Compact operators

Let $AA^*$ be compact I need to show that A is compact, I am quite sure that it needs to be shown that it maps weakly convergent sequence to a sequence that converge in norm but I cant manage to do that. Thanks for help.

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Here is a direct proof. Let $x_n \rightharpoonup x$, that is, $(y,x_n) \to (y,x)$ for all $y \in H$. Then $$(y,Ax_n) = (A^*y,x_n) \to (A^*y,x) = (y,Ax)$$ for all $y \in H$ and therefore $Ax_n \rightharpoonup Ax$. Now $AA^*$ is compact and thus $AA^*x_n \to AA^*x$. This means that $$\|Ax_n\|^2 = (Ax_n,Ax_n) = (AA^*x_n,x_n) \to (AA^*x,x) = (Ax, Ax) = \|Ax\|^2.$$

It follows that $$\|Ax_n - Ax\|^2 = \|Ax_n\|^2 - 2\mbox{Re}(Ax_n, Ax) + \|Ax\|^2 \to 0$$ so that $Ax_n \to Ax$. Therefore $A$ is compact.

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Is this in a Hilbert space? You could try the polar decomposition.

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This seems more of a comment than an answer? – Calvin Lin Jun 23 '13 at 19:47
It's meant as a hint. – Robert Israel Jun 23 '13 at 20:12

Just to elaborate on the answer above, polar decomposition says $A = UP$ where $U$ is a partial isometry and $P$ is the square root of $A^*A$ (defined using functional calculus). So if $A^*A$ is compact, $A=UP$ is also compact. (Same argument applies to $AA^*$ by the corresponding polar decomposition $A=PU$.)

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If $P$ is the square root of $A^*A$, then it is $A=UP$. But $A=QV$ with $Q$ the square root of $AA^*$. – 1015 Jun 23 '13 at 21:08
Thanks for the correction. – Michael Jun 24 '13 at 3:31