# How do norms on $X$ and $Y$ relate to operator norms on $B(X)$ and $B(Y)$?

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?

• "Thus, $||a||_Y≤||a||_X$." You are claiming this is true and not assuming it? Why? – D_S May 3 at 21:05
• 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. – Frederik Ravn Klausen May 4 at 11:56

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