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Let $f$ be a holomorphic, non-constant function on the unit sphere $|z|<1$.

Define the function $g(r)=sup_{|z|leq r}Re(f(z))$ for $rin [0,1]$.

Show that $g$ is an increasing function on $[0,1)$

I am pretty sure this proof is very trivial, but I am missing something really obvious.

This is my thought process so far:

Since $f$ is non-constant on $|z|<1$, then there exists a neighborhood $U$ in $|z|<1$, such that (by the Cauchy Riemann equations) either $frac {partial Re(f(z))}{partial x}neq 0$ or $frac {partial Re(f(z))}{partial y}neq 0$. Suppose without loss of generality that $frac {partial Re(f(z))}{partial x}neq 0$

I want to use the maximum modulus principle. Thus take $r_1in [0,1]$, such that the boundary $|z|=r_1$ intersects $U$. By the maximum modulus principle, since $D=|z|<r_1$ is open, connected and bounded, and $f:overline{D}rightarrowmathbb{C}$ is continuous such that $f_{restriction_D}$ is holomorphic, then $|f|$ attains its maximum over $overline D$ with its maximum on $partial D$ (boundary of D), call this maximum $M_1$

Now pick $r_2in [0,1]$, with $r_2>r_1$, such that the boundary $|z|=r_2$ intersects $U$. By the same argument, $|f|$ now attains a new maximum on the boundary of $|z|=r_2$, call it $M_2$.

Now either $M_1=M_2$, or $M_1<M_2$

Since $frac {partial Re(f(z))}{partial x}neq 0$ on $U$, then clearly $M_1<M_2$.

Does it then follow that $g(r)$ is an increasing function on $[0,1)$?

My apologies if the above attempt at a proof makes no sense, I am not too sure how to go about the problem otherwise. Thanks for any help in advance!

You are correct in trying to use the maximum principle, but notice that you really shouldn't look at the derivative themselves, but rather at the fact that the real part of $f$ is harmonic (this is essentially the meaning of the Cauchy-Riemann equations).

By the (strong) maximum principle for harmonic functions, if $u$ is a non constant harmonic function and $B(0,r)$ is the ball of radius $r$ around $0$, then $u$ has maximas only on the boundary of the ball. In particular, there exists a point $z$ such that $|z| = r$ and $u(z) > u(z')$ for every $z'$ with $|z'| < r$.

So, if $0leq r_1, < r_2 < 1$, there exists a point $z, |z| = r_2$, such that for every other point $z', |z'| leq r_2$ we have, $mathrm{Re}(f(z)) < mathrm{Re}(f(z'))$. This clearly implies
$$suplimits_{|z| leq r_2}mathrm{Re}(f(z)) >suplimits_{|w| leq r_1} mathrm{Re}(f(w)).$$