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Let $T_\varphi$ be an analytic Toeplitz operator (meaning $\varphi \in H^\infty$). Further let $K$ be a compact operator that commutes with $T_\varphi$. Now I want to show that the spectrum of $K$ only consists of $0$.

I'm trying the following things: $T_\varphi K$ is compact so we can apply the spectral theorem for compact operators. So we know that at least $0$ is in the spectrum of this operator (not of $K$) and that the spectrum is countable with the only limit point $0$. Can I somehow use this? I also know that if $C$ is a compact operator $\|T_\varphi - C\| \geq \|T_\varphi \|$ so $\|T_\varphi(1 - K)\| = \|(1 - K)T_\varphi\| \geq \|T_\varphi \|$.

Any suggestions how I can continue?

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    Right, I may have been abusing notation, but that's what I meant by $H^2$. Thanks for clarifying.2010-10-16

1 Answers 1

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I think I got the idea:

Lemma: If $T_\varphi$ is an nonconstant analytic Toeplitz operator then the only invariant finite-dimensional subspace for $T_\varphi$ is $\{0\}$.

Proof: If $M$ is a finite dimensional subspace them $T_\varphi$ restricted to $M$ has an non-zero eigenvector (because it is compact). But this is not possible for an the Toeplitz operator on the whole space.

So now let $K$ be the compact operator that commutes with $T_\varphi$ and let $\lambda \neq 0$. Define $K_\lambda := \textrm{ker}(K - \lambda)$. Now we see that $K_\lambda$ is an invariant subspace for $T_\varphi$ as follows: $f \in T_\varphi K_\lambda$ then $f = T_\varphi g$ for some $g \in K_\lambda$, that is $(K - \lambda)g = 0$. Because they commute we have $(K - \lambda) T_\varphi g = (K - \lambda)f = 0$, so $f \in K_\lambda$. Also note that $K_\lambda$ is finite dimensional by the spectral theorem.

Now, since $\varphi$ is nonconstant $K_\lambda = 0$. So $\lambda$ is not in the spectrum of $K$. $\{0\}$ is in the spectrum of $K$ by the spectral theorem for compact operators.

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    You're right, the spectral theorem is a bit of a hammer in comparison with what we need. But I'm already satisfied that it turns out to work, then I can make it less "fat".2010-10-19