Concerning the fact that weak and strong convergence coincide on {\ell^1} (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 {(x^k)} in {\ell^1} converges weakly iff it converges strongly.

Proof: As always, strong convergence implies weak convergence. To proof the opposite we assume that {x^k\rightharpoonup 0} but {\|x^k\|_1\geq\epsilon} for some {\epsilon>0}. From this assumption we are going to derive a contradiction by constructing a vector {f\in\ell^\infty} with unit norm and a subsequence {x^{k_l}} such that {\langle f,x^{k_l}\rangle} does not converge to zero. We initialize {j_0=0}, set {k_1=1}, choose {j_1\in{\mathbb N}} such that {\sum_{j>j_1}|x^1_j|\leq \epsilon/6} and define the first {j_1} entries of {f} as

\displaystyle  f_j = \text{sign}(x^1_j)\ \text{ for }\ 1\leq j\leq j_1.

Now we proceed inductively and assume that for some {l\geq 1} the numbers {j_1,\dots,j_l} the subsequence {x^1,\dots,x^{k_l}} and the entries {f_1,\dots,f_{j_l}} have already been constructed and fulfill for all {m\leq l}

\displaystyle  \Big|\sum_{j\leq j_{m-1}} f_j x^{k_m}_j\Big| \leq \frac{\epsilon}{6} \ \ \ \ \ (1)

\displaystyle  \sum_{j=j_{m-1}+1}^{j_m} f_j x^{k_m}_j\geq \frac{2\epsilon}{3} \ \ \ \ \ (2)

\displaystyle  \sum_{j>j_m}|x^{k_m}_j|\leq \frac{\epsilon}{6}. \ \ \ \ \ (3)

(Note that for {l=1} these conditions are fulfilled: (1) is fulfilled since the sum is empty, (2) is fulfilled since {\sum_{j=1}^{j_1}f_j x^1_j = \|x\|_1 - \sum_{j>j_1}|x^1_j|>5\epsilon/6} and (3) is fulfilled by definition.) To go from a given {l} to the next one, we first observe that {x^k\rightharpoonup 0} implies that for all {j} it holds that {x^k_j\rightarrow 0}. Hence, we may take {k_{l+1}} such that {\sum_{j\leq j_l} |x^{k_{l+1}}_j|\leq \epsilon/6} and of course we can take {k_{l+1}>k_l}. Since {x^{k_{l+1}}} is a summable sequence, we find {j_{l+1}} such that {\sum_{j>j_{l+1}}|x^{k_{l+1}}_j|<\epsilon/6} and again we may take {j_{l+1}>j_l}. We set

\displaystyle  f_j = \text{sign}(x^{k_{l+1}}_j)\ \text{ for }\ j_l\leq j\leq j_{l+1}

and observe

\displaystyle  \sum_{j = j_l+1}^{j_{l+1}} f_j x^{k_{l+1}}_j = \sum_{j = j_l+1}^{j_{l+1}} |x^{k_{l+1}}_j| > \|x^{k_{l+1}}\|_1 - \frac{\epsilon}{3} > \frac{2\epsilon}{3}.

By construction, the properties (1), (2) and (3) are fulfilled for {l+1}, and we see that we can continue our procedure ad infinitum. For the resulting {f\in\ell^\infty} (indeed {\|f\|_\infty=1}) and the subsequence {(x^{k_l})} we obtain (using the properties (1), (2) and (3)) that

\displaystyle  \begin{array}{rcl}  \langle f,x^{k_l}\rangle &=& \sum_j f_j x^{k_l}_j\\ & = & \sum_{j\leq j_{l-1}} f_j x^{k_l}_j + \sum_{j= j_{l-1}+1}^{j_l} f_j x^{k_l}_j + \sum_{j> j_l} f_j x^{k_l}_j\\ & \geq & -\Big|\sum_{j\leq j_{l-1}} f_j x^{k_l}_j \Big| + \sum_{j= j_{l-1}+1}^{j_l} f_j x^{k_l}_j - \sum_{j> j_l} |x^{k_l}_j|\\ & \geq& - \frac{\epsilon}{6} + \frac{2\epsilon}{3} - \frac{\epsilon}{6}\geq \frac{\epsilon}{3}. \end{array}

\Box

Although this proof is really elementary (you only need to know what convergence means and have a proper {\epsilon}-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 {x^k\rightharpoonup 0} implies {x^k\rightarrow 0} in {\ell^1}.

Proof: We know that a weakly convergent sequence in {\ell^1} is pointwise convergent (by testing with the canonical basis vectors), hence for every {j} it holds that {x^k_j\rightarrow 0} for {k\rightarrow\infty}. For some {J\in{\mathbb N}} we write

\displaystyle  \|x^k\|_1 = \sum_{j< J} |x^k_j| + \sum_{j\geq J} |x^k_j|.

By pointwise convergence of {x^k_j} we know that the first sum converges to zero for every {J}. Hence, we can proceed as follows: For a given {\epsilon>0} first choose {J} so large such that {\sum_{j\geq J} |x^k_j|\leq \epsilon/2} for all {k} and then choose {k} so large that { \sum_{j< J} |x^k_j|\leq \epsilon/2}. This sounds nice but wait! How about the choice of {J}? The tails of the series shall be bounded by {\epsilon/2} uniformly in {k}. 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 {J} is possible by contradiction: Assume that there exists an {\epsilon} such that for all {J} there exists a {k} such that {\sum_{j\geq J}|x^k_j|>\epsilon}. Alas, this is same thing which we have proven in the proof of Markus… \Box

Indeed, the construction Markus made in his proof can be generalized to obtain the Nikodym convergence theorem:

Theorem 2 Let {(\sigma_n)} be a sequence of signed finite measures for which it holds that {\sigma_n(E)\rightarrow 0} for every measurable set {E} (which is equivalent to the weak convergence of {\mu_n} to zero). Then the sequence {(\sigma_n)} is uniformly countably additive, i.e. for any sequence {(E_m)} of disjoint measurable sets the series {\sum_{m=1}^\infty |\sigma_n(E_m)|} converges uniformly in {n}.

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.

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