# $T$ 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. $$\square$$

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.
• For the first part, you are right, I don't know what I was thinking there. But for the second part, I proved that if $p_1$ and $p_2$ hold, then $T$ is bounded. Do you mean that the the statement "Assume $T$ is not bounded. Then $\exists (x_1,x_2,..),x_i \to x$ such that $||T(x_i)||\to \infty$. " is wrong? Dec 20, 2021 at 12:53
• It is indeed true that a linear map between normed spaces is continuous (=bounded) if $x_n\to 0$ always implies that $T(x_n)$ is bounded. Assuming $T$ to be non-bounded there is a sequence $x_n$ with $\|x_n\|\le 1$ such that $\|T(x_n)\|>n$. Then $y_n=\frac 1{\sqrt{n}} x_n\to 0$ and $\|T(y_n)\|\ge \sqrt{n}$. Dec 20, 2021 at 12:55
• How do you say that if $T$ is not bounded then there is a convergent sequence $(x_i)$ with $\|Tx_i\| \to \infty$? Surely this is not at all obvious and you have to provide a proof. That is what I have done in my answer. $T$ is not bounded only says that there is a **bounded ** sequence $(x_i)$ with $(\|Tx_i\| )$ unbounded. Where does your convergent sequence come from? @JustANoob Dec 20, 2021 at 23:15