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Define $g(z)=\frac{1}{2\pi i}\int_{-k}^k \frac{h(\zeta)}{\zeta-z}d\zeta$, where $h$ is continuous and defined on $[-k,k]$.

Let $|h(x)-h(y)|\leq |x-y|^\alpha$ for all $x,y\in[-k,k]$ and for some $\alpha>0$. Prove that $g(x\pm i\epsilon)$ converge uniformly to functions $g_{\pm}$ on $[-k,k]$ as $\epsilon\rightarrow 0$ (where $g_+-g_-=h$).

This was stated as a fact to a question I was working with, where I had to prove that $g$ is the unique function satisfying the above property (we can assume that $g(x+i\epsilon)-g(x-i\epsilon)\rightarrow h(x)$ as $\epsilon\rightarrow 0^+$ for $x\in[-k,k]$, which can be shown through convolutions). However, I do not see why this is the case.

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I'm working on the exact same problem (Stein and Shakarchi, Complex Analysis, Ch. 3, Problem 5), and I also don't see why this can be so easily asserted. It's a shame no one answered this yet. I attempted to show as $\epsilon\rightarrow 0^+$ that $g(x\pm i\epsilon)$ is uniformly Cauchy, but this didn't get me anywhere either. – VF1 Nov 9 '14 at 18:39

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