# Space Of All Compact Operators (Functional Analysis)

I need to prove the following statements: (a)The set $\mathbb{F}$ of compact operators on a Hilbert space $\mathbb{H}$ is a norm closed linear subspace of the space of all bounded linear operators on $\mathbb{H}$ . (b) If $A \in \mathbb{F}$ , and $B$ is a bounded linear operator on $\mathbb{H}$ , then $AB \in \mathbb{F}$ . (c) If $A \in \mathbb{F}$ , then $A^{*} \in \mathbb{F}$ .

I should use the following Lemma: An operator $A$ on a Hilbert Space $\mathbb{H}$ is compact if and only if there exists a sequence of operators $A_n$ of finite rank (that is with range space of finite dimension) which converge in norm to $A$ .

I'll be glad to receive some guidance to it... Maybe someone will be able to explain me the meaning of a norm closed linear subspace.

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Linear subspace: closed under sums and scalar multiples. Norm closed: closed according to the norm topology. So: you must show the sum of two compact operators is compact; a scalar multiple of a compact operator is compact; if a sequence of compact operators converges in the norm topology (to some operator), then the limit is compact. –  GEdgar Jun 1 '12 at 18:10
Thanks a lot !!!! –  joshua Jun 2 '12 at 13:15

• Norm closed means closed with respect to the topology given by the norm defined by $\lVert A\rVert=\sup_{x\neq 0}\frac{\lVert Ax\rVert}{\lVert x\rVert}$. We can show that $\mathbb F$ is sequeantially closed (it's enough since we are in a metric space). If $\{A_n\}$ is a sequence of compact operators which converge to $A$ for the norm, let $A_n'$ of finite rank such that $\lVert A_n-A'_n\rVert\leq n^{-1}$. Show that $\{A_n'\}$ converges to $A$, which gives that $A$ is a norm limit of finite ranked operators.
• Let $\{A_n\}\subset\mathbb F$ which converges to $A$, then show that $A_nB\to AB$, using the fact that $B$ is bounded.