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

Suppose I have a weakly sequentially continuous linear operator T between two normed linear spaces X and Y (i.e. $x_n \stackrel {w}{\rightharpoonup} x$ in $X$ $\Rightarrow$ $T(x_n) \stackrel {w}{\rightharpoonup} T(x)$ in $Y$). Does this imply that my operator T must be bounded?

share|cite|improve this question
Just to clarify, those convergences inside parentheses are both weak? – Jonas Meyer Oct 1 '10 at 2:46
Yes. Sorry about that. I put a w on top now to hopefully clarify things. I usually write it without the w, so I didn't notice that it might've been unclear. – user1736 Oct 1 '10 at 3:01
No problem. In the meantime I noticed that the Wikipedia article on weak topology mentions your original notation, but it was something I wasn't used to. – Jonas Meyer Oct 1 '10 at 3:03
up vote 11 down vote accepted

In my original answer I only mentioned that it works for $Y$ complete, but as Nate pointed out in a comment, I never actually used completeness of $Y$.

The answer is yes. Weakly convergent sequences in a normed space are bounded, as a consequence of the uniform boundedness principle applied to the dual space (which is a Banach space) and the fact that a convergent sequence of real (or complex) numbers is bounded. If $T$ is unbounded, then there is a sequence $x_1,x_2,\ldots$ in $X$ converging in norm (and hence weakly) to 0 such that $\|T(x_n)\|\to\infty$, so by the previous sentence this implies that $T(x_1),T(x_2),\ldots$ does not converge weakly.

share|cite|improve this answer
You don't need $Y$ to be complete; if you check, you are applying the uniform boundedness principle in $Y^*$ which is a Banach space regardless. – Nate Eldredge Oct 1 '10 at 3:20
After a few seconds thought, yes, of course! I'm just so used to thinking about the Banach space case that I had assumed that that is where the completeness is used rather than actually thinking. – Jonas Meyer Oct 1 '10 at 3:22
@Nate: Thanks very much for the correction. – Jonas Meyer Oct 1 '10 at 3:39
Yeah, that seems to work. Thanks for your help Jonas! – user1736 Oct 1 '10 at 3:56
@Svetoslav: Unboundedness of T implies that for each $M>0$ there exists x with norm 1 such that $\|Tx\|>M$. Thus for each positive integer n take $y_n$ with norm 1 such that $\|Ty_n\|>n^2$, and take $x_n=\frac1n y_n$. – Jonas Meyer Feb 15 at 4:23

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.