Holomorphic Functions are Analytic

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Holomorphic Functions are Analytic

02 Jul 2024

Portrait of Weierstrass

This is our sixth post in the series with my notes on complex integration, corresponding to Chapter 4 in Ahlfors’ Complex Analysis.

In this post however, we won’t follow Ahlfors’ book, but rather Terry Tao’s notes on Complex Analysis [2] to show that holomorphic functions are analytic.

The previous posts from the series:

  1. Complex Integration
  2. Path-Independent Line Integrals
  3. Cauchy Integral Theorem
  4. The Winding Number
  5. Cauchy’s Integral Formula

Analytic Functions

An analytic function is a function that can be locally represented by a convergent power series. Definition 1 provides a more formal definition:

Definition 1. Let $f: \Omega \rightarrow \mathbb{C}$ be a function. We say $f$ is analytic in $a$ if there’s an open disk $D$ centered in $a$ such that for $z \in D$:

\[(1) \quad f(z) = \sum_{n = 0}^{\infty} {c_n} (z - a)^n\]

where $c_n \in \mathbb{C}$. For $f(z)$ to be defined, $(1)$ must converge.

Locally meaning

Definition 2. Let $r$ be the radius of disk $D$, i.e. $D = \curly{x \in \Omega \mid \abs{z - a} < r}$. If a power series such as $(1)$ converges when $\abs{z - a} < R$ and diverges when $\abs{z - a} > R$, we call $R$ the radius of convergence.

Lemma 3 provides an explicit formula for the radius of convergence of $(1)$:

Lemma 3. Let $f(z)$ be a power series as in $(1)$. Then the radius of convergence is:

\[(3.1) \quad R = \liminf_{n \rightarrow \infty} \abs{c_n}^{-1/n}\]

assuming that $\abs{c_n}^{-1/n} = \infty$ if $c_n = 0$.

Assume $\abs{z - a} \gt R$. By $(3.1)$, we have $$\lim_{r \rightarrow \infty} \paren{\inf_{n \ge r} \abs{c_n}^{-1/n}} = R$$ So for all $n \ge r$, $\abs{c_n}^{-1/n} \le R \lt \abs{z - a}$, or that $$\abs{c_n}^{-1/n} \lt \abs{z - a}$$ Multiplying by $\abs{c_n}^{1/n}$: $$1 \lt \abs{c_n}^{1/n} \abs{z - a}$$ Since these are all positive terms, we can raise to the $n$-th power: $$1 \lt \abs{c_n} \abs{z - a}^n = \abs{c_n (z - a)^n}$$ Which means as $r$ tends to infinity, $\abs{c_n (z - a)^n}$ is greater than 0, so if $\abs{z - a} \gt R$, then the power series $(1)$ does not converge.

Now suppose $\abs{z - a} \lt R$. Choose $\abs{z - a} \lt R' \lt R$. Then we claim that as $r \rightarrow \infty$, $\abs{c_n}^{-1/n} \gt R'$. Otherwise, we'd have $\abs{c_n}^{-1/n} \le R' \lt R$ which contradicts the hypothesis that $R$ is a lower bound for this term. So we have: $$\abs{c_n}^{-1/n} \gt R'$$ Inverting this expression gives us: $$\abs{c_n}^{1/n} \lt \frac{1}{R'}$$ Multiplying by the positive term $\abs{z - a}$: $$\abs{c_n}^{1/n} \abs{z - a} \lt \frac{\abs{z - a}}{R'}$$ Raising to the $n$-th power: $$\abs{c_n} \abs{z - a}^n \lt \paren{\frac{\abs{z - a}}{R'}}^n$$ Putting everything inside the modulus: $$\abs{c_n (z - a)}^n \lt \paren{\frac{\abs{z - a}}{R'}}^n$$ Because $R' \gt \abs{z- a}$ the geometric series $$\sum_{n = 0}^{\infty} \paren{\frac{\abs{z - a}}{R'}}^n$$ is absolutely convergent and hence $$\sum_{n = 0}^{\infty} \abs{c_n (z - a)}^n$$ is also absolutely convergent. Summarizing, we proved that if $\abs{z - a} < R$, the power series $(1)$ converges and if $\abs{z - a} > R$ it diverges, thus $R$ is the radius of convergence by definition.

