# A linear operator that sends strongly convergent sequences to weakly convergent sequences

So I'm having trouble showing this. Its an old prelim problem. Suppose you have a linear operator $A$ on a Banach space which sends strongly convergent sequences to weakly convergent ones. Show that this operator is continuous. Now my first thought was to use closed graph theorem or hahn banach somehow, but I honestly have no clue what to do here.

Now my first thought was to use closed graph theorem

This!

Hence suppose we have a sequence $(x_n)$ such that $x_n \to x$ and $Ax_n \to y$. Since $A$ maps norm-convergent sequences to weakly convergent sequences, it follows that $Ax_n$ converges weakly to $Ax$ - if that is not clear, consider the sequence $(z_n)$ where $z_{2m} = x_m$ and $z_{2m+1} = x$; then $z_n \to x$, and since $Ax$ occurs infinitely often in the image sequence $(Az_n)$, if that converges weakly to anything, its weak limit must be $Ax$. And since norm-convergence implies weak convergence, $Ax_n$ converges weakly to $y$. Since weak limits are unique - the weak topology is Hausdorff - it follows that $y = Ax$. Hence the graph is closed, and consequently $A$ continuous.

• Ahh so simple. Thanks Daniel! – thegamer Aug 23 '17 at 17:43
• @Fisher Why $lim_nA(x_n)=y$ exists here ? – Tsemo Aristide Aug 23 '17 at 17:51
• @TsemoAristide By assumption. We check that the graph is closed, so we look at a sequence $(x_n, Ax_n)$ in the graph, which converges to some point $(x,y)$. If $(x,y)$ necessarily lies in the graph, it follows that the graph is closed. – Daniel Fischer Aug 23 '17 at 17:54
• The business with $(z_n)$ seems unnecessary. If $x_n\to x$, then $Ax_n\rightharpoonup Ax$ weakly by our assumption on $A$. Since $Ax_n\to y$, also $Ax_n\rightharpoonup y$ weakly. Then continue from "Since weak limits are..." – Jason Aug 24 '17 at 1:11
• @Jason The assumption is only that $Ax_n$ converges weakly to something. A priori, that something could be different from $Ax$. At some point, one has to prove that this cannot happen. If that has been proven before, the business with $(z_n)$ is indeed unnecessary. – Daniel Fischer Aug 24 '17 at 10:10

Recall, that

(1) linear operator is continious iff it is bounded.

(2) if $$x_n \rightharpoonup x$$ then sequence $$\left\Vert x_n \right\Vert$$ is bounded.

Suppose $$A$$ is not bounded and $$\left\Vert Ax \right\Vert = \infty$$. Then we can find sequence $$x_n$$ on a unit spere, such that $$\left \Vert Ax_n \right\Vert \to \infty$$. Now we construct another sequence $$y_n = \frac{x_n}{\sqrt{\left\Vert Ax_n\right\Vert}}.$$ We have now $$y_n \to 0$$ and $$\left\Vert Ay_n\right\Vert \to \infty$$ thus $$Ay_n$$ cannot converge weakly to anything.