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.
Okay, what I'm trying to see is if are Banach spaces and is a net in the space of bounded operators such that is a Cauchy net in for all , then there exists such that in the strong operator topology.
This is what I have so far:
in the strong op. topology iff for all . Every Cauchy net in a Banach space converges. So we have that is a linear operator, but we have yet to prove (or disprove) that 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 are Hilbert spaces, then and by the Hellinger-Toeplitz theorem, since is everywhere defined and self-adjoint, we have that it's bounded. An analogous theorem gives us that is bounded and so is a bounded operator.
Edit: I don't think this previous paragraph works, since as we don't know that is bounded may not exist, or is there a way to define a formal adjoint of a linear operator with the properties ?
Now, I'm at a loss as to how to approach the general case...
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.
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.
Let 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 , 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 to a Hamel basis, and let X be the subspace spanned algebraically by the added elements).
Define a linear map by . Then T is discontinuous.
Let be the set of all finite-dimensional subspaces of H, directed by inclusion, and for let , the restriction of T to F, with on the orthogonal complement of F. Then each is linear and bounded (because it has finite rank). Also, the directed net converges strongly to T and is therefore strongly Cauchy, but it does not converge strongly in B(H).
I believe even that the nice Opalg's method works for the general case,
with infinite dimensional Banach spaces. Given a non bounded linear map , that can be always be constructed, and given a finite dimensional subspace of , there exists always a topological complement of in , i.e . The axiom of choice allows us to take a complement for each finite dimensional subspace . Thus one can define as the restriction of in and 0 on , and is clearly continuous on . The conclusion is now the same as above, the net is Cauchy in for the SOT but not convergent