Prove a function defined by integral is holomorphic

hbghlyj

courses archive Complex Analysis 2017
Theorem 15.27. Let $U$ be an open subset of $\mathbb{C}$ and suppose that $F: U \times[a, b]$ is a function satisfying
(1) The function $z \mapsto F(z, s)$ is holomorphic in $z$ for each $s \in[a, b]$.
(2) $F$ is continuous on $U \times[a, b]$
Then the function $f: U \rightarrow \mathbb{C}$ defined by
$$
f(z)=\int_a^b F(z, s) d s
$$
is holomorphic.
Proof. Changing variables we may assume that $[a, b]=[0,1]$ (explicitly, one replaces $s$ by $(s-a) /(b-a))$. By Theorem 15.25 it is enough to show that we may find a sequence of holomorphic functions $f_n(z)$ which converge of $f(z)$ uniformly on compact subsets of $U$. To find such a sequence, recall from Prelims Analysis that the Riemann integral of a continuous function is equal to the limit of its Riemann sums as the mesh of the partition used for the sum tends to zero. Using the partition $x_i=i / n$ for $0 \leq i \leq n$ evaluating at the right-most end-point of each interval, we see that
$$
f_n(z)=\frac{1}{n} \sum_{i=1}^n F(z, i / n),
$$
is a Riemann sum for the integral $\int_0^1 F(z, s) d s$, hence as $n \rightarrow \infty$ we have $f_n(z) \rightarrow$ $f(z)$ for each $z \in U$, i.e. the sequence $\left(f_n\right)$ converges pointwise to $f$ on all of $U$. To complete the proof of the theorem it thus suffices to check that $f_n \rightarrow f$ as $n \rightarrow \infty$ uniformly on compact subsets of $U$. But if $K \subseteq U$ is compact, then since $F$ is clearly continuous on the compact set $K \times[0,1]$, it is uniformly continuous there, hence, given any $\epsilon>0$, there is a $\delta>0$ such that $|F(z, s)-F(z, t)|<\epsilon$ for all $z \in \bar{B}(a, \rho)$ and $s, t \in[0,1]$ with $|s-t|<\delta$. But then if $n>\delta^{-1}$ we have for all $z \in K$
\begin{aligned}
\left|f(z)-f_n(z)\right| & =\left|\int_0^1 F(z, s) d z-\frac{1}{n} \sum_{i=1}^n F(z, i / n)\right| \\
& =\left|\sum_{i=1}^n \int_{(i-1) / n}^{i / n}(F(z, s)-F(z, i / n)) d s\right| \\
& \leq \sum_{i=1}^n \int_{(i-1) / n}^{i / n}|F(z, s)-F(z, i / n)| d s \\
& <\sum_{i=1}^n \epsilon / n=\epsilon
\end{aligned}Thus $f_n(z)$ tends to $f(z)$ uniformly on $K$ as required.  $_\blacksquare$

Example 15.28. If $f$ is any continuous function on $[0,1]$, then the previous theorem shows that the function $f(z)=\int_0^1 e^{i s z} f(s) d s$ is holomorphic in $z$, since clearly $F(z, s)=e^{i s z} f(z)$ is continuous as a function on $\mathbb{C} \times[0,1]$ and, for fixed $s \in[0,1]$, $F$ is holomorphic as a function of $z$. Integrals of this nature (though perhaps over the whole real line or the positive real axis) arise frequently in many parts of mathematics, as you can learn more about in the optional course on Integral Transforms.

hbghlyj

Remark 15.29. Another way to prove the theorem is to use Morera's theorem directly: if $\gamma:[0,1] \rightarrow \mathbb{C}$ is a closed path in $B(a, r)$, then we have
\begin{aligned}
\int_\gamma f(z) d z & =\int_\gamma\left(\int_0^1 F(z, s) d s\right) d z \\
& =\int_0^1\left(\int_\gamma F(z, s) d z\right) d s=0
\end{aligned}where in the first line we interchanged the order of integration, and in the second we used the fact that $F(z, s)$ is holomorphic in $z$ and Cauchy's theorem for a disk. To make this completely rigorous however, one has to justify the interchange of the orders of integration. Next term's course on Integration proves a very general result of this form known as Fubini's theorem, but for continuous functions on compact subsets of $\mathbb{R}^n$ one can give more elementary arguments by showing any such function is a uniform limit of linear combinations of indicator functions of "boxes" - the higher dimensional analogues of step functions - and the elementary fact that the interchange of the order of integration for indicator functions of boxes holds trivially.

hbghlyj

Prove $f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$ is holomorphic in the open unit disk.
Proof (Using method in 2#) MSE
Use Morera's theorem. First prove that $f$ is continuous.Then, if $\gamma$ is any closed $C^1$ curve in the unit disk, we have
$$
\int_\gamma f(z)\,dz=\int_\gamma\Bigl(\int_{[0,1]}\frac{dw}{1-w\,z}\Bigr)\,dz=\int_{[0,1]}\Bigl(\int_\gamma\frac{dz}{1-w\,z}\Bigr)\,dw=0,
$$
since $1/(1-w\,z)$ is holomorphic on the unit disk as a function of $z$.
In details, see MSE
Case 1: If $w\ne 0$ (**), then $$w \in (0,1] \implies \frac1w \in [1,\infty),$$ so $\frac1w$ is not a singularity of the inner integral below:
$$= \int_{[0,1]} \frac{-1}{w} \int_{\gamma} \frac{dz}{z-\frac{1}{w}} dw \tag{A}$$
Hence $\frac{1}{z-\frac{1}{w}}$ is holomorphic in $D[0,1]$. By Cor 4.20 or Cauchy's Integral Formula 4.27, $$\int_{\gamma} \frac{dz}{z-\frac{1}{w}} dw=0 \to (A) = \int_{[0,1]} \frac{-1}{w} (0) dw = 0$$
$\therefore$ by Morera's Thm 5.6, f is holomorphic in $D[0,1]$.

Case 2: If $w=0$, then we have
$$\int_{[0,1]} \int_{\gamma} \frac{dz}{1-(0)z} dw$$
This is justified because $w$ is fixed in the inner integral above. Now we have
$$ = \int_{[0,1]} \int_{\gamma} \frac{dz}{1} dw$$
Now $\frac11$ is entire, so by Cor 4.20 or Cauchy's Integral Formula 4.27
$$ = \int_{[0,1]} 0 dw= 0$$
$\therefore$, by Morera's Thm 5.6, f is holomorphic in $D[0,1]$.
QED

hbghlyj

Also see complex.pdf Theorem 13.1.