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 $A : \ell_1 \to \ell_1$?

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 '16 at 3:19