Let $H$ be a Hilbert space, and $A:H\rightarrow H$ a linear operator.
Prove that $A$ is compact if and only if $A^*$ is compact.
I saw the following proof in my book -
What I don't understand is, why the original theorem ($A$ is compact iff $A^*$ is compact) is equivalent to $A$ is compact iff $A^*A$ is compact?
Also - when it says that 'if $A$ is compact, then obviously $A^*A$ is compact', if I want to formally prove it:
If $A$ is compact, then for every $\{x_n\}$ that is bounded, we get that $\{Ax_n\}$ has a Cauchy subsequence,$\{x_{n_k}\}$.
Therefore, $A*(f(x_{n_k})) = f(A(x_{n_k}))$ for every $f\in H^*$. Since $A,f$ are bounded, there are both continuous (as it is equivalent to being bounded) then $f(A(x_{n_k}))$ is a Cauchy subsequence as well, meaning that $A^*$ is compact.
Is it correct? is it the way to prove it?
There is maybe a simpler way to show the second assertion. To say that an operator is compact is to say that the image of the unit ball is compact. Hence, if $B$ is the unit ball of $H$, then $K:= A(B)$ is compact and $A^{*}A(B) = A^*(K)$, which is compact since $A$ is continuous, so is $A^*$ and the image of a compact under a continuous map is compact.
Now, to explain the reasoning of the book. Suppose you have proven that "for any operator $A$, $A$ is compact iff $A^*A$ is compact". Call this theorem T1. Let us show that if $A$ is compact then $A^*$ is compact. Suppose $A$ is compact and set $B=A^*$. Since $B^* = A$ is compact, then $B^*B$ is compact (easy it is proven just like before). By Theorem $T1$, this shows that $B$ is compact, i.e. $A^*$ is compact.
I hope this answers your question.