Show that $\|T\|\le\lim\inf_{n\to\infty}\|T_n\|$ under both the weak and strong operator topologies on $B(X,Y)$ I have shown that whenever $x_n\rightharpoonup x$ in $X$, then $\|x\|\le\lim\inf_{n\to\infty}\|x_n\|$ and have also managed to demonstrate the analogous result for when $f_n\overset{\ast}{\rightharpoonup}f$ in $X^*$. In the proof of the former we make recourse to the Hahn Banach theorem, whereas in the latter there is no such need.
I'm interested in showing the corresponding result as well for a sequence of bounded linear operators under the strong and weak operator topologies. 
To clarify, $T_n\overset{SOT}\to T\in B(X,Y)$ means that $\forall x\in X$, $\|T_nx-Tx\|\to0$ as $n\to\infty$, and, $T_n\overset{WOT}\to T\in B(X,Y)$ means that $\forall x\in X$, and $f\in Y^*,$ $|f(T_nx)-f(Tx)|\to0$ as $n\to\infty$.
Here are my attempts so far. We start with the strong operator topology case, and then will do the weak operator topology case.

1.$\,$Suppose that $T_n\overset{SOT}\to T\in B(X,Y)$. Then $\forall x\in X$, $\|T_nx-Tx\|\to0$ as $n\to\infty$. Choose $x\in X:\|x\|=1$, and let $\epsilon\gt0$. 
Now, $\|Tx\|\le\|T\|\implies\|Tx\|\gt\|T\|-\epsilon$.
But $\|T_nx\|\to\|Tx\|$ by strong operator convergence. So that, for large enough $n\in\mathbb N$, $\|T_nx\|\gt\|T\|-\epsilon.$ But, $\|T_nx\|\le\|T_n\|$ for all $n\in\mathbb N$, so that furthermore, $\|T_n\|\ge\|T\|-\epsilon.$
But then, this impliex that $\|T\|-\epsilon\le\lim\inf_{n\to\infty}\|T_n\|$, on taking lim infs of both sides. Which further implies that $\|T\|\le\lim\inf_{n\to\infty}\|T_n\|$, since $\epsilon\gt0$ was arbitrary. Thus showing the result we required.

2.$\,$Suppose that $T_n\overset{WOT}\to T\in B(X,Y)$. Then $\forall x\in X$, and $f\in Y^*,$ $|f(T_nx)-f(Tx)|\to0$ as $n\to\infty$. Now the Hahn Banach theorem tells us that $\exists f\in X^*:\|f\|=1$ and that $f(x)=\|x\|$. 
With this in mind, consider that $|f(T_nx)|\le\|T_n\|$, using the boundedness of $f$, $T_n,\,\forall n\in \mathbb N$ and the above observations from Hahn Banach.
But $|f(T_nx)|\to|f(Tx)|\le\|T\|$, by the weak operator convergence and then the boundedness and Hahn Banach deductions.
On putting both of these observations together, we see that $\|T\|\le\|T_n\|$, and so on taking lim infs, we get that $\|T\|\le\lim\inf_{n\to\infty}\|T_n\|$, as required. 

Are both of these attempts correct? Or are there any ways in which I could improve?
 A: In the SOT attempt, you are first choosing $x$. But then you cannot guarantee that $\|Tx\|\geq\|T\|-\epsilon$. What you need to do is to choose $x$ so that $\|x\|=1$ and $\|Tx\|\geq\|T\|-\varepsilon$. Then, by the convergence, you find that for $ n$ big enough , $\|T_nx\|\geq\|Tx\|-\epsilon$. Thus, for $n$ big enough, 
$$
\|T_n\|\geq\|T_nx\|\geq\|Tx\|-\epsilon\geq\|T\|-2\epsilon. 
$$
Now you conclude that $\liminf_n\|T_n\|\geq\|T\|-2\epsilon$. 
In the WOT attempt, you are glossing so much that it is hard to see what you are doing. There is no inequality whatsoever with $\|T\|$ on the low side, so it is hard to see how you conclude; you don't say how you choose $x$. The way I would do it to first notice that, by Hahn-Banach, 
$$
\|Tx\|=\sup\{|f(Tx)|:\ f\in Y^*,\ \|f\|=1\}.
$$
So, fix $x$. Given $\varepsilon>0$, there exists $f\in Y^*$ with $\|f\|=1$ and 
$$
\|Tx\|\leq|f(Tx)|+\varepsilon=\lim_n|f(T_nx)|+\varepsilon. 
$$
As $|f(T_nx)|\leq\|T_n\|$, we get $\lim_n|f(T_nx)|\leq\liminf_n\|T_n\|$. Thus
$$
\|Tx\|\leq\liminf_n\|T_n\|+\varepsilon.
$$
As this works with any $x$ with $\|x\|=1$, taking supremum we get 
$$
\|T\|\leq\liminf_n\|T_n\|+\varepsilon, 
$$
and we finally can remove the $\varepsilon$ since its aritrarily small. 
A: Assume $T_n \xrightarrow{SOT} T$. For every $x \in X$ we have
$$\|Tx\| = \lim_{n\to\infty} \|T_nx\| \le \liminf_{n\to\infty} \|T_n\|\|x\| = \left(\liminf_{n\to\infty} \|T_n\|\right)\|x\|$$
Therefore $\|T\| \le \liminf_{n\to\infty} \|T_n\|$.

Assume $T_n \xrightarrow{WOT} T$ and take $x \in X$. The Hanh-Banach theorem implies that there exists $f \in Y^*$ such that $\|f\| = 1$ and $f(Tx) = \|Tx\|$. We have

$$\|Tx\| = |f(Tx)| = \lim_{n\to\infty} |f(T_n x)| \le \liminf_{n\to\infty} \overbrace{\|f\|}^{=1}\|T_nx\| \le \liminf_{n\to\infty} \|T_n\|\|x\| = \left(\liminf_{n\to\infty} \|T_n\|\right) \|x\|$$
Therefore $\|T\| \le \liminf_{n\to\infty} \|T_n\|$.
