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?