Proof. Consider \(\quad \displaystyle {\lim \limits _{h \rightarrow 0}\left |\frac {F(z + h) - F(z)}{h} - f(z)\right |}\)

We want to show that the limit of the above expression is 0 for all \(z \in D\).

We have that \begin {align*} \frac {F(z + h) - F(z)}{h} & = \frac {1}{h} \int ^{z + h}_{z_0} f(\zeta ) d\zeta - \frac {1}{h}\int ^z_{z_0} f(\zeta )d\zeta \\\\ & = \frac {1}{h}\int ^{z + h}_zf(\zeta ) d\zeta \quad \cdots \quad (*) \end {align*}

Choose the arc joining \(z + 0\, z + h\) to be straight line \(\quad \displaystyle {\int ^{z+h}_zd\zeta = h}\)

\begin {align*} \text {Now, write}\quad f(z) & = \frac {f(z)}{h}\cdot h = \frac {f(z)}{h}\int _z^{z + h} d\zeta \\\\ & = \frac {1}{h}\int ^{z + h}_z f(z)d\zeta \quad \cdots \quad (**) \end {align*}

Thus by \((*)\) and \((**)\) we have \begin {align*} \left |\frac {F(z + h) - F(z)}{h} - f(z)\right | & = \left |\frac {1}{h}\displaystyle {\int ^{z +h}_z f(\zeta ) d\zeta - \frac {1}{h}\int ^{z + h}_z f(z)d\zeta }\right |\\\\ & = \frac {1}{\left |h\right | }\left |\displaystyle {\int ^{z + h}_z \big ( f(\zeta ) - f(z)\big ) d\zeta }\right |\\\\ & \leq \frac {1}{\left |h\right |} \max \limits _{\zeta \in \gamma }\left |f(\zeta ) - f(z)\right |\left |\displaystyle {\int ^{z + h}_z d\zeta }\right |\\\\ & = \frac {1}{\left |h\right |}\,\max \limits _{\zeta \in \gamma } \left |f(\zeta ) - f(z)\right |\,\left |h\right |\\\\ & = \max \limits _{\zeta \in \gamma }\left |f(\zeta ) - f(z)\right |\\ \end {align*}

But \(f(z)\) is continuous in \(D\). So that \(\, \displaystyle {\lim \limits _{h\rightarrow 0} \,\max \limits _{\zeta \in D} \left |f(\zeta ) - f(z)\right | = 0}.\quad \)
Hence \(F'(z) = f(z)\) □