0
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ "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
2
$\begingroup$

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 $\|A\|_{B(X)} = n$, $\|A\|_{B(Y)} = \sqrt n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.