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 $A : \ell\_1 \to \ell\_1$?



Monotone convergence theorem for operators in the weak operator topology
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.


