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.
 A: 
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.
A: 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.
