# An inequality involving a holomorphic function

Suppose $f \in H(\mathbb{D})$ and $f(0)=0 , |Ref(z)|<1$ . Then prove that : $$|f(z)|\le \frac{2}{\pi}\log\left(\frac{1+r}{1-r}\right)$$ for $|z|\le r<1$ and $$|Ref(z)|\le \frac{4}{\pi}\tan^{-1}(|z|)$$. I tried showing this by using the fact the unit disc is a convex domain so it retains the hyperbolic metric. But i think that it can be solved by using that f takes the unit disc to the strip $|Ref|<1$

• probably it should be $$|\operatorname{Im} f(z)|\le \frac{2}{\pi}\log\left(\frac{1+r}{1-r}\right)$$ Commented Nov 29, 2016 at 5:43
• There os no mistake in the problem's statement i am sure about that. Commented Nov 29, 2016 at 7:06
• OK. I will post an answer. Commented Dec 2, 2016 at 11:34
• I would be grateful, thanks ! Commented Dec 2, 2016 at 11:35

Let $D_r=\{z : |z|\le r\}$ and $\varphi (z)=\frac{2}{\pi i}\log \left(\frac{1+z}{1-z}\right)$, which maps $\mathbb{D}$ to the strip $|\operatorname{Re}\, z|<1$ bijectively. First we prove \begin{align} \left|\log \left(\frac{1+z}{1-z}\right)\right|\le \log \left(\frac{1+r}{1-r}\right) \quad (|z|\le r<1).\tag{1} \end{align}

In other words $$|w|\le \frac{2}{\pi}\log \left(\frac{1+r}{1-r}\right)$$ holds for all $w\in \varphi (D_r)$.

Proof.\begin{align} \log \left(\frac{1+z}{1-z}\right)&=\log(1+z)-\log(1-z)\\ &=\int_\gamma \frac{d\zeta }{1+\zeta }+\int_\gamma \frac{d\zeta }{1-\zeta } \quad (\gamma : \zeta =zt\,\,(0\le t\le 1))\\ &=\int_0^1 \frac{zdt}{1+zt}+\int_0^1 \frac{zdt}{1-zt}=2z\int_0^1 \frac{dt}{1-z^2t^2}. \end{align}

Taking absolute values we have \begin{align} \left|\log \left(\frac{1+z}{1-z}\right)\right|&\le 2|z|\int_0^1 \frac{dt}{1-|z|^2t^2}\\ &=\int_0^1 \frac{|z|dt}{1+|z|t}+\int_0^1 \frac{|z|dt}{1-|z|t}\\ &=\log(1+|z|)-\log(1-|z|)=\log\left(\frac{1+|z|}{1-|z|}\right)\\ &\le \log\left(\frac{1+r}{1-r}\right). \end{align} The proof is complete.

Now we consider $$g(z)=(\varphi ^{-1}\circ f)(z).$$ Since $g$ satisfies $g(0)=0$ and $|g(z)|<1$, by Schwarz's lemma we see that $|g(z)|\le |z|,$ that is, $g(D_r)\subset D_r.$ This implies $$f(D_r)=(\varphi \circ g)(D_r)\subset \varphi (D_r).$$ Therefore we see that$$|f(z)|\le \frac{2}{\pi}\log \left(\frac{1+r}{1-r}\right)\quad (|z|\le r<1)$$ holds.

Next we prove that $$|\operatorname{Re}\, f(z)|\le \frac{4}{\pi}\tan^{-1} r$$ for $z$ with $|z|\le r$. Let $\phi(z)=\frac{1+z}{1-z}.$ Then $\phi(D_r)$ is the disk centered at $z=\frac{1+r^2}{1-r^2}$ with radius $\frac{2r}{1-r^2}$. Therefore we see that $$|\arg \phi(z)|\le \sin^{-1} \left(\frac{2r}{1+r^2}\right)=2\tan^{-1}r$$ for $z$ with $|z|\le r$.

Thus we know that $$|\operatorname{Re}\varphi (z)|\le \frac{4}{\pi}\tan^{-1}r \quad (|z|\le r),$$ which implies that $$|\operatorname{Re}f(z)|\le \frac{4}{\pi}\tan^{-1}r \quad (|z|\le r),$$ since $f(D_r)\subset \varphi (D_r).$

• It seems this works properly! Thank you very much good sir! Commented Dec 2, 2016 at 11:42