integration over sphere for complex variable Given a function $u: \mathbb C^n \to \mathbb R$ and $a \in \mathbb C^n$, suppose that
$u(a) \leq \frac{1}{2 \pi} \int_{0}^{2 \pi} u(a + re^{i \theta}\zeta) d\theta$, for any $\zeta \in \mathbb C^n$ such that $|\zeta| = 1$ and want to prove that
$u(a) \leq \frac{1}{2 \pi} \int_{|\zeta'| = 1} u(a + r\zeta') d\sigma$ where $d\sigma$ is the volume form of $S^{2n-1}$
I want to do this by integrating both sides of the first inequality with respect to $d\sigma$, but I am having a hard time calculating
$\int_{|\zeta'| = 1} \int_{0}^{2 \pi} u(a + re^{i \theta} \zeta') d\theta d\sigma$
 A: By the Mean Value Theorem, there exists $\overline{\theta}(a,r, \zeta)$ such that
$ \int_{0}^{2 \pi} u(a + re^{i \theta} \zeta) d\theta =2 \pi u(a+re^{i\overline{\theta}})$.
If $a$ and $r$ are fixed, one can assume $\overline{\theta} = \overline{\theta}(\zeta)$. Therefore,
$ |S_{n-1}| u(a) \leq \frac{1}{2 \pi} \int_{|\zeta| = 1} \int_{0}^{2 \pi} u(a + re^{i \theta} \zeta) d\theta d\sigma =  \int_{|\zeta| = 1} u(a + re^{i\overline{\theta}}\zeta) d\sigma$.
One can show that the dependency of $\overline{\theta}$ on $\zeta$ is (at least locally) $C^1$. Thus, making the change of variables $\zeta' = e^{i\overline{\theta}(\zeta)}\zeta$, one sees that
$
|S_{n-1}| u(a) \leq   \int_{|\zeta'| = 1} u(a + r\zeta') d\sigma, 
$
which is, up to a constant factor (in my opinion, unavoidable), the inequality you desire.
A: We are given that for any $\zeta \in S=S^{2n-1},$
$$u(a)\le \frac{1}{2\pi}\int_0^{2\pi}u(a+re^{it}\zeta)\,dt.$$
This implies
$$ u(a)= \frac{1}{\sigma (S)}\int_S u(a)\,d\sigma \le \frac{1}{\sigma (S)}\int_S\frac{1}{2\pi}\int_0^{2\pi}u(a+re^{it}\zeta)\,dt\,d\sigma (\zeta)$$ $$\tag 1 =  \frac{1}{2\pi}\int_0^{2\pi}\frac{1}{\sigma (S)}\int_S u(a+re^{it}\zeta)\,d\sigma (\zeta)\,dt.$$
We have used Fubini here.
Claim: The integral $\int_S u(a+re^{it}\zeta)\,d\sigma (\zeta)$ is independent of $t.$
Suppose the claim is true. Then $(1)$ equals
$$\frac{1}{2\pi}\int_0^{2\pi}\frac{1}{\sigma (S)}\int_S u(a+r\zeta)\,d\sigma (\zeta)\,dt = \frac{1}{\sigma (S)}\int_S u(a+r\zeta)\,d\sigma (\zeta).$$
This shows $u(a)\le \dfrac{1}{\sigma (S)}\int_S u(a+r\zeta)\,d\sigma (\zeta)$ as desired.
So we are done if we prove the claim.
Proof of claim: The measure $d\sigma$ is rotation invariant. This means that if $T$ is an orthogonal transformation on $\mathbb C^n=\mathbb R^{2n},$ then for any continuous $f$ on $S,$
$$\int_S f(\zeta)\,d\sigma(\zeta) = \int_S f(T(\zeta))\,d\sigma(\zeta).$$
Now for any fixed $t\in \mathbb R ,$ the map $T(z)= e^{it}z$ is orthogonal. This proves the claim.
Ask if you have questions.
