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Problem :

C is a smooth simple closed curve in $\mathbb C$

$g(x)$ is a continuous function on C

$f(z)=\int_C\frac {g(x)}{x-z}dx\quad (z\notin C)$

Show that $f(z)$ is holomorphic in $\mathbb C\setminus C$

My try :

$\lim_{h\to0}\frac{f(z+h)-f(z)}{h} =\lim_{h\to0}\int_C\frac{g(x)}{[x-(z+h)][x-z]}dx=\int_C\lim_{h\to0}\frac{g(x)}{[x-(z+h)][x-z]}dx=\int_C\frac{g(x)}{[x-z]^2}dx$

So $f(z)$ is holomorphic in $\mathbb C\setminus C$.

Is it right? I think I didn't use the fact that C is a smooth simple closed curve and $g(x)$ is continuous.

And I'm not confident about the way I prove $\lim_{h\to0}\int_C\frac{g(x)}{[x-(z+h)][x-z]}dx=\int_C\lim_{h\to0}\frac{g(x)}{[x-(z+h)][x-z]}dx$

I think the interchange of integral and limit is possible by uniformly convergence $\sup_{x\in C}|\frac{g(x)}{[x-(z+h)][x-z]}-\frac{g(x)}{[x-z]^2}|=\sup |g(x)\frac{h}{[x-z]^2[x-(z+h)]}|\le|\frac{\max g(x)}{\min [x-z]^3}||h|$ . So it is uniformly convergent. (I think I used the fact $g(x)$ is continuous here around $\max g(x)$ but, still I think I didn't use the fact that C is a smooth simple closed curve.)

Is my proof correct? Thanks for your reading.

Arbitrary
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  • Problem I see is you haven't showed me anything about $\int_C\frac{g(x)}{(x-z)^2}dx$. How do I know that thing exists and makes sense? In answer to this question https://math.stackexchange.com/q/779176/254075 they use Morera's Theorem to solve the problem. – sharding4 Jun 08 '17 at 02:08
  • @sharding4 Thanks. Do you mean the integral can be infinite value? $g(x)$ and $[x-z]^2(\neq0)$ is bounded so the integral is also bounded. Is it right? (I want not to use fubini's theorem in your link because I don't know it well. Sorry.) – Arbitrary Jun 08 '17 at 03:09
  • I don't think that you know the integral exists for an arbitrary closed curve $C$ and continuous function $g$. I think that needs to be justified. If it were a circle, you could parametrize it, and that would probably be ok because everything is bounded away from $C$. I agree with you about having to appeal to Fubini's theorem, but... – sharding4 Jun 08 '17 at 03:16
  • @sharding4 Thanks. Maybe I don't understand. But, isn't it enough because $\frac{g(x)}{[x-z]^2}$ is continuous and bounded? – Arbitrary Jun 08 '17 at 03:32
  • Maybe since $C$ is smooth you don't need to worry about it. For an arbitrary simple closed curve it would not be. – sharding4 Jun 08 '17 at 03:36

1 Answers1

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The interchange of the limit and the integral is not justified, as you note, so you need to try something else. The following is the usual wey to get round your problem, and consists of expanding the integral kernel $(z-\xi)^{-1}$ into a uniformly convergent powerseries, so that you can interchange summation and integration: in general, it is much harder to prove one can interchange limits and integrals, but uniform convergence of series does the trick here.

Let $C$ be any piecewise smooth curve and define $f : \mathbb C\smallsetminus C\longrightarrow \mathbb C$ so that $$ f(z)=\int_C\frac {g(\xi)}{\xi-z}d\xi$$

To show that $f$ is holomorphic, take a $z_0\in C$ and pick a ball $B = B(z_0,r)$ strictly missing $C$. I will show that $f$ admits a powerseries development in $B$, so that $f$ is analytic at $z_0$. Because $z_0$ is arbitrary, this shows $f$ is analytic and hence holomorphic throughout its domain.

For $z\in B$ and $\xi \in C$ we have
$$\left|\frac{z-z_0}{\xi-z_0}\right|<1-\delta$$ for some $0 < \delta <1$. Thus we may consider the powerseries development

$$\frac{1}{z-\xi} =\frac{1}{\xi-z_0} \sum_{n\geqslant 0 } \left(\frac{z-z_0}{\xi-z_0}\right)^n $$

that is valid in $B$. This converges uniformly over $B$ by the estimate made above, and thus it is valid to interchange the order of integration and summation to obtain that for $z\in B$ we have

$$ f(z)=\sum_{n\geqslant 0} a_n (z-z_0)^n$$

where $a_n = \displaystyle \int_C \frac{g(\xi)}{(\xi-z_0)^{n+1}}d\xi$ for each $n\in \mathbb N$.

Pedro
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