Let $(\xi_n)_{n\ge 1}$, $(\eta_n)_{n\ge 1}$ be independent mean-zero random variables with values in a Banach space $X$ such that $$\sum_n\mathbb P(\xi_n\in A)\le\sum_n\mathbb P(\eta_n\in A)$$for any Borel set $A\subset X\setminus\{0\}$.

Let $1\le p<\infty$. Is there a constant $C$ (perhaps depending on $p$ and $X$) such that

$$\textstyle\mathbb E\|\sum_n\xi_n\|^p\le C\,\mathbb E\|\sum_n\eta_n\|^p?$$

This problem was posed by Ivan Jaroslavtsev 28.06.2019 on page 133 of Volume 2 of the Lviv Scottish Book.


2 Answers 2


Certainly not always. The most trivial example seems to be $X=\ell^\infty$, $\eta_n=\pm e_n$ (with probability $1/2$ for each sign), and $\xi_n$ being uniformly distributed on $\pm e_1,\dots,\pm e_N$ with large $N$ for $n=1,\dots,N$ (as usual, $e_n$ is the vector with the $n$-th coordinate $1$). The rest of $\xi_n$ and $\eta_n$ can be put to $0$, say.

Then the sums of probabilities in question are equal (and equal to one half times the number of vectors $\pm e_n, n=1,\dots,N$ lying in $A$). However, $\|\sum\eta_n\|$ is always $1$ while $\|\sum\xi_n\|$ is typically (i.e., with probability $\ge \frac12$, say) at least of order $\sqrt{\frac{\log N}{\log\log N}}$ (that is just the classical balls into bins problem combined with the random choice of signs for the balls in the maximal bin, and in this computation I ignore the fact that the bin with nearly maximal number of balls is typically not unique, which may drive the estimate up even more).

Next, the "perhaps, depending on $X$" construct looks very fishy in the question as it is posed now (nobody prevents us from taking the appropriate sum of spaces with large constants to get a space without any constant, say; also, since the question is essentially finite-dimensional, it is clear that $\ell^\infty$ must be a counterexample if anything at all is).

Are you sure that the problem is not "Describe (in some familiar terms) the Banach spaces $X$ such that..."?

  • $\begingroup$ Thank you very much for the answer. Concerning the proper form of this question, I do not know what to answer as I even cannot decrypt the name of the person that posed this problem, see math.lviv.ua/szkocka/viewpage.php?vol=2&page=133 $\endgroup$ Commented Nov 7, 2019 at 11:24
  • $\begingroup$ @LvivScottishBook As I understand the name is Ivan Yaroslavtsev (written as Iv. Yar.) $\endgroup$ Commented Nov 10, 2019 at 5:53
  • $\begingroup$ @MikhailOstrovskii Thank you very much for the suggestion. I will try to contact with Iv.Yar. Probably this will be a kind of surprise for him. $\endgroup$ Commented Nov 11, 2019 at 11:48

First of all, many thanks to @LvivScottishBook for putting the problem online (that was indeed very surprising for me) and to @fedja for finding a counterexample.

While writing the problem I was sure that it has a positive answer for any Banach space (for some reason this was kind of intuitive for me) with $C$ independent of $X$, so @fedja answered the original question. Nonetheless, as @fedja noticed, it is not clear for which Banach spaces such an inequality holds true. The only thing that I can guarantee so far is that the necessary assumption on the Banach space is having a finite cotype thanks to @fedja counterexample and the Maurey–Pisier theorem (see Corollary 7.3.14 in Analysis in Banach spaces, Volume II by Hytönen, van Neerven, Veraar, and Weis).

It remains open whether finite cotype is a sufficient condition as well.

  • $\begingroup$ P.S. I guess I better make a comment out of this text but I could not as I do not have enough reputation. $\endgroup$
    – Iv Yar
    Commented Nov 11, 2019 at 14:26
  • 6
    $\begingroup$ You should ask a separate question out of this and make a link to the old one. This way no one finds your comment/question. $\endgroup$ Commented Nov 11, 2019 at 14:46
  • $\begingroup$ I think it merits a new question rather. $\endgroup$
    – Amir Sagiv
    Commented Nov 11, 2019 at 16:17
  • 1
    $\begingroup$ Thanks! Here is the new question. $\endgroup$
    – Iv Yar
    Commented Nov 11, 2019 at 17:04

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