# Upper bound for coefficients of a power series

I am doing the following problem.

Suppose $f(z)=\sum_{n=0}^{\infty}a_nz^n$ is an analytic function on the unit disc $|z|<1$. Let $0<r<1$. Prove that $$|a_n|r^n\leq \max\{4A(r),0\}-2Ref(0),$$ where $A(r)=\max_{|z|=r}Ref(z)$.

Here is my approach (but fail). Let $u(z)=Ref(z)$, then I have computed that (hope my computation is correct) $$Re(a_n)=\frac{1}{\pi r^n}\int_0^{2\pi}u(r,\theta)\cos(n\theta)d\theta.$$ $$Im(a_n)=\frac{1}{\pi r^n}\int_0^{2\pi}u(r,\theta)\sin(n\theta)d\theta.$$ Then, I used these to get the estimate \begin{align*} |a_n|r^n\leq&r^n(|Re(a_n)|+|Im(a_n)|) \\\leq&\frac{2}{\pi}\int_0^{2\pi}|u| \\=&\frac{2}{\pi}\bigg(\int_{u\geq 0}u-\int_{u\leq 0}u\bigg) \\=&\frac{2}{\pi}\bigg(2\int_{u\geq 0}u-\int_0^{2\pi}u\bigg) \\\leq&\frac{2}{\pi}\bigg(4\pi A(r)-\frac{2\pi}{2\pi}\int_0^{2\pi}u\bigg) \\=&8A(r)-4u(0), \end{align*} where I have used the mean value formula $$u(0)=\frac{1}{2\pi}\int_0^{2\pi}u(r,\theta)d\theta$$ for harmonic function. But it is seems that which is still over estimated (a factor of 2). Is the question wrong or there is a better way to obtain the required estimate? Thanks a lot!

• For any estimate containing the max of the real part try to examine $g(z)=\exp(f(z))$, since then $|g(z)|=\exp(Re(f(z)))$. – LutzL Dec 27 '13 at 8:18

One of the oldest tricks in the complex variables book is the identity $$|a|=\max_{\theta\in \mathbb R} \operatorname{Re}(e^{i\theta}a) \tag{1}$$ which holds for every complex number $a$. Thus, one can obtain an estimate of absolute value by estimating the real part of something, which has the advantage of being linear.
Fix $n$. Choose $\theta$ so that the $n$th coefficient of $g(z)=f(e^{i\theta}z)$ is real (you can even make it positive). The function $g$ satisfies the same assumptions as $f$. Applied to $g$, your inequalities produce the desired estimate, since there is no imaginary part of the coefficient.