I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,
Let $X,Y,Z$ be real Banach spaces with corresponding norms $|\cdot|_X, |\cdot|_Y, |\cdot|_Z$. Assume that $X \subset Y$ with compact injection and that $Y \subset Z$ with continuous injection.
- Prove that for every $\varepsilon>0$ there is $C_\varepsilon > 0$ such that $$ |u|_Y \le \varepsilon |u|_X + C_\varepsilon |u|_Z \quad \forall u \in X. $$
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
Assume the contrary that there is $\varepsilon>0$ such that for each $n \in \mathbb N^*$ there is $u_n \in X$ such that $$ |u_n|_Y > \varepsilon |u_n|_X + n |u_n|_Z. $$
Then $u_n \neq 0$ for all $n$. Let $v_n := \frac{u_n}{|u_n|_X}$. Then $|v_n|_X=1$ and $|v_n|_Y > \varepsilon + n |v_n|_Z$ for all $n$. Because $X \subset Y$ with compact injection, we assume WLOG that there is $v \in Y$ such that $|v_{n}-v|_Y \to 0$. In particular, $(|v_n|_Y)_n$ is bounded. It follows that $|v_n|_Z \to 0$. Because $Y \subset Z$ with continuous injection, we get $|v_{n}-v|_Z \to 0$. Then $v=0$ and thus $|v_n|_Y \to 0$. On the other hand, $$ \lim_n |v_n|_Y \ge \varepsilon + \lim_n n |v_n|_Z. $$
Then we get a contradiction $0 \ge \varepsilon >0$. This completes the proof.
