I am working on the proof of the following statement;
Suppose $f$ is analytic on a a rectangle $R$ and $|f'(z)|\leq 1$ for all $z \in R$. Then $f$ is a contraction on $R$, that is $$ |f(b)-f(a)|\leq |b-a| $$ for all $a,b \in R$. Now I have the following proof;
Since $f'$ is the derivative of some analytic function $f$ on $R$, for any two $a,b \in R$ we can construct a smooth curve $C\subset R$ from $a$ to $b$ and then the fundamental theorem of calculus applies to $f'$ on this curve. Now fix $a,b \in R$ and let $C$ be a smooth curve from $a$ to $b$. Then we have $$ |f(b)-f(a)| = |\int_Cf'(z)\text{d}z| \leq \int_C|f'(z)|\text{d}z \leq \int_C1\text{d}z. $$ Now I know that the next step is to show that $$ \int_C1\text{d}z \leq |b-a| $$ How ever here I am a little confused, as it seems to me that we have $$ \int_C1\text{d}z =b-a $$ But then how do we get the inequality, as this seems to be saying that a real number is less than a complex number, which does not make sense. So obviously somewhere I have a mistake or I am misunderstanding something.
Cheers for the help.
You can take your curve to be straight line. Let $\gamma (t)=(1-t)a+tb, 0 \leq t \leq 1$. Then $f(b)-f(a)=\int_0^{1} f'((1-t)a+tb)(b-a)\, dt$ because $f'((1-t)a+tb)(b-a)$ is the derivative of $f(1-t)a+tb)$. Hence $|f(b)-f(a)| \leq (b-a)$. The problem with what you have done is that $|\int_C f'(z)\, dz| \leq \int_C |f'(z)|\, dz$ is wrong. The right sdie is a complex number so the inequality does not even make sense. Here is the correct way of getting an upper bound for an integral over a path: $|\int_C g(z)\, dz| \leq ML$ where $M$ is an upper bound for $|g(z)|$ and $L$ is the length of $C$.