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Suppose $\phi$ is an invertible function in $L^\infty(T)$ where $T$ is the unit circle such that the essential range of $\phi$, $R(\phi)$ is contained in the open right half-plane. Show there exist an $\epsilon>0$ such that $\epsilon R(\phi)\subset \{z\in\mathbb{C}:|z-1|<1\}$.

This is part of a proof. I cannot see why it is true. I know $R(\phi)$ is compact and does not contain 0 as $\phi$ is invertible.

For this, we have $R(\phi)=\{\lambda\in\mathbb{C}:\mu(\{x\in X:|f(x)-\lambda|<\epsilon\})>0 \ \forall \epsilon>0\}$. Here $\mu$ is the Lebesgue measure and $X$ is the unit circle.

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The essential range is the smallest closed subset $R\subset\mathbb{C}$ so that $\phi(z)\in R$ for almost every, $z\in T$. Is that right? –  Harald Hanche-Olsen Dec 17 '12 at 20:06

1 Answer 1

The essential range is compact, and the sets of the form $\varepsilon^{-1}\{z\in\mathbb{C}\colon\lvert z-1\rvert<1\}$ with $\varepsilon>0$ is an open cover (of the open right half plane, and hence of $R(\phi)$). So $R(\phi)$ is contained in a finite union of these sets, and therefore in one of them since these sets are nested.

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