# If $f$ is complex analytic on $S=\{x+iy : |x|<1, |y|<1\}$, continuous on $\bar{S}$ and bounded by $1,2,3,4$ on each side, then is $|f(0)|>2$ possible?

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.

• To the question author: Which tools do you have at your disposal? Are you familiar with harmonic functions and the Dirichlet problem? That can be used to show that there is a solution. – Martin R Oct 7 '20 at 11:26
• @MartinR We were taught that a harmonic function is one that satisfies the Laplace differential equation, and how to use the Cauchy-Riemann equations to try and compute the harmonic conjugate (when it exists). I am not familiar with the Dirichlet problem, though. – user833460 Oct 7 '20 at 11:29
• @anonymous_user: There is a solution, but I currently do not see how to show that with elementary means. On the other hand, it is quite easy to show that $|f(0)|$ cannot exceed $\sqrt{24} \approx 2.21$. Just asking: did you quote the problem correctly? – Martin R Oct 7 '20 at 11:34
• @MartinR Here's a screenshot of the question from my laptop — the problem is quoted correctly in the body now (unless my eyes have overlooked something again). I might encounter the Dirichlet problem in my upcoming differential equations course, so could you please tell me how one would go about it using that? Even if it's too advanced, I might still learn something! – user833460 Oct 7 '20 at 11:41

## 2 Answers

I think that actually one can construct such an example using a little harmonic function theory.

First, notice that by RMT and by symmetry there is a conformal map that extends to a homeomorphism from the closed unit disc to the closed square (and which is actually conformal everywhere outside the vertices), $$F:\mathbb D \to S, F(0)=0, F(\pm 1, \pm i)$$ being the vertices of the square in counterclockwise order corresponding to the circle ordering (so if one fixes the image of one vertex say $$F(1)=(-1,-1)$$, the others are fixed eg $$F(i)=(1,-1)$$ etc).

(One can write a formula for this as $$F(z)=c\int_0^z\frac{dz}{\sqrt{1-z^4}}$$ but that is not needed)

Consider the finite non-negative measure on the unit circle given by $$0, \log 2, \log 3, \log 4$$ on the four open arcs $$(1,i), (i-1), (-1,-i), (-i,1)$$ and by zero in the four points or if you want the absolute continuous one given by $$d\mu=qdt$$ where $$q$$ takes the given values on the four open arcs and is unimportant what finite value we give it at the $$4$$ roots of unity of order $$4$$

Let $$u_1(re^{i\theta})=\frac{1}{2\pi}\int_0^{2\pi}\frac{1-r^2}{1-2r\cos (\theta-t)+r^2}d\mu(t)$$ the Poisson transform of $$d\mu$$ which is harmonic, bounded and positive in the open disc and satisfying $$u_1(re^{it}) \to q(t)$$ non-tangentially outside the four special points; actually note that $$u_1(0)=\frac{1}{2\pi}\int_0^{2\pi}d\mu(t)=\frac {\log 2+\log 3+\log 4}{4} >2$$.

So considering $$g=u+iv$$ holomorphic in the unit disc, $$h(z)=e^{g(z)}$$ almost satisfies the required properties on the four arcs since $$|h(z)|=e^{u(z)}$$ and $$|h(0)|=e^{\frac{\log 24}{4}}=24^{1/4}>2$$ (however $$h$$ is not continuous on the boundary, though it is "almost" so) and then clearly $$f(w)=h(F^{-1}(w))=e^{g(F^{-1}(w))}$$ almost satisfies the requires properties on the square and $$|f(0)|>2$$;

But now it is clear that $$h_1(z)=(1-\epsilon)h(rz)$$ for $$r$$ close enough to $$1$$ and $$\epsilon>0$$ small will do (will satisfy the boundness properties on the arcs and will be continuous, even holomorphic on the closed disc) and then $$|h_1(0)|>2$$ if $$1-r$$ hence $$\epsilon$$ are small enough, so taking $$f_1(w)=h_1(F^{-1}(w))$$ solves the problem.

Note that since $$F, F^{-1}$$ are not conformal at the vertices, $$f_1$$ is only continuous on the closed square (though holomorphic outside the vertices) despite that $$h_1$$ is holomorphic on the closed unit disc

Yes, it can happen that $$|f(0)|>2$$. The existence of an example implies that some polynomial works, hence it's at least theoretically possible to write down an example "explicitly". Here's a sketch of a not-quite-elementary construction:

Let $$\psi$$ be a smooth function on the boundary such that $$\psi=0$$ on $$\gamma_1$$ and $$\log(j-1)\le\psi\le\log j$$ on $$\gamma_j$$, $$j=2,3,4$$, and also such that $$\psi=\log j$$ on "most of" $$\gamma_j$$, for example on all of $$\gamma_j$$ except two small subintervals at the ends.

Let $$u$$ be the solution to the Dirichlet problem with boundary data $$\psi$$. Note that $$u$$ is smooth up to the boundary, and hence so is the harmonic conjugate $$v$$. By symmetry the harmonic measure of $$\gamma_j$$ at the origin is $$1/4$$; hence $$u(0)$$ is "close to" $$\frac14(\log1+\log2+\log3+\log4).$$ Note that $$\frac14(\log1+\log2+\log3+\log4)>\frac14(\log2+\log2+2\log2)=\log2.$$Let $$f=e^{u+iv}$$.