# For functions $f$,$g$ analytic on nbhd. of the unit disk, $|f|\leq |g|$ it's boundary, and $g \neq 0$ on its closure, is $|f| \le |g|$ on interior?

This is an old qualifying exam question in Complex Analysis:

If $f$ and $g$ are analytic on a neighborhood of the closure of the unit disk $\mathbb{D}$ (denoted $\overline{\mathbb{D}}$), with $g$ never vanishing on $\overline{\mathbb{D}}$ and $|f(z)|\le|g(z)|$ for all $z \in \partial \mathbb{D}$, is $|f(z)|\le|g(z)|$ for all $z \in \overline{\mathbb{D}}$?

I've been stuck on this for a while with no progress. What's hurting is the fact that $g$ doesn't vanish; if it were allowed to vanish, then the statement would be false (e.g. $f\equiv z$, $g \equiv 1/2$). Any ideas?

• Maximum modulus principle for $f/g$. – WimC May 11 '18 at 3:36
• @N.S. What about $f=1,g(z)=z?$ – zhw. May 11 '18 at 14:45
• @N.S. You do on the boundary. – zhw. May 11 '18 at 16:43

Since $g$ is nonvanishing in $\overline{\mathbb{D}}$, we can apply the maximum modulus principle to $f/g$.
More precisely, either $g=cf$ for some constant $c\geq 1$ or $\vert f/g\vert$ must attain its maximum on the boundary of $\mathbb{D}$. Hence, for some $z_0\in \partial \mathbb{D}$ $$\frac{\vert f(z)\vert}{\vert g(z)\vert}< \frac{\vert f(z_0)\vert}{\vert g(z_0)\vert}\leq 1$$ for all $z\in \mathbb{D}$.
• You could shorten it by just saying $f/g$ is holomorphic on a neighborhood of $\overline D$ and $|f/g|\le 1$ on $\partial D.$ Therefore $|f/g|\le 1$ in $D$ my the MMT. – zhw. May 11 '18 at 14:42