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It is known that the only Banach space that satisfies the von-Neumann inequality is the Hilbert space:

Theorem (see e.g. Pisier, "Similarity Problems and Completely Bounded Maps", p 27) For a Banach space $F$ the following conditions are equivalent:

  • for every linear operator $T$ on $F$ with $\|T\|\le 1$ and any algebraic polynomial $p$ we have $\|p(T)\|\le\sup_{|z|<1}|p(z)|$;

  • $F$ is isometric to a Hilbert space.

If we allow $F$ to be merely isomorphic to a Hilbert space, the norm on $L(F)$ is replaced with an equivalent one and we get the following weaker von-Neumann inequality:

  • There is $C>0$ such that for every linear operator $T$ on $F$ with $\|T\|\le 1$ and any algebraic polynomial $p$ we have $\|p(T)\|\le C\sup_{|z|<1}|p(z)|$.

What are the Banach spaces for which the weaker von-Neumann inequality holds?

Clearly, if the inequality holds for $F$ it holds for any isomorphic space, and so it is not a geometric property. Could it be an isomorphic characterization of Hilbert spaces? Or perhaps, something like super reflexivity would imply this property? Or maybe it just holds for all Banach spaces?

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    $\begingroup$ I am almost certain that it doesn't hold for $F=\ell^1$, and I have a suspicion that it won't hold for $\ell^p$ for $p\neq 2$ $\endgroup$
    – Yemon Choi
    Commented Mar 4, 2020 at 7:35
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    $\begingroup$ Small remark: If the weaker von Neumann inequality holds on a Banach space $F$; then every linear contraction on $F$ admits a $A(D)$-functional calculus (where $A(D)$ denotes the disk algebra). This seems to be a rather strong property. $\endgroup$
    – Jochen Glueck
    Commented Mar 4, 2020 at 10:29
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    $\begingroup$ I have just taken a quick look at the proof in Pisier's book (the result is attributed to Foias) and I wonder if one could follow the argument, replacing $1$ by $C$ where necessary, to end up with a purely Banach-space condition on the two-dimensional real subspaces of $F$. Have you checked if this works? It would then reduce the problem to a question about Banach space geometry. $\endgroup$
    – Yemon Choi
    Commented Mar 4, 2020 at 15:10
  • $\begingroup$ @YemonChoi yes, I thought about it, but I don't know what to do with the condition the condition that there is $C$ such that $\frac{\|y-\lambda x\|}{\|x-\overline{\lambda} y\|}\le C$, if $\|x\|=\|y\|=1$. $\endgroup$
    – erz
    Commented Mar 4, 2020 at 21:37
  • $\begingroup$ This paper seems to be closely related On polynomialy bounded operators acting on Banach spaces. At least it gives examples of a space on which the implication won't hold. $\endgroup$
    – an_ordinary_mathematician
    Commented Apr 4, 2020 at 20:47

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