# Prove the operator on hilbert space is compact

My question is actually the same as the first part of this one,

Prove that T is compact

I am thinking about two ways, 1) use a bounded sequence $\{g_n\}$, and try to pick a convergent subsequence from $\{Tg_n\}$; 2) use Arzela-Ascoli. Uniform-boundedness is obvious. I got stuck in proving equicontinuity.

I know how to show the second part: Suppose $T$ has an eigenvector $g\neq 0$, so that $Tg=\lambda g$. With given orthonormal basis, $g=\sum_{k\in K}c_k\phi_k$, where $K=\{k|c_k=\langle \phi_k,g\rangle\neq 0\}$. Let $k_0=\min K$, then $\langle \phi_{k_0},\lambda g\rangle=\lambda c_{k_0}\neq 0$, but, \begin{align*} \langle \phi_{k_0},Tg\rangle &= \langle f_{k_0},\sum_{k\in K}c_kT\phi_k\rangle \\ &= \langle \phi_{k_0},\sum_{k\in K}\frac{c_k}{k}\phi_{k+1}\rangle \\ &= \sum_{k\in K}\frac{c_k}{k}\langle \phi_{k_0},\phi_{k+1}\rangle \\ &= 0 \end{align*} Contradiction! So $T$ cannot have eigenvectors.

Thank you, T.A.E.

I think of another way, \begin{equation} \sum_k||Af_k||^2=\sum_k||\frac{1}{k}f_{k+1}||^2=\sum_k\frac{1}{k^2}=\frac{\pi^2}{6}<\infty \end{equation} So $A$ is Hilbert-Schmidt and thus compact.

Is that correct?

Assume that $T\varphi_{k}=\frac{1}{k}\varphi_{k+1}$ where $\{ \varphi_{k}\}_{k=1}^{\infty}$ is an orthonormal basis of $\mathcal{H}$. For a general $x$, $$Tx = T\sum_{k=1}^{\infty}(x,\varphi_{k})\varphi_{k} = \sum_{k=1}^{\infty}\frac{1}{k}(x,\varphi_{k})\varphi_{k+1}$$ The operator $T$ is approximated by $T_{N}$, where $$T_{N}x = \sum_{k=1}^{N}\frac{1}{k}(x,\varphi_{k})\varphi_{k+1}.$$ The operator $T_{N}$ is a finite-rank operator, and \begin{align} \|(T-T_{N})x\|^{2} & = \left\|\sum_{k=N+1}^{\infty}\frac{1}{k}(x,\varphi_{k})\varphi_{k+1}\right\|^{2} \\ & = \sum_{k=N+1}^{\infty}\frac{1}{k^{2}}|(x,\varphi_{k})|^{2} \\ & \le \frac{1}{(N+1)^{2}}\sum_{k=N+1}^{\infty}|(x,\varphi_{k})|^{2} \\ & \le \frac{1}{(N+1)^{2}}\|x\|^{2}. \end{align} Therefore, $\|T-T_{N}\| \le 1/(N+1)\rightarrow 0$ as $N\rightarrow\infty$. So $T$ is the norm limit of a sequence of finite rank operators, which makes $T$ compact.
To show $T$ has no eigenvalues. First, notice that $Tx=0$ iff $x=0$. So $\lambda=0$ is not an eigenvalue. Suppose $Tx=\lambda x$ for some $\lambda \ne 0$. Then $$\lambda^{n}x = T^{n}x = \sum_{k=1}^{\infty}\frac{1}{k(k+1)(\cdots)(k+n-1)}(x,\varphi_{k})\varphi_{k+n}.$$ The smallest expression in $k$ is where $k=1$, which gives $1/n!$. Therefore, $$|\lambda|^{n}\|x\| \le \frac{1}{n!}\|x\|,\\ \|x\| \le \frac{(1/|\lambda|)^{n}}{n!}\|x\|.$$ Summing the left side over $n=1$ to $n=\infty$ is infinite unless $x=0$. But summing the right side over $n$ is finite and equal to $(e^{1/|\lambda|}-1)\|x\|$. So the above inequality forces $x=0$. Therefore $T$ has no eigenvalues.