Leibniz's Rule in the Complex Domain

Question

Let $G$ be an open subset of $\mathbb{C}$ and $\gamma$ a rectifiable curve in $\mathbb{C}$. Suppose that $\varphi:\{\gamma\}\times G\to\mathbb{C}$ is continuous. Define $g:G\to\mathbb{C}$ by $$g(z)=\int_{\gamma}\varphi(w,z)\,dw$$ Then $g$ is continuous. Show that if $\frac{\partial \varphi}{\partial z}$ exists for each $(w,z)$ in $\{\gamma\}\times G$ and is continuous, then $g$ is analytic and $$g'(z)=\int_{\gamma}\frac{\partial \varphi}{\partial z}(w,z)\,dw$$

Answer 1

Probably the easiest is to use a Morera's theorem or one of its variants. That is show that for each sufficiently small triangle $T$ entirely contained (with its interior) in $G$, we have $$\int_{\partial T}g(z)\,dz=0\;.$$

We have $$\int_{\partial T}g(z)\,dz=\int_{\partial T}\int_\gamma\varphi(w,z)\,dw\,dz=\int_\gamma\int_{\partial T}\varphi(w,z)\,dz\,dw\;.$$ For each fixed $w$ the function $\varphi(w,z)$ is analytic in $z$, so the inner integral in the last double integral is zero, and $\int_{\partial T}g(z)\,dz=0$. This shows that $g(z)$ is analytic.

I realize this does not prove the formula for the derivative. For proving the formula in detail, see for example these two links:

• Rigorous Proof of Leibniz's Rule for Complex
• Analogue of Leibniz's Rule

Here is a different proof for the formula of $g'(z)$ that does not require low level estimations, but uses the notion of antiderivative for complex functions.

Proposition: Let $f(z)$ and $F(z)$ be continuous complex functions defined in $G$ (open subset of $\Bbb C$). $F$ is holomorphic in $G$ with $F'=f$ if and only if for any points $z_0,z_1\in G$ and any path $\eta$ in $G$ going from $z_0$ to $z_1$, $$\int_\eta f(\zeta)\,d\zeta=F(z_1)-F(z_0)\;.$$

Now define $$h(z)=\int_{\gamma}\frac{\partial \varphi}{\partial z}(w,z)\,dw\;.$$ This is a continuous function of $z$ and from the proposition above, to show $g'(z)=h(z)$ it is equivalent to show $$\int_\eta h(\zeta)\,d\zeta=g(z_1)-g(z_0)$$ for any $z_0,z_1\in G$ and any path $\eta$ in $G$ going from $z_0$ to $z_1$.

We have $$\begin{align*}\int_\eta h(\zeta)\,d\zeta & =\int_\eta\int_{\gamma}\frac{\partial \varphi}{\partial z}(w,\zeta)\,dw\,d\zeta \\ & =\int_{\gamma}\int_\eta\frac{\partial \varphi}{\partial z}(w,\zeta)\,d\zeta\,dw \\ & =\int_{\gamma}(\varphi(w,z_1)-\varphi(w,z_0))\,dw \\ & = g(z_1) - g(z_0) \end{align*} $$ where the second equality is changing the order of integration, and the third equality is applying the same above proposition to the function $\varphi(w,z)$ which is analytic in $z$ with derivative $\frac{\partial \varphi}{\partial z}(w,z)$. It's basically the antiderivative result for complex functions. That proves the desired formula for $g'(z)$.