How to show that $|f(z_{1}) − f(z_{2})| ≤ M|z_{1} − z_{2}|$ for any $z_{1}, z_{2} ∈ U$?

Question

Let $U = B(c, R)$ denote the open ball of radius $R$ centered at $c ∈ \mathbb{C}$. Let $f$ be an analytic function on $U$ such that there exists $M > 0$, $$|f'(z)| ≤ M. \quad \forall z ∈ U$$ Prove that $$|f(z_{1}) − f(z_{2})| ≤ M|z_{1} − z_{2}|. \quad ∀z_{1}, z_{2} ∈ U$$

I was thinking that

$$|f'(z_0)|\leq {M}$$ as we know that $$f'(z_0)=\lim_{z\rightarrow z_0} \frac{f(z_1)-f(z_2)}{z_1-z_2}.$$

Here I don't know how to proceed…

Please help me.

Answer 1

Let $C$ the path given by $C(t)=z_1+t(z_2-z_1)$ for $t \in [0,1]$. Then we have

$f(z_1)-f(z_2)= \int_C f'(w)dw=\int_0^1 f'(C(t))C'(t)dt$.

Can you take it from here ?