least if we assume $f$ is continuous).

So say $f(x)=x^\alpha$. Then $\int_0^1 f^p<\infty$ if and only if $\alpha p>-1$, while $\int_1^\infty f^p<\infty$ if and only if $\alpha p<-1$. So, if you promise not to tell anyone I put it this way, setting $\alpha=-1/p$ makes $\int_0^\infty f^p$ "as close as possible to finite at both endpoints".

(Hmm. An actual true fact is that $\alpha=-1/p$ is the only choice that makes $\int_{1/A}^Af^p=O(\log A)$.)

Better yet, start over and look at it this way: In fact $H$ is just a convolution operator on the multipicative group $(0,\infty)$. This is really the right way to look at it, for example it makes both Hardy's inequality and the fact that almost-maximizers are what they are completely transparent.

We wimp out and make a change of variables so we can consider convolutions and Fourier transforms on $\Bbb R$ instead of on that group:

Define a surjective isometry $T:L^p((0,\infty))\to L^p(\Bbb R)$ by $$Tf(x)=e^{x/p}f(e^x).$$Define $\tilde H:L^p(\Bbb R)\to L^p(\Bbb R)$ by $$\tilde H=THT^{-1}.$$

You can calculate that $$\tilde Hf=f*K,$$where $$K(x)=e^{-((p-1)/p)x}\chi_{(0,\infty)}(x).$$Hence $$||\tilde Hf||_p\le||K||_1||f||_p=\frac p{p-1}||f||_p.$$(Since $T$ is a surjective isometry this is exactly Hardy's inequality, made totally obvious/motivated.)

And at least formally $$\widehat{\tilde Hf}=\hat K\hat f.$$Since $K\ge0$ it's clear that $$||\hat K||_\infty=||K||_1=\hat K(0).$$ This makes it at least very plausible that the almost-maximizers for $\tilde H$ should be $f$ such that $\hat f$ is supported near the origin (for $p=2$ that's not only plausible it's even true, by Plancherel). But if $f=1$ then $\hat f$ is literally supported on $\{0\}$. And $$T^{-1}1=x^{-1/p}.$$

Final Edit: No, that's wrong.

I was unable to prove that the last paragraph above actually works, even for $p=2$, which was supposed to be clear. In fact the plausibility argument was too fuzzy. Say $p=2$.. It's true that if $f_n$ is a maximizing sequence then $\widehat {f_n}$ must in some sense have most of its mass concentrated near the origin, but it does not follow that $f_n$ must be close to $1$ on a large set. If $\widehat {f_n}$ were $L^1$ with integral $1$ that would be true, but $\widehat {f_n}$ is not integrable. What's true is that $|\widehat {f_n}|^2\to\delta_0$ weakly, but that says nothing about $f_n$ tending to $1$.

And in fact there are jillions of other maximizers. Temporarily define $$\phi_n(x)=n^{-1/p}\phi(x/n),$$and note that $$||\phi_n||_p=||\phi||_p.$$

It turns out that if $f\in L^p(\Bbb R)$ and $f_n$ is defined as above then $(f_n)$ is a maximizing sequence for $\tilde H$. If it happens that $f$ is continuous at the origin then $f_n$ is approximately constant on compact sets, consistent with our wrong conjecture about a maxmizing sequence being essentially unique. But if $f$ oscillates suitably near the origin then every $f_n$ does a lot of oscillation on $[-1,1]$. So "$f=1$, except truncated to lie in $L^p$" is far from the only sort of thing that maximizes $\tilde H$, hence similarly for $x^{-1/p}$ and $H$.

Proof: If $p=2$ then it's easy to see from Plancherel that $$||f_n*K||_2\to||K||_1||f||_2=||\tilde H||\,||f||_2.$$This was the first thing I noticed indicating that the conjecture was wrong, further evidence that looking at $H$ in terms of convolutions is the right way to look at it. It's not hard to give a direct proof valid for $p>1$:

First, a change of variable shows that $$f_n*K=(f*K^n)_n,$$where $(.)_n$ is defined as above and $$K^n(x)=nK(nx);$$hence $$||f_n*K||_p=||f*K^n||_p.$$But $(K_n)$ is an approximate identity, except for the mis-normallization $\int K_n=p/(p-1)$. So $$\left|\left|f*K_n-\frac p{p-1}f\right|\right|_p\to0,$$hence $$||\tilde Hf_n||_p=||f*K^n||_p\to\frac p{p-1}||f||_p.$$

