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!

  • $\begingroup$ For any estimate containing the max of the real part try to examine $g(z)=\exp(f(z))$, since then $|g(z)|=\exp(Re(f(z)))$. $\endgroup$ – 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.

If the above hint is not enough, read further.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.