Why is $\int_{[0,1]} \frac{dw}{1-wz}$ is holomorphic in unit disc?

Question

Expanding on the question and answer in: Prove $f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$ is holomorphic in the open unit disk.

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka

Pf: Let $\gamma \subset D[0,1]$ be piecewise smooth and closed. Supposing f is continuous on $D[0,1]$. and we can switch integrals, we have:

$$\int_{\gamma} f(z) dz := \int_{\gamma} \int_{[0,1]} \frac{dw}{1-wz} dz$$

$$= \int_{[0,1]} \int_{\gamma} \frac{dz}{1-wz} dw \tag{*}$$

Case 1: If $w\ne 0 (**)$, then $$w = Re(w) \in [0,1] \implies \frac1w \in [1,\infty),$$ 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

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1. Where have I gone wrong?

2. To show $f$ is continuous on $D[0,1]$, I will argue that $f$ is an antiderivative of some function on $D[0,1]$. By definition, antiderivatives are holomorphic on the regions they're defined on and thus differentiable on said region and thus continuous on said region.

Define $g_z: D[0,1] \to \mathbb C, z \in D[0,1]$ s.t. $g_z(t) := \frac{1}{1-tz}$. Observe $g_z$ is continuous on $D[0,1]$ and that $\int_{\gamma} g_z(t) dt = 0, \forall \gamma \subset D[0,1]$, closed and piecewise smooth is actually what we argued earlier (so, I'm not using Cor 4.13, which would be circular!). Let $w_0 \in D[0,1].$ By Thm 4.15, the function $F_{\gamma_w}: D[0,1] \to \mathbb C$ where $\gamma_w$ is any piecewise smooth path from $w_0$ to $w$ defined by $F(w) := \int_{\gamma_w} g_z(t) dt$ is an antiderivative for g on $D[0,1]$.

Now I'm stuck. I guess I can't pick $\gamma_w = [0,1]$ to turn $F(w)$ into $f(z)$. How could I adjust this approach? I have a feeling there's some change of variable I'm missing or that I made a mistake with variables. I asked about this in another question: Find radius of $\mathbb C$-power series

3. Alternatively to show $f$ is continuous, is there some way to justify switching limit and integral as follows $$\lim_{z \to z_0} \int_{[0,1]} \frac{dw}{1-wz} = \int_{[0,1]} \lim_{z \to z_0} \frac{dw}{1-wz}$$

I think this limit switching is precisely equivalent to $f$'s continuity in $D[0,1]$.

?

4. (*) How do we justify switching the two integrals please?

5. (**) What do we do about $w=0$? I have a feeling we can exclude like how $\int_{[2,3]} 3x dx = \int_{(2,3]} 3x dx$. Update: Added answer for $w=0$ based on nextpuzzle's answer.

6. Another route for continuity of $f$: Closed form of $\int_{[0,1]} \frac{dw}{1-wz}$ involving Ln?

P.S. Starting Ch8, I guess we can say that $f$ is holomorphic because it is analytic: Find a power series for $f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$.

Answer 1

You can interchange the integrals due to Fubini's theorem. Since the integrand is continuous on $[0,1]$ and the image of $\gamma$, the integral of its absolute value is finite.

For $w=0$, you just don't do (A). Keep it as $$\int_{[0,1]}\int_\gamma\frac{dz}{1}$$ As before the integral $\int_\gamma dz=0$ by Cauchy's theorem, or if you like by using that $\frac11$ has $z$ as anti-derivative. The step (A) wasn't really necessary. You can argue that the inner integral in (*) is zero by Cauchy's theorem, without taking the $w$ out.

In (2), if you are going to show that $f$ has a derivative, then why do anything else? You can prove continuity directly. The function $g(w,z)=\frac{1}{1-wz}$ is continuous for $w\in[0,1]$ and $z$ in the open unit disc, and therefore $g$ is uniformly continuous on $[0,1]\times K$ for $K$ a compact subset of the unit disc. Therefore, integrating with respect to $w$ in $[0,1]$ results in a continuous function on $K$.

Answer 2

No need to justify switching integrals. Just obtain a closed form of $f(z)$, which allows to prove continuity and that $\int_{\gamma} f =0$:

$$f(z) := \int_{[0,1]} \frac{dw}{1-wz} = -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} + 1_{z=0},$$

1. $f(z)$ is continuous on $D[0,1]$ because $$\lim_{z \to 0}-\frac1z\operatorname{Ln}(1-z) = 1 = f(0)$$

2. $$\int_{\gamma} f(z) dz = \int_{\gamma} -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} + 1_{z=0} dz$$

$$ = \int_{\gamma} -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} dz + \int_{\gamma} 1_{z=0} dz = 0$$

Now all conditions of Morera's Thm 5.6 are satisfied without (explicit) reference to antiderivatives or Fubini's theorem, measure spaces, uniform continuity, etc