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Does a robust version of Schur's lemma exist? Specifically, I was wondering about something like this:

Let $B$ be a bounded operator over a vector space $V$, with underlying field $\mathbb{C}$ and let $\rho$ be an irreducible representation of a group $G$ over $V$. Taking $\epsilon \geq 0$, if for all $g \in G$ it is the case that $\| [ B, \rho(g) ] \| \leq \epsilon$, then there exists a constant $\lambda \in \mathbb{C}$ such that $\| B - \lambda I \| \leq f(\epsilon)$, where $f$ is some ''nice'' function of $\epsilon$ (obviously we want that $f(\epsilon) \rightarrow 0$ as $\epsilon \rightarrow 0$).

Basically I'm asking: if $B$ almost commutes with the irrep does that imply that $B$ is close to a multiple of the identity (in operator norm)?

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    $\begingroup$ What's $G$? I think that there is a fairly easy positive answer if $G$ is a compact topological group acting continuously. If you want something more general then you may have trouble. $\endgroup$
    – Neil Strickland
    Jul 8, 2018 at 12:12
  • $\begingroup$ I was actually thinking of $G$ as a finite group :) $\endgroup$
    – dimquasar
    Jul 8, 2018 at 12:17
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    $\begingroup$ If $G$ is finite then the map $\overline{B}=|G|^{-1}\sum_{g\in G}gBg^{-1}$ will act as a scalar, and $B-\overline{B}$ can be bounded using the bounds on $[g,B]$. $\endgroup$
    – Neil Strickland
    Jul 8, 2018 at 12:26
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    $\begingroup$ On the other hand, if $\rho(G)$ contains the scalars (or even a single scalar with absolute value not equal to $1$) then the commutator identity would only hold for $\epsilon=0$, so the statement reduces to ordinary Schur's lemma. $\endgroup$
    – Victor Protsak
    Jul 8, 2018 at 12:40
  • $\begingroup$ @NeilStrickland thanks! I guess for a compact topological group you'd define $\bar{B}$ using the Haar measure instead? $\endgroup$
    – dimquasar
    Jul 8, 2018 at 13:03

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