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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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Are your operators acting on $L^2$ of the circle, or on $H^2$? – Jonas Meyer Oct 16 '10 at 22:25
On the space $\tilde{H}^2$, that is the subspace of $L^2(S^1)$ of functions whose Fourier coefficients for negative indices are 0. – Jonas Teuwen Oct 16 '10 at 23:19
Right, I may have been abusing notation, but that's what I meant by $H^2$. Thanks for clarifying. – Jonas Meyer Oct 16 '10 at 23:25
up vote 4 down vote accepted

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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Nice proof, +1. – Jonas Meyer Oct 18 '10 at 22:44
A sort of nitpick: My preference would be not to invoke the spectral theorem in those places. $K_\lambda$ is finite-dimensional because the restriction of $K$ to $K_\lambda$ is both compact and a nonzero constant multiple of the identity. 0 is in the spectrum of $K$ because compact operators on infinite dimensional normed spaces are not invertible; the compacts form a proper ideal, which again follows from the fact that the identity on an infinite dimensional normed space is not compact. These are basic facts used in the spectral theorem, but the theorem itself seems a bit much here. – Jonas Meyer Oct 18 '10 at 22:49
Or you could apply another big theorem: The spectrum of an operator on a complex Banach space (or of an element of a complex Banach algebra) is nonempty. Since you showed using the Fredholm alternative that the spectrum of $K$ contains no nonzero elements, it must be {0}. – Jonas Meyer Oct 18 '10 at 22:56
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". – Jonas Teuwen Oct 19 '10 at 14:11

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