Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $X$ and $Y$ are banach spaces and T map from $X$ to $Y$. If every sequence $x_n$ in weak topology in $X$ converges to $0$ , and $T(x_n)$ converges to $0$ in weak topology , does that imply that $T$ is bounded ?

From this question i would like to understand how convergence in weak topology differs from other topology. What would be the case if the topology defined was not weak ?

Thank you for your kind help . Keenly Looking forward to get some good ideas about weak topologies .

share|cite|improve this question
Do you mean to require that for every sequence $x_n$ which converges weakly to 0, that $T(x_n)$ converges weakly to 0? – Nate Eldredge Nov 20 '12 at 20:18
@NateEldredge : Yes Sir . – Theorem Nov 20 '12 at 20:23
The fact that $T$ is bounded follows from closed graph theorem. Let $\{x_n\}$ a sequence which converges strongly to $0$, and $\lVert Tx_n-y\rVert\to 0$. Let $f\in Y^*$. Then we have that $\{Tx_n\}$ converges weakly to $0$. As $\{Tx_n\}$ converges strongly hence weakly to $y$, $y=0$. Now let $x_n\to x$, $Tx_n\to y$. Then $x_n-x\to 0$ and $T(x_n-x)\to y-Tx$ so $y=Tx$. – Davide Giraudo Nov 20 '12 at 20:30
Now I guess the question is, for example when $X=Y$, what are the topologies $\cal T$ on $X$ such that if $T\colon (X,\cal T)\to (X,\cal T)$ is continuous, then $T$ is strongly continuous (and conversely)? – Davide Giraudo Nov 20 '12 at 20:40
@DavideGiraudo : Sir, thats exactly what i want to know. And also how to connect the continuity and other properties of norm with the weak topologies . I am kind of mixing up everything and not able to see clearly how to think connect norm and weak topology . I hope i am asking a sensible question :P – Theorem Nov 21 '12 at 8:53

Hint: If $T$ is unbounded then there is a sequence $x_n$ such that $x_n \to 0$ in norm (hence also weakly) but $T x_n$ is unbounded. However, it follows from the uniform boundedness principle that any weakly convergent sequence is bounded.

share|cite|improve this answer
Sir, But the thing here i don't understand is , i am given convergence in weak topology , how can i consider convergence in norm and connect these two things ?? – Theorem Nov 21 '12 at 8:55
@Theorem: Any sequence that converges in norm also converges weakly to the same limit (try proving this if you haven't already). So if your condition holds, you can apply it to $x_n$ and learn that $T x_n$ converges weakly, and hence is bounded. – Nate Eldredge Nov 21 '12 at 13:50
Yes sir , but i have been given weak convergence and not the norm convergence . Or have i misunderstood the question ?? – Theorem Nov 21 '12 at 14:12
@Theorem: It isn't that you have "been given weak convergence". You have been given a statement which applies to sequences that converge weakly. Therefore it also applies to sequences that converge in norm. – Nate Eldredge Nov 21 '12 at 14:56
Aha, now something is clear . I had been thinking that convergence in weak topology means weak convergence . Thank you , i will come back if i have some more doubts. – Theorem Nov 21 '12 at 17:15

A Banach space is by definition a normed vector space $X$ which is complete as a metric space, with the distance function induced by the norm. The norm induces a topology on $X$, with respect to which you can consider topological questions such as continuity, compactness, etc.

A linear map $T : (X, ||\;||) \rightarrow (Y, ||\;||)$ between normed vector spaces (not neccesarily Banach) is continuous if and only if it is bounded. It is implicit, that you are talking about continuity with respect to the topologies induced by the norms on $X$ and $Y$.

Now, there are a lot of different useful topologies in functional analysis one can put on a normed vector space $(X, ||\;||)$ other than the topology that is induced from the norm. One such topology is called the weak topology. How is it defined?

The norm structure on $(X, ||\;||)$ allows us to define when a linear functional $\phi : X \rightarrow \mathbb{C}$ is continuous. The collection of all continuous linear functionals on $X$ is denoted by $X^*$, and it by itself is a normed vector space (even Banach), with the norm being the operator norm $$ ||\phi|| = \sup_{||x|| = 1} |\phi(x)|. $$ The weak topology on $X$ is the weakest topology with respect to which all the maps $\phi \in X^*$ are continuous. Obviously, the original topology, induced by the norm on $X$ is a topology on $X$ under which all maps $\phi \in X^*$ are continuous, but it is quite possible you can throw away some open sets, and still remain with a topology for which all $\phi \in X^*$ are still continuous. Indeed, the weak topology on $X$, which we will denote by $\tau$, is the minimal topology which contains the obvious subsets which must be open for the functionals to be continuous - $\phi^{-1}(U)$ for $U \subset \mathbb{C}$ open and $\phi \in X^*$.

Assume that $X$ is infinite dimensional. Some properties of this topology you should get used to and prove:

  1. The weak topology $\tau$ has less open subsets than the norm topology on $X$.
  2. The weak topology $\tau$ is not first countable. It is not induced by any norm on $X$ nor by any distance function.
  3. The identity map $\mathrm{id} : (X, ||\;||) \rightarrow (X, \tau)$ is continuous. The inverse of the identity map is not.
  4. A map $T : X \rightarrow Y$, which is continuous as a map between the normed vector spaces $T : (X, ||\;||_1) \rightarrow (Y, ||\;||_2)$ will also be continuous as a map between the topological vector spaces $T : (X, \tau_1) \rightarrow (Y, \tau_2)$ when $\tau_i$ are the weak topologies on $X$ and $Y$. The other direction is generally false.

Hmm, it seems I misunderstood the question, but this still might be useful...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.