# Compact operator between Hilbert spaces: range and orthogonal complement of the kernel are separable

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.

• 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.) Jul 9, 2013 at 23:07
• Oh, I didn't know that. I will clarify the definitions. Thanks for your remark! Jul 9, 2013 at 23:10

• $T$ 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.
• If $T_n \to T$ in operator norm, then $\mathrm{ran}(T) \subset \overline{ \bigcup_n \mathrm{ran}(T_n)}$.
• 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 Feb 20 at 14:17