$f$ is entire without any zeros, then for $ r>0$ $m(r) = \inf\left \{ \left|f(z)\right |:\left|z\right|=r\right \}$, $m$ is non increasing

Question

$f$ is entire function without any zeros, then for $ r>0$ $m(r) = \inf\left \{ \left|f(z)\right |:\left|z\right|=r\right \}$, $m$ is not increasing function.

I have an idea of how to solve this problem, but I´ve came across a problem.

If $f$ is constant, then it´s trivial.

If $f$ is non-constant with $f(z)\neq 0$ for every $z \in \mathbb{C}$, and $f$ is holomorphic, then we define $$g(r) := \frac{1}{m(r)}$$ then $$ = \frac{1}{\inf\left \{ \left|f(z)\right |:\left|z\right|=r\right \}}$$ $$ = \sup \left \{ \frac{1}{f(z)} : \left|z\right|=r \right \} $$

Because $\left \{ \frac{1}{f(z)} : \left|z\right|=r \right \}$ is closed and bounded, then $g$ takes its maximum values on boundary.

Question is: I assumed, that $g$ increases with $r$, and from that $g$ is an increasing function, but unfortunately I cannot find right arguments for this statemnt and I would like if someone can explain this?

Answer 1

Hint: Let $g=\frac 1f$. Then $\sup \{|g(z)|: |z|=r\}=\sup \{|g(z)|: |z|\leq r\}$ by MMP and hence $\sup \{|g(z)|: |z|=r\}$ is non-decreasing.

Answer 2

The result follows from the "minimum" version of the maximum principle applied to analytic functions. (Since $f(z)\neq 0,$ the function $g(z)=1/f(z)$ is analytic.)

Let $R_2>R_1$. We find that the minimum modulus on $B_{R_2}=\{z: |z|\leq R_2 \}$ is achieved when $|z|=R_2.$ But then $B_{R_2}\supset B_{R_1}$. Therefore, $m(R_2)\leq m(R_1).$

Answer 3

If $g$ takes its maximum on the boundary, then $$ \sup \left\{ \frac{1}{f(x)} : |z| = r \right\} = \sup \left\{ \frac{1}{f(x)} : |z| \leq r \right\} \text{.} $$

Since the sequence of disks $|z| \leq r$ are nested, this means the suprema on these disks are a not decreasing function of $r$. So we know $g'(r) \geq 0$ for all $r$.

Further, $0 \leq g'(r) = \frac{-1}{(m(r))^2} \cdot m'(r)$. Since the first factor is nonpositive, the second factor is too. That is, $m'(r) \leq 0$ and so $m$ is a not increasing function.