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?)
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$g:D \rightarrow \mathbb{C}$ 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.
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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
$$ \frac{\partial}{\partial x} \int_a^b u(t,z) \, dt = \frac{\partial}{\partial y} \int_a^b v(t,z) \, dt , \quad \frac{\partial}{\partial y} \int_a^b u(t,z) \,dt = – \frac{\partial }{\partial x} \int_a^b v(t,z) \, dt $$ -
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 $[a,b] \times [-1,1]$.
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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 $$ |g_n(t)| \le \Big| \frac{\partial}{\partial x} U(t,x') \Big| \le C \chi_{[a,b]}$$
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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
$$ \lim_{x \rightarrow 0} G(x) = \int_{[a,b]} \frac{\partial }{\partial x} U(t,0) \, dt $$ -
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 $D$.
Best Answer
The "real case" was maybe proven using the MVT, and we don't have that in the complex environment. We therefore have to start afresh.
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 $(1)$.