Consider a complex function $f(z)$. Suppose it's analytic in $|z|<1$ and continuous in $|z|\leq 1$. If $f(z)\in\mathbf{R}$ on the boundary $|z|=1$, how to prove it's a real constant in $|z|\leq 1$?
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Since $f$ is uniformly continuous we have that for all $\varepsilon>0$ there exists $\delta > 0$ such that for every point $w \in \mathbb C$ with $|w| = 1- \delta$ we have that the imaginary part of $f(w)$ is at most $\varepsilon$. Compute $$\text{im }f(z) = \text{im }\frac{1}{2\pi i}\int_{|w|=1-\delta}\frac{f(w)}{w-z}\,dw<\varepsilon.$$ Letting $\varepsilon$ go to $0$ we obtain that $f(z) \in \mathbb R$. Since holomorphic functions with image in $\mathbb R$ are constant (e.g., by the Cauchy-Riemann equations), $f$ is constant.
Levi
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This is wrong: $f$ need not admit a holomorphic continuation to some larger ball. See e. g. https://en.wikipedia.org/wiki/Ostrowski%E2%80%93Hadamard_gap_theorem – Bruno Krams Nov 09 '19 at 18:02
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@BrunoKrams fixed it. – Levi Nov 09 '19 at 18:09