$T$ is continuous $\iff \forall (x_1, x_2,..), x_i \in X, x_i \to^w x \implies T(x_i) \to^w T(x)$

I have solved (I think..) the following problem. The thing is that in my proof, I have not needed to use the fact that $X$ is complete. Which makes me wonder if my proof is correct.

Problem

Let $X$ be a Banach space and let $T: X \to X$ be a linear map. Show that:

$T$ is continuous $\iff \forall (x_1, x_2,..), x_i \in X, x_i \to^w x \implies T(x_i) \to^w T(x)$

By "$\to^w$" I mean "converges weakly". $X'$ denotes the dual of $X$.

Solution $\implies$

Assume $T$ is continuous. Then, since $T$ is linear as well, $T \in X'$. $x_i \to^w x \implies T(x_i) \to T(x) \implies T(x_i) \to^w T(x) \ \ \square$

Solution $\impliedby$

Assume $p_1 \iff \forall (x_1, x_2,..), x_i \in X, x_i \to^w x \implies T(x_i) \to^w T(x)$.

We know that every weakly convergent sequence is bounded, i.e.:

$p_2 \iff x_i \to^w x \implies \exists C > 0: \forall i: ||x_i|| \le C$

I claim that $p_1 \wedge p_2 \implies T$ is bounded: Assume $T$ is not bounded. Then $\exists (x_1, x_2, ..), x_i \to x$ such that $||T(x_i)|| \to \infty$. But $x_i \to x \implies x_i \to^w x \implies T(x_i) \to^w T(x) \implies \exists C>0 :\forall i: ||T(x_i)|| \le C$.

Hence $T$ is bounded, i.e. continuous. $\square$

Solution 1:

Both parts have errors. $T\in X'$ is not correct since $T$ takes values in $X$, not in the scalar field. Suppose $T$ is continuous and $x_i \to^{w} x$. To show that $Tx_i \to^{w} Tx$ pick any $g \in X'$. Then $g\circ T \in X'$ so $g(T(x_i))=(g\circ T) (x_i) \to (g\circ T) (x_i)=g(Tx)$ and this proves that $Tx_i \to^{w} Tx$.

In the converse part you assumed that $x_i \to x$ (in the norm) and proved that $(T(x_i))$ is bounded. That does not prove that $T$ is bounded. Suppose $T$ is not bounded. Then there exists a sequence $(x_i)$ with $\|x_i\| \leq 1 $ for all $i$ and $\|Tx_i\| >i$. Let $y_i=\frac {x_i} {\sqrt i}$ Then $y_i \to 0$. By your argument $T(y_i)$ is bounded. But $\|Ty_i\|.\sqrt i \to \infty$ This is a contradiction.