I was thinking about the following problem:
Let $(A_t)_{t \geq 0}$ be a family of bounded operators on a Banach space $X$ which is uniformly bounded and let $(B_{t,\alpha})$ be a net in the Banach algebra of bounded operators on $X$ such that $B_{t,\alpha}$ converges in weak operator topology to $A_t$ for all $t\geq 0$, that is $f(B_{t,\alpha}x) \to f(A_tx)$ for all $x \in X$, $f \in X^*$ and $t \geq 0$.
Is it true that $(B_{t,\alpha})$ has to be uniformly bounded?
What if we assume the uniform boundedness of the net, does it imply the uniform boundedness of the limit?
Thank you for any help.
A convergent net in a Banach space need not be bounded. A trivial (and admittedly uniteresting) example is
$$(x_\alpha)_{\alpha \in (0,\infty)};\quad x_\alpha = \alpha^{-1}\cdot x$$
for some $x \neq 0$, where $(0,\infty)$ is a directed index set with the usual order. Of course for every norm-convergent net $x_\alpha \to x$, there is an $\alpha_0 \in A$ such that the "tail" $(x_\alpha)_{\alpha \geqslant \alpha_0}$ is bounded, so for the norm topology, there are no really interesting examples of unbounded convergent nets. For the weak topology, there are (in the infinite-dimensional case) more interesting examples of unbounded convergent nets, although explicitly constructing one is not easy.
If the system of weak neighbourhoods of $0$ has the same cardinality as $(0,\infty)$, choose a bijection $\nu \colon \mathscr{V}_w(0) \to (0,\infty)$, and for every $U \in \mathscr{V}_w(0)$ choose an $x_U \in U$ with $\lVert x_U\rVert = \nu(U)$. Since every weak neighbourhood of $0$ contains an infinite-dimensional subspace, such an $x_U$ exists. Then the net $(x_U)_{U \in \mathscr{V}_w(0)}$, where $\mathscr{V}_w(0)$ is directed by reverse inclusion, is a net converging to $0$ in the weak topology, but it is unbounded, and most likely it has no bounded "tails" $(x_U)_{U \subset W}$.
So the answer to the first question is no, the family $(B_{t,\alpha})$ has no reason to be uniformly bounded.
For the second question, the answer is yes, if $\lVert B_{t,\alpha}\rVert \leqslant M$ for all $t,\alpha$, then $\lVert A_t\rVert \leqslant M$ for all $t$, so the family $(A_t)_{t\geqslant 0}$ of limits is uniformly bounded too. For every $x$ and $f$, we have for every $t$
$$\lvert f(A_t x)\rvert = \lim_{\alpha} \lvert f(B_{t,\alpha} x)\rvert \leqslant \lVert f\rVert \cdot M\cdot \lVert x\rVert,$$
and hence $\lVert A_t x\rVert \leqslant M\cdot \lVert x\rVert$.