# Problem of the Week #255 - Aug 15, 2017

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Let $\Bbb D$ be the open unit disc in the complex plane, and let $f$ be a continuous complex function on $\partial\Bbb D$. Consider the function

$$F(re^{i\phi}) \,\dot{=}\, \frac1{2\pi}\int_0^{2\pi} f(e^{i\theta})\frac{1-r^2}{1-2r\cos(\theta-\phi) + r^2}\, d\theta\quad (re^{i\phi}\in \Bbb D)$$

Prove $F$ is harmonic on $\Bbb D$, and that for all $z_0\in \partial \Bbb D$, $\lim\limits_{z\to z_0} F(z) = f(z_0)$.
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#### Euge

##### MHB Global Moderator
Staff member
I'll give one more week for users to attempt a solution. Keep in mind, the definition of $F$ involves the Poisson kernel. It's key to use the fact that the Poisson kernel is an approximate identity.

#### Euge

##### MHB Global Moderator
Staff member
No one answered this week's problem. You can read my solution below.

Note that $F(z)$ is the real part of the holomorphic function $$\frac{1}{2\pi}\int_{0}^{2\pi} f(e^{i\theta})\frac{e^{i\theta}-z}{e^{i\theta} + z}\, d\theta$$ whence $F$ is harmonic. Now fix $z_0 = e^{i\phi_0}\in \partial \Bbb D$. By continuity of $u$, given $\epsilon > 0$ there corresponds an $\eta > 0$ such that for all $\theta$, $\lvert \theta - \phi_0\rvert < \eta$ implies $\lvert f(e^{i\theta}) - f(e^{i\phi_0})\rvert < 0.5\epsilon$. Fix $\phi$ such that $\lvert \phi - \phi_0\rvert < 0.25\eta$, and note

$$F(re^{i\phi}) - f(re^{i\phi_0}) = \frac{1}{2\pi} \int_{-\pi}^\pi [f(re^{i\theta}) - f(re^{i\phi_0})]P_r(\phi-\theta)\, d\theta$$ where $P_r$ is the Poisson kernel. Break up the latter integral as $J_1 + J_2$, where $J_1, J_2$ are integrals over regions $\lvert \theta - \phi_0\rvert < \eta$ and $\lvert \theta - \phi_0\rvert \ge \eta$, respectively. Then $\lvert J_1 \rvert< 0.5\epsilon$ and $\lvert J_2 \rvert < 2\max_{\lvert \phi - \theta\rvert \ge 0.75\eta} P_r(\phi - \theta) \cdot \max_{-\pi \le \theta \le \pi} \lvert u(re^{i\theta})\rvert$. Since, $\lim_{r\to 1} \max_{\lvert \phi - \theta\rvert \ge 0.75\eta} P_r(\phi - \theta) = 0$, then $|J_2| < 0.5 \epsilon$ for all $r$ sufficiently close to $1$. Hence, $\lvert F(re^{i\phi}) - F(re^{i\phi_0})\rvert < 0.5 \epsilon + 0.5 \epsilon = \epsilon$ whenever $\lvert \phi - \phi_0\rvert < 0.25\eta$ and $r$ is sufficiently close to $1$. Consequently, $\lim_{z \to z_0} F(z) = f(z_0)$.

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