Intuitively speaking, for the series to converge, each term $c_n \abs{z - a}^n$ must tend towards 0, so $\abs{z - a}$ must be smaller than $1/c_n^{1/n}$, otherwise the series either goes to infinity or “oscilates” around some value. So the necessity of this makes sense. The sufficiency is less obvious though!

Holomorphic Functions

Theorem 4 proves one of the most important results in complex analysis, that holomorphic functions are analytic:

Theorem 4. Let $f: \Omega \rightarrow \mathbb{C}$ be a holomorphic function. Let $r \gt 0$ such that the closed disk $D$, $\abs{z - a} \le r$, is contained in $\Omega$. Let $C$ be a circle of radius $r$ centered in $a$. Define $c_n$ as:

\[(4.1) \quad c_n = \frac{1}{2\pi i} \int_C \frac{f(z)}{(z - a)^{n+1}} dz\]

Then the power series

\[(4.2) \quad \sum_{n = 0}^{\infty} c_n (z - a)^n\]

converges to $f(z)$ if $\abs{z - a} \lt r$ and hence is analytic in $a$.

Outline of the proof. First we prove that the radius of convergence of $c_n$ is $r$. Then we pick a point $w$ inside the circle $\abs{z - a} \lt r$ and show that $(4.2)$ converges to $f(w)$.

To do so, we express $f(w)$ as a function of $f(z)$, $z \in C$, using Cauchy’s integral formula [6],

\[f(w) = \frac{1}{2\pi i} \int_{C} \frac{f(z)}{z - w} dz\]

The most clever step on the proof is to expand the denominator $1/(z - w)$ into an infinite geometric series. This makes $f(w)$ almost look like $(4.2) + (4.1)$:

\[f(w) = \frac{1}{2\pi i} \int_{C} \sum_{n = 0}^\infty \frac{f(z) (w - a)^n}{(z - a)^{n+1}} dz\]

Except that the integral and sum are in a different order. Sums and integrals are commutative when they’re finite, but since we have an infinite sum, proving we can swap them is non-trivial, and is actually the most complicated part of the proof! But once we do, rearranging terms gives us what we want:

\[f(w) = \sum_{n = 0}^\infty \paren{\frac{1}{2\pi i} \int_{C} \frac{f(z)}{(z - a)^{n+1}} dz} (w - a)^n\] Since $D$ is a closed disk, it's a compact set [4] and because holomorphic functions are continuous, it can be shown that $f$ itself is also bounded in $D$ [4]. Thus, let $\abs{f(z)} \le M$ for some finite $M \in \mathbb{R}$. For points $z \in C$, $\abs{z - a} = r$, so $$(4.3) \quad \abs{\frac{f(z)}{(z - a)^{n+1}}} \le \frac{M}{r^{n + 1}}$$ Taking the modulus of $(4.1)$: $$\abs{c_n} = \frac{1}{2\pi} \abs{\int_C \frac{f(z)}{(z - a)^{n+1}} dz}$$ Using Lemma 5 (Appendix) and the fact that $\abs{C}$ is the circumference of the circle of radius $r$, that is $2\pi r$, $$\abs{c_n} \le \frac{1}{2\pi} 2\pi r \frac{M}{r^{n + 1}} = \frac{M}{r^n}$$ From Lemma 3, the radius of convergence $R$ is: $$R = \liminf_{n \rightarrow \infty} \abs{c_n}^{-1/n} = \liminf_{n \rightarrow \infty} \frac{r}{M^{1/n}}$$ As $n$ approaches infinity, $M^{1/n}$ tends to 1, and since $r$ is constant, we have $R = r$. So we proved the radius of convergence of $(4.2)$ is $r$. It remains to show it converges to $f(z)$ when $\abs{z - a} \lt r$, i.e. for any point inside $D$.

