Problem: Suppose $f(z) = \sum_{n=0}^\infty a_nz^n$ when $|z|<1$. Show that if $|a_4|=2^4\max\limits_{|z|=1/2}|f(z)|$, then $f(z) = a_4z^4$.
I have a proof when $f$ has no zeros in $0<|z|<1$ but not in general. I also have a proof which uses Parseval's theorem. I would like to see a proof that uses only tools from a standard complex analysis course.
Proof (assuming $f$ has no zeros in $0<|z|<1$)
Using the Cauchy's formula, we can get
$$ |a_4| = \dfrac{|f^{(4)}(0)|}{4!} \le \dfrac{1}{2\pi}\int_{|w|=1/2}\dfrac{|f(w)|}{|w^5|}|dw| \le 2^4\max\limits_{|z|=1/2}|f(z)| = |a_4|. $$
So all the inequalities above are actually equalities and in particular we get that for all $z$ with $|z|=1/2$ one has $|f(z)| = \max\limits_{|z|=1/2}|f(z)| = |a_4|/2^4$. So $f$ maps the circle of radius $1/2$ to another circle. If $f$ does not have zeros in $0<|z|<1/2$, then we can reflect $f$ about the circle $|z|=1/2$ to obtain an entire function by defining
$$ \overline{f(z)} = \dfrac{(|a_4|/2^4)^2}{{f\left( \dfrac{1}{4\overline{z}} \right)}} \quad\text{for}\quad |z|\ge 1/2. $$
The order of the pole of this entire function (say $k$) is same as the order of the zero of $f$ at the origin. Then the function
$$ g(z) = \dfrac{f(z)}{z^k} $$
has a removable singularity at the origin and no zeros in $0<|z|<1/2$. Hence we can similarly reflect $g$ to obtain an entire function which is bounded near $\infty$. Hence by Liouville's theorem $g$ is constant and we get
$$ f(z)=cz^k \quad\text{for some constant }c. $$
If $k<4$, then $a_4=0$ and we get that $f$ vanish on $|z|=1/2$ (see the second sentence) and it is identically zero (by the identity theorem). Hence $f(z) = 0 = 0z^4$ is trivially true then. If $a_4\ne 0$, then we get that $k=4$ and $c=a_4$ and we are again done.
Proof using Parseval's theorem
Note that
\begin{equation}
\dfrac{1}{2\pi}\int_0^{2\pi} |f\left( e^{i\theta}/2 \right)|^2 d\theta \le \max_{|z|=1/2}|f(z)|^2 = \left( \dfrac{|a_4|}{2^4} \right)^2.
\end{equation}
Also note that the Fourier series of $f$ is
$$ f\left( e^{i\theta}/2 \right) = \sum_{n=0}^\infty a_n \left( \dfrac{e^{i\theta}}{2} \right)^n = \sum_{n=0}^\infty \dfrac{a_n}{2^n}e^{in\theta} $$
and hence the Parseval's identity tells us
\begin{equation}
\dfrac{1}{2\pi}\int_0^{2\pi} |f\left( e^{i\theta}/2 \right)|^2 d\theta = \sum_{n=0}^\infty\left(\dfrac{|a_n|}{2^n}\right)^2.
\end{equation}
Substituting the last equation in the first displayed formula gives us
$$ \sum_{n=0}^\infty\left(\dfrac{|a_n|}{2^n}\right)^2 \le \left( \dfrac{|a_4|}{2^4} \right)^2 $$
and this tells us that $a_n$ must be zero for all $n$ other than $4$.
Best Answer
As you pointed out, we have the chain of inequalities $$ |a_4| = \dfrac{|f^{(4)}(0)|}{4!} \le \dfrac{1}{2\pi}\int_{\{|w|=1/2\}}\dfrac{|f(w)|}{|w^5|}|dw| \le 2^4\max\limits_{|z|=1/2}|f(z)| = |a_4|, $$ and thus both inequalities are in fact equalities. The fact that the second inequality is an equality implies that $|f|$ is constant on the circle centered at $0$ of radius $1/2$. The first inequality being an equality gives, after parametrization $w=(1/2)e^{it}$, becomes
$$\left| \frac{1}{2\pi i} \int_{0}^{2\pi} \frac{f((1/2)e^{it})}{((1/2)e^{it})^5} \, \frac{i}{2}e^{it} \, dt \right| = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{|f((1/2)e^{it})|}{(1/2)^5} \frac{1}{2} \, dt.$$
This implies that the integrand on the left must be a unimodular constant multiple of the integrand on the right, i.e., $$\frac{f((1/2)e^{it})}{((1/2)e^{it})^5} \, \frac{i}{2}e^{it} = \alpha \frac{|f((1/2)e^{it})|}{(1/2)^5} \frac{1}{2}$$ for all $t \in [0,2\pi]$, for some $\alpha$ with $|\alpha|=1$. Recall that the right-hand side is constant, and we get $$f(z)=Cz^4$$ for all $z$ with $|z|=1/2$, for some constant $C$. This remains valid for all $z$ with $|z|<1$, by the identity theorem. The constant $C$ must be $a_4$ by uniqueness of power series coefficients.