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Note: This is a generalisation of an earlier problem as suggested by user Jochen Glueck in the comments.

Let $1 \leq p < q \leq \infty$, and $f_n: [0, 1] \to \mathbb R$ be a sequence of functions in the closed unit ball of $L^q$.

Question: Is it true that there exists a constant $C < 2$, depending only on $p$ and $q$ such that

$$\inf_{n_k} \sup_{i,j \in\mathbb N} \|f_{n_i} - f_{n_j}\|_{L^p} \leq C?$$

Where the infimum is taken over all increasing sequences of natural numbers $n_k$.

If so, what is the sharpest such constant for each $p, q$?

Remark: In the linked problem, the sharp constant $C = 1$ for $p = 1, q = \infty$ is obtained in the answer by Yuval Peres.

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    $\begingroup$ Isn't the question equivalent to ask about the Istratescu measure of noncompactness of the unit ball $M$ of $L^q$ in the space $L^p$? (Literature in that formulation might be easier to search for.) Recall that the Istratescu measure of noncompactness of $M$ is defined as the supremum of $\inf_{n\ne m}d(x_n,x_m)$ over all sequences $x_n\in M$. $\endgroup$ Nov 20, 2021 at 4:06
  • $\begingroup$ I suppose it is! Are there any references on this? $\endgroup$
    – Nate River
    Nov 20, 2021 at 4:07
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    $\begingroup$ I don't know, but there is quite an active community around the Istratescu measure of noncompactness. It might be a well-known result for them. $\endgroup$ Nov 20, 2021 at 4:10
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    $\begingroup$ I looked at that literature, yet could not find such a result there. $\endgroup$ Nov 25, 2021 at 21:46
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    $\begingroup$ BTW, the equivalence of the problem with the Istratescu measure of noncompactness is not as trivial as it might seem. I did not recall that when I wrote the above comment, but now I recalled that I had once given a rigorous proof of this equivalence myself (formula (3.6) in my monograph on topological analysis). There are quite some publications on the Istratescu mnc which are not translated into English and not electronically accessible; it is probably necessary to contact one of the experts (e.g. Nina A. Erzakova). $\endgroup$ Nov 27, 2021 at 13:38

2 Answers 2

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Let me prove that such constant $C$ always exists.

It is not hard to find such $\alpha$, $\beta$ that the inequality $$x^p\leqslant \alpha x^q+\beta$$ holds for all positive $x$ and turns into equality if and only if $x=2$. Then $$|f-g|^p+|g-h|^p+|f-h|^p\leqslant \alpha (|f-g|^q+|g-h|^q+|f-h|^q)+3\beta.$$ Since all three differences $|f-g|$, $|g-h|$, $|f-h|$ can not be equal to $\pm 2$, we actually have $$|f-g|^p+|g-h|^p+|f-h|^p\leqslant \alpha (|f-g|^q+|g-h|^q+|f-h|^q)+3\tilde{\beta}, \quad \text{with some}\,\,\tilde{\beta}<\beta$$ Then integrating against $[0,1]$ we get $$\|f-g\|_p^p+\|g-h\|_p^p+\|f-h\|_p^p\leqslant \alpha (\|f-g\|_q^q+\|g-h\|_q^q+\|f-h\|_q^q)+3\tilde{\beta}.$$ If all $f,g,h$ are in the closed unit ball in $L^q$, this yields $$\|f-g\|_p^p+\|g-h\|_p^p+\|f-h\|_p^p\leqslant 3(2^q\alpha+\tilde{\beta}).$$ Since $2^q\alpha+\beta=2^p$, we see that at least one of expressions $\|f-g\|_p, \|g-h\|_p, \|f-h\|_p$ is at most $(2^q\alpha+\tilde{\beta})^{1/p}=:C<2$.

Now join $n$ and $m$ by a red edge if $\|f_n-f_m\|\leqslant C$ and by a blue edge otherwise. By infinite Ramsey theorem, there exists either a blue triangle or an infinite red clique. The second case is impossible, the first case is what we need.

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  • $\begingroup$ Very nice! Have you any intuition on when this constant can be expected to be sharp? $\endgroup$
    – Nate River
    Nov 26, 2021 at 23:51
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Here are just a few observations to discard some trivial cases. I'll talk directly about the Istratescu formulation and will prefer the probabilistic language, but that shouldn't be a problem after everything that has been posted already.

First of all, the problem is equivalent to asking what is the best constant in the inequality $E[|X-X'|^p]^{1/p}\le C_{p,q}E[|X|^q]^{1/q}$ where $X$ is an arbitrary real random variable and $X'$ is an independent copy of $X$.

