# 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.

• How are being bounded and being weakly bounded related? – Daniel Fischer Dec 8 '15 at 14:20
• I dont know...i dont know how to do that relation – pipita Dec 8 '15 at 14:21
• The relation is not particular to this specific situation. It's a general relation, it should have been at least mentioned before this exercise was posed. – Daniel Fischer Dec 8 '15 at 14:26
• But it was not...i was just checking the exercices before this one. Is this that you mean? en.wikipedia.org/wiki/Bounded_set_(topological_vector_space) – pipita Dec 8 '15 at 14:29
• I would expect it to be mentioned in the lecture/book, not in the exercises. Yes, that's what I mean, bounded sets in topological vector spaces. – Daniel Fischer Dec 8 '15 at 14:33

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.