Concerning the fact that weak and strong convergence coincide on (also know as Schur’s Theorem) I asked about an elementary proof in my last post. And in fact Markus Grasmair send me one which I’d like to present here. It is indeed elementary as it does not use any deep mathematics – however, it is a bit tricky.
Theorem 1 (Schur’s Theorem) A sequence in converges weakly iff it converges strongly.
Proof: As always, strong convergence implies weak convergence. To proof the opposite we assume that but for some . From this assumption we are going to derive a contradiction by constructing a vector with unit norm and a subsequence such that does not converge to zero. We initialize , set , choose such that and define the first entries of as
(Note that for these conditions are fulfilled: (1) is fulfilled since the sum is empty, (2) is fulfilled since and (3) is fulfilled by definition.) To go from a given to the next one, we first observe that implies that for all it holds that . Hence, we may take such that and of course we can take . Since is a summable sequence, we find such that and again we may take . We set
By construction, the properties (1), (2) and (3) are fulfilled for , and we see that we can continue our procedure ad infinitum. For the resulting (indeed ) and the subsequence we obtain (using the properties (1), (2) and (3)) that
Although this proof is really elementary (you only need to know what convergence means and have a proper -handling) I found it hard to digest. I think that this is basically not avoidable – Schur’s Theorem seems to be one of these facts which easy to remember, hard to believe and hard to prove.
Let’s try a direct proof that implies in .
Proof: We know that a weakly convergent sequence in is pointwise convergent (by testing with the canonical basis vectors), hence for every it holds that for . For some we write
By pointwise convergence of we know that the first sum converges to zero for every . Hence, we can proceed as follows: For a given first choose so large such that for all and then choose so large that . This sounds nice but wait! How about the choice of ? The tails of the series shall be bounded by uniformly in . That sounds possible but not obvious and indeed this is the place where the hard work starts! Indeed, one can prove that this choice of is possible by contradiction: Assume that there exists an such that for all there exists a such that . Alas, this is same thing which we have proven in the proof of Markus…
Indeed, the construction Markus made in his proof can be generalized to obtain the Nikodym convergence theorem:
Theorem 2 Let be a sequence of signed finite measures for which it holds that for every measurable set (which is equivalent to the weak convergence of to zero). Then the sequence is uniformly countably additive, i.e. for any sequence of disjoint measurable sets the series converges uniformly in .
One may compare this alternative proof with Exercise 11 (right after Theorem 5 [Nikodym convergence theorem]) in this blog post by Terry where one shall take a similar path to proof Schur’s Theorem.