Indeed, our constant cannot be any better because we can just take a sequence of independent copies of $X$ normalized by $E[|X|^q]=1$. On the other hand, suppose that $X_j$ is any sequence of random variables satisfying $E[|X_j|^q]\le 1$. Take large integer $N$ and consider the random variable $X$ taking real values $x_j$ ($j=1,\dots,N$) with probability $1/N$ each. Then we conclude that $$ \left[\frac 1{N^2}\sum_{1\le i,j\le N}|x_i-x_j|^p\right]^{q/p} \le C_{p,q}^q\frac 1N\sum_{1\le j\le N}|x_j|^q\,. $$ In particular, for each fixed $\omega$ in the probability space, we have $$ \left[\frac 1{N^2}\sum_{1\le i,j\le N}|X_i(\omega)-X_j(\omega)|^p\right]^{q/p} \le C_{p,q}^q\frac 1N\sum_{1\le j\le N}|X_j(\omega)|^q $$ Taking the expectations of both sides, we conclude that $$ \left[\frac 1{N^2}\sum_{1\le i,j\le N, i\ne j}E|X_i(\omega)-X_j(\omega)|^p\right]^{q/p}\\ \le E\left[\frac 1{N^2}\sum_{1\le i,j\le N, i\ne j}|X_i(\omega)-X_j(\omega)|^p\right]^{q/p} \le C_{p,q}^q\,, $$ so $$ \left[\frac {N-1}{N}\min_{1\le i,j\le N, i\ne j}E|X_i(\omega)-X_j(\omega)|^p\right]^{1/p} \le C_{p,q}\,. $$ This shows that $C_{p,q}$ can be used in the original problem (by Fedor Petrov's remark about infinite Ramsey, say).

The most trivial case now is $p=2$. It boils down to the observation that $$ E[|X-X'|^2]^{1/2}=\sqrt 2(E[|X|^2]-E[X]^2)^{1/2}\le\sqrt 2E[|X|^2]^{1/2}\le \sqrt 2E[|X|^q]^{1/q} $$ for all $q\ge 2$ and independent Rademachers give the identity.

This can be easily generalized to $p\ge 2$. All we need is the following elementary inequality: $$ |x-y|^p\le 2^{p-2}\left||x|^{p/2}\operatorname{sgn} x-|y|^{p/2}\operatorname{sgn} y\right|^2\,,\quad x,y\in\mathbb R\,, $$ which reduces to 2 inequalities $$ (1-t)^{p/2}\le 2^{\frac p2-1}(1-t^{p/2})\quad\text{and}\quad (1+t)^{p/2}\le 2^{\frac p2-1}(1+t^{p/2}) $$ for $t\in[0,1]$, the first of which holds because the LHS is convex, the RHS is concave, and the endpoints $0$ and $1$ are fine, while the second one is just the convexity of $s\mapsto s^{p/2}$.

Once we have it, we just write $$ E[|X-X'|^p]^{1/p}\le 2^{1-\frac 2p}E\left[\left||X|^{p/2}\operatorname{sgn}X-|X'|^{p/2}\operatorname{sgn}X'\right|^2\right]^{1/p}= \\ 2^{1-\frac 1p}(E[|X|^p]-E[|X|^{p/2}\operatorname{sgn}X]^2)^{1/p}\le 2^{1-\frac 1p}E[|X|^p]^{1/p}\le 2^{1-\frac 1p}E[|X|^q]^{1/q}\,, $$ and the Rademachers give an equality again, so $$ C_{p,q}=2^{1-\frac 1p} $$ for any $q\ge p\ge 2$.

The interesting case is thus $p\in[1,2)$. I will do $p=1$ now, which can be pulled through to a final closed formula.

The heuristic is simple. Let $\lambda$ be the density of $X$. Then we want to maximize $\iint |x-y|\lambda(x)\lambda(y)\,dx\,dy$ subject to the constraints $\lambda\ge 0$, $\int\lambda(x)\,dx=1$, $\int|x|^q\lambda(x)\,dx=\operatorname{const}$. It is tempting to use the Lagrange multipliers technique and to write $$ I(x)=\int |x-y|\lambda(y)\,dy=\alpha+\beta|x|^q\,, $$ from where $\lambda(x)\asymp |x|^{q-2}$ (the second derivative of the LHS is $2\lambda(x)$).

This doesn't seem to make much sense until you realize that the Lagrange equation doesn't really need to hold on the entire $\mathbb R$, just on the support of $\lambda$. This prompts one to consider the probability densities $$ \lambda_M(x)=\frac{(q-1)|x|^{q-2}}{2M^{q-1}}\chi_{[-M,M]}\,. $$ The corresponding value of the potential $I_M(x)$ can be determined from $$ I_M(0)=\int|y|\lambda_M(y)\,dy=\frac{q-1}q M\,, $$ so $\alpha=\frac{q-1}{q}M$, and $I_M''(x)=2\lambda_M(x)$, so $\beta=\frac {1}{q}M^{1-q}$. Note also that for $|x|>M$, $I_M$ is just linear with correct slope at $\pm M$, so we have $I_M(x)\le \alpha+\beta|x|^q$ there.

