Jtdylktuy

Let $$g(z)=frac{1}{2 pi i} int_{-M}^M frac{h(x)}{x-z} ,dx$$
where $h$ is continuous and supported in $[-M,M]$ .

a)Prove that the function $g$ is holomorphic in $C-[-M,M]$ , and vanishes at infinity , that is $lim_{|z| to infty} |g(z)|=0$ . Moreover , the jump of $g$ across $[-M,M]$ is $h$ , that is $$h(x)=lim_{epsilon to 0 ,epsilon gt o} g(x+i epsilon )-g(x-i epsilon)$$
b)If $h$ satisfies a mild smoothness condition , for instance a Holder condition with exponent $alpha$ , then $g(x+i epsilon)$ and $g(x- i epsilon)$ converge uniformly to functions $g_+(x) $ and $g_-(x)$ as $epsilon to 0$ . then $g$ can be characterized as the unique holomorphic function that satisfies :

($1$) $g$ is holomorphic outside $[-M,M]$ ,

($2$) $g$ vanishes at infty ,

($3$) $g(x+ i epsilon)$ and $g(x-i epsilon)$ converge uniformly as $epsilon to 0$ to $g_+(x)$ and $g_-(x)$ with $$g_+(x)-g_-(x)=h(x)$$
[Hint:If $G$ is another function satisfying these conditions , $g-G$ is entire.]

This is an exercise in Stein's complex analysis on Page $110$ . And for a), notice that $$g(x+i epsilon )-g(x-i epsilon)= frac{1}{pi}int_{-M}^M frac{epsilon}{(x-x_0)^2+epsilon ^2} h(x) , dx$$ Then , after proving that $frac{1}{pi} frac{epsilon}{(x-x_0)^2+epsilon ^2}$ is a good kernel we complete the proof of a) , and by the similiar way we can prove for b) $$g_+(x)=frac{1}{2}(h(x)-ih(x))$$ $$g_-(x) = frac12 (-h(x)-ih(x))$$ and then with Holder condition we can prove the uniform converges .

The question is : If $h$ satisfied the Holder condition , and function $G_1 , G_2 $ satisfied the condition (1),(2),(3) , then we have $$G_1(z) = G_2(z) = g(z)$$ To prove this , I want to use the hint , let $f=g-G$ . If $f$ is entire , $f$ must be continuous on some open set contains $[-M,M]$ , for $xin [-M,M]$ ,if it is continuous , then $$f(x)=lim_{epsilon to 0} f(x+i epsilon)=0$$ So we reset $f(z)=0$ if $z in [-M,M]$ and $f(z)= g(z)-G(z)$ otherwise . Then it suffice to prove $f$ is continuous and differentiable on the interval $[-M,M]$ . For $z in (-M,M)$ and with the equality above , we have $$lim_{a,b to 0} |f(z+ai+b)-0|=lim_{a,b to 0} |f(z+ai+b)|=0$$ But there might be some trouble on two points $z=-M$ and $z=M$ , I can not prove $f$ is continuous on these two points . However , I think we only need to prove $f$ is continuous and holomorphic on some small interval $(a,b) subset [-M,M]$ , then $f$ is identity $0$ on that interval so $f=0$ . WLOG , we set $f_0(z) = 0 $ , $z in (-M,M)$ and $f(z)=g(z)-G(z)$ otherwise . Then $f_0$ is continuous . To complete our conclusion , it is suffice to prove $f_0$ is differentiable and $f'(z)=0$ on $(-M,M)$ , $$f_+'(z)=lim_{h to 0, Im(h) gt 0} frac{f(z+h)-f(z)}{h}=lim_{h to 0, Im(h) gt 0} frac{g_(z+h)-G_(z+h)}{h}$$ My question:

For the last equation , as $h to 0$ , $g(z+h) to g_+(z) $ and $G(z+h) to G_+(z)$ , and $G_+(z)=g_+(z)$ , but how to deal with this next ? And If we set $f(z)=0$ on $[-M,M]$ , can we prove that $f$ is continuous on $-M$ and $M$ ?