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I found the next exercise in Haim Brezis's book Functional Analysis, Sobolev Spaces and Partial Differential Equations. I feel like I solved the problem, but I'm not sure.

The problem is:

Let $E,F$ be two Banach spaces with norms $\|\cdot \|_E,\| \cdot \|_F$. Assume that $E$ is reflexive. Let $T :E \to F$ be a compact operator. Consider on $E$ another norm $| \cdot |$ weaker than $\|\cdot \|_E$, i.e. $|u|\leq C \|u\|_E$. Prove that for every $\varepsilon >0$ ther exists $C_\varepsilon>0$ such that $$ \|Tu\|_F \leq \varepsilon \|u\|_E +C_\varepsilon |u| $$

My approach is the following: Assume that the conclusion does not hold. Then there is an $\varepsilon>0$ and a sequence $u_n$ in $E$ such that $$ \|Tu_n\|_F > \varepsilon \|u_n\|_E+n|u_n|$$

We can normalize the sequence such that $\|Tu_n\|=1$. Then $u_n$ is bounded in $E$, and because $E$ is reflexive then without loss of generality we can assume that $u_n$ converges weakly to $u \in E$. Since compact operators map weakly convergent sequences onto strongly convergent sequences it follows that $Tu_n \to Tu$ in $F$, so $\|Tu\|_F=1$, which means that $u\neq 0$.

On the other hand $|u_n|<1/n$ so that $u_n$ converges to $0$ in the weaker norm $|\cdot |$. Is this enough to prove that $u=0$ and reach a contradiction?

If my approach does not lead to a good end then what else should I try?

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up vote 1 down vote accepted

Your approach looks good and almost complete. To finish the proof, note that the weak convergence of $u_n$ to $u$ in $(E,\|\cdot\|_E)$ implies weak convergence in $(E,|\cdot|)$ – the assumption $|u|\le C \|u\|_E$ says that $|\cdot|$-continuous functionals are $\|\cdot\|_E$-continuous. Then $u=0$ since the weak topology of $(E,|\cdot|)$ is Hausdorff and $|u_n|\to0$.

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I was thinking something on these lines, but I didn't think of the weak topology given by $|\cdot|$. It is a nice idea indeed. Thanks –  Beni Bogosel May 31 '12 at 12:46
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