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I'm working on an old qualifying exam problem, which asks me to show that if $f:\mathbb{D}\rightarrow \mathbb{C}$ is holomorphic and injective and $f'(0)=1$, then the area of $f(\mathbb{D})$ is at least $\pi$. I know that the area of $f(\mathbb{D})$ is $$\int \int _\mathbb{D} |f'(z)|^2 dx dy. $$ At $z=0$, we get that the area of $f(\mathbb{D})$ is the area of $\mathbb{D}$, which is $\pi$. How can I show that at any other point, z, $|f'(z)|\geq 1$?

qualsqualsquals
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  • You cannot show that $|f'(z)|\geq 1$ for all $z \in \Bbb C$. If $f$ is not linear then there will always be points with $|f'(z)| < 1$. – Martin R Aug 06 '24 at 18:29

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Note that $f'(z)$ is never $0$ on $\mathbb D$, so $\log |f'(z)|$ is a harmonic function on $\mathbb D$ and $$\frac{1}{\pi} \int \int_D \log |f'(z)|\; dx\; dy = \log |f'(0)| = 0$$ By Jensen's inequality,

$$ \eqalign{\frac{1}{\pi} \int \int_{\mathbb D} |f'(z)|^2 \; dx\; dy &= \frac{1}{\pi} \int \exp(2 \log |f'(z)|) \; dx \; dy\cr &\ge \exp\left( \frac{2}{\pi} \int \int_{\mathbb D} \log |f'(z)|)\; dx\right) = 1}$$

Robert Israel
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