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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.

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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.

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  • $\begingroup$ One point I would like to point out: For a metric space, it is compact if and only if it is sequentially compact. I used this fact implicitly. $\endgroup$ – Danny Pak-Keung Chan Jul 26 '18 at 21:13

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