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.


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 $\|B_N - A^{1/2}\| \le \lambda_N^{1/2} \to 0$.

  • $\begingroup$ But why $A^{1/2}$ equals $\sum_{n\ge 1} \lambda_n^{1/2}(\cdot, x_n)x_n$ $\endgroup$ – Aram Malkhasyan May 22 '12 at 19:43
  • $\begingroup$ @user31919: Did you try squaring it? $\endgroup$ – Jonas Meyer May 22 '12 at 20:21
  • 1
    $\begingroup$ @user31919 Let $B = \sum_{n\ge 1} \lambda_n^{1/2}(\cdot, x_n)x_n$. Then $B$ is self-adjoint and non-negative, moreover for $x\in H$ we have $$ B(Bx) = \sum_{n\ge 1} \lambda_n^{1/2}(Bx, x_n)x_n = \sum_{n,m\ge 1} \lambda_n^{1/2}\lambda_m^{1/2}(x, x_m)(x_m, x_n)x_n = \sum_{n\ge 1}\lambda_n (x,x_n)x_n = Ax $$ and so $B = A^{1/2}$. $\endgroup$ – martini May 22 '12 at 20:28
  • $\begingroup$ @martini I see,i got it already! Thanks a lot! $\endgroup$ – Aram Malkhasyan May 22 '12 at 21:02
  • $\begingroup$ So this argument shows also that if A is compact and positive, then it has the same range as $A^{1/2}$, right? $\endgroup$ – Axesilo Jul 3 '14 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.