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 $Kh=0$?

Question

$\newcommand{\span}{\operatorname{span}}\newcommand{\im}{\operatorname{Im}}$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.

Answer 1

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.