Sequence of strongly convergent operators $\{T_n\}$ satisfies $\sup_{x\in K }\| T_n(x) - T(x)\| = 0$ for compact $K$ Exercise: Let $E,F$ be Banach Spaces and $T,T_1,T_2$,... operators in $L(E,F)$ such that $T_n(x)\rightarrow T(x)$, $\forall x\in E$ . prove that for all compact $K\subset E$
\begin{equation}
\sup_{x\in K}{||T_n(x)-T(x)||}\rightarrow 0
\end{equation}
the book gives the following suggestion:
proceed by contradiction and use the  Banach-Steirhauss theorem to guarantee that $\sup\{||T||,||T_1||, ||T_2||,...\}< \infty $.
Proof:
Suppose that $\sup_\limits{x\in K}^{}{||T_n(x)-T(x)||}\nrightarrow 0$ then we have that exist an $\varepsilon>0$  such that $\forall n\in N$ exist $T_n$ such that
\begin{equation*}
||T_n-T||=\sup_\limits{x\in K}^{}{||T_n(x)-T(x)||}>\varepsilon
\end{equation*}
By hypothesis $T_n(x) \rightarrow T(x)$  for all $x \in E$, i.e,  $\forall \epsilon=1$ exists $N \in \mathbb{N}$ s.t $n \geq N$
\begin{equation*}
 ||T_n(x)-T(x)|| < 1
\end{equation*}
then $||T_n(x)||< 1 +||T(x)|| \leq 1+c$, i.e, $\sup\{||T(x)||,||T_1(x)||, ||T_2(x)||,...\}< c_x $.  By  Banach-Steirhauss theorem
$$sup\{||T||,||T_1||, ||T_2||,...\}< \infty.$$
In this moment I have two ideas to continue.
1.Since $K$ is compact, we know that  all sequences $\{T_n\}\in K$ admits a subsequence convergent.I don't know how to relate this fact with the negation and how to get to the contradiction.
2.On the other hand, I was thinking in prove that the space generated by the sequence of operators is Cauchy, but since $L(E,F)$ is Banach then all Cauchy sequences converge. However, I don't know how I can apply in this prove that $K\subset E$ is compact.
I would be very grateful if someone could help me.
 A: Suppose to the contrary that $$\lim_n\sup_{x \in K}\|T_nx - Tx \| \neq 0.$$
By going to a subsequence, we may assume that $$c_n:=\sup_{x \in K}\|T_n x- Tx \|\geq \epsilon$$
for some $\epsilon > 0$ and all $n \geq 1$.
Choose a sequence $(x_n)_n$ in $K$ with $$\|T_n x_n- Tx_n \| = c_n. \quad \quad(*)$$ This exists by continuity and compactness of $K$. Again, by choosing a subsequence, we may assume that $(x_n)_n$ converges to some $x\in K$.
Note that $\lim_n T_n x_n = Tx$. Indeed
$$\|T_n x_n - Tx\| \leq \|T_n x_n - T_n x\| + \|T_n x -Tx\| \leq \|T_n\| \|x_n- x \| + \|T_n x  - Tx\|$$
and by an application of the uniform boundedness principle, $\sup_n \|T_n\| < \infty$.
Letting $n \to \infty$ in $(*)$, we thus obtain
$$\lim_n c_n=0$$
which is impossible since $c_n \geq \epsilon$ for all $n \geq 0$. This is the desired contradiction.
A: $\newcommand{\F}{{\mathscr{F}}}$
This post already has an excellent answer by @MathQED. Nevertheless, since this question pops up in many situations in Analysis, I
think it is interesting to highlight some of the ingredients involved as this might  help to see things from a better
perspective and, most importantly, to identify the phenomenon when it shows up in other situations.
Let me phrase this in terms of one definition and three exercises for the interested reader:
1. Definition.  Given a metric (or topological space) $X$, a metric space $M$, and a set $\F$ of functions from $X$ to $M$,  one says
that $\F$ is equicontinuous at a given point $x_0\in  X$  if, for all $\varepsilon >0$,  there exists an open set $U\subseteq X$ containing
$x_0$,  such that
$$
  d(f(x), f(x_0))<\varepsilon ,  \quad \forall x\in  U,  \quad \forall f\in  \F.
  $$
(The crucial point here is that the same $U$ can be chosen for all functions $f$ in $\F$).
If $\F$ is equicontinuous at every point of $X$, we simply say that $\F$ is equicontinuous.
2. Exercise.
A collection $\F$ of linear transformations from a normed space $E$ to a normed space $F$ is equicontinuous if and
only if $\F$ is equicontinuous at $0$, if and only if
$$
  \sup_{T\in \F}\|T\|<\infty .
  $$
3. Exercise.  Given $X$ and $M$ as in (1), given an equicontinuous sequence  $\{f_n\}_n$ of functions from $X$ to $M$, and given another
function $f$ from $X$ to $M$,  TFAE:

*

*$f_n\to f$ pointwise,


*$f_n(x)\to f(x)$, for every $x$ in some dense subset $D\subseteq X$,


*$f_n|_K\to f|_K$, uniformly, for every compact subset $K\subseteq X$.
4. Exercise.  Answer the question posed by the OP based on the above exercises!
