# weak sequential continuity of linear operators

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?

• 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

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.
• 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
• @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 '16 at 4:23