I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,
Let $E$ and $F$ be two real Banach spaces with corresponding norms $|\cdot|_E$ and $|\cdot|_F$. Assume that $E$ is reflexive. Let $T:E \to F$ be a compact (bounded linear) operator. Consider another norm $|\cdot|$ on $E$, which is weaker than the norm $|\cdot|_E$, i.e., $|u| \le C|u|_E$ for all $u \in E$.
- Prove that for every $\varepsilon>0$ there is $C_\varepsilon > 0$ such that $$ |Tu|_F \le \varepsilon |u|_E + C_\varepsilon |u| \quad \forall u \in E. $$
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
Let $\tau_1$ be the weak topology on $E$ induced by $|\cdot|_E$. Let $\tau_2$ be the weak topology on $E$ induced by $|\cdot|$.
Assume the contrary that there is $\varepsilon>0$ such that for each $n \in \mathbb N^*$ there is $u_n \in E$ such that $$ |Tu_n|_F > \varepsilon |u_n|_E + n |u_n|. $$
Then $u_n \neq 0$ for all $n$. Let $v_n := \frac{u_n}{|u_n|_E}$. Then $|v_n|_E=1$ and $|Tv_n|_F > \varepsilon + n |v_n|$ for all $n$. Because $E$ is reflexive, we assume WLOG that there is $v \in E$ such that $v_n \to v$ in $\tau_1$. I proved that weaker norm induces weaker weak topology, so $\tau_2 \subset \tau_1$. Then $v_n \to v$ in $\tau_2$.
Because $T$ is compact, we get from exercise 6.7.4 (in the same book) that $|Tv_n-Tv|_F \to 0$. In particular, $(|Tv_n|_F)_n$ is bounded. It follows that $|v_n| \to 0$ and thus $v_n \to 0$ in $\tau_2$. Because $\tau_2$ is Hausdorff, we get $v=0$. Then $|Tv_n|_F \to 0$. On the other hand, $$ \lim_n |Tv_n|_F \ge \varepsilon + \limsup_n n |v_n|. $$
Then we get a contradiction $0 \ge \varepsilon >0$. This completes the proof.