Dominated convergence theorem and Cauchy's integral formula

Question

Let $U\subseteq \mathbb{C}$ be open and $\bar B(a,r) \subseteq U$. Let $\gamma(t) =a+ re^{it}$ with $t \in [0,1]$ be the boundary path of $B(a,r)$. By Cauchy's integral formula $f(w) = \frac{1}{2 \pi i}\int_{\gamma} \frac{f(z)}{(z-w)} dz$, where $w \in B(a,r)$.

I want to prove $\frac{d f(w)}{dw} = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-w)^2}dz$.

The usual argument is to interchange the order of differentiation and integration and this is justified by uniform convergence.

Is it possible to justify this interchange with the DCT?

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My attempt:

For DCT to apply, I need to check that $\frac{d}{dw}(\frac{f(\gamma(t))}{\gamma(t)-w}\gamma'(t)) = \frac{d}{dw}(\frac{f(re^{it})}{re^{it}-w}ire^{it}) = \frac{f(re^{it})}{(re^{it}-w)^2}ire^{it} $ is dominated by some function which is integrable over $[0,1]$. Because $f$ is continous over a compact it is bounded by some $M$ and therefore $\frac{M}{(r+ |w-a|)^2}$ should be the desired dominating function.

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For the bounty:

I am happy with the accepted answer. I would just like to know if my attempt is wrong and if it is necessary to consider the real and imaginary parts of $w$. Many thanks!

Answer 1

I think the only possible issue with the proposed solution in the OP is a careful proof that the proposed dominating function is a dominating function and a reference to the appropriate version of the DCT (or bounded convergence theorem, see below) that the OP would like to use.

There are at least two ways to approach this problem with the DCT. In each approach, we must identify which parameter serves as the parameter that we take a limit in, and identify an appropriate dominating function, prove it is a dominating function, and then quote the appropriate version of the DCT.

Note that the DCT is most commonly stated in terms of sequences of functions, so in any application of the "sequential" DCT to problems involving limits with a continuous parameter, we must use a characterization of limits in terms of sequences—see the second approach below. (Also see this old answer of mine regarding DCT with respect to continuous and discrete parameters for more on this.)

Now, we want to justify the equation: \begin{align*} \frac{\partial}{\partial w} \oint_{\partial B(a,r)}\frac{f(z)}{z-w}\,dz= \oint_{\partial B(a,r)}\frac{\partial }{\partial w}\frac{f(z)}{z-w}\,dz. \end{align*}

The first approach using real parameters:

We will use that $\frac{\partial}{\partial w} = \frac12\left(\frac\partial{\partial w_1}-i\frac\partial{\partial w_2}\right)$, and can show that \begin{align*} \frac{\partial}{\partial w_j} \oint_{\partial B(a,r)}\frac{f(z)}{z-w}\,dz= \oint_{\partial B(a,r)}\frac{\partial }{\partial w_j}\frac{f(z)}{z-w}\,dz\qquad(j=1,2). \end{align*} Because then by linearity and the definition of $\partial/\partial w$, we will have the equality we are after. With this approach, the parameters $w_j$ are the parameters we take limits in, and they are real parameters, which has the advantage that we can use the version of differentiating under the integral sign quoted here:

Differentating under the integral sign. Suppose that $F(x,t)$ is integrable as a function of $x \in \mathbb{R}^d$ for each value of $t \in \mathbb{R}$ and differentiable as a function of $t$ for each value of $x$. Assume also that $$\bigg| \frac{\partial}{\partial t} F(x,t) \bigg| \le G(x),$$ for all $x,t$, where $G(x)$ is an integrable function of $x$. Then $\frac{\partial}{\partial t} F(x,t)$ is integrable as a function of $x$ for each $t$ and $$\frac{d}{dt} \int F(x,t)\, dx = \int \frac{\partial}{\partial t} F(x,t)\,dx.$$

To prove this, you can mimic the second approach we will use to the problem in the involving the characterization of limits I mentioned (to prove this theorem I quoted above, the mean value theorem is also useful). To apply this, write out $\frac{f(z)}{z-w}$ as a function $F_j = F(t,w_j)$ where $t$ can be the parameter for $\partial B(a,r)$ for each $j = 1,2$ and apply this result to each of $F_1$ and $F_2$ separately.

