# Prove that $f$ is a contraction.

Suppose $f$ is analytic in a convex region $D$ and $|f'|\le 1$ throughout $D$. Prove that $f$ is a contraction, that is, show that $|f(b)-f(a)|\le|b-a|$ for all $a,b$ in $D$.

Since $f$ is analytic, it is differentiable in $D$. This seems like we would use the mean-value theorem, namely there exists $c$ such that $f'(c)=(f(b)-f(a))/(b-a)$. Then we take the absolute value of both sides and $|f'|<1$. So $|f(b)-f(a)|<|b-a|$, but I'm not sure if I can use the Mean Value Theorem applies, since we do not know if $f$ is real-valued. Also I haven't used that the region is convex. Any help would be appreciated. Thanks!

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If it's a complex-valued function of a complex variable, and $f'$ is its complex derivative rather than some sort of matrix-valued derivative of a real-vector-valued function of a real-vector-valued variable, then the mean value theorem for real-valued functions of a real variable would not be directly applicable. – Michael Hardy Feb 14 '12 at 23:12
For $a,b \in D$ you can consider $f$ as a map on the segment $[a,b] = \{ta + (1-t)b: t \in [0,1]\}$. Now this is basically a differentiable map $[0,1] \to \mathbb{R}^2$ and the vector valued MVT applies. – user12014 Feb 14 '12 at 23:12
....but could somewhat indirectly applicable as in "PZZ"'s comment. – Michael Hardy Feb 14 '12 at 23:13
To refrain oneself from wrongly applying the MVT to complex valued functions, one can keep in mind the case $f(z)=\mathrm e^{z}$: for $a=0$ and $b=2\mathrm i\pi$, $f(b)-f(a)\ne(b-a)f'(c)$ for any complex number $c$, since $f(b)=f(a)$ but $f'(c)=\mathrm e^c\ne0$. – Did Feb 14 '12 at 23:24

Let $[w,z]$ denote the directed line segment from $w$ to $z$. Note that if $w$ and $z$ are in $D$, then so is $[w,z]$ (exactly the definition of convexity). If we fix $a\in D$, then for all $z\in D$, $f(z)=f(a)+\int\limits_{[a,z]}f'(w)dw$ (an application of a version of the Fundamental Theorem of Calculus for analytic functions integrated over curves). Subtract $f(a)$, apply absolute values, and bound the integral using the hypothesis on the derivative to finish (I have seen physicists call the relevant bound on the integral "Darboux's inequality" (q.v.)).
\begin{aligned} |f(b)-f(a)| = &|\int_0^1 \frac{d}{dt}f((1-t)a+bt)\,dt| \leq \\ &\int_0^1 |\frac{d}{dt}f((1-t)a+bt)|\,dt = \\ &\int_0^1 |f'((1-t)a+bt)(b-a)||\,dt \leq \\ &\int_0^1|f'((1-t)a+bt)|b-a|\,dt \leq |b-a|.\\ \end{aligned}