Can this way show that the $A$ is a compact operator?

Question

Let $\left\{ {{e_n}} \right\}_{n = 1}^\infty $are orthonormal basis in a Hilbert space H,A is a bounded linear operator on H; if ${M_n} = {\{ {e_1},{e_2}, \cdots ,{e_n}\} ^ \bot }$,when $x\in M_n,\mathop {\sup }\limits_{\left\| x \right\| = 1} \left\| {Ax} \right\| \to 0(n \to \infty )$. Show that the $A$ is a compact operator

my proof: $H=M_n + {M_n}^{\bot}$,so $A$ is a compact on ${M_n}^{\bot}$,because it's a finite rank operator on ${M_n}^{\bot}$;$\forall y \in H,\Vert y \Vert \leq B$,$y=y_1+y_2$,where $y_1 \in {M_n}^{\bot},y_2\in M_n$;it's clear that $\{Ay_1\}$has a subsequence$\{Ay_{n_k}\}$.on the other hand, for $y_2,\frac{1}{{\left\| y \right\|}}\left\| {A{y_2}} \right\| = \left\| {A\frac{{{y_2}}}{{\left\| y \right\|}}} \right\| \le \sup \left\| {Ax} \right\| \to 0$,$\Vert Ay_2 \Vert=\Vert y\Vert \Vert Ay_2 \Vert \to 0$.so $\{ Ay_2\}$ has a subsequence $\{Ay_{n_s}\}$; then $\{Ay_{n_k}+Ay_{n_s}\}$ is a subsequence of $\{ Ay\}$ and is convergence;A is a compact operator.

Answer 1

because $$sup\Vert Ax \Vert \to 0\quad (x\in {M_{n}}^\bot,\Vert x \Vert =1 )\quad (n\to \infty)$$ so$$\forall \epsilon >0,\exists N, \Vert Ax \Vert<\epsilon \quad(x\in {M_{N}}^\bot,\Vert x \Vert =1 )$$ define $$A_Nx=A[<x,e_1>e_1+<x,e_2>e_2+\cdots+<x,e_N>e_N]$$so $A_N$ is a finite rank operator,then is a compact operator.then $$\Vert Ax-A_Nx\Vert=\Vert (A-A_N)x \Vert=\Vert A P_Nx\Vert\leq\epsilon \Vert x\Vert$$ where $P_N$is the projection onto orthogonal complement of ${M_{N}}^\bot$; $(x=x_1+x_2,x_1\in M_N,x_2\in {M_{N}}^\bot,P_Nx=x_2\in{M_{N}}^\bot)$ so $$\Vert A-A_N\Vert\leq \epsilon$$ $$\mathop {\lim }\limits_{N \to \infty } \left\| {A - {A_N}} \right\|{\rm{ = 0}}$$ so A is a compact operator.