Assume that f is entire, $f(0)=1$, and $\int_{0}^{2\pi} \lvert f(e^{i\theta}) \rvert d\theta = 2 \pi$. Prove that f is constant.
Now, I tried to bound the integral by $2\pi$ times the max of $\lvert f(e^{i\theta}) \rvert$ where $0 \leq \theta \leq 2\pi$ and then using the max-modulus principle tired to get a contradiction but we get an obvious inequality. Next, I tried to make use of the power series of f to evaluate the integral and use the properties of integral, but then we again get obvious inequalities. Any hint would be helpful.
Best Answer
Since $1=f(0) = {1 \over 2 \pi i}\int_C {f(z) \over z} dz = {1 \over 2 \pi }\int_0^{2 \pi}f(e^{i \theta}) d \theta $ we see that the integral is real and ${1 \over 2 \pi }\int_0^{2 \pi}f(e^{i \theta}) d \theta = {1 \over 2 \pi }\int_0^{2 \pi}|f(e^{i \theta})| d \theta $. In particular, $\operatorname{re} f(e^{i \theta}) = | f(e^{i \theta}) | $ everywhere and so $f(e^{i \theta}) = | f(e^{i \theta}) | $ is real everywhere.
Consider $h(z) = e^{ i f(z)}$ which is entire, and $|h(e^{i \theta})| = 1$, hence $|h(z)| \le 1$ on the unit disk. Repeating with $g(z) = e^{ -i f(z)}$ shows that $| e^{ i f(z)}| = 1$ on the unit disk and hence $f$ is constant.
Addendum:
Using the curve $C(t) = e^{it}$ we have
$f(0) = {1 \over 2 \pi i}\int_C {f(z) \over z} dz = {1 \over 2 \pi i}\int_0^{2 \pi} {f(e^{i \theta}) \over e^{i \theta}} i e^{i \theta} d \theta = {1 \over 2 \pi}\int_0^{2 \pi} f(e^{i \theta})d \theta $.