TITLE: $\lambda_k \to 0$ implies $T$ is compact?
QUESTION [3 upvotes]: I am doing an exercise which asks to show that if $\{\varphi_k\}$ is an orthonormal basis in a Hilbert space with $T$ a bounded operator satisfying $T\varphi_k = \lambda_k \varphi_k$, then $\lambda_k \to 0$ implies that $T$ is compact. Now I will be done if I can show that $TP_n \to T$ in norm for $P_n$ is a compact operator. Here $P_n$ is the orthogonal projection onto the linear span $\{\varphi_1\ldots,\varphi_n\}$. 
However the hint given is that $\lambda_k\to 0$ already implies this. Why is this so? I guess if I can understand why $||T|| = \sup_k |\lambda_k|$ then I would be done. For I believe $||P_nT - T|| = \sup_{k\geq n} |\lambda_k|$ and then $\lim_{n\to \infty} ||P_nT - T|| = \limsup_{n\to \infty} |\lambda_n| = 0.$


My question is: Why is $||T|| = \sup_{k} |\lambda_k|$?

REPLY [3 votes]: Any vector $x$ can be expanded in the orthonormal basis as $x = \sum_k x_k \varphi_k$, and any vector expressed in such a way has norm given by $\Vert x \Vert^2 = \sum_k |x_k|^2$. Thus $$ \Vert T(x)  \Vert^2 = \Vert  \sum_k  x_k T(\varphi_k) \Vert^2  =  \Vert \sum_k x_k\lambda_k \varphi_k\Vert^2 \\= \sum_k |x_k \lambda_k|^2  \le \sum_k\sup_j|\lambda_j|^2x_k^2 = \sup_k|\lambda_k|^2 \lVert x \rVert^2.$$
This shows $\Vert T \Vert \le \sup_k |\lambda_k|$. For the other direction, note that $\Vert T(\varphi_k)\Vert= \Vert \lambda_k \varphi_k \Vert = |\lambda_k|$ implies $\Vert T\Vert\ge |\lambda_k|$ for every $k$; i.e. $\Vert T\Vert \ge \sup_k |\lambda_k|$.
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