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Suppose that $f$ is holomorphic in $D_{R_0}(0) = \{|z| < R_0\}$. Show that whenever $0 < R < R_0$ and $|z| < R$, then $f(z) = \frac{1}{2\pi}\int_0^{2\pi} f(Re^{i\theta})\text{Re}(\frac{Re^{i\theta}+z}{Re^{i\theta}-z})d\theta$.

It seems as if they want you to use Cauchy's Integral formula and I set $\zeta = Re^{i\theta}$, and so $f(z) = \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta - z}d\zeta =\frac{1}{2\pi}\int_0^{2\pi} \frac{f(Re^{i\theta})}{Re^{i\theta} - z}(Re^{i\theta})d\theta$. I'm not quite sure if this is where I should go with it. I've also tried using analytic-ness for $f$ and replacing $\zeta = Re^{i\theta}-z$ but I don't know how to get that real part it says in the answer. Any hints appreciated, thanks.

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  • $\begingroup$ It is much easier if you use Poisson kernels and the fact that harmonic function in 2 variables are locally real parts of holomorphic functions. $\endgroup$ – Christian Sep 24 '18 at 21:25
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Consider the linear fractal transform $ T(\xi) = \frac{R(R\xi+z)}{\overline{z}\xi + R} $ . Because $ T $ has its singularity at $ \xi = -\frac{R}{\overline{z}} $ the function $ f \circ T $ is holomorphic in a $ B_r(0) $ where $ r $ is choosen slightly larger then $ 1 $. Further we have

$$ T^ {-1}(\omega) = \frac{R\omega - Rz}{-\overline{z}\omega+R^2} $$ and $$ (T^{-1})'(\omega) = R \frac{R^2-|z|^2}{(R^2-\overline{z}\omega)^2} $$ and the substitution formula $$ \int_{T^{-1} \circ (T \circ \partial B_1(0))} {\frac{(f \circ T)(\xi)}{\xi} d\xi = \int_{T \circ \partial B_1(0)} {\frac{f(x)}{T^{-1}(x)}\cdot {(T^{-1})'(x)}dx} } = \int_{T \circ \partial B_1(0)} {\frac{f(x)\cdot(R^ 2-|z|^2)}{(x-z)\cdot(R^2 - \overline{z}x)}dx} $$ which gives in Combiation with Cauchy's Integral formula $$ f(z) = f(T(0)) = \frac{1}{2 \pi i} \int_{\partial B_1(0)} { \frac{f(T(\xi))}{\xi} d\xi } = \frac{1}{2 \pi i} \int_{T \circ \partial B_1(0)} {\frac{f(x)\cdot(R^ 2-|z|^2)}{(x-z)\cdot(R^2 - \overline{z}x)}dx} $$

Now we calculate $ T \circ \partial B_1(0) $. A direct computation shows that $|T(e^{i\theta})|=R$. We conclude this linear fractional transformation carries $ \partial B_1(0) $ into $ \partial B_R(0) $. Further we have

$$ \text{Ind}_{T \circ \partial B_1(0)}(0) = \frac{1}{2 \pi i} \int_{\partial B_1(0)} {\frac{T'(x)}{T(x)}}dx = 1 $$

because the integral is equal to the number of zeros of $ T $ in $ B_1(0) $ which can be calculated directly. This show that $ T $ carries $ B_1(0) $ onto $ B_R(0) $. Otherwise there would be a "gab" on $ \partial B_R(0) $ which connects $ 0 $ to the unbounded component. Now it is easy to conclude that $ T \circ \partial B_1(0) $ and $ \partial B_R(0) $ are homotopic. So we can calculate

$$ f(z) = \frac{1}{2\pi i} \int_{0}^{2 \pi} {f(Re^{i\theta}) \frac{R^2-|z|^2}{(Re^{i\theta}-z)\cdot(R^2 - \overline{z} Re^{i \theta})} Rie^{i\theta} d \theta} = \frac{1}{2\pi} \int_{0}^{2 \pi} {f(Re^{i\theta}) \frac{R^2-|z|^2}{(Re^{i\theta}-z)\cdot(Re^{-i\theta} - \overline{z} e^{i \theta - i \theta})} d \theta} = \frac{1}{2\pi} \int_{0}^{2 \pi} {f(Re^{i\theta}) \frac{R^2-|z|^2}{(Re^{i\theta}-z)\cdot(Re^{-i\theta} - \overline{z})} d \theta} = \frac{1}{2\pi} \int_{0}^{2 \pi} {f(Re^{i\theta}) \frac{R^2-|z|^2}{|Re^ {i\theta}-z|^ 2} d \theta} = \frac{1}{2\pi} \int_{0}^{2 \pi} {f(Re^{i\theta}) \Re{\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right)} d \theta} $$ because of the following equalities $$ \Re{\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right)} = \frac{1}{2 i}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z} + \frac{\overline{Re^{i\theta} + z}}{\overline{Re^{i\theta} - z}}\right) = \frac{1}{2 i}\left( \frac{ \left(Re^{i\theta}+z \right) \cdot \overline{\left( Re^{i\theta} - z\right)} + \overline{\left(Re^{i\theta} + z\right)} \cdot \left( Re^{i\theta} - z\right) }{|Re^{i\theta}-z|^2} \right) = \frac{1}{2 i}\left( \frac{2 |Re^{i\theta}| + 2 |z|^ 2}{|Re^ {i\theta} - z|^2} \right) = \frac{R-|z|^2}{|Re^{i\theta}-z|^2} $$

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