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I'm trying to solve a functional analysis problem

A self-adjoint non-negative operator $A$ on a Hilbert space $H$ is compact if and only if its $\sqrt{A}$ is compact.

1 Answers 1

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If $A^{1/2}$ is compact, $A = A^{1/2}A^{1/2}$ is as a composition of a compact and a bounded operator is compact.

If, on the other hand, $A$ is compact, we can write $A = \sum_{n\ge 1} \lambda_n(\cdot, x_n)x_n$ for some $\lambda_n \to 0$ and an orthonormal sequence (the $\lambda_n$ are the eigenvalues of $A$ and $x_n$ are the corresponding eigenvectors. Then $A^{1/2} = \sum_{n\ge 1} \lambda_n^{1/2}(\cdot, x_n)x_n$ is compact as it is the limit (in the norm topology) of the finite dimensional operators $B_N = \sum_{1\le n \le N} \lambda_n^{1/2}(\cdot, x_n)x_n$ since $\|B_N - A^{1/2}\| \le \lambda_N^{1/2} \to 0$.