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Let $\mathcal{H}_1$ and $\mathcal{H}_2$ be Hilbert spaces and $T: \mathcal{H}_1 \rightarrow \mathcal{H}_2$ a compact operator.

I want to show that $(\ker T)^\perp$ and $\text{ran}\ T$ are separable.

Since $T$ maps to a Hilbert space, there is a sequence $(T_n)_{n \in \mathbb{N}}$ of finite-dimensional operators that converges to $T$ in the operator norm. Unfortunately, I was unable to make proper use of this fact. Or do I have to use another approach?

I don't know much about compact operators between Hilbert spaces...

Can someone help me to get started on this?

[Update on the Definitions I use here:]

A Hilbert space is an inner product space $(\mathcal{H}, \langle .,.\rangle)$ that is complete with respect to the norm that is induced by $\|x\|=\langle x,x\rangle^{1/2}$.

A linear space $\mathcal{X}$ is separable iff there exists a dense countable subset of $\mathcal{X}$, where countable means finite or countably infinite.

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  • $\begingroup$ Some people include separable in the definition of a Hilbert space. I take it you are not one of those people? (I just want to be clear the statement on the problem.) $\endgroup$
    – Potato
    Jul 9, 2013 at 23:07
  • $\begingroup$ Oh, I didn't know that. I will clarify the definitions. Thanks for your remark! $\endgroup$
    – Amarus
    Jul 9, 2013 at 23:10

1 Answer 1

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Here are some remarks to get you started.

  • $T$ is compact iff $T^*$ is compact.
  • $\ker(T)^\bot = \overline{ \mathrm{ran}(T^*)}$.
  • $\mathcal{K} \subset \mathcal{H}$ is separable iff $\overline{ \mathcal{K}}$ is separable.

With knowledge of these facts, you should be able to reduce one of your statements to the other. You might also consider trying to prove a lemma along the lines of

  • If $T_n \to T$ in operator norm, then $\mathrm{ran}(T) \subset \overline{ \bigcup_n \mathrm{ran}(T_n)}$.
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  • $\begingroup$ Doesn’t this argument assume that there is an orthonormal basis, ie the hilbert spaces are separable? Otherwise you cannot obtain $T$ as the norm limit of the finite rank projective sequence $\endgroup$
    – FShrike
    Feb 20 at 14:17

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