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Let $\phi$ be a nonzero function in $L^\infty(T)$ where $T$ is the unit circle. Let $M_\phi$ be the multiplication operator and $T_\phi$ be the Toeplitz operator. Show $T_\phi$ and $M_\phi$ have no eigenvalues in common.

I keep getting that $\phi$ must be a constant if $T_\phi$ has an eigenvalue.

Recall: if $\phi\in L^\infty(T)$ and $P$ is the projection onto the Hardy space $H^2$, then $T_\phi(f)=P(\phi f)$ for $f\in H^2$.

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Correction concerning the eigenvalues of $M_\phi$: the equation $\phi f = \lambda f$ is equivalent to $(\phi-\lambda)f\equiv 0$, which means that $\phi$ is constant on the support of $f$. Therefore, $M_\phi$ has eigenvalues for any function $\phi$ that is constant on some set of positive measure. – user53153 Jan 3 '13 at 2:12
That being said, I am confused by this question. If $\phi\equiv 1$ then both operators are the identity, so they have $1$ as an eigenvalue. – user53153 Jan 3 '13 at 3:18
i was thinking the same thing. i think the quesrion is wrong – john Jan 3 '13 at 12:37

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