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.

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