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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?

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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. en.wikipedia.org/wiki/Weak_topology#Weak_convergence –  Jonas Meyer Oct 1 '10 at 3:03

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up vote 9 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.

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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
    
Looking back at this, it's more obvious than I realized that this answer has nothing to do with $Y$ being complete, because weak convergence and boundedness obviously are independent of being complete. –  Jonas Meyer Oct 1 '10 at 15:25

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