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

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.

• (d) Suppose $A$ is of finite rank. Then $\mathcal{R}(A)\oplus\mathcal{R}(A)^{\perp}=H$. If $e \in \mathcal{R}(A)^{\perp}$, then $F_e(f) = \langle f,e\rangle$ is a linear functional in $H^*$ such that $F_e \circ A = A^*F_e=0$. So the range of $A^*$ is spanned by $\{ F_e : e \in \mathcal{R}(A) \}$, which is finite-dimensional. Jun 16, 2018 at 3:52

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 $H$.