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I was following this problem,

Let $D$ be the open unit disc and $C$ be the unit circle. Suppose $$f:C\to\mathbb{C}$$ is continuous. Show that $$g(w)=\dfrac{1}{2\pi i}\int_{C} \dfrac{f(z)}{z-w} \rm{d}z$$

is an analytic function of $w$ for $w\in D$.

and the answer seemed fair, but it uses fubini's theorem. But my complex analysis class doesn't cover this theorem so either I have to prove it or I should find another proof that these integrals can be done opposite ways.

Here is the snippet from the link: $$\int_\gamma \left( \frac{1}{2\pi i} \int_C \frac{f(z)}{z-w}\,dz \right)\,dw \\= \frac{1}{2\pi i} \int_C \left( \int_\gamma \frac{f(z)}{z-w}\,dw \right)\,dz$$

Is there a way of showing this to be true?(not necessarily general, it is sufficient to prove for this case only) I have tried to do it by myself but couldn't find solution.

Also, can you give some hint on showing that $g$ is continuous in the case above? This link lacks of it but I can't think of way of showing it.

1 Answers1

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Let $w\in \mathbb{D}$, and let $d=\text{dist}(w,C)$. As $f$ is continuous on $C$, $|f|$ has a maximum $M$ on the circle.

$$g(w_1)-g(w_2)=\frac{1}{2\pi i}\int_C\frac{f(z)}{(z-w_1)(z-w_2)}(w_1-w_2)\\|g(w_1)-g(w_2)|\le \frac{1}{2\pi}\int_C\left| \frac{f(z)}{(z-w_1)(z-w_2)}(w_1-w_2)\right|dz\ \le \frac{M}{d(d-|w_1-w_2|)}|w_1-w_2|\\ \lim_{w_2\to w_1}|g(w_1)-g(w_2)|=2\pi\frac{M}{d^2}\lim_{w_2\to w_1}|w_1-w_2|=0$$

To prove that $g$ is analytic without using Fubini, we better avoid interchanging integrals (there are some results on that for Riemann multiple integrals which can be proved without measure theory, but they are technical and usually not covered in calculus courses) and instead use some form of the Leibniz rule, which is usually discussed in calculus courses.

The application of Leibniz's rule is straightforward:

$$\frac{dg}{dz}=\int_C\frac{\partial}{\partial w} \frac{f}{z-w}dz=\int_C \frac{f}{(z-w)^2}dz$$

I leave it to you to check that the hypotesis to apply your favorite form of Leibniz rule are met.

One can also write a direct and self contained proof:

$$\lim_{w_2\to w_1}\frac{g(w_2)-g(w_1)}{w_2-w_1}=\lim_{w_2\to w_1}\int_C\frac{f(z)}{(z-w_1)(z-w_2)}dz$$

It remains only to prove that the convergence is uniform: the limit of the integrand is clearly

$\frac{f}{(z-w)^2}$. To show that the convergence is in fact uniform we estimate

$$\left|\frac{f(z)}{(z-w_1)(z-w_2)}-\frac{f(z)}{(z-w_1)^2}\right|=\left|\frac{f(z)}{z-w_1}\right|\left|\left(\frac{1}{z-w_2}-\frac{1}{z-w_1}\right)\right|=\\= \frac{|f(z)|}{|z-w_1|^2|z-w_2|}|w_1-w_2|\le \frac{M}{d^2(d-|w_2-w_1|)}|w_2-w_1|$$ Since this bound is independent of $z$, the convergence is uniform in $w$ and so we can interchange limit and integral.