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

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.

• Please use proper sentences. In current state, it is close to unreadable.
– daw
Jan 17, 2022 at 13:25
• In what sense is it known that "it" ($A$? some operator related to $A$?) is finite rank? I agree that it may be possible to show that $A$ is a norm limit of finite rank operators, but this attempt needs clarification. It may be helpful to consider $P_n A P_n$, where $P_n$ is orthogonal projection onto $M_n^{\perp}$. Jan 17, 2022 at 16:34
• the below is a new answer Jan 18, 2022 at 8:24

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 $$A_Nx=A[e_1+e_2+\cdots+e_N]$$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.