# Rigorous Proof of Leibniz's Rule for Complex

Let $$f(t,z):[a,b] \times D \rightarrow \mathbb{C}$$, $$D \subseteq \mathbb{C}$$ open, a continuous function analytic in $$D$$ for all $$t \in [a,b]$$. Also, $$\frac{\partial f }{\partial z} (t,z) : [a,b] \times D \rightarrow \mathbb{C}$$ is continuous. Then $$g(z) = \int_a^b f(t,z) \, dt$$ is analytic on $$D$$ with $$g'(z) = \int_a^b \frac{\partial f}{\partial z} (t,z) \, dt.$$ (P97, Complex Analysis, Freitag)

The author did not provide a proof, and said it is "direct" from the real case. I am really interested in how one writes this out completely. (probably in a much more neater way than mine's).

My attempt: (Correct?)

1. $$g:D \rightarrow \mathbb{C}$$ is analytic in $$D$$ iff $$g$$ is frechet differentiable at $$z$$ (in the reals) and $$g$$ satisfies the Cauchy Riemann equations, i.e, $$\frac{\partial g }{ \partial x} = \frac{1}{i} \frac{\partial g} {\partial y}$$. It is sufficient if we show the partial derivatives are continuous for frechet differnetiability.

2. Let $$f(t,z) = u(t,z) + iv(t,z)$$, then, $$g(z) = \int_a^b u(t,z) \, dt + i \int_a^b v(t,z) \, dt$$ and it suffices to show $$\frac{\partial}{\partial x} \int_a^b u(t,z) \, dt = \frac{\partial}{\partial y} \int_a^b v(t,z) \, dt , \quad \frac{\partial}{\partial y} \int_a^b u(t,z) \,dt = - \frac{\partial }{\partial x} \int_a^b v(t,z) \, dt$$

3. Fix $$z_0 \in D$$ and consider $$i: [a,b] \times [c,d] \rightarrow [a,b] \times D$$ given by $$(t,x) \mapsto (t,z_0+x)$$ which defines continuous function $$U = u \circ i :[a,b] \times [c,d] \rightarrow \mathbb{C}.$$ We may wlog assume $$[c,d] = [-1,1]$$ as $$D$$ is open. We have $$\frac{\partial }{\partial x} U (t,0) = \frac{\partial}{\partial x}u (t,z_0),$$ and $$\frac{\partial}{\partial x} U (t,s) = \mathfrak{R} \frac{\partial f}{\partial z} (t,z_0+s)$$ is continuous on $$[a,b] \times [-1,1]$$ as $$\frac{\partial f}{\partial z}$$ exists and is continuous on $$[a,b] \times D$$. So $$\frac{\partial U}{\partial x}$$ is bounded by some constant $$C$$ on $$[a,b] \times [-1,1]$$.

4. Set $$g_n (t) := \frac{U(t,x_n) - U(t,0)}{x_n }$$ for any real seqence $$x_n \rightarrow 0$$. Existence of partial derivative implies $$\lim_{n \rightarrow \infty} g_n(t) = \frac{\partial}{ \partial x} U(t,0).$$ $$g_n:[a,b] \rightarrow \mathbb{C}$$ is a continuous function; by MVT, exists some $$x'$$ between $$x_n$$ and $$0$$ such that $$|g_n(t)| \le \Big| \frac{\partial}{\partial x} U(t,x') \Big| \le C \chi_{[a,b]}$$

5. Applying DCT to the sequence of measurable functions $$\{g_n(t)\}$$, \begin{align*} \lim_{n \rightarrow \infty} \int_{[a,b]} \frac{U(t,x_n) -U(t,0)}{x_n} \, dt & = \lim_{n \rightarrow \infty} \int_{[a,b]} g_n(t) \, dt \\ & = \int_{[a,b]} \lim_{n \rightarrow \infty} g_n(t) \, dt \\ & = \int_{[a,b]} \frac{\partial }{\partial x} U(t,0) \, dt \end{align*} As this holds for any sequence $$x_n \rightarrow 0$$, the function $$G(x):= \int_{[a,b]} \frac{U(t,x) - U(t,0)}{x}\, dt$$, is sequentially convergent at $$0$$, hence continuous at $$0$$, with $$\lim_{x \rightarrow 0} G(x) = \int_{[a,b]} \frac{\partial }{\partial x} U(t,0) \, dt$$

6. By definition, the above equality is equivalent to, $$\frac{ \partial }{\partial x} \Big| _{z_0} \int_a^b u(t,z) \, dt = \int_a^b \frac{\partial }{\partial x} \Big| _{z_0} u(t,z) \,dt .$$ We obtain, by the same argument, an equation for the partials with respect to $$y$$. As $$f$$ is holomorphic, we have $$\frac{\partial u}{\partial x}(t,z_0) = \frac{\partial v}{\partial y}(t,z_0) .$$ As the choice $$z_0 \in D$$ was arbitrary, Cauchy Riemann Equations are satisfied on whole of $$D$$, and $$g$$ is holomorphic on whole of $$D$$.

Fix a point $z_0\in D$. We have to prove that $$\lim_{h\to0}{g(z_0+h)-g(z_0)\over h}=\int_a^b f_z(t,z_0)\>dt\ .\tag{1}$$ By assumption there is a closed disc $B$ of radius $r_0>0$ around $z_0$ such that $f_z$ is continuous, hence uniformly continuous, on the set $R:=[a,b]\times B$. For complex $h$ with $0<|h|<r_0$ we then have $$f(t,z_0+h)-f(t,z_0)=\int_{z_0}^{z_0+h}f_z(t,z)\>dz=h\int_0^1 f_z(t,z_0+\tau h)\>d\tau$$ and therefore $$\Phi(t,h):={f(t,z_0+h)-f(t,z_0)\over h}-f_z(t,z_0)=\int_0^1 \bigl(f_z(t,z_0+\tau h)-f_z(t,z_0)\bigr)\>d\tau\ .$$ The uniform continuity of $f_z$ on $R$ now allows to conclude the following: Given an $\epsilon>0$ there is a $\delta>0$ such that $$\bigl|\Phi(t,h)\bigr|<\epsilon\qquad\bigl(a\leq t\leq b, \ 0< |h|<\delta)\ .$$ This means that $\lim_{h\to0}\Phi(t,h)=0$ uniformly in $t$. Since $${g(z_0+h)-g(z_0)\over h}-\int_a^b f_z(t,z_0)\>dt=\int_a^b \Phi(t,h)\>dt$$ we may infer $(1)$.
• Maybe this is obvious, but why does $f_z$ being continuous on $B$ mean that it's uniformly continuous on $R$? Mar 9, 2018 at 22:57
• @Arbutus because $B$ is closed and bounded it is compact and so is $[a, b] \times B$. A continuous function on a compact set is uniformly continuous. Jul 15, 2018 at 18:37
• I don't see how the uniform continuity of $f_z$ implies that we can make $|\Phi(t,h)|$ small. Can you please clarify this? Thanks in advance. Sep 14, 2019 at 23:18
• @EpsilonDelta: Because $|f_z(t,z_0+\tau h)-f_z(t,z_0)|$ is small when $h$ is small. Sep 15, 2019 at 8:45