I'm trying to solve an exercise in Brezis' Functional Analysis
Let $E,F$ be Banach spaces where $\dim E = \infty$. Let $T:E \to F$ be a compact bounded linear operator. Then there is a sequence $(u_n)$ in $E$ such that $|u_n|_E=1$ and $|Tu_n|_F \to 0$.
The author's proof is by contradiction and more complex than mine, i.e.,
If the conclusion fails, there exists some $\delta>0$ such that $|T u|_F \geq \delta |u|_E \quad \forall u \in E$. Hence $R(T)$ is closed. Consider the operator $T_0: E \rightarrow R(T)$ defined by $T_0=T$. Clearly $T_0$ is bijective. By Corollary 2.6, $T_0^{-1} \in \mathcal{L}(R(T), E)$. On the other hand, $T_0 \in \mathcal{K}(E, R(T))$. Hence $B_E$ is compact and $\dim E<\infty$
There are possibly subtle mistakes that I could not recognize in below attempt. Could you have a check on it? Thank you so much for your help!
My attempt Fix $\varepsilon \in (0, 1)$. Because $E$ is infinite-dimensional and by Riesz's lemma, there is a sequence $(u_n)$ in $E$ such that $|u_n|_E=1$ and $|u_n-u_m|_E \ge \varepsilon$ for all $m \neq n$. Because $T$ is compact, there is a subsequence $(u_{n_k})_k$ and $v \in F$ such that $Tu_{n_k} \xrightarrow{k \to \infty} v$. Let
$$
x_k := \frac{u_{n_{k+1}}-u_{n_k}}{|u_{n_{k+1}}-u_{n_k}|_E}
\quad \forall k \in \mathbb N.
$$
Then $|x_k|_E=1$. We have
$$
|Tx_k|_F = \frac{|Tu_{n_{k+1}}-Tu_{n_k}|_F}{|u_{n_{k+1}}-u_{n_k}|_E} \le\frac{ |Tu_{n_{k+1}}-v|_F + |Tu_{n_k}-v|_F}{\varepsilon}.
$$
Taking the limit $k \to \infty$, we see that $(x_k)$ satisfies our requirement.
Best Answer
I would just say that your proof is perfectly valid :)
You did not have to edit since convergence implies de Cauchy