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$f$ is continuous on $\gamma$ which is smooth & bounded. Let F be a function such that $$F=\int_\gamma \frac{f(\beta)}{\beta-z} d\beta$$ for all $z$ not in $\gamma$

Show F is holomorphic at $z$ not on the curve and that

$$F^'(z)=\int_\gamma \frac{f(\beta)}{(\beta-z)^2} d\beta$$

I used the basic definition:

$$F^'(z)=\lim_{h\to0}\frac{F(z+h)-F(z)}{h}$$ $$=\lim_{h\to0}\frac{\int_\gamma \frac{f(\beta)}{\beta-(z+h)} d\beta-\int_\gamma \frac{f(\beta)}{\beta-z} d\beta}{h}$$ $$=\lim_{h\to0}\int_\gamma \frac{(f(\beta)(\beta-z)-f(\beta)(\beta-z-h)}{h(\beta-z-h)(\beta-z)} d\beta$$ $$=\lim_{h\to0}\int_\gamma \frac{f(\beta)}{(\beta-z-h)(\beta-z)} d\beta$$ $$=\int_\gamma \frac{f(\beta)}{(\beta-z)^2} d\beta$$

Just wondering is what I done correct (since it seems a bit trivial)? If so is it necessary that the curve is smooth & bounded, and that $f$ is continuous on it?

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You're commuting the limit with the integral, so you want to say something about uniform continuity. – Dylan Moreland Mar 2 '12 at 20:59
Thanks for the reply - but I haven't really learnt about communting limits and intergrals, I just thought you were allowed. Though I have learnt about uniform continuity. Could you please explain more? or is there another way to to this? – Rcwt Mar 2 '12 at 21:03
Usually commuting limits and integrals is allowed when uniform convergence is satisfied. That is, if you a (2 dimensional) series of functions $f_{n,m}(x)$ then if $f_{n,m}$ converges uniformly we have $\lim_{n\rightarrow \infty}\lim_{m\rightarrow\infty}f_{n,m}(x)=\lim_{m \rightarrow \infty}\lim_{n \rightarrow \infty}f_{n,m}(x)$, but without uniform convergence, this result is not true in general. – ldog Mar 2 '12 at 22:34
Another approach might be to use Fubini's theorem as follows. First integrate over a closed curve in $\mathbb C\setminus \gamma$, interchange the order of integration, and apply Morera's theorem to conclude that $F$ is holomorphic. Then use Cauchy's integral formula to express the derivative of $F$ as an integral, and interchange the order of integration once more. The applicability of Fubini's theorem in this context is easy to justify if you are familiar with it. – Nick Strehlke Mar 3 '12 at 4:12
up vote 1 down vote accepted

Here we know what would be the derivative. We can write \begin{align*} \left|\frac{F(z+h)-F(z)}h-\int_{\gamma}\frac{f(\beta)}{(\beta-z)^2}dz\right|&=|\int_{\gamma}\frac{f(\beta)}{h(\beta-(z+h))}dz-\int_{\gamma}\frac{f(\beta)}{h(\beta-z)}dz\\&-\int_{\gamma}\frac{f(\beta)}{(\beta-z)^2}dz|\\ &=|\int_{\gamma}\frac{f(\beta)}{h(\beta-z)^2(\beta-(z+h))}((\beta-z)^2-\\ &(\beta-z-h)(\beta-z)-h(\beta-(z+h)))dz|, \end{align*} and computing \begin{align*} (\beta-z)^2-(\beta-z-h)(\beta-z)-h(\beta-(z+h))&=(\beta-z)h-h(\beta-z-h)\\ &=h^2 \end{align*} so $$\left|\frac{F(z+h)-F(z)}h-\int_{\gamma}\frac{f(\beta)}{(\beta-z)^2}dz\right|\leq |h|^2\int_{\gamma}\frac{|f(\beta)}{|\beta-z|^2(|h|-d(z,\gamma))}dz$$ (if we assume that $\gamma$ is closed).

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