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Reading some article a while ago I read the following: (here $H^2$ represents the Hardy space)

Let $f\in H^2$ be such that $f(0)=1$, and let $0<\lvert\lambda\rvert<1$, then $$\lVert f(\lambda z)\cdots f(\lambda^{n-1}z)\rVert _\infty \lVert f\rVert _2\leq \prod_{m\geq 1} \lVert f(\lambda^{m-1}z)\rVert_2$$

for every $n\geq 1$.

This seems a bit 'weird' to me, as usually the $2$-norm is bounded above by the infinity norm, not the other way as here.

I would greatly appreciate any help.

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  • $\begingroup$ Is there a typo in your inequality? The LHS depends on $n$, while the RHS doesn't. $\endgroup$
    – Christian Remling
    Commented Jul 8 at 17:46
  • $\begingroup$ @ChristianRemling No, there is no typo. $\endgroup$
    – Tomas smith Smith
    Commented Jul 8 at 17:59
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    $\begingroup$ I would suggest changing $n$ to $m$ as the index of the product on RHS since in the statement you fix $n \ge 1$ while you use it also as an index of an infinite product and that is confusing $\endgroup$
    – Conrad
    Commented Jul 8 at 20:21
  • $\begingroup$ @Conrad Ok, will do. $\endgroup$
    – Tomas smith Smith
    Commented Jul 8 at 20:47

1 Answer 1

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This is false (and immediately suspicious, as you already pointed out). Recall that $\|\sum a_n z^n\|_{H^2}=\|a\|_{\ell^2}$. So if $f(z)=1+\epsilon z$, then $\|f(\lambda^k z)\|^2_{H^2}=1+\lambda^{2k}\epsilon^2$ and $f(\lambda^kz)$ assumes its maximum at $z=1$ and $f(\lambda^k)=1+\lambda^k\epsilon$.

This makes the LHS of your inequality $\simeq 1+c\epsilon$, while the RHS is $\simeq 1+d\epsilon^2$ as $\epsilon\to 0+$.

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  • $\begingroup$ I cannot see how something published in a paper (revised multiple times) can be false, so... (plus, we have the infinity norm of the product of the functions, not the product of the infinity norms) $\endgroup$
    – Tomas smith Smith
    Commented Jul 8 at 21:33
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    $\begingroup$ Indeed, how something published can be false. That is so funny. On a more serious note, in Christian's example, all functions attained maximum at the same point, so the product of infinity norms is the norm of the product. $\endgroup$
    – Oleg Eroshkin
    Commented Jul 8 at 21:58
  • $\begingroup$ This is from a very prestigious journal, so no need for jokes. I still cannot see how this is a counterexample $\endgroup$
    – Tomas smith Smith
    Commented Jul 8 at 22:05
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    $\begingroup$ @TomassmithSmith How about you just tell us what is the article you are referring to? Perhaps you forgot an assumption or wrote the inequality wrong... $\endgroup$
    – Malik Younsi
    Commented Jul 10 at 18:32

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