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A second approach from first principles:

Without separating the integral into real and imaginary parts and quoting the theorem on differentiating under the integral that we are familiar with from real variables, we can choose to write the integral in a form that lets us apply the DCT for sequences of functions $([0,2\pi],\mathrm{Borel},dt)\to(\mathbb C,\mathrm{Borel})$ from first principles. We still would like to show \begin{align*} \frac{\partial}{\partial w} \oint_{\partial B(a,r)}\frac{f(z)}{z-w}\,dz= \oint_{\partial B(a,r)}\frac{\partial }{\partial w}\frac{f(z)}{z-w}\,dz. \end{align*} The DCT is stated for sequences of functions, so recall the following characterization of limits in a metric space: \begin{align*} \lim_{h\to a}g(h) = L \iff \text{for all sequences $h_j\to a$,}\ \lim_{j\to\infty}g(h_j) = L. \end{align*} (Cf. Rudin's Principles of Mathematical Analysis p. 84.) Thus, let $h_j\to 0$ be an arbitrary sequence of complex numbers and write the difference quotient corresponding to the left-hand side as (after skipping some algebra): \begin{align*} \int_0^{2\pi}\frac{f(re^{it})}{(re^{it}-w)^2-h_j(re^{it}-w)}ire^{it}\,dt. \end{align*} By continuity, $f$ is bounded by $M$ say on $\partial B(a,r)$. To bound the expression in the denominator, we use the reverse triangle inequality, \begin{align*} |(re^{it}-w)^2-h_j(re^{it}-w)| &\ge |re^{it}-w|\big(|re^{it}-w|-|h_j|\big). \end{align*} Because the distance $\delta$ from $w$ to the boundary of the disk is positive, we have $|re^{it}-w|\ge \delta > 0$ for all $t$, so if $j$ is so large that $|h_j|<\frac\delta2$, then the right-hand side of the last inequality is bounded below by $$ |re^{it}-w|\big(|re^{it}-w|-\frac\delta2\big)\ge \delta\big(\frac\delta2\big). $$ Hence we see that for $j\gg1$, $$ \bigg|\frac{f(re^{it})}{(re^{it}-w)^2-h_j(re^{it}-w)}ire^{it}\bigg| \le \frac{2M}{\delta^2}r, $$ which is bounded and hence belongs to $L^1([0,2\pi],dt)$. By the DCT (in fact, merely bounded convergence theorem will do here), \begin{align*} \lim_{j\to\infty}\int_0^{2\pi}\frac{f(re^{it})}{(re^{it}-w)^2-h_j(re^{it}-w)}ire^{it}\,dt &= \int_{0}^{2\pi}\lim_{j\to\infty}\frac{f(re^{it})}{(re^{it}-w)^2-h_j(re^{it}-w)}ire^{it}\,dt \\ &= \int_{0}^{2\pi}\frac{f(re^{it})}{(re^{it}-w)^2}ire^{it}\,dt\\ &= \oint_{\partial B(a,r)}\frac{f(z)}{(z-w)^2}\,dz\\ &= \oint_{\partial B(a,r)}\frac{\partial }{\partial w}\frac{f(z)}{z-w}\,dz. \end{align*} As the sequence $h_j\to 0$ we chose was arbitrary, we have the desired conclusion by the characterization of limits we stated.

Answer 2

The answer to your first question is yes, dominated convergence can be used to justified the change of order of integration and differentiation in Complex analysis; however, in the setting you described it is an overkill since you are integrating over piecewise differentiable maps, and there are Calculus theorems (see Tom Apostol's Analysis in the section about Riemann integration) that can handle the change of order of differentiation and integration in your setting.

In ant event, what you are attempting do, you are missing what the limiting process is: For a fixed $z_0$, one should try to dominate $\frac{1}{z-z_0}\Big(\frac{f(\gamma(t)}{\gamma(t)-z}-\frac{f(\gamma(t)}{\gamma(t)-z_0)}\Big)\gamma'(t)$ in a neighborhood of of $z_0$.

