Ideal of compact operators

Remarks

Example: on $H = \ell^2$, the diagonal operator with entries $1, 1/2, 1/3, \ldots$ belongs to $\mathcal{K}(H)$, since it is the norm limit of its finite-rank truncations. For any bounded $A$, the products $AT$ and $TA$ are again in $\mathcal{K}(H)$. Indeed, a bounded operator maps bounded sets to bounded sets and relatively compact sets to relatively compact sets.

Non-example: if $H$ is infinite-dimensional, the identity $I$ is not in $\mathcal{K}(H)$, since an orthonormal sequence has no convergent subsequence; so $\mathcal{K}(H)$ is a proper ideal. If $H$ is finite-dimensional, every operator is compact and $\mathcal{K}(H) = \mathcal{B}(H)$.

Other notations are $\mathcal{K}$ and $\mathcal{B}_0(H)$. On a Hilbert space, $\mathcal{K}(H)$ equals the norm closure of the finite-rank operators.