Hölder Condition Implying Uniform Convergence
I'm also working on Problem 5(b) from Chapter 3 in Stein's Complex Analysis(Page 110). In the discription of the problem, it assert that when some mild smoothness condition (such as Hölder Condition) is imposed on $ h(x)$ , which is continuous and supported in $ [-k,k]$ , then $g(x_0+i\epsilon)=\frac{1}{2\pi i}\int_{-k}^k \frac{h(x)}{x-(x_0+i\epsilon)}dx$ can converge to some function $ g_+(x_0) $ as $ \epsilon\rightarrow 0^+$ uniformly on $ x_0\in[-k,k]$.
I may find a counterexample to the above argument. If assuming that $ h(x)=1$ on $ [-k,k]$ , it certainly satisfies Hölder condition. However by definition of $g(z)$,
$g(z)=\frac{1}{2\pi i}\int_{-k}^k \frac{h(x)}{x-z}dx=\frac{1}{2\pi i}\int_{-k}^k \frac{1}{x-z}dx=\frac{1}{2\pi i}log\left(\frac{k-z}{k+z}\right)$, and it seems that $log\left(\frac{k-z}{k+z}\right)$ has singularities at $z=\pm k $,
$g(x_0+i\epsilon)$ can not converge to a unbouded function uniformly on $ x_0\in[-k,k]$.
Do I miss something? any help would be greatly appreciated.