# An analytic function $f=u+iv$ on a unit open disk obeying $|u|+|v|<=1$

Given an analytic function $f=u+iv$ on an open unit disk $\mathbb D(0,1)$ obeying $|u|+|v|\le1$. Prove that either $|u|+|v|<1$ everywhere or $f$ is a constant.

I am stuck with this problem. It is clear that $|u|+|v|=1$ is a closed set and $f$ can't map an open set from the disk to it but can't make the steps showing that this implies $|u|+|v|<1$. Or I may be on the wrong path.... Any help will be appreciated .

• Use Open mapping theorem. Oct 26, 2016 at 14:36

Since $u$ is a harmonic function, $|u|$ is a subharmonic function. Also $|v|$ is subharmonic.
Then we see that $|u|+|v|$ is subharmonic.
Suppose that $|u|+|v|$ has the maximum value $1$ in $\mathbb D(0,1)$. Then $|u|+|v|$ is constant by the maximum principle for subharmonic functions. Therefore we have $$|u|+|v|=1$$ for all $z\in \mathbb D(0,1)$.
But this is a contradiction because $f$ maps $\mathbb D(0,1)$ to the closed set $|u|+|v|=1$.
Thus if $f=u+iv$ is not a constant ($f=a+ib$ with $|a|+|b|=1$) we see that $$|u|+|v|<1$$ for all $z\in \mathbb D(0,1)$.