Hilbert–Schmidt operator

Remarks

The condition and the value of the sum do not depend on the orthonormal basis. Indeed, for an orthonormal basis $(f_j)_{j \in J}$ of $K$, Parseval's identity gives $\sum_i \norm{A e_i}^2 = \sum_i \sum_j \abs{\ip{A e_i}{f_j}}^2 = \sum_j \norm{A^* f_j}^2$, and the right side does not involve $(e_i)$.

Every Hilbert–Schmidt operator is compact and satisfies $\norm{A} \le \norm{A}_{\mathrm{HS}}$. A compact operator is Hilbert–Schmidt if and only if its singular values $\sigma_n$ satisfy $\sum_n \sigma_n^2 < \infty$, and then $\norm{A}_{\mathrm{HS}}^2 = \sum_n \sigma_n^2$. The Hilbert–Schmidt operators form the Schatten class $S_2$; every trace class operator is Hilbert–Schmidt. On $L^2$ spaces of $\sigma$-finite measure spaces, they are the integral operators with square-integrable kernels, and $\norm{A}_{\mathrm{HS}}$ is the $L^2$ norm of the kernel.

Example: on $\ell^2$, the operator $A(x_1, x_2, \dots) = (x_1, x_2/2, x_3/3, \dots)$ is Hilbert–Schmidt. For the standard basis, $\norm{A e_n} = 1/n$ and $\sum_n 1/n^2 = \pi^2/6$.

Non-example: on $\ell^2$, the operator $B(x_1, x_2, \dots) = (x_1, x_2/\sqrt{2}, x_3/\sqrt{3}, \dots)$ is compact but not Hilbert–Schmidt, since $\sum_n \norm{B e_n}^2 = \sum_n 1/n$ diverges. The identity of an infinite-dimensional Hilbert space is not Hilbert–Schmidt either.