Diagonalizable operator is compact [duplicate]

Question

Possible Duplicate:
How to prove that an operator is compact?

Let $H$ be a separable Hilbert space with bases $\left\{ e_{n}\right\} $, $% \left\{ \alpha _{n}\right\} \subset \mathbb{F} $ with $M=\sup \left\{ \left\vert \alpha _{n}\right\vert :n\geq 1\right\} $. If $Ae_{n}=\alpha _{n}e_{n}$ for all $n$, would you help me to show that $A$ extends by linearity to a bounded operator on $H$ with $\left\Vert A\right\Vert =M$. Furthermore, prove that $A$ is compact operator iff $\alpha _{n}\rightarrow 0 $ as $n\rightarrow \infty $.

Answer 1

This kind of problem can be very difficult for general operators but for diagonal ones it is easy. Just let $x=\sum x_j e_j$ be a vector in $\mathcal{H}$ such that \begin{equation} \|x\|=\sqrt{\sum|x_j|^2}=1. \end{equation} By definition of $A$, \begin{equation} Ax=\sum a_j x_j e_j, \end{equation} with norm \begin{equation} \|Ax\|=\sqrt{\sum |a_jx_j|^2}\le \sqrt{\sup |a_j|^2\sum |x_j|^2}=\sup |a_j| = M, \end{equation} hence $\|A\|\le M$.

Now for any $\epsilon>0$, you find $|a_j|>M-\epsilon$, then take $x=a_je_j$ whould give you $\|A\|\ge M-\epsilon$, and this implies $\|A\|=M$.

For the second problem we use the fact that in $B(\mathcal{H})$, $A$ is compact if and only if $A$ is the limit of finite rank operators. And of course this is equivalent to $\operatorname{lim}a_n=0$, because you can just approximate $A$ by $AP_n$, where $P_n$ is the projection onto the space spaned by the first $n$ basis elements.