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Let $I \colon X \to Y$ a non-isometric isomorphism between Banach spaces. Let $T \colon X \to X$ be a bounded operator. Do we have the equality $$ \|T\|_{B(X)} =\|ITI^{-1}\|_{B(Y)}\ ? $$ Here $B(X)$ is the space of bounded linear operators on $X$. Here "isomorphism" means bounded bijective linear map with bounded inverse.

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  • $\begingroup$ Must $I$ or $ITI^{-1}$ be bounded? $\endgroup$ – mechanodroid Jul 18 '18 at 19:37
  • $\begingroup$ $I$ and $I^{-1}$ are bounded. $\endgroup$ – Zouba Jul 18 '18 at 19:44
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In general, we don't have $\lVert T\rVert_{B(X)} = \lVert ITI^{-1}\rVert_{B(Y)}$. Let $X = \mathbb{R}^2$ endowed with the Euclidean norm, $Y = \mathbb{R}^2$ endowed with an $\ell^p$-norm for $p \neq 2$, $I$ the identity and $T$ a rotation by an angle not an integer multiple of $\pi/2$. On $X$, $T$ is an isometry, but on $Y$ its norm is greater than $1$.

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The answer is no.

Consider the Banach space $X = (\mathbb{R}^2, \|\cdot\|_\infty)$ and let $T : X \to X$ be given as $T = \begin{bmatrix}2 & 3 \\ 0 & 1\end{bmatrix}$.

Then

$$\|T(x,y)\|_\infty = \|(2x + 3y, y)\|_\infty = \max\{|2x+3y|, |y|\} \le 2|x| + 3|y| \le 5\|(x,y)\|_\infty$$

and $T(1,1) = (5,1)$ so we conclude $\|T\| = 5$.

Now we shall diagonalize $T$. Define $I : X \to X$ as $I = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ which is an isomorphism, but it is not isometric because $I(1,1) = (2,1)$.

Then $$ITI^{-1} = \begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}$$

so $\|ITI^{-1}\| = 2$.

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