Strongly convergence and Uniformly convergence in Banach space Let $X,Y$ be Banach spaces, $T_n:X\to Y$ are bounded linear operators and $S:X\to Y$ is compact operator. Suppose for all $x\in X$, $||T_nx||\leq ||Sx||$  and $T_n$ strong convergent to $T$(bounded linear operator) . Then $T_n$ convergent uniformly.
My idea: If $T_n$ not convergent uniformly, there exist $\epsilon$ and $x_n$ s.t.,$||x_n||=1$,  $||(T_n-T)x_n|| \geq \epsilon$
Then, there are subsequence s.t., $Sx_{n_k}$ convergent.
But I don't know the relation $T_n$ to $S$, so I can't prove.
 A: Firstly, observe that we may assume that $T=0$.
Reason: For any $x\in X$, we have $||T_{n}x||\leq||Sx||$. Letting
$n\rightarrow\infty$, we have $||Tx||\leq||Sx||$. Now 
\begin{eqnarray*}
||(T_{n}-T)x|| & \leq & ||T_{n}x||+||Tx||\\
 & \leq & ||Sx||+||Sx||\\
 & = & ||(2S)x||.
\end{eqnarray*}
$\{T_{n}-T\mid n\in\mathbb{N}\}$ is a sequence of bounded linear
map from $X$ into $Y$, with $(T_{n}-T)x\rightarrow0$ for each $x\in X$
and that $||(T_{n}-T)x||\leq||(2S)x||$. Note that $2S$ is still
a compact linear map from $X$ into $Y$. Now replace the original
$\{T_{n}\}$ with $\{T_{n}-T\}$.
Let us rephrase the question: Let $X$ and $Y$ be Banach spaces.
Let $T_{n}:X\rightarrow Y$ be a bounded linear map, $S:X\rightarrow Y$
be a compact linear map. Suppose that $T_{n}x\rightarrow0$ for all
$x\in X$ and that $||T_{n}x||\leq||Sx||$ for all $x\in X$. Prove that
$||T_{n}||\rightarrow0$.
Proof: Prove by contradiction. Suppose the contrary that $||T_{n}||\not\rightarrow0$.
By passing to a subsequence, without loss of generality, we may assume
that there exists $\varepsilon_{0}>0$ such that $||T_{n}||>\varepsilon_{0}$
for all $n$. Let $B=\{x\in X\mid||x||\leq1\}$. For each $n$, choose
$x_{n}\in B$ such that $||T_{n}x_{n}||>\varepsilon_{0}$. Since $\overline{S(B)}$
is a compact subset in $Y$ and the sequence $\{Sx_{n}\mid n\in\mathbb{N}\}\subseteq\overline{S(B)}$.
The sequence $\{Sx_{n}\mid n\in\mathbb{N}\}$ has a convergent subsequence.
By passing to a suitable subsequence, without loss of generality,
we may assume that $\{Sx_{n}\mid n\in\mathbb{N}\}$ is convergent.
Choose $N\in\mathbb{N}$ such that $||Sx_{n}-Sx_{m}||<\frac{\varepsilon_{0}}{4}$
whenever $m,n\geq N$. Since $T_{n}x_{N}\rightarrow0$ as $n\rightarrow\infty$,
there exists $n_{0}>N$ such that $||T_{n_{0}}x_{N}||<\frac{\varepsilon_{0}}{4}$.
Note that $||T_{n_{0}}(x_{n_{0}}-x_{N})||\leq||S(x_{n_{0}}-x_{N})||<\frac{\varepsilon_{0}}{4}$.
It follows that 
\begin{eqnarray*}
||T_{n_{0}}x_{n_{0}}|| & \leq & ||T_{n_{0}}(x_{n_{0}}-x_{N})||+||T_{n_{0}}x_{N}||\\
 & < & \frac{\varepsilon_{0}}{2}
\end{eqnarray*}
which is a contradiction.