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In attempting to clarify the issues with your attempt, I will present some instances where Lebesgue integration is a great tool to solve Complex variable problems with three examples, which I believe, will illustrate that one may have to go beyond Calculus tools. To achieve that I consider only three. The first is slightly more than what you asked; the second one deals with what you asked; the third is to try to show how Lebesgue integration can be used to give an extension of Cauchy's theorem and estimates.

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Example 1: Let $\mu$ be a complex measure on a measurable space $(X,\mathscr{B})$ and let $D\subset\mathbb{C}$ be open. Suppose $\varphi$ is a bounded complex valued function in $D\times X$ such that $\varphi(\cdot,x)$ is holomorphic in $D$ for each $x\in X$, and that $\varphi(z,\cdot)$ is $\mathscr{B}$--measurable for each $z\in D$. Define $$ f(z):=\int_X \varphi(z,x)\mu(dx),\qquad z\in D. $$ Then $f$ is holomorphic in $D$.

There are several ways to go about this. Here is one that makes use of Cauchy's theorem along with dominated convergence.

The conditions of the statement above imply that for $z\in D$ fixed $$\partial_z\varphi(z,x)=\lim_{w\rightarrow z}\frac{\varphi(w,x)-\varphi(z,x)}{w-z}$$ is measurable in $x$. Let $M:=\sup_{(z,x)\in D\times X}|\varphi(z,t)|$. For $z_0\in D$, choose $r>0$ small enough so that the closed ball $\overline{B(z_0;2r)}\subset D$. For any $z\in B(z_0;r)$, $$ \begin{align} \frac{\varphi(z,x)-\varphi(z_0,x)}{z-z_0}&=\frac{1}{z-z_0}\frac{1}{2\pi i}\int_\gamma\Big(\frac{\varphi(\xi,x)}{\xi-z}-\frac{\varphi(\xi,x)}{\xi-z_0}\Big)\,d\xi\\ &=\frac{1}{2\pi i}\int_\gamma\frac{\varphi(\xi,x)}{(\xi-z)(\xi-z_0)}\,d\xi \end{align} $$

where $\gamma$ is the path $\gamma(t)=z_0+2re^{it}$, $0\leq t\leq 2\pi$. Then $$ \begin{align} \Big|\frac{\varphi(z,x)-\varphi(z_0,x)}{z-z_0}\Big|&=\frac{1}{2\pi}\left|\int^{2\pi}_0\frac{\varphi(z_0+2r e^{it},x)}{(z-z_0-2re^{it})2r e^{it}}i2r e^{it}\,dt\right|\leq \frac{M}{r} \end{align} $$ since $|z-z_0-2r e^{it}|\geq r$. Being a complex measure, $|\mu|(X)<\infty$ and so, we can apply dominated convergence to obtain first that $x\mapsto \partial_z\varphi(z_0,x)$ is integrable with respect to $\mu$ (with respect to $|\mu|$ rather, which in terns implies integrability with resect to positive and negative parts of the real and imaginary parts of $\mu$), and second that

$$ \begin{align} \lim_{z\rightarrow z_0}\frac{f(z)-f(z_0)}{z-z_0}&=\lim_{z\rightarrow z_0}\int_X \frac{\varphi(z,x)-\varphi(z_0,x)}{z-z_0}\,\mu(dx)\\ &=\int_X \lim_{z\rightarrow z_0}\frac{\varphi(z,x)-\varphi(z_0,x)}{z-z_0}\,\mu(dx)=\int_X\partial_z\varphi(z,x)\,\mu(dx) \end{align} $$

Comment:

• Dominated convergence for the complex measure $\mu$ can be understood in the sense that dominated convergence is being applied with respect to the positive and negative parts of the real and imaginary parts of $\mu$. Alternatively, one can consider the variation measure $|\mu|$ of $\mu$ and apply dominated convergence with respect to $|\mu|$; this will imply dominated convergence with respect t the positive and negative parts of the real and imaginary parts of $\mu$.

• The machinery from Lebesgue integration theory can be used along with classical complex analysis results to solve many problems in complex analysis. In Example 1 for instance we can furnish another solution, which is much more elegant, by apply Morera's theorem (integration over triangles) together with Fubini's theorem to obtain that $f$ is holomorphic in $D$. I leave the details to you.

