Show that the function $\phi(z)=\sum_{j=1}^n |f_j(z)|^2 $ has no local max Suppose $f_j(z) \in H(\Omega)(j=1,2, \ldots,n) $. Show that the function $$\phi(z)=\sum_{j=1}^n |f_j(z)|^2 $$ has no local maximum in the region $\Omega$ unless all the functions $f_j(z)(j=1,2, \ldots,n)$ reduce to a constant function.
My first instinct was to prove that the function is unbounded proving it would have not local maximum. but I am not sure if this is sufficient or not and I am not sure about the function reducing to a constant.
 A: If one knows some theory of subharmonic functions, one can argue that $z \mapsto \lvert f(z)\rvert^2$ is subharmonic if $f$ is a holomorphic function, and strictly subharmonic unless $f$ is constant. Since the sum of subharmonic functions is constant, it follows that $\phi$ is subharmonic, and strictly subharmonic unless all of the $f_j$ are constant. But a strictly subharmonic function has no local maxima.
Without recourse to the theory of subharmonic functions, we can prove the assertion in various ways using diverse tools of complex analysis. One way:
From Cauchy's integral formula, we obtain the mean value formula for holomorphic functions,
$$f(z) = \frac{1}{2\pi} \int_0^{2\pi} f(z + re^{i\varphi})\,d\varphi\tag{1}$$
if $r$ is small enough that the closed disk $\overline{D_r(z)}$ is contained in the domain of $f$. Applying the absolute value to $(1)$, we obtain the inequality
$$\lvert f(z)\rvert \leqslant \frac{1}{2\pi} \int_0^{2\pi} \lvert f(z + re^{i\varphi})\rvert\,d\varphi\tag{2}$$
under the same conditions, and the inequality is strict unless $f$ is constant on the circle $\{w : \lvert w-z\rvert = r\}$. By the identity theorem that would imply that $f$ is constant. Setting $f = f_j^2$ for $j = 1,\dotsc,n$ and summing, we obtain that
$$\phi(z) \leqslant \frac{1}{2\pi} \sum_{j = 1}^n \int_0^{2\pi} \lvert f_j(z+ re^{i\varphi})\rvert^2\,d\varphi, \tag{3}$$
and the inequality is strict unless we have the equality
$$\lvert f_j(z)\rvert^2 = \frac{1}{2\pi} \int_0^{2\pi} \lvert f_j(z+re^{i\varphi})\rvert^2\,d\varphi$$
for all $j$, i.e. unless all $f_j$ are constant. But if $\phi$ has a local maximum at $z_0\in \Omega$, then for small enough $r > 0$ we have
$$\phi(z) \geqslant \frac{1}{2\pi} \int_0^{2\pi} \phi(z + re^{i\varphi})\,d\varphi,\tag{4}$$
and $(4)$ is compatible with $(3)$ only if we have equality in both, that is, if all $f_j$ are constant.
A: If $f(z) = \sum_{k=0}^{\infty}c_k(z-a)^k$ in $D(a,R),$ then the orthogonality of the exponentials shows
$$\tag 1\int_0^{2\pi}|f(a+re^{it})|^2\,dt = \int_0^{2\pi}|\sum_{k=0}^{\infty}c_kr^ke^{ikt}|^2 = 2\pi\sum_{k=0}^{\infty}|c_k|^2r^{2k}$$
for $0\le r <R.$ Note that $(1)$ is an increasing function of $r.$ Furthermore it is strictly increasing unless $c_k = 0$ for all $k>0,$ i.e., unless $f$ is constant, in $D(a,R).$
You can apply this to the integrals of $|f_1|^2 + \cdots +|f_n|^2$ to get the desired result for the problem at hand.
