Let f be a holomorphic function on the closed unit disk such that : $$ | f(e^{i\pi t})| \leq e^{t}, \forall t \in [0,2] \, .$$
Show that $$ | f(0)| \leq e \, .$$
I tried to use this relation to use the average value of $f$ in $0$
$$ f(0)=\frac{1}{2\pi}\int_{0}^{2\pi}f(e^{it})dt= \frac{1}{2}\int_{0}^{2}f(e^{i\pi t})dt$$
so $$ | f(0)| \leq \frac{e^2-1}{2}$$ which is far from the desired inequality.
The function $$ g(z) = f(z) \cdot \overline{f(\bar z)} $$ is holomorphic in the closed unit disk and satisfies $$ |g(e^{i\pi t})| = |f(e^{i\pi t})| \cdot |f(e^{i\pi (2-t)})| \le e^t \cdot e^{2-t} = e^2 \, . $$ for $0 \le t \le 2$. The maximum modulus principle then gives $$ |f(0)|^2 = |g(0)| \le e^2 \, . $$
Alternatively one can use that $\log|f(z)|$ is subharmonic and therefore satisfies the mean-value inequality $$ \log |f(0)| \le \frac 12 \int_0^2 \log |f(e^{i\pi t})| \, dt \le \frac 12 \int_0^2 t \, dt = 1 \, . $$
The same estimate also follows from Jensen's formula.
