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?
weak sequential continuity of linear operators
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functional-analysis
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0Just to clarify, those convergences inside parentheses are both weak? – 2010-10-01
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0Yes. 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. – 2010-10-01
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0No 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. http://en.wikipedia.org/wiki/Weak_topology#Weak_convergence – 2010-10-01
1 Answers
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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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0You 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. – 2010-10-01
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0After 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. – 2010-10-01
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0@Nate: Thanks very much for the correction. – 2010-10-01
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0Yeah, that seems to work. Thanks for your help Jonas! – 2010-10-01
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0Looking 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. – 2010-10-01
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0@JonasMeyer : is it not that $T(x_n)$ is in double dual of $X$ ? I am quite confused here . $x_n$ are in $X$, and as operator $T(x_n)$ takes value from $X'$ say $f$and gives me a value $f(x_n)$. What is my misunderstanding here ? – 2012-12-08
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0@Theorem: $T:X\to Y$, as in says in the question. – 2012-12-08
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0It is a long time since you posted that answer, but I would like to ask you something. Why the assumption of unboundedness of the operator $T$ (or equivalently not continuous at $0$ ) implies the existence of such a sequence $x_n\to 0$ such that $\|Tx_n\|\to\infty$ > Isn't it that there exists a sequence $x_n\to 0$ in norm and some $\epsilon>0$ such that $\|Tx_n\|\ge \epsilon,\,\forall n\in\mathbb N$ ? – 2016-02-15
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3@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 $x_n=\frac1n y_n$. – 2016-02-15