If a linear operator is strong-weak continuous, then it is bounded 
$X$ and $Y$ are normed spaces and $L: X\to Y$ is a linear operator from $X$ to $Y$. Show that if $L$ is a continuous operator from $X$ with the strong (norm) topology to $Y$ with the weak topology, so $L: X\to Y$ (here $X,Y$ have both the strong topologies) is continuous. 

My try:
The hypothesis is equivalent to say that: for all $f \in Y'$ (ie, dual of $Y$) we have $f\circ  L$ is continuous.
So I want to prove that $L$ is continuous in the strong topologies, ie, $L$ takes bounded sets to bounded sets. I don't know what to do next.
 A: 
So I want to prove that $L$ is continuous in the strong topologies, ie, $L$ takes bounded sets to bounded sets.

Using bounded sets is a good plan. We have the fact that a continuous linear map between topological vector spaces maps bounded sets to bounded sets: Let $E,F$ be topological vector spaces (real or complex), and $T \colon E \to F$ a continuous linear map. Further, let $B \subset E$ be bounded. If $W$ is any neighbourhood of $0$ in $F$, by continuity there is a neighbourhood $V$ of $0$ in $E$ with $T(V) \subset W$. By boundedness, there is a $\delta > 0$ such that $\alpha B \subset V$ for all scalars $\alpha$ with $\lvert\alpha\rvert < \delta$. But then $\alpha T(B) = T(\alpha B) \subset T(V) \subset W$ for $\lvert\alpha\rvert < \delta$, so $T(B)$ is bounded.
Thus in our situation, we know that $L$ maps norm-bounded subsets of $X$ to weakly bounded subsets of $Y$, in particular $L(B_X)$ is weakly bounded in $Y$, where $B_X$ is the unit ball in $X$. A theorem of Mackey tells us that in locally convex spaces every weakly bounded subset is bounded in the original topology, so in fact $L(B_X)$ is norm-bounded, but that means precisely that
$$\lVert L\rVert = \sup \{ \lVert Lx\rVert_Y : \lVert x\rVert_X \leqslant 1\} < +\infty,$$
i.e. $L$ is continuous with respect to the norm topologies.
In normed spaces, the assertion of Mackey's theorem follows easily from the Banach-Steinhaus theorem (aka the uniform boundedness principle): Let $S\subset Y$ be weakly bounded. Via the canonical isometric embedding $\Phi_Y \colon Y \to Y''$ of $Y$ into its bidual, we can view $S$ as a family of linear functionals on the Banach space $Y'$, and that $S$ is weakly bounded means precisely that this family is pointwise bounded,
$$\sup \{ \lvert\Phi_Y(y)(\lambda)\rvert : y \in S\} = \sup \{ \lvert\lambda(y)\rvert : y \in S\} < +\infty$$
for all $\lambda \in Y'$. By the Banach-Steinhaus theorem it follows that
$$\sup \{ \lVert \Phi_Y(y)\rVert_{Y''} : y \in S\} = \sup \{ \lVert y\rVert_Y : y \in S\} < +\infty,$$
i.e. $S$ is norm-bounded.
