# Function holomorphic in the units disk with different bound

Suppose $$f$$ is continuous in the closed unit disk $$\bar{D}(0,1)$$ and holomorphic over its interior $$D(0,1)$$. Moreover suppose that for $$|z|=1$$ we have:

$$\Re(z)\leq0\Rightarrow |f(z)|\leq\ 1$$

$$\Re(z)>0\Rightarrow |f(z)|\leq 2$$

and I Need to prove $$|f(0)|\leq \sqrt{2}$$ I know from the maximum modulus principle we have that:

$$1\leq \max_{|z|=1}|f|=\max_{\bar{D}(0,1)}|f|\leq 2$$

but I can't really see where the square root come from so I cannot go any further.

• out of curiosity, is the bound tight? – AccidentalFourierTransform Dec 10 '18 at 18:11
• @AccidentalFourierTransform As far as the text of my exercises says no – Renato Faraone Dec 11 '18 at 10:04

First try. By Cauchy integral $$f(0) = \frac{1}{2\pi i} \int_{\lvert z\rvert = 1} \frac{f(z)}{z}\,dz =\frac{1}{2\pi} \int_0^{2\pi} f(e^{i\varphi})\,d\varphi.$$ Hence $$|f(0)|\leq \frac{1}{2\pi} \int_0^{2\pi} |f(e^{i\varphi})|\,d\varphi\leq\frac{2\pi+1\pi}{2\pi}=\frac{3}{2}.$$ But unfortunately $$\sqrt{2}<3/2$$.
Second try. Consider the function $$F(z)=f(z)f(−z)$$ which is continuous in the closed unit disk $$\bar{D}(0,1)$$ and holomorphic over its interior $$D(0,1)$$. Then, $$\text{Re}(z)\leq 0$$ iff $$\text{Re}(-z)\geq 0$$ and therefore, for $$|z|=1$$ we have that
$$|F(z)|\leq |f(z)||f(−z)|\leq 2\cdot 1.$$ Now apply the Cauchy integral to $$F$$: $$|f(0)|^2=|F(0)|\leq \frac{1}{2\pi} \int_0^{2\pi} |F(e^{i\varphi})|\,d\varphi\leq 2\implies |f(0)|\leq \sqrt{2}.$$
For a slightly different proof than the one RobertZ gave, note that $$\log\lvert f\rvert\colon\overline{D}\to\bar{\mathbb{R}}$$ is subharmonic (it is actually harmonic with poles), since it is $$\Re\log f$$ away from the zeros of $$f$$, and if $$f(z)=0$$ then $$\log\lvert f(z)\rvert=-\infty$$. Now the mean value property of harmonic function gives $$\log\lvert f(0)\rvert$$ is at most the average value of $$\log\lvert f\rvert$$ on the unit circle, and the latter is bounded by $$\frac12\log 2$$.