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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

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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


  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

  1. 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 $D[0,1]$.

?

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

  2. (**) 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.

  3. 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 $f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$.

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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 $K$.

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  • $\begingroup$ Thanks nextpuzzle. About Fubini's, what exactly is the elementary complex analysis statement please? I learned the general Fubini's in Stochastic Calculus. It wasn't discussed, proved or even stated in my stochastic calculus classes or advanced probability class, but I skimmed the chapter in Probability with Martingales. Sooo, to someone who knows elementary real analysis and doesn't know measure theory or even advanced real analysis, what is 'Fubini's theorem'? $\endgroup$ – BCLC Jul 23 '18 at 10:14
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    $\begingroup$ @BCLC In the statement in the link, forget about the part that says $\sigma$-finite measure. The integration that you are doing is over such a measure. That just means that the real line can be covered by countable many sets of finite length. For example, $[n,n+1]$. The other hypothesis is that one of the three three integrals listed is finite. In each, you are integrating a continuous function over a compact set. Therefore, any of them is finite. $\endgroup$ – user578878 Jul 23 '18 at 10:20
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    $\begingroup$ @BCLC You are trying to prove that $f$ is holomorphic by using Morera. Holomorphic on the unit disc means having complex derivative on in the unit disc. If you are going to show that $f$ has complex derivative, then that is the whole point. You either do it directly or use Morera or other method to do it indirectly. $\endgroup$ – user578878 Jul 23 '18 at 10:28
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    $\begingroup$ @BCLC You should read and think before "what?". You are the one that is in the process of learning this. First understand what you are being told before all that show. I said, you don't need to prove that $f$ is an antiderivate directly. That will be a consequence of Morera. If you were to prove that it is an antiderivative directly you would be done, and wouldn't need to use Morera. You do need to prove that it is continuous, a much weaker condition. I gave an argument above. $\endgroup$ – user578878 Jul 23 '18 at 10:57
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    $\begingroup$ @BCLC No, that is not what I said either. You can indeed prove that $f$ is an antiderivative. It will involve a bit of computation. If you do, you are done with the problem. The point of using Morera is maybe obtaining a proof without getting your hands dirty in calculus. $\endgroup$ – user578878 Jul 23 '18 at 11:36
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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

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