Suppose that $X \subset Y$ are both Banach spaces with norms $\vert \vert \cdot \vert \vert_X$ and $\vert \vert \cdot \vert \vert_Y$ respectively. Thus, $\vert \vert a \vert \vert_Y \leq \vert \vert a \vert \vert_X$. Suppose further, that $X$ is dense in $Y$ (in the norm of $Y$).

Now, consider $B(X)$ and $B(Y)$ with respective operator norms $\vert \vert \cdot \vert \vert_{B(X)} $ and $\vert \vert \cdot \vert \vert_{B(Y)}$.

If an operator $A \in B(X)$ is also bounded in the norm $\vert \vert \cdot \vert \vert_{B(Y)}$ we know that they it has a unique extension to $Y$ and thus we can compare the norms $\vert \vert A \vert \vert_{B(Y)}$ and $\vert \vert A \vert \vert_{B(X)}$. But as far as I can see it might not be bounded.

Do we know whether there exists a constant $C>0$ such that for those operators $A$ where we can compare the norms in this way then either $\vert \vert A \vert \vert_{B(X)} \leq C \vert \vert A \vert \vert_{B(Y)}$ or $\vert \vert A \vert \vert_{B(Y)} \leq C \vert \vert A \vert \vert_{B(X)}$?

If not what are counter examples to those such inequalites?

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    $\begingroup$ "Thus, $||a||_Y≤||a||_X$." You are claiming this is true and not assuming it? Why? $\endgroup$ – D_S May 3 at 21:05
  • $\begingroup$ Yes, good point. I think of $X$ being the set of all elements that have finite $X$-norm and $Y$ as being the set of all elements that have finite $Y$-norm. $\endgroup$ – Frederik Ravn Klausen May 4 at 11:56

This is not possible.

First, let $X=L^2(0,1)$, $Y=L^1(0,1)$. For $n\in \mathbb N$, define $(Ax)(s):=x(s/n)$. Then $\|A\|_{B(X)} = \sqrt n$, $\|A\|_{B(Y)} = n$.

Second, let $X=l^1$, $Y=l^2$. For $n\in \mathbb N$, define $$ Ax=(\underbrace{x_1,\dots,x_1}_{n\text{ times}},\underbrace{x_2,\dots,x_2}_{n\text{ times}},\dots). $$ In this case $\|A\|_{B(X)} = n$, $\|A\|_{B(Y)} = \sqrt n$.


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