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.

Thanks in advance

  • 6
    $\begingroup$ 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. $\endgroup$ – GEdgar Jun 1 '12 at 18:10

Once you have the lemma, the problem is simpler.

  • 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.
  • Use the fact that the norm of a bounded operator is the same as the norm of its adjoint.

An interesting exercise would be the following: show the statements starting from the definition of a compact operator (that is, the closure of the image of the unit ball is compact).


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