# Compactness of operators, and a lot of projections

Let $\mathcal{H}$ be an infinite-dimensional Hilbert space. Claim: an operator $T\colon\mathcal{H}\to\mathcal{H}$ is compact if and only if $Te_n\to 0$ for every orthonormal sequence $\lbrace e_n\rbrace$.

Proof: Suppose $T$ is compact. Given an orthonormal sequence, extend it to an orthonormal basis $e_n$. Let $P_n$ be the projection onto the span of $e_1,\dots, e_n$. We know that $\|P_nT-T\|\to 0$ for any compact operator. Since $T^*$ is compact as well, we have $\|P_nT^*-T^*\|\to 0$, hence $\|TP_n-T\|\to 0$. Finally, note that $Te_n=(T-TP_{n-1})e_n$, and therefore $\|Te_n\|\le \|T-TP_{n-1}\|\to 0$ as $n\to\infty$.

Conversely, suppose $T$ is not compact. Pick an orthonormal basis $\lbrace e_n\rbrace$ and let $P_n$ be as above. The non-compactness of $T$ implies that there exists $\epsilon>0$ such that $\|T-TP_n\|\ge \epsilon$ for all $n$ (because $TP_n$ has finite rank). Let $n_0=1$.

Pick a unit vector $\|u\|$ such that $\|(T-TP_{n_0})u\| > \epsilon/2$. Since $\lbrace e_n\rbrace$ is a basis, for any vector $h$ we have $P_n h\to h$ as $n\to\infty$. Thus, there exists $n_1>n_0$ such that $\|(T-TP_{n_0})P_{n_1}u\| > \epsilon/2$. Let $v_1$ be the vector $(I-P_{n_0})P_{n_1}u$ normalized and note that $\|Tv_1\| > \epsilon/2$.

Continuing in this way, we form a sequence of unit vectors $v_k$ such that $\|Tv_k\| > \epsilon/2$ and $v_k$ lies in the range of projection $Q_k:=(I-P_{n_{k-1}})P_{n_k}$ (this is a projection because all $P_m$ commute). These ranges are mutually orthogonal, because for any $j>k$ we have $(I-P_{n_{j-1}})P_{n_k}=0$, which implies $Q_jQ_k=0$ and therefore $\langle Q_jx,Q_ky\rangle=0$ for all $x,y$. Thus, $\lbrace v_k\rbrace$ is the desired sequence.

This site uses Akismet to reduce spam. Learn how your comment data is processed.