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Suppose that $K:C([0,1])\to C([0,1])$ is a continuous operator both with respect to $L^2$ and $L^\infty$ norms. Consider the following operator norm $$\sup_{\|f\|_2\leq 1}\|Kf\|_\infty$$ where $\|.\|_2$ is $L^2$ norm. Is it possible to compare up to a constant the following two operator norms $$\sup_{\|f\|_2\leq 1}\|Kf\|_\infty$$

and

$$\sup_{\|f\|_\infty\leq 1} \|Kf\|_\infty$$

Since $\|f\|_2\leq \|f\|_\infty$, the unit ball under $\|.\|_\infty$ norm is contained in the unit ball under the $\|.\|_2$ norm and $$\sup_{\|f\|_\infty\leq 1} \|Kf\|_\infty\leq \sup_{\|f\|_2\leq 1} \|Kf\|_\infty.$$

In other direction, since $K$ is continuous, we can restrict the supremum to some countable dense subset of the unit ball under $\|.\|_2$ norm. But I'm not sure if it gives me anything.

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When you say $K$ is continuous I assume it is with respect the norm on $C([0,1])$, that is, $|f\|_\infty$. The two norms are not comparable. Let $K$ be the identity. Then $$\sup_{\|f\|_\infty\leq 1} \|Kf\|_\infty=1$$ but $$\sup_{\|f\|_2\leq 1} \|Kf\|_\infty=\infty.$$ To see it consider a sequence $\{f_n\}$ of functions such that $f_n$ is supported on $[0,1/n]$ and $\int_0^1|f_n|^2\,dx=1$, which implies $\|f_n\|\infty\ge\sqrt n$.

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  • $\begingroup$ Thank you for this nice example. I didn't write that the assumption is that $K$ is bounded with respect to $L^2$ and $L^\infty$ norms. $\endgroup$ – Kiril Apr 20 '16 at 17:34
  • $\begingroup$ The identity of course is bounded with respecto to both norms. $\endgroup$ – Julián Aguirre Apr 20 '16 at 18:20

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