[Math] Rigorous proof of higher derivatives form Cauchy Integral formula.

complex-analysis

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

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

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

  4. Why Lemma 3 is required for a rigorous proof of equation 23 and 24 ?

Thank you for your kind help.

Best Answer

  1. 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).

  2. 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.

  3. 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 $F_1$.

  4. 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 $nF_{n+1}$.

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