1. ## Functional Analysis

Okay, what I'm trying to see is if $\displaystyle X, Y$ are Banach spaces and $\displaystyle (T_a) \subset \mathcal{B} (X,Y)$ is a net in the space of bounded operators such that $\displaystyle (T_ax)$ is a Cauchy net in $\displaystyle Y$ for all $\displaystyle x\in X$, then there exists $\displaystyle T\in \mathcal{B} (X,Y)$ such that $\displaystyle T_a \rightarrow T$ in the strong operator topology.

This is what I have so far:

$\displaystyle T_a \rightarrow T$ in the strong op. topology iff $\displaystyle T_ax \rightarrow Tx$ for all $\displaystyle x\in X$. Every Cauchy net in a Banach space converges. So we have that $\displaystyle Tx:= \lim_{a} T_ax$ is a linear operator, but we have yet to prove (or disprove) that $\displaystyle T$ is bounded. The case where the net is a sequence follows immediately from the Banach-Steinhaus theorem because every convergent sequence in bounded, but this need not happen for an arbitrary net.

Let's assume for a moment that $\displaystyle X, Y$ are Hilbert spaces, then $\displaystyle T= \frac{T+T^*}{2} +\frac{T-T^*}{2} =T_1+T_2$ and by the Hellinger-Toeplitz theorem, since $\displaystyle T_1$ is everywhere defined and self-adjoint, we have that it's bounded. An analogous theorem gives us that $\displaystyle T_2$ is bounded and so $\displaystyle T$ is a bounded operator.

Edit: I don't think this previous paragraph works, since as we don't know that $\displaystyle T$ is bounded $\displaystyle T^*$ may not exist, or is there a way to define a formal adjoint of a linear operator with the properties $\displaystyle \langle x,Ty \rangle = \langle T^*x,y\rangle$?

Now, I'm at a loss as to how to approach the general case...

2. I believe you actually can use the banach-steinhaus principle, as a net is surely also a family.

In fact, what you are asked to show is so general, that I bet it is equivalent to the principle in certain unexpected cases.

3. Originally Posted by Rebesques
I believe you actually can use the banach-steinhaus principle, as a net is surely also a family.
I don't think I can since I can't gurantee that it is pointwise bounded: Take for example $\displaystyle a_t=e^{-t}$ with $\displaystyle t\in \mathbb{R}$ then the net converges as $\displaystyle t\rightarrow \infty$ but it is not bounded.

4. That's not linear.

Now, use the B-S principle to obtain $\displaystyle \sup_a\vert\vert T_a\vert\vert<\infty$. For x in the unit ball of X, $\displaystyle \vert\vert Tx\vert\vert\leq\vert\vert T-T_a\vert\vert+\sup_a\vert\vert T_a\vert\vert$, therefore $\displaystyle \vert\vert T\vert\vert$ is bounded.

5. Originally Posted by Rebesques
That's not linear.
Take $\displaystyle (T_a)(x)=ax$ with $\displaystyle a\in (-\infty ,0)$ then in converges at every point to $\displaystyle 0$ (we take the directed set with $\displaystyle <$ relation), but it is not pointwise bounded since $\displaystyle |T_a(x)|=|a||x|\rightarrow \infty$ if $\displaystyle a\rightarrow -\infty$ so the theorem does not apply.

6. The question is in fact if L(X,Y) is complete (in the locally convex sense) under the SOT. It is sequentially complete, because the Banach Steinhauss principle, but, as you have argued, it is not clear for nets.

I don't know the answer, but I guess it is negative and not at all trivial, but sure it is known!. The Strong Operator Topology is not metrisable but locally convex, try to look for the answer in Schaefer or Köethe book "Topological vector spaces" (perhaps in Jarchow's book "Locally convex spaces")

sorry, I can't do any more but provide some references! but I am curious also about the answer, I will try also to check it.

7. Originally Posted by Jose27
Take $\displaystyle (T_a)(x)=ax$ (...) if $\displaystyle a\rightarrow -\infty$ so the theorem does not apply.
The index set does not follow the defined increasing order.

it is not clear for nets.

I think it is straightforward for cauchy nets in banach spaces.

8. Originally Posted by Rebesques
The index set does not follow the defined increasing order.

I think it is straightforward for cauchy nets in banach spaces.
A convergent sequence in a Banach space is bounded because all excepte finitely many are contained in a (bounded) neighbourhood. In a net this is simply not true, and José27 argument is correct. What he means, is that for any positive x, if $\displaystyle a>-\varepsilon /x$ then $\displaystyle T_a(x)\subset ]-\varepsilon,\varepsilon[$, hence the net is convergent to 0 as a goes to 0 (for negative x is the analogous, of course). But $\displaystyle (T_a(x))_{a<0}$ is not bounded.

Anyway here there are countable cofinal subsets in the index set, and hence Banach Steinhauss works because you can reduce it to a problem of sequences. It is hard for me to think about one net with no cofinal subsets in the index set and convergent at every point!!. If the answer were negative, the example would be quite sophisticated I suppose.

9. Originally Posted by Enrique2
The question is in fact if L(X,Y) is complete (in the locally convex sense) under the SOT. It is sequentially complete, because of the Banach–Steinhaus principle, but, as you have argued, it is not clear for nets.

I don't know the answer, but I guess it is negative
I agree with that. I think the following example works, but it's weird enough that I may have overlooked something.

Let $\displaystyle \{e_n\}_{n\in\mathbb{N}}$ be an orthonormal basis for a Hilbert space H (or it could be any Banach space with the right sort of basis, but I'll stick to a Hilbert space). Let E be the non-closed subspace of H spanned algebraically by the $\displaystyle e_n$, in other words the set of finite linear combinations of the basis elements. Let X be a complementary subspace (which must exist by the axiom of choice: extend the set $\displaystyle \{e_n\}$ to a Hamel basis, and let X be the subspace spanned algebraically by the added elements).

Define a linear map $\displaystyle T:H\to H$ by $\displaystyle Tx=x\ (x\in E),\ \ Tx=0\ (x\in X)$. Then T is discontinuous.

Let $\displaystyle \mathcal{F}$ be the set of all finite-dimensional subspaces of H, directed by inclusion, and for $\displaystyle F\in\mathcal{F}$ let $\displaystyle T_F = T|_F$, the restriction of T to F, with $\displaystyle T_F=0$ on the orthogonal complement of F. Then each $\displaystyle T_F$ is linear and bounded (because it has finite rank). Also, the directed net $\displaystyle \{T_F\}_{F\in\mathcal{F}}$ converges strongly to T and is therefore strongly Cauchy, but it does not converge strongly in B(H).

10. I believe even that the nice Opalg's method works for the general case,
$\displaystyle L(X,Y)$ with $\displaystyle X,Y$ infinite dimensional Banach spaces. Given a non bounded linear map $\displaystyle T:X\to Y$, that can be always be constructed, and given a finite dimensional subspace $\displaystyle F$ of $\displaystyle X$, there exists always a topological complement $\displaystyle W$ of $\displaystyle F$ in $\displaystyle X$, i.e $\displaystyle X=F\oplus W$. The axiom of choice allows us to take a complement $\displaystyle W_F$ for each finite dimensional subspace $\displaystyle F$. Thus one can define $\displaystyle T_F$ as the restriction of $\displaystyle T$ in $\displaystyle F$ and 0 on $\displaystyle W_F$, and $\displaystyle T_F$ is clearly continuous on $\displaystyle X$. The conclusion is now the same as above, the net $\displaystyle (T_F)_F$ is Cauchy in $\displaystyle L(X,Y)$ for the SOT but not convergent