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Show that if $e_n$ is an orthonormal basis for a Hilbert space $H$, and $T:H\to H$ is a bounded linear operator that satisfies $$\sum_{n=1}^{\infty}\lvert\lvert Te_n\rvert\rvert^2<\infty$$, then $T$ is compact. I need to do this by defining a sequence of finite rank operators $T_k$, and showing that $\lvert\lvert T-T_k\rvert\rvert\to 0$, so I know my $T_k$ will probably be of the form $\sum_{n=1}^k...$ and need to end up with something like $\lvert\lvert T-T_k\rvert\rvert\leq\sum_{n=k+1}^{\infty}\lvert\lvert Te_n\rvert\rvert^2\to0$ but I've tried loads of things for $T_k$ such as $T_k(x)=\sum_{n=1}^k\langle x,e_n\rangle Te_n$ but couldn't get anywhere.

I have that $\lvert\lvert (T_k-T)x\rvert\rvert^2=...=\langle\sum_{k+1}^\infty\langle x,e_n\rangle Te_n,\sum_{k+1}^\infty\langle x,e_n\rangle Te_n\rangle$ and I tried to simplify this much more but since $Te_n$ aren't orthogonal it's more difficult than usual

Loobear23
  • 1,160

1 Answers1

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Your choice of $T_k$ is correct.

As $\{e_n\}$ is an orthonormal basis we have $x = \sum_{n=1}^\infty \langle x,e_n\rangle e_n$ for all $x$ and $\|x\|^2 = \sum_{n=1}^\infty |\langle x,e_n\rangle|^2$.

We have \begin{align} \|T(x)-T_k(x)\| &= \|T(x) - \sum_{n=1}^k \langle x,e_n \rangle T e_n \|\\ &= \| T(x - \sum_{n=1}^k \langle x,e_n \rangle e_n) \| \quad \text{(linearity of $T$)}\\ &= \| T(\sum_{n=k+1}^\infty \langle x,e_n \rangle e_n)\| \quad \text{(expansion series of $x$)}\\ &= \| \sum_{n=k+1}^\infty \langle x,e_n \rangle T e_n\| \quad \text{(continuity of $T$)}\\ & \leq \sum_{n=k+1}^\infty |\langle x,e_n\rangle| \|Te_n\| \quad \text{(triangle inequality)}\\ & \leq \left( \sum_{n=k+1}^\infty |\langle x,e_n\rangle|^2 \right)^\frac{1}{2} \left( \sum_{n=k+1}^\infty \|Te_n\| \right)^\frac{1}{2} \quad \text{(Cauchy-Schwarz inequality in $\ell^2$)}\\ & \leq \|x\| \left( \sum_{n=k+1}^\infty \|Te_n\| \right)^\frac{1}{2}\\. \end{align} Therefore $$ \|T-T_k\| \leq \left( \sum_{n=k+1}^\infty \|Te_n\| \right)^\frac{1}{2} \to 0 $$

Gonzalo A. Benavides
  • 993