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I know from the spectral theorem for compact self-adjoint operators that $K$ can be written as $$K = \sum_{n} \lambda_n P_n$$ where $\lambda_n$ are eigenvalues of $K$ and $P_n$ are orthogonal projections. By the spectral theorem, the image of each $P_n$ is finite dimensional, but there can be a countably infinite many of them.

Given that, I'm not sure how to proceed. My instinct is that I need to show that $K$ cannot have infinitely many eigenvalues due to the condition that its image is closed.

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Let $M$ be the kernel of $K$. Then $S:H/M \to K(H)$ defined by $S(x+M)=Kx$ is a well defined bounded operator which is also bijective. By Open Mapping Theorem its inverse is also continuous. Since $K$ is compact this implies that the closed unit ball of $K(x)$ is compact and hence $K(X)$ is finite dimensional.

Some details for the last part: suppose $(y_n)$ is a bounded sequence in $K(X)$. Let $x_n=S^{-1}y_n$. Then $\|x_n\| \leq \|S^{-1}\| \|y_n\|$ so $(x_n)$ is bounded. By the definition of the norm in $H/M$ we can find a sequence $(z_n)$ in $M$ such that $(x_n+z_n)$ is bounded. Since $K$ is compact $K(x_n+z_n)$ has a convergent subsequence. But $Kz_n=0$ so $(K(x_n))$ has a convergent subsequence. By definition of $x_n$ we have $y_n=S(x_n)=Kx_n$. Hence $(y_n)$ has a convergent subsequence. Thus every bounded sequence in $K(H)$ has a convergent subsequence. This implies that $K(H)$ is finite dimensional.

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  • $\begingroup$ Where you wrote $H \vert M$, do you mean $S$ is defined on the quotient space $H/M$ or that $S$ is defined on $H$ when restricted to $M$? $\endgroup$ – kkc Aug 14 at 18:07
  • $\begingroup$ I meant the quotient space. $\endgroup$ – Kavi Rama Murthy Aug 14 at 23:12
  • $\begingroup$ could you please explain your answer with a bit more detail. I am having trouble understanding it $\endgroup$ – kkc Aug 14 at 23:34
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    $\begingroup$ @kkc I have added some details. $\endgroup$ – Kavi Rama Murthy Aug 14 at 23:44
  • $\begingroup$ Did you read my answer? Isn't $(y_n)$ any bounded sequence in $K(H)$ in my answer? $\endgroup$ – Kavi Rama Murthy Aug 15 at 23:11
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The other answer is correct and helpful to know because it is quite general.

However, it is probably instructive to choose a more constructive path in your simple situation. For this, show that there exist $(a_n)\in \ell^2$ such that $(a_n \lambda_n^{-1})\not\in \ell^2$. Finally choose a unit vector $x_n$ in the image of each $P_n$ and conclude that $\sum_n x_n a_n $ is not in the image of your operator but in the closure of the image.

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