Lemma 10.8 - Rudin's Functional Analysis, simple inequality.

Question

From a question asked by myself long ago apparently:

All the notation is in that question. There's the inequality

$$ |g(\lambda)| \leq |2r - g(\lambda)| $$

Which I am not sure where it comes from. Here $g$ is an entire function with the following properties: $g(0) = g'(0) = 0$ and $Re\left[ g(\lambda) \right] \leq |\lambda|$. This last inequality should imply the one I am asking.

Can you help to figure out?

Answer 1

Suppose $g(\lambda)=a+ib$. In the argument in the posting you quote it is established that $g(\lambda)$ is in the region $A_\lambda:=\{x+iy: x\leq |\lambda|\}$. If $|\lambda|\leq r$ then $a\leq|\lambda|\leq r$

Then

1. $|g(\lambda)|^2=a^2+b^2$

2. $|2r-g(\lambda)|^2=(2r-a)^2+b^2=4r^2-4ra+a^2+b^2=4r^2-4ra+|g(\lambda)|^2$

3. $4r^2-4ra=4r(r-a)\geq0$

Putting things together, you obtain the desired conclusion.