Let $u$ be a subharmonic on open set $\Omega$. Let $a\in\Omega,R>0$ such that $B(a,r)\subset \Omega$. Prove $$v(\rho)=\int_0^{2\pi}u(a+\rho e^{it})dt$$ is a monotone increasing function on $(0,R)$.
For a sub harmonic function we know that $u(a)\le v(\rho)$ for any value of $\rho<R$. I thought maybe using Harnack's inequality but I don't know how.
How can we use the fact that $u$ is sub harmonic to prove monotonicity of the integral?
A couple of approaches:
Majorization
Let $h$ be a harmonic function in $B(a,r)$ such that $h=u$ on $\partial B(a,r)$. Then $u\le h$ in $B(a,r)$ (hence the name, sub-harmonic). For harmonic functions, averages on concentric circles are the same. Hence, the average of $u$ on $\partial B(a,s)$ for $s<r$ does not exceed its average on $\partial B(a,r)$.
Differentiation
Assume $u$ is smooth for now. Differentiate its average on $\partial B(a,r)$ with respect to $r$. This gives the integral of normal derivative of $u$, which is the flux of the gradient $\nabla u$ out of the disk. By the divergence theorem, this flux is equal to the integral of Laplacian over the disk, which is nonnegative. This proves monotonicity for smooth subharmonic functions.
The general case is then handled by approximation: every subharmonic function is the decreasing limit of smooth subharmonic functions, namely its mollifications.