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For real numbers, we know that any monotonic bounded sequence converges to a finite limit. Does this generalize to sequences of operators?

More formally, I have a sequence of operators $\{A_n\}_{n=1}^{\infty}$ where each $A_n: \ell_1 \to \ell_1$ and $\|A_n\|_1 \leq 1$. I know from the Banach-Alaoglu theorem that the unit ball $\{T : \ell_1 \to \ell_1 $ such that $ \|T\|_1 \leq 1\}$ is compact when the space of operators is endowed with the weak operator topology.

By weak operator topology I mean that $\{A_n\}$ converges to $A$ if and only if for all $x \in \ell_1^*$ and $y \in \ell_1$ we $$ \langle x ,A_n y \rangle \to \langle x, A y \rangle .$$

Does the Banach-Alaoglu theorem imply that if we have a monotonicity property $A_1 \leq A_2 \leq A_3 \leq ...$ that the sequence converges (in the weak topology) to some operator $A : \ell_1 \to \ell_1$?

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    $\begingroup$ From your second paragraph you must mean by "weak topology" the topology of pointwise convergence in the weak topology on $\ell_1$ rather than the Banach space weak topology. But even with that interpretation it is clear that the answer is no. Take the projection $P_n$ onto the first $n$ unit vectors of $\ell_1$. The sequence $(P_N)$ converges in the pointwise weak$^*$ topology to the identity operator, but not pointwise in the weak operator topology. This is basically obvious. $\endgroup$
    – Bill Johnson
    Oct 11, 2016 at 1:10
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    $\begingroup$ What does $\le$ mean here? $\endgroup$
    – Nate Eldredge
    Oct 11, 2016 at 2:32
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    $\begingroup$ So, is that equivalent to saying that $B-A$ is positivity preserving? $\endgroup$
    – Nate Eldredge
    Oct 11, 2016 at 2:52
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    $\begingroup$ Then don't we conclude that for every nonnegative $x$, the sequence $A_n x$ is pointwise increasing and $\ell^1$ bounded. So monotone convergence says it converges in $\ell^1$ to some $y$. By considering positive and negative parts, we get the same without the nonnegativity assumption. So this would seem to imply that $A_n$ converges in the strong operator topology. $\endgroup$
    – Nate Eldredge
    Oct 11, 2016 at 2:57
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    $\begingroup$ I said strong operator topology, not the operator norm topology. And I wish you would stop saying "weak topology" when you mean "weak operator topology". They are not the same and the definitions are crucial here. Note the limiting operator $y$ will be a bounded operator by the uniform boundedness principle. $\endgroup$
    – Nate Eldredge
    Oct 11, 2016 at 3:19

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You've clarified in comments that $\le$ means that the operators $A_{n+1} - A_n$ are positivity preserving. So for any nonnegative $x \in \ell^1$, we have $A_1 x \le A_2 x \le \dots$ pointwise. Since $A_1 x \in \ell^1$ and the sequence $\{A_n x\}$ is $\ell^1$-bounded, the monotone convergence theorem implies that $A_n x$ converges in $\ell^1$-norm to some $Ax \in \ell^1$. For general $x \in \ell^1$, writing $x = x^+ - x^-$ shows that again $A_n x$ converges in $\ell^1$-norm to some $Ax$. In particular, $\|A_n x\|_1 \to \|Ax\|_1$ which implies that $A$ is bounded with operator norm at most 1. We have thus shown $A_n \to A$ in the strong operator topology, which is even more than requested.

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