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.


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. $\square$


1 Answer 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.

  • $\begingroup$ @justANoob I was a bit careless in my earlier comments. Sorry about that. I have given a detailed answer now. $\endgroup$ Dec 20, 2021 at 11:34
  • $\begingroup$ 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? $\endgroup$
    – JustANoob
    Dec 20, 2021 at 12:53
  • $\begingroup$ 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 $\|T(y_n)\|\ge \sqrt{n}$. $\endgroup$
    – Jochen
    Dec 20, 2021 at 12:55
  • $\begingroup$ @KaviRamaMurthy Just to send you a notification :) see my answer above $\endgroup$
    – JustANoob
    Dec 20, 2021 at 17:14
  • $\begingroup$ 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 $\endgroup$ Dec 20, 2021 at 23:15

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