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$X$ 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.

pipita
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  • How are being bounded and being weakly bounded related? – Daniel Fischer Dec 08 '15 at 14:20
  • 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 08 '15 at 14:26
  • But it was not...i was just checking the exercices before this one. Is this that you mean? https://en.wikipedia.org/wiki/Bounded_set_(topological_vector_space) – pipita Dec 08 '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 08 '15 at 14:33
  • Not the definition. The relation between being bounded in the norm topology and being bounded in the weak topology. – Daniel Fischer Dec 08 '15 at 14:39
  • Let us continue this discussion in chat. – pipita Dec 08 '15 at 14:42
  • The relation is (i think): if a set is bounded in strong topology so it is bounded in the weak topology. Right? But how this help me? – pipita Dec 08 '15 at 14:50
  • That's one part. How about the other direction, which weakly bounded sets are strongly bounded? – Daniel Fischer Dec 08 '15 at 14:52
  • That is impossible to determinate because the strong topology has more open sets than the weak toppology. – pipita Dec 08 '15 at 15:02
  • It is nevertheless possible. Mackey's theorem gives a simple relation. – Daniel Fischer Dec 08 '15 at 15:09
  • http://mathworld.wolfram.com/MackeysTheorem.html

    Ok, but this dont help me nowhere...and I've never heard about this in my classes Thanks anyaway

     – pipita Dec 08 '15 at 15:16
  • Does someone has some simple hints? – pipita Dec 08 '15 at 15:17
  • Uh, that Wolfram link is about another theorem by Mackey. Seems wikipedia doesn't have an article about it (mildly surprising). So let's state it: Mackey's theorem (the one I meant) says that in a locally convex TVS $E$, a set $B\subset E$ is weakly bounded if and only if it is bounded in the original topology. So for our situation with normed spaces, weakly bounded = strongly bounded. Can you see how that helps? – Daniel Fischer Dec 08 '15 at 15:24
  • But for that happen, the set must be locally convex...i dont know if it is or not What is TVS? – pipita Dec 08 '15 at 15:28
  • But i cannot use that theorem because i have never learnt in the classes – pipita Dec 08 '15 at 15:29
  • The space must be locally convex. A normed space is locally convex. TVS is the abbreviation for "topological vector space". – Daniel Fischer Dec 08 '15 at 15:29
  • There must be another way to do this because I have never heard about that and my teacher would do not tell me to this if I had that to use something that he didnt teach – pipita Dec 08 '15 at 15:34
  • There are other ways, one is taken here. Without knowing what theory you have available, I could only guess which ways are accessible. – Daniel Fischer Dec 08 '15 at 15:50
  • Have you met the uniform boundedness principle? See here. – Josh Keneda Feb 21 '16 at 06:45
  • @DanielFischer, you've already answered the question. Why don't post it as answer? – Norbert Feb 23 '16 at 09:19

1 Answers1

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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.

Daniel Fischer
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