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Let $T$ be an operator $l^2({\mathbb{Z}_{\geq 0}}) \to l^2({\mathbb{Z}_{\geq 0}})$, $e_n \mapsto \sqrt{1 - q^{2(n+1)}}e_{n+1} $, where $0<q<1$. I want to compute $\|f(T,T^{*}) \|$ (operator norm) for any $f \in \mathbb{C} [z, \bar z]$. Operator $T^{*}$, of course, is the Hilbert conjugate and $T^{*}e_0 = 0$, $T^{*}e_n = \sqrt{1-q^{2n}}e_{n-1}$ for $n >0$.

I am not sure if that computation even possible and I would be happy just to show that $\|f(T,T^{*})\| \to \sup\limits_{|z| \leq 1} f(z, \bar z)$ as $q$ goes to 1 because it is actually why I need to compute norms in the first place.

I think that there is some generalization of the von Neumann inequality like $q$-analogue or something (there are a lot of generalizations) because the usual inequality proves this for polynomials $g \in \mathbb{C}[z]$ (without $\bar z$).

Question: does anyone know some useful facts or inequalities related to my question? Something that might help.

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    $\begingroup$ $T$, $T^*$ don't commute, so there's no unique obvious interpretation of $f(T,T^*)$. $\endgroup$
    – Christian Remling
    Commented Oct 12, 2019 at 16:08
  • $\begingroup$ Well, may be my notation is not the best. I meant any “noncomutative polynomial”. $\endgroup$
    – Invincible
    Commented Oct 12, 2019 at 18:49

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