I'm a second-year undergraduate taking an introductory course in complex analysis. I am stuck on this problem from one of the previous year's exam:
True or False: For a function $f$ analytic on $S = \{ x + iy : x \in \mathbb{R}, y \in \mathbb{R}, |x| < 1, |y| < 1 \}$ and continuous on $\bar{S} = \{ x + iy : x \in \mathbb{R}, y \in \mathbb{R}, |x| \leq 1, |y| \leq 1 \}$, and satisfying that $|f|$ is bounded on the four sides $\gamma_1, \gamma_2, \gamma_3, \gamma_4$ of the square $\bar{S}$ respectively by $1, 2, 3, 4$, it is possible to have $|f(0)| > 2$.
I'm not able to disprove the existence of such a function or construct an example of such a function, but my guess is that it should be false. We have learnt about the Maximum Modulus Theorem, which says that
A non-constant holomorphic function on an open connected domain never attains its maximum modulus at any point in the domain.
Maybe by shifting the function $f$ by some constant or linear function I can show that it violates this Theorem, and so $f$ cannot exist, but I am not able to come up with a proof. Another result that we were taught that seems relevant is the Schwarz Lemma, which says that:
Let $\mathbb{D} = \{ z : |z| < 1 \}$ be the open unit disk and let $f \colon \mathbb{D} \to \mathbb{C}$ be a holomorphic map such that $f(0) = 0$ and $|f(z)| \leq 1$ on $\mathbb{D}$. Then $|f(z)| \leq |z|$ $\forall\ z \in \mathbb{D}$ and $|f'(0)| \leq 1$. Moreover, if $|f(z)| = |z|$ for some non-zero $z$ or $|f'(0)| = 1$, then $f(z) = az$ for some $a \in \mathbb{C}$ with $|a| = 1$.
Maybe by considering the restriction of $f$ to the unit disk and rescaling I could apply Schwarz Lemma, but I'm not sure how to go about this either.
Of course, I could be wrong and there is indeed such a function $f$, but in that case, I don't know how to go about constructing it.
How can I solve this problem? Any useful hints are also fine, a complete solution is not necessary.