Prove that $F(z)=\int_Xf(x,z)d\mu(x) $ is holomorphic

Question

I'm trying to figure out this exercise:

Let $(X,\mathbb{E},\mu)$ be a measureable space and let $G\subseteq\mathbb{C}$ be open. Assume that $f:X\times G\to\mathbb{C}$ satisfies:

(1): $\forall x\in X: f(x,\cdot)\in H(G)$

(2): $\forall z\in G: f(\cdot, z)$ is measurable on $X$.

(3): There exists a measurable function $g:X\to [0,\infty]$ satisfying $\int gd\mu<\infty$ such that $$|f(x,z)|\leq g(x)$$ for alle $x\in X,z\in G$.

Prove that $$F(z)=\int_Xf(x,z)d\mu(x) \quad z\in G$$ is holomorphic in $G$ and that $$F'(z)=\int_X \frac{\partial f}{\partial z}(x,z)d\mu(x) \quad z\in G$$ So my idea was to use Cauchy's integral formula for the n'th derivative we then get:

$$\frac{F(z_0+h)-F(z_0)}{h}=\int_{X}\frac{1}{h}(f(x,z_0+h)-f(x,z_0))d\mu(x)=\int_{X} \left(\int_{0}^{1}\frac{\partial f}{\partial z}(x,z_0+th)dt\right)d\mu(x)\\ =\int_{0}^{1}\left(\int_{X}\frac{\partial f}{\partial z}(x,z_0+th)d\mu(x)\right)dt$$

I have shown the second equality in a previous exercise, but I'm unsure if I'm the third holds(interchange the integrals?). And also am I allowed to take the limit for $h\to 0$ to conclude the stated?

Answer 1

In your attempt, I'm not sure how you used Cauchy's integral formula for derivatives, but yes, the first two inequalities are correct. The problem is that we only have an estimate on $|f(x,z)|, $ not $|\partial f(x,z)/\partial z|.$ The following lemma will take care of that problem.

Lemma: Suppose $f\in H(D(a,2r))$ and $|f|\le M$ in this disc. Then

1. $|f'(z)| \le \dfrac{M}{r}$ for $z\in D(a,r),$ and
2. $ |f(z) - f(w)|\le \dfrac{M}{r}|z-w|$ for $z,w\in D(a,r).$

Proof: I'll be brief; hopefully this is familiar to you. 1. follows from Cauchy's estimates, and 2. follows from 1.

Fix $z_0\in G.$ Choose $r>0$ such that $D(z_0,2r) \subset G.$ For $x\in X,$ we have $|f(x,z)|\le g(x)$ for all $z\in D(z_0,2r).$ So by 2. of the lemma,

$$ \left |\frac{f(x,z_0+h) - f(x,z_0)}{h}\right| \le \frac{g(x)}{r}\,\, \text { for }0<|h|<r\,\text { and } x\in X.$$

So now let's look at

$$\frac{F(z_0+h) - F(z_0)}{h} = \int_X \frac{f(x,z_0+h) - f(x,z_0)}{h}\, d\mu(x)$$

as $h\to 0.$ The difference quotients on the right are measurable and converge pointwise on $X$ to $\partial f (x,z_0)/\partial z.$ Moreover they are bounded on $X$ in absolute value by $g(x)/r$ for $0<|h|<r.$ Since $\int_x g(x)/r\,d\mu(x) < \infty,$ the dominated convergence theorem gives the desired result:

$$F'(z_0) = \int_X \frac{\partial f (x,z_0)}{\partial z}\,d\mu(x).$$

Answer 2

W.l.o.g. we may assume that $G$ is simply connected (or just it is a disk).

Let $u(x,z)=\frac{\partial f}{\partial z}(x,z)$ and $v(x,z)=\frac{\partial^2 u}{\partial z^2}(x,z)$, and define $$ U(z) = \int_X u(x,z) d\mu(x), \qquad V(z) = \int_X v(x,z) d\mu(x). $$ The function $u(x,z)=\lim\frac{f(x,z+1/n)-f(x,z)}{1/n}$, being limit of measurable functions, is $\mu$-measurabe for every fixed $z$. By Cauchy's estimates, if $B(z,r)\subset G$ then $\big|u(x,z)\big|\le\frac{g(x)}{r}$, so $U(z)$ is well-defined. Similary, $v$ is $\mu$-measurable and $\big|v(x,z)\big|<\frac{2g(x)}{r^2}$, so $V$ is well-defined.

We show that $F$ is an integral function of $U$, and $U$ is an integral function of $V$; that is, for any line segment $[a,b]\subset G$, $$ \int_{[a,b]} U(z) dz = F(b)-F(a) \quad\text{and}\quad \int_{[a,b]} V(z) dz = U(b)-U(a). \tag{*} $$ This imply that $U$ is continuous, and then $F'=U$.

It suffices to prove one part from (*), say $\int_{[a,b]} V(z) dz = U(b)-U(a)$. Suppose that the integral does not exist or is not equal to $U(b)-U(a)$. Then there exists some $\varepsilon>0$ such that for every $n$, the segment $[a,b]$ can be subdivided into some segments $[z^{n}_0,z^{n}_1],\ldots,[z^{n}_{N_n-1},z^{n}_{N_n}]$ with $|z^{n}_k-z^{n}_{k-1}|<\frac1n$, and some sample points $\xi^{n}_k\in[z_{k-1},z_k]$ can be selected such that $$ \left|\sum_{k=1}^{N_n} V(\xi^{n}_k)(z^{n}_k-z^{n}_{k-1}) -(U(b)-U(a))\right| \ge\varepsilon . \tag{**}$$

For every fixed $x\in X$ we have $$ \lim_{n\to\infty}\sum_{k=1}^{N_n} v(x,\xi^{n}_k)(z^{n}_k-z^{n}_{k-1}) = \int_{[a,b]} v(x,z) \mathrm{d}z = u(b)-u(a). $$

Let $r$ be the distance between the segment $[a,b]$ and the complement of $G$. Then $$ \left|\sum_{k=1}^{N_n} v(x,\xi^{n}_k)(z^{n}_k-z^{n}_{k-1})\right| \le \frac{2g(x)}{r^2} \cdot |b-a| $$ So, by the dominated convergence theorem, $$ \lim_{n\to\infty} \sum_{k=1}^{N_n} V(\xi^{n}_k)(z^{n}_k-z^{n}_{k-1}) = \lim_{n\to\infty} \sum_{k=1}^{N_n} \int_X v(x,\xi^{n}_k)(z^{n}_k-z^{n}_{k-1})d\mu(x) =\\= \int_X \lim_{n\to\infty} \sum_{k=1}^{N_n} v(x,\xi^{n}_k)(z^{n}_k-z^{n}_{k-1})d\mu(x) = \int_X (u(x,b)-u(x,a)) d\mu(x) = U(b)-U(a). $$ That contradicts (**), done.

$F'=U$ shows that the function $F(z)=\int_X f(x,z)d\mu(x)$ is holomorphic and $F'(z)=\int_X u(x,z)d\mu(x)=\int_X \frac{\partial f(x,z)}{\partial z}d\mu(x)$.