# If $K$ is a compact, self-adjoint linear operator on a separable Hilbert space with closed image, prove that the image of $K$ is finite dimensional

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.

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.

• 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 $M$? – kkc Aug 14 at 18:07
• I meant the quotient space. – Kavi Rama Murthy Aug 14 at 23:12
• could you please explain your answer with a bit more detail. I am having trouble understanding it – kkc Aug 14 at 23:34
• @kkc I have added some details. – Kavi Rama Murthy Aug 14 at 23:44
• Did you read my answer? Isn't $(y_n)$ any bounded sequence in $K(H)$ in my answer? – Kavi Rama Murthy Aug 15 at 23:11

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.