Let $A$ be an invertible opeartor, $B$ being another operator, is |ABA^{-1}| \leq |B|?
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Seems unlikely: https://math.stackexchange.com/questions/525970/norm-of-an-inverse-operator-t-1-t-1 – Eric Towers Mar 28 '24 at 20:42
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Have you tried anything ? A random choice of two noncommuting $2\times 2$ matrices give counterexamples. – Ryszard Szwarc Mar 28 '24 at 20:43
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When $A = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, we have $ABA^{-1} = \begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}$. So, $\|ABA^{-1}\| = 2 \not\leq 1 = \|B\|$.
David Gao
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