A problem from Stein and Shakarchi complex analysis (problem 5, Chapter 3) The origin question is below:
Let
$$g(z)=\frac{1}{2\pi i}\int_{-M}^M\frac{h(x)}{x-z}dx$$
where $h$ is continuous and supported in $[-M,M]$.

*

*Prove that the function $g$ is holomorphic in $\mathbb{C}\backslash[-M,M]$, and vanished at infinity, that is $\lim_{|z|\to\infty}|g(z)|=0$. Moreover, the "jump" of $g$ across $[-M,M]$ is $h$, that is
$$h(z)=\lim_{\varepsilon\to0+}g(x+i\varepsilon)-g(x-i\varepsilon).$$

*If $h$ satisfies a mild smoothness condition, for instance a Hölder condition with exponent $\alpha$, then $g(x+i\varepsilon)$ and $g(x-i\varepsilon)$ converge uniformly to functions $g_+(x)$ and $g_-(x)$ as $\varepsilon\to0$. Then, $g$ can be characterized as the unique holomorphic function that satisfies:

*

*$g$ is holomorphic outside $[-M,M]$,

*$g$ vanished at infinity,

*$g(x+i\varepsilon)$ and $g(x-i\varepsilon)$ converges uniformly to the functions $g_+(x)$ and $g_-(x)$ with
$$g_+(x)-g_-(x)=h(x)$$.



The first problem is easy as long as one notice that $g(x+i\varepsilon)-g(x-i\varepsilon)=h*K_\varepsilon$, where $K_\varepsilon(x)=\frac{\varepsilon}{\pi(x^2+\varepsilon^2)}$ is a good kernel. I found it difficult to prove that $g(x\pm i\varepsilon)\rightrightarrows g_\pm(x)$. As a matter of fact, the real part of $g_\varepsilon^\pm$ is nothing but $\pm\frac{1}{2}(g_\varepsilon^+-g_\varepsilon^-)$ which had been proved to converge uniformly to $\pm\frac{1}{2}h(x)$, but the imaginary part is propotional to
$$\int_{-M}^Mh(t)\frac{t-x}{(t-x)^2+\varepsilon^2}dt.$$
I even cannot prove that it does converge to some function. and I didn't know how to use the Hölder condtion as well. I saw that follows from this condition, given $\varepsilon>0$, one have $g_\varepsilon^\pm(x):=g(x\pm i\varepsilon)$ also satisfies the Hölder condition. But what can I do with it?
 A: One can notice that $\mathrm{Re}(g(x\pm i\varepsilon))=\pm(g(x+i\varepsilon)-g(x-i\varepsilon))$ uniformly converges to $h(x)/2$ as $\varepsilon\to0$. The imaginary part of $g(x\pm i\varepsilon)$ are all proportional to
$$\int_\mathbb{R}\frac{th(t-x)}{t^2+\varepsilon^2}dt.$$
For any $\varepsilon>0$, $t/(t^2+\varepsilon^2)$ is odd, so for all $R>0$,
$$\int_{-R}^R\frac{t}{t^2+\varepsilon^2}dt=0.$$
Now as $h$ is supported in $[-M,M]$, for $R$ large enough
$$\int_\mathbb{R}\frac{th(t-x)}{t^2+\varepsilon^2}dt=\int_{-R}^R\frac{t(h(t-x)-h(-x))}{t^2+\varepsilon^2}dt.$$
As $h$ satisfies Hölder condition
$$\bigg|\frac{t(h(t-x)-h(-x))}{t^2+\varepsilon^2}\bigg|\leq C|t|^{\alpha-1}\in L^1([-R,R]).$$
By LDC,
$$\lim_{\varepsilon\to0+}\int_\mathbb{R}\frac{th(t-x)}{t^2+\varepsilon^2}dt=\int_{-R}^R\frac{h(t-x)-h(-x)}{t}dt:=\widetilde{h}(x).$$
Next
\begin{align*}
    \bigg|\int_\mathbb{R}\frac{t(h(t-x)-h(-x))}{t^2+\varepsilon^2}dt-\widetilde{h}(x)\bigg|&\leq\int_{-R}^R\bigg|\frac{\varepsilon^2(h(t-x)-h(-x))}{t(t^2+\varepsilon^2)}\bigg|dt\\
    &\leq C\left(\int_{-\varepsilon}^\varepsilon|t|^{\alpha-1}dt+\int_{\varepsilon\leq|t|\leq R}\varepsilon^2|t|^{\alpha-3}dt\right)\\
    &\leq 2C\left(\frac{\varepsilon^{\alpha}}{\alpha}+\frac{R^{\alpha-2}\varepsilon^2-\varepsilon^{\alpha}}{\alpha-2}\right)\to0.
\end{align*}
So $g(x\pm i\varepsilon)\rightrightarrows g_{\pm}(x)$. Now if $\widetilde{g}$ satisfies these three conditions too, $f:=g-\widetilde{g}$ is holomorphic outside $[-M,M]$, vanishes at infinity and $f(x\pm i\varepsilon)\rightrightarrows0$ on $[-M,M]$. By symmetry principle, $f$ can be extended to an entire function, then by Liouville's theorem, $f\equiv0$. So $g$ is unique.
Remark:The Hölder condition guarantees that $g(x\pm i\varepsilon)$ does converges to some $g_\pm(x)$. The uniqueness of $g$ still holds when generalizing the condition (iii) to $g(x\pm i\varepsilon)$ converges to $g_\pm(x)$.
