If we have a function $\tilde{f}\in L^p(T)$ on the unit circle $T$ with $p \geq 1$ we can regain a harmonic function $f$ on the unit disk using the Poisson kernel $P_r$:
$$
f\left(re^{i\theta}\right)=\frac{1}{2\pi} \int_0^{2\pi} P_r(\theta-\phi) \tilde f\left(e^{i\phi}\right) \,\mathrm{d}\phi, \quad r < 1
$$
It looks somewhat similar to the Cauchy's Integral formula. The latter states that a holomorphic function defined on a disk is completely determined by its values on the boundary $\gamma$ of the disk.
$$
f(a)={\frac {1}{2\pi i}}\oint _{\gamma }{\frac {f(z)}{z-a}}\,dz
$$
Questions:
1) Could you please explain the logic behind the integral transformation with the Poisson kernel? Why does the kernel have the form $P_{r}(\theta )=\operatorname {Re} \left({\frac {1+re^{i\theta }}{1-re^{i\theta }}}\right),\ 0\leq r<1$ ?
2) What is the difference between the integral transformation with the Poisson kernel and the Cauchy's integral formula?
Thank you for your help in advance.
Best Answer
The usual Cauchy kernel is a member of a wider family: we can add an arbitrary holomorphic function. If $h$ is holomorphic on the closed disk then $$ f(a) = \frac1{2\pi i}\oint_{|z|=1} f(z) \left(\frac1{z-a}+h(z)\right) \mathrm{d}z = \frac1{2\pi}\oint_{|z|=1} f(z) \left(\frac1{1-a\bar{z}}+zh(z)\right) \frac{\mathrm{d}z}{iz}. $$ $$ = \frac1{2\pi}\oint_{|z|=1} f(z) \left(1+\frac{a\bar{z}}{1-a\bar{z}}+zh(z)\right) \frac{\mathrm{d}z}{iz}. $$
From this point we can look for a function $zh(z)$ (which must have a root at $0$) such that the modified kernel $1+\frac{a\bar{z}}{1-a\bar{z}}+zh(z)$ is real along the unit circle. (Notice that $\frac{\mathrm{d}z}{iz} = \mathrm{d}(\arg z)$ is real.) This can be done by choosing $$ zh(z) = \overline{\left(\frac{a\bar{z}}{1-a\bar{z}}\right)} = \frac{\bar{a}z}{1-\bar{a}z}, $$ so $$ h(z) = \frac{\bar{a}}{1-\bar{a}z}, $$ that is holomorphic in the disk $|z|<\frac1{|a|}$.
Hence, the modified kernel is $$ 1+\frac{a\bar{z}}{1-a\bar{z}} + \overline{\left(\frac{a\bar{z}}{1-a\bar{z}}\right)} = \mathrm{Re}\left(1+2\frac{a\bar{z}}{1-a\bar{z}}\right) = \mathrm{Re}\left(\frac{1+\bar{a}z}{1-\bar{a}z}\right). $$
Therefore, the difference between the Cauchy and the Poission kernels is a holomorphic function. :-)