Dominated Convergence Theorem and Holomorphic Functions

Question

This is exercise 133Xc in Fremlin Volume 1:

Let $(X,\Sigma,\mu)$ be a measure space and let $G\subset\mathbb{C}$ be open. Let $f:X\times G\to\mathbb{C}$ be a function and suppose that the derivative $\frac{\partial f}{\partial z}$ of $f$ with respect to the second variable exists for all $x\in X$, $z\in G$. Suppose also that $(i)$ $F(z)=\int f(x,z) dx$ exists for every $z\in G$ and $(ii)$ there is an integrable function $g$ such that $\lvert \frac{\partial f}{\partial z} (x,z) \rvert \leq g(x)$ for every $x\in X$, $z\in G$. Show that the derivative $F'$ of $F$ exists everywhere in $G$, and $F'(z)=\int \frac{\partial f}{\partial z} (x,z) dx$ for every $z\in G$. (Hint: you will need to check that $\lvert f(x,z)-f(x,w) \rvert \leq \lvert z-w \rvert g(x)$ whenever $x\in X, z\in G$ and $w$ is close to $z$.)

My attempt:

Let $w\in G$ and let $(w_n)$ be a sequence in $G\setminus\{w\}$ converging to $w$. For each $n$ we have

$$\frac{F(w_n)-F(w)}{w_n-w}=\int \frac{f(x,w_n)-f(x,w)}{w_n-w} dx$$ Consider the sequence of functions $(f_n)$ defined by

$$f_n(x)=\frac{f(x,w_n)-f(x,w)}{w_n-w}$$ for all $n$ and all $x\in X$. Suppose I can show that $\lvert f_n(x)\rvert\leq g(x)$ for all $n$ and all $x\in X$. Then, since $f_n(x)\to \frac{\partial f}{\partial z}(x,w)$ as $n\to\infty$ for all $x\in X$, I can invoke the DCT (complex version) to deduce that

$$\lim_{n\to\infty}\frac{F(w_n)-F(w)}{w_n-w}= \int \frac{\partial f}{\partial z}(x,w) dx$$

Since $(w_n)$ is arbitrary, the conclusion follows.

My problem is how to show the inequality $\lvert f_n(x)\rvert\leq g(x)$. In the real case we can invoke the MVT to obtain this inequality but in the complex case I don't know how to proceed. It is supposed to be a basic exercise but somehow I don't have the right tool.

Thanks a lot for your help.

EDIT: I posted a solution attempt below based on DavidC.Ulrich's comments. But I have the problem that $\alpha_z$ is a function of $z$ and I can't prove that this function is differentiable.

Answer 1

"Advanced calculus" details, on request:

Lemma. Suppose $G\subset \Bbb C$ is open, $f:G\to\Bbb C$ is complex-differentiable, and $|f'(z)|\le c$ for all $z$. If the segment $[z,w]$ lies in $G$ then $|f(z)-f(w)|\le c|z-w|$.

Proof: Fix $z,w$ with $[z,w]\subset G$. Choose $\alpha\in\Bbb C$ with $|\alpha|=1$ and $$|f(z)-f(w)|=\alpha(f(z)-f(w).$$Define $$u=\Re(\alpha f)$$and define $\phi:[0,1]\to\Bbb R$ by $$\phi(t)=u(tw+(1-t)z).$$Then MVT shows there exists $s\in[0,1]$ with $$\begin{aligned}|f(w)-f(z)|&=u(1)-u(0)=u'(s)\\&=\Re\alpha(w-z)f'(sw+(1-s)z)\\&\le|(w-z)f'(sw+(1-s)z)|\le c|z-w|.\end{aligned}$$

Answer 2

Following DavidC.Ulrich's advice here is my solution.

Since $w\in G$ and $G\subset\mathbb{C}$ is open, there exist $r>0$ such that $B(w,r)\subset G$, where $B(w,r)$ denotes the open ball of radius $r$ centered at $w$. Since $w_n\to w$, we may assume WLOG that $(w_n)\subset B(w,r)$, for otherwise we can start the sequence at a larger $n$ without affecting the limit.

Fix $x\in X$ and $n$. For each $z \in G$ there exist a complex number $\alpha_z$ such that $$\alpha_z f(x,z)=\lvert f(x,z) \rvert $$

Indeed $\alpha_z$ can be taken to be $e^{-i\theta}$, where $\theta$ is an argument of the complex number $f(x,z)$. Note that the RHS of the equality shows that $\alpha_z f(x,z)$ is real for all $z \in G$.

Because the line $\{(1-t)w_n+tw : t\in[0,1]\}$ lies in $G$, we can define the function $h:[0,1] \to \mathbb{R}$ by $$h(t)=\alpha_z f(x,(1-t)w_n+tw)$$

Moreover, because $\frac{\partial f(x,z)}{\partial z}$ exists for all $z \in G$ and $\frac{\partial (1-t)w_n+tw }{\partial t} =w-w_n$ for all $t\in [0,1]$, the chain rule implies that $h$ is differentiable on $[0,1]$ with

$$h'(t)=\alpha_z \frac{\partial f(x,(1-t)w_n+tw)}{\partial z} (w-w_n)$$

Now, we can apply the MVT to $h$ to obtain obtain $c \in (0,1)$ such that

$$ h(1)-h(0)=h'(c)$$

Translating back in terms of $f$ we get

$$ \alpha_z f(x,w)-\alpha_z f(x,w_n) = \alpha_z\frac{\partial f(x,(1-c)w_n+cw)}{\partial z} (w-w_n) $$

Taking absolute values on both sides, cancelling out $\lvert \alpha_z \rvert$ and dividing by $\lvert w-w_n \rvert$ we get

$$ \lvert f_n(x) \rvert = \Bigg \lvert \frac{ f(x,w)- f(x,w_n) }{ w-w_n } \Bigg \rvert = \Bigg \lvert \frac{\partial f(x,(1-c)w_n+cw)}{\partial z} \Bigg \rvert \leq g(x)$$

Since $x\in X$ and $n$ were arbitrary we are done.

PROBLEM : I DIDN'T TAKE INTO ACCOUNT THE FACT THAT $\alpha_z$ IS A FUNCTION of $z$!