I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,
Let $E, F$ be Banach spaces and $T:E \to F$ a bounded linear operator. Consider the following properties:
- (P) If $(u_n)$ is a sequence in $E$ such that $u_n \to u$ in the weak topology $\sigma(E, E^*)$, then $Tu_n \to Tu$ in norm topology of $F$.
- (Q) $T$ is continuous from $E$ equipped with $\sigma(E, E^*)$ to $F$ equipped with norm topology.
- Prove that (Q) $\iff$ $T$ is a finite-rank operator.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
- Let $R(T)$ be the range of $T$.
- Assume $T$ is a finite-rank operator. Then $\dim R(T) < \infty$. Then the norm topology of $R(T)$ coincides with its weak topology $\sigma(R(T), R(T)^*)$. By Theorem 3.10 in the same book, $T$ is continuous from $E$ equipped with $\sigma(E, E^*)$ to $F$ equipped with $\sigma(F, F^*)$. Then $T$ is continuous from $E$ equipped with $\sigma(E, E^*)$ to $R(T)$ equipped with $\sigma(R(T), R(T)^*)$. Then $T$ is continuous from $E$ equipped with $\sigma(E, E^*)$ to $R(T)$ equipped with norm topology. Then $T$ is continuous from $E$ equipped with $\sigma(E, E^*)$ to $F$ equipped with norm topology.
- Assume (Q) holds. Let $B$ be the closed unit ball of $F$. Let $U:=T^{-1} (B)$. Then $U$ is a neighborhood of $0$ in $\sigma(E, E^*)$. Then $U$ contains a subspace $X$ of $E$ such that $\operatorname{codim} X < \infty$. Then there is a subspace $Y$ of $E$ such that $\dim Y = \operatorname{codim} X$ and that $E = X+Y$. Because $B$ is bounded and $f(X) \subset B$, we get $X \subset \ker T$. Then $R(T) \subset T(Y)$ and thus $\dim R(T) \le \dim Y < \infty$. This completes the proof.