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Example 2:. For simplicity, suppose $f$ is analytic in an open convex set $D$. Cauchy's theorem (in a convex set) states that for any closed path $\gamma$ in $D$

$$ f(z)\operatorname{Ind}_\gamma(z)=\frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{\xi -z}\,d\xi $$ where $z\in D\setminus\gamma^*$ ($\gamma^*$ is the range in $\mathbb{C}$ of the the path $\gamma$). All that can be proved without resorting to Lebesgue integration of course. In any event, suppose $\gamma$ is the path and $\gamma(x)= z_0+ Re^{ix}$, $0\leq x\leq 2\pi$. We can apply the result of Example 1 with $\varphi: B(0;R/2)\times[0,2\pi]\rightarrow\mathbb{C}$ given by $$ \varphi(z,x)=\frac{f(z_0+R e^{xi})}{\gamma(x)-z} $$ and $$ \mu(dx)=\frac{1}{2\pi i}\gamma'(x)\,dx $$ since $$|\phi(z,x)|\leq 2\frac{\sup_{w:|w-z_0|=R}|f(w)|}{R}=M<\infty$$ All this gives you $$ f'(z)=\frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{(\xi -z)^2}\,d\xi $$

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Example 3: Let $\mu$ be a complex measure on a measurable space $(\Omega,\mathscr{F})$ and let $\varphi$ be a complex--valued measurable function on $\Omega$. Suppose $D\subset\mathbb{C}$ is an open set which does not intersect $\varphi(\Omega)$.

Then, the map $f:D\rightarrow\mathbb{C}$ given by $$ \begin{align} f(z)= \int_\Omega\frac{\mu(d\omega)}{\varphi(\omega)-z}\tag{1}\label{one} \end{align} $$ is analytic. Moreover, if the closed ball $\overline{B}(a;r)\subset D$, then $$ \begin{align} f(z)=\sum^\infty_{n=0}c_n(z-a)^n,\qquad z\in B(a;r)\tag{2}\label{two} \end{align} $$ where $$ \begin{align} c_n=\int_\Omega\frac{\mu(d\omega)}{(\varphi(\omega)-a)^{n+1}},\qquad |c_n|\leq \frac{\|\mu\|_{TV}}{r^{n+1}},\qquad n\in\mathbb{Z}_+.\tag{3}\label{three} \end{align} $$ If $R$ is the radius of convergence of $\eqref{two}$, then $r\leq R$.

Here is a short proof to illustrate how dominated convergence may be used (see also notes at the bottom of this answer).

If $\overline{B}(a;r)\subset D$, then $q:=\inf_{\omega\in\Omega}|\varphi(\omega)-a|>r$, and so $$ \begin{align} \frac{|z-a|}{|\phi(\omega)-a|}\leq \frac{|z-a|}{q}\leq\frac{r}{q}<1,\qquad \omega\in\Omega,\quad z\in B(a;r). \end{align} $$ Hence, for any $z\in B(a;r)$ fixed, the series $$ \omega\mapsto \sum^\infty_{n=0}\frac{(z-a)^n}{(\varphi(\omega)-a)^{n+1}} =\frac{1}{\varphi(\omega)-z} $$ converges absolutely and uniformly in $\Omega$. By dominated convergence (to justify change of order of summation and integration) $$ f(z)=\int_\Omega \frac{\mu(d\omega)}{\varphi(\omega)-z}= \int_\Omega \sum^\infty_{n=0}\frac{(z-a)^n}{(\varphi(\omega)-a)^{n+1}} \,\mu(d\omega) = \sum^\infty_{n=0}c_n(z-a)^n, $$ where the $c_n$ satisfy $\eqref{three}$. The last statement follows from the estimate $$\begin{align} \limsup_{n\rightarrow\infty}\sqrt[n]{|c_n|}\leq\lim_{n\rightarrow\infty}\frac{1}{r}\sqrt[n]{\frac{\|\mu\|_{TV}}{r}}=\frac{1}{r} \end{align} $$

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Notes:

• Examples 1 and 2 cover the typical situation of integration along paths. In such cases you may consider the complex measure $\mu(dt)=\gamma'(t)\,dt$ where $\gamma:[a,b]\rightarrow\mathbb{C}$ is a piecewise differentiable function.

• Example 3 is reminiscent of Cauchy's estimates.