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  • $\begingroup$ Very nice. This is a connection with the dilation group. Thank you for this beautiful answer! It is probably the case that this train of reasoning yields the uniqueness of maximizers. $\endgroup$ – Giuseppe Negro Dec 4 '18 at 14:57
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    $\begingroup$ Yes it is a connection with that. In fact $G=(0,\infty)$ is a locally compact abelian group, and in fact $Hf=K*f$ if you look at it right. Hence one could for example use Plancherel to prove something about almost-maximizers in $L^2(G)$. It's not clear to me whether that's really relevant to your question, since $L^2(G)\ne L^2((0,\infty)$ (the measure implicit in $L^2(G)$ being $dt/t$ instead of $dt$.) $\endgroup$ – David C. Ullrich Dec 4 '18 at 15:09
  • $\begingroup$ @DavidCUllrich: I have been thinking a little... As you show here, formally, maximizers to the Hardy inequality should be powers; $f(x)=Cx^a$, for some $a>-1$. Indeed, these are the unique solutions to the functional equation $f_t=c(t)f$. Of course this is not correct, since powers are not $L^p$ functions, as you clearly remark. Now, among all powers, the only one that "almost" maximizes is $x^{-1/p}$; if we redo this limiting process with another power, we get another constant. What makes $x^{-1/p}$ so special? $\endgroup$ – Giuseppe Negro Dec 29 '18 at 9:38
  • $\begingroup$ @GiuseppeNegro Edit... $\endgroup$ – David C. Ullrich Dec 29 '18 at 14:26
  • $\begingroup$ @GiuseppeNegro Don't miss the second edit, "Better yet...". $\endgroup$ – David C. Ullrich Dec 29 '18 at 15:40
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SUMMARY OF DAVID C. ULLRICH's ANSWER.

  1. The transformation $Tf(x)=e^\frac{x}{p} f(e^x)$ is an isomorphism of $L^p(\mathbb R)$ onto $L^p(0, \infty)$ and it conjugates the Hardy operator to a convolution; $$THT^{-1} g = K_p\ast g,\qquad \text{where }\,K_p(x):=e^{-\frac{p-1}{p} x} \mathbf 1_{\{x>0\}}.$$ In particular, the Hardy inequality $\|Hf\|_{L^p(0, \infty)}\le C \|f\|_{L^p(0,\infty)}$ is equivalent to $\|K_p\ast g\|_{L^p(\mathbb R)}\le C\|g\|_{L^p(\mathbb R)}$, and a sequence $f_n$ saturates the Hardy inequality if and only if the corresponding sequence $g_n=T f_n$ saturates the latter inequality; $$ \frac{\|Hf_n\|_{L^p(0, \infty)}}{\|f_n\|_{L^p(0,\infty)}}\to \frac{p}{p-1} \quad \iff\quad \frac{\|K_p\ast g_n\|_{L^p(\mathbb R)}}{\|g_n\|_{L^p(\mathbb R)}}\to \frac{p}{p-1}.$$

    With respect to the transformation $T$, the function $x^{-1/p}$ corresponds to the constant $1$; $$ T(x^{-1/p})=1.$$ Thus, my question is rephrased as the conjecture that $1$ is the "essential unique maximizer" for the convolution operator $g\mapsto K_p\ast g$.

  2. Since $K_p\in L^1(\mathbb R)$, by the Young inequality $$\|K_p\ast g\|_{L^p(\mathbb R)}\le \frac{p}{p-1}\|g\|_{L^p(\mathbb R)}, $$ where we used that $p/(p-1)=\|K_p\|_{L^1(\mathbb R)}$. This gives an alternative proof of the Hardy inequality.
  3. By a scaling argument, for any $g\in L^p(\mathbb R)$ the sequence $g_n(x):=g(x/n)$ is such that $$\frac{\|K_p\ast g_n\|_{L^p(\mathbb R)}}{\|g_n\|_{L^p(\mathbb R)}} \to \frac{p}{p-1}.$$ (Actually, David C.Ullrich considered the differently normalized sequence $g_n(x):=n^{-1/p}g(x/n)$).
  4. Conclusion. If $g$ is continuous at $0$, then the sequence $g_n$ constructed in the point 3 converges pointwise to a constant, and therefore the corresponding sequence $f_n\in L^p(0, \infty)$ converges pointwise to $x^{-1/p}$. But these are not the only maximizing sequences. In conclusion, the conjecture of point 1 is disproved and the answer to my original question is negative.
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  • $\begingroup$ Great. I considered cleaning things up myself, decided to leave my answer as is because the heuristics that turned out to be leading me astray could be nonetheless instructive. Detail The $g_n$ above presumably tend pointwise to zero, certainly so if $g$ is continuous at $0$. That's not the constant you meant - what tends to $g(0)$ if $g$ is continuous at $0$ is the differently normalized $n^{1/p}g_n$ (which of course also give a norming sequence.) $\endgroup$ – David C. Ullrich Jan 5 '19 at 15:00
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Given

$$\mathcal H(f(x))=\dfrac1x\int\limits_0^xf(y)\,\mathrm dy,\quad \left\|f(x)\right\|_p = \left(\int\limits_0^\infty f^p(x)\,\mathrm dx\right)^{\frac1p},\quad p>1.\tag1$$

