Weak operator topology convergence of sequence of bounded operators to a bounded operator I need to prove the following theorem.
Let $X, Y$ be Banach spaces, with $Y$ weakly sequentially complete.  Let $\{T_n\} \subset \mathscr{L}(X,Y)$ with $\Lambda(T_nx)$ converges for all $x \in X$, $\Lambda \in Y^*$.  Prove there is a $T \in \mathscr{L}(X,Y)$ such that $T_n \to T$ in the weak operator topology.
Proof so far: $\Lambda(T_nx)$ converges for all $x \in X$, $\Lambda \in Y^*$.  Since $Y^*$ is weakly sequentially complete, there is a $y \in Y$ such that $\Lambda(T_nx) \to \Lambda y_x$ for all $\Lambda \in Y^*$.  Define $Tx :=y_x$.  We need to show $T$ is bounded.  Note that because $\Lambda(T_nx)$ converges, it is most certainly true that $\{\Lambda(T_nx) : T_n \in \{T_n\}\} < \infty$.  Hence, by a Corollary of the PUB, $\|T_n\|$ is uniformly bounded by say $M > 0$. 
So here now I want to prove that $\|T\|$ is bounded.  Using reverse triangle inequality, we can get that $|\Lambda(Tx)| \leq M \|\Lambda\| \|x\|$, but the fact that we have $\Lambda$ on the left hand side does not seem to allow me to prove that $T$ is bounded. Reed and Simon do a similar proof with $X = Y$ Hilbert spaces, and they refrain from defining $T$ until later, but they do it using a very clever result of the Riesz Representation Theorem.  I'm not sure maybe how to construct an analogous $T$ here instead of choosing the seemingly obvious choice from the fact that $Y$ is weakly sequentially complete.  Any hints/help would be much appreciated.
 A: As you said, we have 
$$|\Lambda(Tx)| = \left|\lim_{n\to\infty} \Lambda(T_n x)\right|= \lim_{n\to\infty}|\Lambda(T_n x)| \le  \limsup_{n\to\infty} \|\Lambda\| \|T_n\|\|x\| \le  M \|\Lambda\| \|x\|$$
so for $\widehat{Tx} \in Y^{**}$, the element of $Y^{**}$ represented by $Tx$ we have
$$\left|\widehat{Tx}(\Lambda)\right| = |\Lambda(Tx)| \le M \|\Lambda\| \|x\|$$
so $\|Tx\| = \left\|\widehat{Tx}\right\| \le M\|x\|$.
Therefore $T$ is bounded and $\|T\| \le M$.
Finally, for any $x \in X$ we have $\Lambda(T_n x) \to \Lambda(Tx), \forall \Lambda\in Y^*$ so $T_nx \xrightarrow{w} Tx$. We conclude $T_n \to T$ in the weak operator topology.

A counterexample when $Y$ is only a Banach space:

Consider $T_n : c_0 \to c_0$ given by $T_n(x_k)_k = (\overbrace{x_1, x_1, \ldots, x_1}^n, 0, 0\ldots)$.
We have $(c_0)^* = \ell^1$ so for any $x \in c_0, \Lambda = (\lambda_n)_n \in \ell^1$ we have
$$\Lambda(T_n x) = \Lambda(\overbrace{x_1, x_1, \ldots, x_1}^n, 0, 0\ldots) = x_1\cdot \sum_{k=1}^n \lambda_n \xrightarrow{n\to\infty} x_1\cdot \sum_{k=1}^\infty \lambda_n$$
so $(\Lambda(T_n x))_n$ certainly converges. But $(T_n)_n$ doesn't converge in the weak operator topology. Indeed, for $x = (1, 0, 0, \ldots) \in c_0$ we have $T_nx = (\overbrace{1, \ldots, 1}^n, 0, 0\ldots)$. The coordinate-wise limit of this sequence is $(1, 1, \ldots) \notin c_0$ so $(T_nx)_n$ cannot converge weakly in $c_0$.
