TITLE: Banach-Steinhaus variant
QUESTION [5 upvotes]: Let $T_n$ be a sequence of continuous linear operators from a Banach
  space $X$ to a normed linear space $Y$. Now, for all $x \in X$,
  $\lim_{n \rightarrow \infty} T_n(x)$ exists in $Y$. Define $T(x) = \lim_{n \rightarrow \infty} T_n (x)$ on $X$. Show that $$\|T\| \leq \liminf \|T_n\|.$$

Here's my proof:

Convergence of $T_n(x)$ implies that for all $x \in X$, $\|T_n(x)\|\leq M_x$ for some $M_x$. The uniform boundedness principle implies $T_n$ is uniformly bounded, in particular, $\liminf  \|T_n\| < \infty.$ Now for all $z \in X$ with $\|z\|=1$, $$\|T(z)\|=\lim \|T_n(z)\| \leq \liminf  \|T_n\| \|z\| = \liminf  \|T_n\|.$$

I'm not too sure about that last line. How can I improve this proof?

REPLY [3 votes]: You are correct. Let me just add a tiny intermediate step to convince you you are right.
Fix $x$.
For all $n$
$$
\|T_{n}x\|\leq \|T_{n}\|\|x\|.
$$
So 
$$
\|Tx\|=\lim \|T_nx\|=\liminf\|T_{n}x\|\leq \liminf \|T_{n}\|\|x\|
$$
for all $x$.
Hence
$$
\|T\|\leq \liminf \|T_n\|.
$$
Note: we have $\lim \|T_nx\|=\|Tx\|$ because $\lim T_nx=Tx$ and $|\|T_nx\|-\|Tx\||\leq \|T_nx-Tx\|$.