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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.

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  • $\begingroup$ out of curiosity, is the bound tight? $\endgroup$ – AccidentalFourierTransform Dec 10 '18 at 18:11
  • $\begingroup$ @AccidentalFourierTransform As far as the text of my exercises says no $\endgroup$ – Renato Faraone Dec 11 '18 at 10:04
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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 $F$: $$|f(0)|^2=|F(0)|\leq \frac{1}{2\pi} \int_0^{2\pi} |F(e^{i\varphi})|\,d\varphi\leq 2\implies |f(0)|\leq \sqrt{2}.$$

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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 $\frac12\log 2$.

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