$A$ has finite rank implies $A^*$ has finite rank. Also $A$ compact implies $A^*$ compact.

Question

I want to solve the following problem

Let $C$ be a closed and bounded subset of a Hilbert space $H$.

a) Suppose that $C$ is compact. Show that, for every $\epsilon>0$, there exists a finite-dimensional subspace $F\subset H$ such that $$\sup_{x\in C}d(x,F)<\epsilon.$$

b) Inversely, we suppose that for every $\epsilon>0$, there exists a finite-dimensional subspace $F\subset H$ such that $d(x,F)<\epsilon$ for all $x\in C$. Prove that $C$ is compact.

c) Let $A:H\to H$ be a bounded operator. Show that $A$ is compact if and only if $A$ is the limit of a sequence of bounded operators with finite rank.

d) Let $A:H\to H$ be a bounded operator with finite rank. Show that $A^*$ has finite rank.

e) Let $A:H\to H$ be a compact operator. Show that $A^*$ is also a compact operator.

I already solved the first 3 itens of this problem. However I can't seem to figure out how to use the first 3 itens to solve the other two. If anyone can give me a hint of how I could solve those 2 problems, that would be very helpful.

Answer 1

For solving part d), take a basis $\{e_1,\ldots,e_n\}$ of the range of $A$. Then write $Ax=\sum_{j=1}^nF_j(x)e_j$ for some $F_j\in H^*$. Certainly you know what $H^*$ looks like. Now calculate $\langle Ax_1,x_2\rangle$ for some $x_1,x_2\in H$ to see what $A^*$ looks like.

To solve part e), just use c), d), and the fact that $\|A*\|=\|A\|$ for any operator $A$ on $H$.