Let us denote $$\mathcal H_M(f(x)) = \dfrac1M\int\limits_0^M f(x)\,\mathrm dx,\tag2$$ then $$\mathcal H(f(x)) = \lim\limits_{M\to x}\mathcal H_M(f(x)),\quad \left\|f(x)\right\|_p = \lim\limits_{M\to \infty} \left(M\cdot \mathcal H_M(f^p(x))\right)^{\frac1p}.\tag3$$

Let us consider the ratio $$R_p(f(x)) = \dfrac{\|\mathcal H(f(x))\|_p}{\left\|f(x)\right\|_p} = \dfrac{\lim\limits_{z\to \infty} \left(z\cdot \mathcal H_z(\mathcal H^p(f(x)))\right)^{\frac1p}} {\lim\limits_{z\to \infty} \left(z\cdot \mathcal H_z(f^p(x))\right)^{\frac1p}} =r_p^{1/p}(f(x)),\tag{4}$$ where $$r_p(f(x))=\lim\limits_{z\to \infty}\dfrac{\mathcal H_z(H^p(f(x)))}{\mathcal H_z(f^p(x))}.\tag{5}$$

On the class of power functions, $$\begin{align} &\mathcal H(x^q) = \dfrac1x\int\limits_{0}^{x}y^q\,\mathrm dy= \dfrac1{q+1}x^q,\quad q\ge -1, \\[4pt] &\mathcal H_z(\mathcal H^p(x^q)) = \dfrac1{(q+1)^p}\dfrac1z\int\limits_{0}^{x}x^{pq}\,\mathrm dx= \dfrac1{(q+1)^p}\mathcal H_z(x^{pq}), \\[4pt] &\mathcal H_z(x^{pq})=\dfrac1z\int\limits_{0}^{z}x^{pq}\,\mathrm dx= \dfrac1{pq+1}z^{pq},\quad pq\ge -1,\\[4pt] &r_p(x^q)=\dfrac1{(q+1)^p}\\ &R_p(x^q)=\dfrac1{q+1}. \end{align}$$ If $pq+1 < 0,$ then $\mathcal H_z(x^{pq}) < 0.$

So the optimization task contains additional constraint and can be presented in the form of $$\textrm{maximize}\quad R=\dfrac1{q+1},\quad\text{where}\quad (p>1)\wedge(pq \ge -1),\tag{6}$$ with the solution $$R_p=\dfrac p{p-1}\quad\text{at}\quad q=-\dfrac1p,\quad p>1.\tag{7}$$

This means that $f_{opt}(x)=x^{-\frac1p}$ provides required maximum of $R_p.$

Thus, all attempts to find another function $f_{opt}$ must be unsuccessful.

Due to the Weierstrass theorem, any monotonic functional sequence in the form of $$\{f_n(x)\},\quad f_n(x)=x^{q_n},\quad \lim\limits_{n\to\infty} q_n = -\dfrac1p,\quad q_1 \ge -\dfrac1p,$$ satisfies the OP task requirements.

If $\{f_n(x)\}$ is not monotonic, it must contain limited quantity of members with $q<-\dfrac1p.$

In partial, the sequence of $$f_n=x^{-\frac1p-\frac1n}$$ does not satisfy the requuirements.

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  • $\begingroup$ It seems to me that this proves that $x^{-1/p}$ is the unique maximizer among power functions. Which is surely true, of course; I studied this here. But there are other maximizers that are not power functions; see the answer by David C.Ullrich (and my answer, which is a summary to it). $\endgroup$ – Giuseppe Negro Jan 8 '19 at 9:45
  • $\begingroup$ @GiuseppeNegro I had more reach results, but deleted drafts. The interest in task is lost. There were some inequalities. $\endgroup$ – Yuri Negometyanov Jan 8 '19 at 10:04
  • $\begingroup$ I am sorry you lost interest, but I had to assign the bounty or it would have gone to waste. Thank you for your interest in any case. $\endgroup$ – Giuseppe Negro Jan 8 '19 at 10:07

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