Now we have $$ \int|x|^q\lambda_M(x)\,dx=\frac{q-1}{2q-1}M^q $$ and $$ \iint |x-y|\lambda_M(x)\lambda_M(y)\,dx\,dy=\int I_M(x)\lambda_M(x)\,dx \\ =M\left[\frac{q-1}q+\frac1q\frac{q-1}{2q-1}\right]=\frac{2(q-1)}{2q-1}M\,. $$ Thus, $$ C_{1,q}\ge 2\left(\frac{q-1}{2q-1}\right)^{1-\frac 1q}\,. $$

Now we want to show that $\lambda_M$ is, indeed, optimal. Let $\lambda$ be any compactly supported probability density (I'll skip the mumbo-jumbo about how to pass to the limit to get arbitrary probability measures from here). Choose $M$ so that $\int|x|^q\lambda(x)\,dx=\int|x|^q\lambda_M(x)\,dx$. Then we can write $$ \iint|x-y|\lambda(x)\lambda_M(y)\,dx\,dy=\int I_M(x)\lambda(x)\,dx\le \int(\alpha+\beta|x|^q)\lambda(x)\,dx \\ =\int(\alpha+\beta|x|^q)\lambda_M(x)\,dx=\int I_M(x)\lambda_M(x)\,dx=\iint|x-y|\lambda_M(x)\lambda_M(y)\,dx\,dy\,. $$ On the other hand, $|x|$ is negative definite in the sense that for every zero integral compactly supported real function $\psi$, $$ \iint|x-y|\psi(x)\psi(y)\,dx\,dy\le 0\,. $$ Applying it with $\psi=\lambda-\lambda_M$, we conclude that $$ \iint |x-y|\lambda_M(x)\lambda_M(y)\,dx\,dy+\iint |x-y|\lambda(x)\lambda(y)\,dx\,dy \\ \le 2\iint |x-y|\lambda(x)\lambda_M(y)\,dx\,dy\le 2\iint |x-y|\lambda_M(x)\lambda_M(y)\,dx\,dy\,, $$ so $$ \iint |x-y|\lambda(x)\lambda(y)\,dx\,dy\le \iint |x-y|\lambda_M(x)\lambda_M(y)\,dx\,dy\,, $$ and we are done.

What happens for $1<p<2$? One can try to run the same argument, but to carry it out one would need to show that we can find a probability density $\lambda_1$ on $[-1,1]$ (going to $M$ is just trivial scaling) such that $$ I(x)=\int_{[-1,1]}|x-y|^p\lambda_1(y)\,dy=\alpha+\beta|x|^q $$ on $[-1,1]$ and $I(x)\le \alpha+\beta|x|^q$ for $|x|>1$. Then this density will be the extremizer ($|x|^p$ is still negative definite, so the rest of the argument will sail through). However, currently I cannot even prove the existence of $\lambda_1$, forget about a decent expression for it and related quantities. Any ideas?

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    $\begingroup$ a comment from V. Petrov: what you get for $\lambda_1$ is a generalized Abel's integral equation, it may be solved explicitly, see (54.23') in Gakhov's Краевые задачи (we still have something to do of course to assure that the solution is non-negative and so on) $\endgroup$ Nov 27, 2021 at 13:54
  • $\begingroup$ @FedorPetrov Interesting. I knew only how to solve the "one-sided" Abel. Unfortunately, the last equation in the copy of Gahov I have an access to is (52.16) on page 528. It looks like you cite some substantially newer edition I'm not aware of. Can you just post the formula for the explicit solution in our case? $\endgroup$
    – fedja
    Nov 27, 2021 at 18:54
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    $\begingroup$ I am afraid to reproduce such formulae, so here is a screenshot from Gakhov: disk.yandex.ru/i/es6uitvE4iqndA disk.yandex.ru/i/IHG47tiOC8sc5Q $\endgroup$ Nov 27, 2021 at 19:46
  • $\begingroup$ @FedorPetrov Thanks a lot and, erm, brr-r-r-r. It looks a bit scary, I should confess, though it may still turn out to be useful. As far as I understand, we have to use it for the second derivative of the solution (because it is $p-2$ that is between $-1$ and $0$) and there may be issues with possible singularity at $0$ of the RHS, which is $|x|^{q-2}$ (technically this is a milder singularity than the one in the kernel, which gives some hope because otherwise we would have no chance). OK, I will take a closer look. It would be nice to finish this problem off :-) $\endgroup$
    – fedja
    Nov 28, 2021 at 2:05
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    $\begingroup$ @NateRiver Just ask questions if anything is unclear :-) $\endgroup$
    – fedja
    Nov 30, 2021 at 4:31

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