# If $\{\varphi_n\}$ is an eigenbasis of $K^\ast K$'s support and $h\perp K^m\varphi_n$ for all $n,m\ge1$, then how do you show $Kh=0$?

$$\newcommand{\span}{\operatorname{span}}\newcommand{\im}{\operatorname{Im}}$$EDIT: According to the below answer, Royden's construction is wrong. However, the key detail which was omitted by the resources available to me is that because $$K$$ has precompact image on the unit ball, this image is separable and thus by scaling one can see that all of $$\im K$$ is separable, hence the below answer's construction with:

$$H_0=\overline{\im K+\im K^\ast}$$

Indeed works.

Op:

This is following an exercise from Royden. To clarify, Royden does not a priori define Hilbert spaces as separable, and nor do I. I’ve seen some posts on this site attempting to prove the same lemma (any compact operator has a separable invariant subspace for which it is zero on the complement) but these posts assume separability and argue that a compact operator is the limit of finite rank operators which as far as I know only holds in a separable context.

Let $$H$$ be a real Hilbert space, and $$K$$ a compact operator of infinite rank. Then $$K^\ast$$ also has infinite rank and is compact, so $$K^\ast K$$ is a self-adjoint compact operator of infinite rank and by the Hilbert-Schmidt theorem there is an orthonormal eigenbasis $$\{\varphi_n:n\in\Bbb N\}$$ of $$[\ker K^\ast K]^\perp$$, so $$K^\ast Kh=0\iff h\perp\varphi_n,\,\forall n\in\Bbb N$$. Then note that if $$K^\ast Kh=0$$ or equivalently $$h$$ is orthogonal to the basis, then $$\|Kh\|^2=\langle Kh,Kh\rangle=\langle K^\ast Kh,h\rangle=0$$ so $$Kh=0\iff K^\ast Kh=0\iff h\perp\{\varphi\}$$.

Let $$H_0$$ be the closed linear span of $$\{K^m(\varphi_n):n,m\ge 1\}$$, which is a closed separable subspace and $$K(H_0)\subseteq H_0$$.

I need to show that $$K=0$$ on $$H_0^\perp$$, which is equivalent to $$K^\ast K=0$$ on $$H_0^\perp$$, which is equivalent to $$H_0^\perp\subseteq\span\{\varphi_n:n\in\Bbb N\}^\perp$$.

I've managed to verify that $$H_0^\perp\subseteq\ker K^\ast$$, since $$\langle h,K\varphi_n\rangle=0\implies K^\ast h\perp\varphi_n\implies K^\ast h\in\ker K$$ if it holds for all $$n$$, and the only intersection of $$\im K^\ast$$ and $$\ker K$$ is at $$0$$, so the result follows.

I also know that:

$$H=H_0\oplus H_0^\perp=\overline{\im K}\oplus\ker K^\ast=\overline{\im K^\ast}\oplus\ker K=[\ker K]^\perp\oplus\ker K$$

And so on, and have thought for a very long time but I've not managed to take that anywhere. The exercise also feels like it should be easy, and I've just been overthinking for a very long time.

I feel quite stuck in a rut with this one - how does:

$$\forall n\in\Bbb N,\,\langle h,K\varphi_n\rangle=0$$

Imply:

$$\forall n\in\Bbb N,\,\langle h,\varphi_n\rangle=0$$

I can get a sort of reverse implication:

$$\forall n\in\Bbb N:\langle Kh,K\varphi_n\rangle=0\iff\langle h,\lambda_n\varphi_n\rangle=0\iff Kh=0$$

Where $$\lambda_n$$ is the eigenvalue of $$K^\ast K$$. That is, $$Kh\in H_0^\perp\implies Kh=0$$, but it is not necessarily the case that $$h\in H_0^\perp\implies Kh\in H_0^\perp$$... Yes, $$K(H_0)\subseteq H_0$$, but as $$K$$ is not symmetric this doesn't imply $$K(H_0^\perp)\subseteq H_0^\perp$$.

What can I do? I feel as if I'm missing something really straightforward.

• @ Shrike - would you mind stating the problem as its given in the literature and refer to one or two of these posts you mention? Commented Feb 20, 2022 at 14:22
• @undefined I am trying to show this without assuming separability. Royden phrases it as “If $K:H\to H$ is a compact operator on a (real) Hilbert space (with infinite rank) then there is a closed, separable subspace $H_0$ where $K(H_0)\subseteq H_0$ and $K=0$ on $H_0^\perp$”. Commented Feb 20, 2022 at 14:28

Consider a matrix $$K=\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}.$$ Then $${\rm Im}K=\ker K= \left\{\begin{pmatrix}x\\ 0\end {pmatrix}\,:\, x\in \mathbb{R}\right \}.$$ In this case $$H_0= \ker K$$ and $$K$$ does not vanish on $$H_0^\perp\setminus{\{0\}}=\left\{\begin{pmatrix}0\\ y\end {pmatrix}\,:\, 0\neq y\in \mathbb{R}\right \}.$$
If one insists on a compact operator with infinite-dimensional range we can consider $$K$$ being the direct sum of the matrices $$\begin{bmatrix} 0 & 2^{-n}\\ 0 & 0\end{bmatrix}.$$ The same effect occurs: $$K$$ does not identically vanish on $$H_0^\perp.$$
If $$K$$ is a compact operator then so is $$K^*.$$ Therefore the subspace $$H_0=\overline{{\rm Im}K+{\rm Im}K^*}$$ is separable and invariant for $$K.$$ Moreover $$K$$ vanishes on $$H_0^\perp,$$ as it vanishes on $$({\rm Im}K^*)^\perp .$$ So the claim in Royden's book is correct.
• Not really, as in my example the whole space is separable and invariant. I just wanted to indicate that $K$ does not necessarily vanish on $H_0^\perp$ with $H_0$ defined as in your question. Commented Feb 20, 2022 at 14:41
• Why should $H_0$ as defined in your answer be separable if I don’t assume that $H$ is separable? Commented Feb 20, 2022 at 17:30
• I have now learnt elsewhere why $H_0$ is separable. Thank you for providing that example Commented Feb 20, 2022 at 18:06