Prove that $\int_{C_r} {f(z)}\,\mathrm dz→0$ if $r \rightarrow 0$, where $C_r$ is the circle $|z-z_0|=r<R$

Question

Let the function $f(z)$ is continuous on $0<|z-z_0|<R$ and $M(r)$ is the maximum of $|f(z)|$ on the circle $0<|z-z_0|=r<R$. Suppose that $rM(r) \rightarrow 0$.

Prove that $\int_{C_r} {f(z)} \,\mathrm{d}z \rightarrow 0$ if $r \rightarrow 0$. Here $C_r$ is the circle $|z-z_0|=r<R$.

I tried to use the Cauchy inequality $\left|\dfrac{f^{(k)}(0)}{k!}\right| \le \dfrac{Me^r}{r^k}$. But it does not help.

Could someone give a hint, please.

Answer 1

Note that

$$ \big|\int_{C_r}f(z)dz\big| \leq \int_{C_r}|f(z)||dz| $$

Can you find a bound for this integral in terms of a bound for $|f|$ and a property of $C_r$? Does this bound get small when $r \to 0$?

Final step:

Conclude that $\int_{C_r}|f(z)||dz| \leq L(C_r)\max_{C_r}|f| = 2\pi rM(r) \to 0$