Rigorous proof of higher derivatives form Cauchy Integral formula. I am studying the Cauchy Integral Formula from Ahlfors' book. I can not understand some points. If you can discuss them a little, it will be very much helpful to me.
I have added this portion from the book.  



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*Why equation (24) is not sufficient to say that derivative of all orders of the analytic function $f(z)$ are not analytic ?

*In Lemma 3 we are first showing $F_1(z)$ is continuous. Why? For analyticity we want continuous derivative of the function $F_1$.

*Can you expand a little the expression $F_n(z) - f_n(z_0)$ in Lemma 3.

*Why Lemma 3 is required for a rigorous proof of equation 23 and 24 ?
Thank you for your kind help.  
 A: *

*Equation (24) is based on the assumption (which you need to prove) that we can differentiate under the sign of integral the equation preceding (23).

*We prove the continuity of $F$ for any $\varphi$, so that, when we compute
$$G(z)=\int_{\gamma}\frac{\varphi/(\zeta-z_0)}{(\zeta-z)}d\zeta$$
we know that this function $G$ is continuous, but
$$G(z)=\int_{\gamma}\frac{\varphi/(\zeta-z_0)}{(\zeta-z)}d\zeta=\frac{1}{z-z_0}\int_\gamma\left(\frac{\varphi(\zeta)}{\zeta-z}-\frac{\varphi(\zeta)}{\zeta-z_0}\right)d\zeta=\frac{1}{z-z_0}(F_1(z)-F_1(z_0))$$
So, $\lim_{z\to z_0}G(z)=G(z_0)$ and $G(z_0)=F_2(z_0)$ by definition.

*The reasoning behind the formula for $F_n(z)-F_n(z_0)$ is the same:
$$F_n(z)-F_n(z_0)=\int_\gamma\left(\frac{\varphi(\zeta)}{(\zeta-z)^n} - \frac{\varphi(\zeta)}{(\zeta-z_0)^n}\right)d\zeta=\int_\gamma\varphi\left(\frac{1}{(\zeta-z)^{n-1}(\zeta-z_0)} + \frac{z-z_0}{(\zeta-z)^n(\zeta-z_0)}-\frac{1}{(\zeta-z_0)^n}\right)d\zeta$$
Now, the term
$$\int_{\gamma}\frac{\varphi}{(\zeta-z_0)^n}d\zeta$$
is constant in $z$, the term
$$\int_{\gamma}\frac{\varphi}{(\zeta-z)^{n-1}(\zeta-z_0)}d\zeta$$
is $F_{n-1}$ where we used $\varphi(\zeta)/(\zeta-z_0)$ in place of $\varphi(\zeta)$, so by inductive hypothesis is continuous and the last term
$$\int_{\gamma}\frac{(z-z_0)\varphi(\zeta)}{(\zeta-z)^n(\zeta-z_0)}d\zeta$$
is estimated like we did for $F_1(z)-F_1(z_0)$, when proving continuity of $F_1$.

*I think it's the same answer I gave at point 1. Lemma 3 says that the derivative of $F_n$ is $nF_{n+1}$, but $nF_{n+1}$ is obtained exactly by differentiating with respect to $z$ under the sign of integral. So, the Lemma actually says that the derivative of $F_n$, which is
$$\frac{d}{dz}\int_{\gamma}\frac{\varphi(\zeta)}{(\zeta-z)^n}d\zeta$$
can be computed by exchanging the differentiation and the integration, i.e. by
$$\int_{\gamma}\frac{d}{dz}\frac{\varphi(\zeta)}{(\zeta-z)^n}d\zeta$$
which gives exactly $nF_{n+1}$.
