Suppose $u$ is a twice continuously differentiable function on $a< |z|<b, \ z\in \mathbb C,$ which is harmonic that is, it satisfies $u_{rr}+\frac{1}{r}u_r + u_{\theta \theta}=0.$ (If we put $z=x+iy,$ then $x=r \cos \theta, y =r \sin \theta, r=|z|$ (polar coordinates)).
Define $f(r):= \frac{1}{2\pi} \int_{0}^{2\pi} u(re^{i\theta}) d\theta$
My Question is: Is $f(r)$ diffrentiable? Can we expect $\frac{d^{2}f}{dr^{2}} +\frac{1}{r} \frac{df}{dr}=0$?
My vague attempt: $f'(r)= \frac{df(r)}{dr}= \frac{1}{2\pi} \int_{0}^{2\pi} \frac{d}{dr}[u(re^{i\theta})] d\theta;$ (can we interchanged derivative and integrals; do I need to use chain rule;$\frac{d}{dr}[u(re^{i\theta})]$ = ? )
Edit: Does there exist $g\in L^{1}(\mathbb T)$ such that $|\frac{\partial} {\partial r}h(r, \theta)| \leq g(\theta)$? (If this this the case then we $f(r)$ is diffrentiable and we can take derivative inside the integral sign [Folland Real Analysis: page 56. Theorem 2.27])
You can interchange the integral over $\theta$ and the differential by $r$. Then you obtain:
$f'(r) = \frac{1}{2 \pi} \int_0^{2 \pi} \frac{du}{dr}(re^{i \theta}) d \theta$.
Now you can substitute the harmonicity condition $u_r = -r(u_{rr}+u_{\theta \theta})$:
$f'(r) = - \frac{r}{2 \pi} \int_0^{2 \pi} (u_{rr}(re^{i \theta}) + u_{\theta \theta}(re^{i \theta}))d \theta$.
The integration over $u_{\theta \theta}$ will vanish due to the periodicity of $u_{\theta}$ and the $u_{rr}$ term can be evaluated with the chain rule (it is easy). Now consider $|f'(r)|$ and use the identity $|\int_0^{2 \pi} A(\theta) B(\theta,r)d \theta| \leq \int_0^{2 \pi} |A(\theta)| d\theta\int_0^{2 \pi} |B(\theta,r)| d \theta$.
You will get a finite value for the maximum estimation since $u$ is assumed to be twice differentiable.