Let $b_{n,j}\in \mathbb{C}$ for each $n,j\in \mathbb{N}$. I was wondering if there is some characterization of those $b_{n,j}$ such that for all bounded sequences $s_j\in \mathbb{C}, j\in \mathbb{N}$, we have that $$\sigma_n := \sum_{j=1}^{+\infty} b_{n,j} s_j$$ converges. Thus we want that the sequence $(\sigma_n)_n$ converges for all bounded sequences.

Note that I do not ask that $s_j\rightarrow s$ implies $\sigma_n\rightarrow s$ as is usual in summability methods.

So can we characterize exactly which $b_{n,j}$ satisfy this?

  • $\begingroup$ How about: Show that the linear map sending $(s_j) \in l^\infty$ to $\lim \sigma_n$ has closed graph, and is therefore a bounded linear functional by the closed graph theorem. The general principle is: you cannot actually "write down" a discontinuous linear map defined on a Banach space. $\endgroup$ – Gerald Edgar May 28 '17 at 23:37

Your condition is equivalent to the convergence in $\ell^1$ of $b_n=(b_{nj})_{j\ge 1}$.

First of all, we indeed must insist that $b_n=(b_{nj})_{j\ge 1}\in\ell^1$ for each fixed $n$, or otherwise we couldn't even define $\sum b_{nj}s_j$ for general bounded $s$.

Recall that $(\ell^1)^*=\ell^{\infty}$, and observe that if $(b_n,s)\to L(s)$ for all $s\in\ell^{\infty}$, then $\|b_n\|_1$ must be bounded by the uniform boundedness principle. (Or a direct argument, by hand, could be given, and you could check an earlier version of this answer for this if interested.)

This implies that the pointwise limit $b_j=\lim b_{nj}$ defines an $\ell^1$ sequence. We want to show that $\|b_n-b\|_1\to 0$. Originally, I thought this would just follow from the well known fact that weakly convergent sequences in $\ell^1$ are norm convergent. However, it's not so clear at this point that $b_n\to b$ weakly.

Instead, we make use of the argument from the usual proof of the fact just quoted. It suffices to show that $b_n$ is uniformly summable in the sense that for every $\epsilon>0$, there exists $J\ge 1$ so that $\sum_{j\ge J}|b_{nj}|<\epsilon$ for all $n$. Suppose this were not the case, so there is an $\epsilon>0$ such that given any $J\ge 1$, we can find an $n_J$ so that $\sum_{j\ge J}|b_{n_Jj}|\ge 3\epsilon$. In fact, we can then also find arbitrarily large $n_J$'s with this property.

Let's use this now to pick an $n_1$ and an interval $I_1$ such that: (1) $\sum_{j\in I_1}|b_{n_1j}|\ge 2\epsilon$; (2)$\sum_{j\in I_1}|b_j|<2^{-1}\epsilon/10$; (3) $\sum_{j>\max I_1}|b_{n_1j}|<\epsilon$.

Now we take $s_j$'s of absolute value $1$ on $I_1$ such that $b_{n_1j}s_j=|b_{n_1j}|$. Note that this ensures that $\sum_{j\ge 1} b_{n_1j}s_j$ will be close to a positive number of size $\ge\epsilon$, and this is true for any choice of $s_j$'s of absolute value $1$ for larger $j$'s at later stages.

Now we just continue in this style: we pick an index $n_2>n_1$ and an interval $I_2$. Note that if $n_2$ is sufficiently large, then what we did on $I_1$ will not have much of an effect on $\sum b_{n_2j}s_j$ because for $j\in I_1$, we will have $b_{n_2j}$ very close to $b_j$, and then property (2) from above strikes. We now arrange matters so that $\sum b_{n_2j}s_j$ is close to a negative number $\ge\epsilon$ in absolute value.

Of course this oscillating pattern is incompatible with convergence, so it follows that $b_n$ is uniformly summable, and thus $b_n\to b$ in $\ell^1$, as claimed.

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    $\begingroup$ I was writing exactly the same answer :) $\endgroup$ – Pietro Majer May 28 '17 at 19:50
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    $\begingroup$ @PietroMajer: Actually, I think the argument had a pretty serious gap (or how do we know that $b_n\to b$ weakly?). I think I got it fixed, but it's messy now. If you have a better idea, by all means post an alternative answer please. $\endgroup$ – Christian Remling May 28 '17 at 23:00
  • $\begingroup$ Yes, I was writing the same answer with the same gap :) I posted a comment in an answer. $\endgroup$ – Pietro Majer May 29 '17 at 6:49
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    $\begingroup$ I would be lazy and use the fact that every operator from $\ell_\infty$ to a separable space is weakly compact. $\endgroup$ – Bill Johnson May 29 '17 at 14:49
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    $\begingroup$ @BillJohnson: Thanks! I would like to be lazy too, but I don't know enough for that. $\endgroup$ – Christian Remling May 29 '17 at 18:26

A comment on Christian Remling's answer. As proved there, we know that $(b_n)$ is a sequence in $\ell_1$, and by assumption we have that $\langle b_n, s\rangle$ converges in $\mathbb{C}$ for any $s\in\ell_\infty$. This defines an element $\tilde b$ of $\ell_\infty^*=\ell_1^{**}$ such that $b_n\to \tilde b$ in the weak$^*$ topology of $\ell_\infty^*$, that is $\sigma(\ell_\infty^*,\ell_\infty)$. The key point is that $\ell_1$ is weakly sequentially complete (essentially proven in C.R. answer), so that $\tilde b$ actually is in $\ell_1$ (not just $\ell_1^{**}$) and the convergence is also in the $\ell_1$ norm as said. (Reference for $L^1$ spaces are weakly sequentially complete: Dunford-Schwarz, Linear Operators, Part I, IV.8.6).

rmk. In fact Schur's theorem on the equivalence of strong and weak convergence in $\ell_1$, that is the fact that a weak convergent sequence is also a norm convergent sequence, extends to a more complete statement : a weak Cauchy sequence is also a norm Cauchy sequence (whence the weak sequential completeness). Indeed, if $b_n$ is a weak-Cauchy sequence, for any pair of diverging sequences of integers $n_k$ and $m_k$, one has $b_{n_k}- b_{m_k}\to0$ weakly as $k\to0$, therefore also $\|b_{n_k}- b_{m_k}\|\to0$.

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  • $\begingroup$ This is very useful, to clean things up. $\endgroup$ – Christian Remling May 29 '17 at 18:27
  • $\begingroup$ Thanks a lot Pietro. It seems that weak sequentially completeness was exactly the concept I was missing. Thank you for the amazing answer. $\endgroup$ – Onion Dip Carlip May 29 '17 at 19:53
  • $\begingroup$ I'm glad I contributed ;) $\endgroup$ – Pietro Majer May 29 '17 at 20:08

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