Let $w$ be such a point inside $D$. Since the circle $C$ is $n(C, w) = 1$ we can apply Cauchy's integral formula [6] to $w$: $$(4.4) \quad f(w) = \frac{1}{2\pi i} \int_{C} \frac{f(z)}{z - w} dz$$ Since $w$ is a point in the interior of $C$ and $z$ is on the circumference of $C$, we have $\abs{w - a} \lt \abs{z - a}$ and we can use Lemma 6 (Appendix) to conclude: $$\frac{1}{z - w} = \sum_{n = 0}^\infty \frac{(w - a)^n}{(z - a)^{n+1}}$$ Replacing this in $(4.4)$: $$f(w) = \frac{1}{2\pi i} \int_{C} \sum_{n = 0}^\infty \frac{f(z) (w - a)^n}{(z - a)^{n+1}} dz$$ We are allowed to move $f(z)$ inside the sum because it's invariant with $n$. From $(4.3)$, we have that the integrant is bounded, more precisely: $$\abs{\frac{f(z)(w - a)^n}{(z - a)^{n+1}}} \le \frac{M \abs{(w - a)}^n}{r^{n+1}}$$ Since $\abs{w - a} \lt r$, $\abs{(w - a)}^n / r^{n+1} \lt 1$ and thus the geometric series: $$\sum_{n = 0}^{\infty} \frac{M \abs{(w - a)}^n}{r^{n+1}}$$ is finite. Then, by *Lemma 9 (Appendix)* we can swap the integral and the sum we get: $$f(w) = \frac{1}{2\pi i} \sum_{n = 0}^\infty \int_{C} \frac{f(z) (w - a)^n}{(z - a)^{n+1}} dz$$ Since $(w - a)^n$ is independent of $z$ we can move it out of the integral and move the outer constant inside: $$f(w) = \sum_{n = 0}^\infty (w - a)^n \paren{\frac{1}{2\pi i} \int_{C} \frac{f(z)}{(z - a)^{n+1}} dz}$$ From $(4.1)$, the second factor on the summand is $c_n$: $$f(w) = \sum_{n = 0}^\infty (w - a)^n c_n$$

Taylor series

In Cauchy’s Integral Formula, we showed that a holomorphic function is infinitely differentiable and from Theorem 4 so are analytic functions. The explicit formula is:

\[f^{(n)}(a) = \frac{n!}{2\pi i} \int_\gamma \frac{f(z)}{(z - a)^{n+1}} dz\]

It looks very similar to the definition $c_n$ $(4.1)$ except for the factorial factor. Indeed we can see that:

\[f^{(n)}(a) = n!c_n\]

or

\[c_n = \frac{f^{(n)}(a)}{n!}\]

Replacing this in $(4.2)$ we get:

\[f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (z - a)^n\]

Which is the exact definition of the Taylor series!

Conclusion

My goal was to write about removable singularities but I got stuck in understanding the proofs in Ahlfors. I found Tao’s notes more clear, but they follow different approaches. Tao builds the concept of removable singularities on top of analytic functions, while Ahlfors delays discussing them until much later.

Thus, in order to continue my progression on Complex Analysis, I decided to write about analytic functions first, which builds on top of Cauchy’s Integral Formula, so I think it’s a reasonable time to do so.

I found that the proofs from Tao’s notes rely a lot more on results from real analysis applied to the complex domain. An indication of this is the number of items in References, I had to refresh a bunch of concepts, especially regarding series and convergence.

Appendix

Here we prove general results used by the main proofs in the post, but that would get in the way if inline with the proof itself.

Lemma 5. Let $\gamma$ be a curve and $\abs{f(z)} \le M$ for $z \in \gamma$. Then

\[\abs{\int_{\gamma} f(z) dz} \le M \abs{\gamma}\] This is almost a direct consequence of Theorem 4 in [3], which claims: $$\abs{\int_{\gamma} f(z) dz} \le \int_{\gamma} \abs{f(z)} \abs{dz}$$ Using the hypothesis $\abs{f(z)} \le M$, $$\le \int_{\gamma} M \abs{dz} = M \int_{\gamma} \abs{dz}$$ The last integral corresponds to the length of the curve, $\abs{\gamma}$.

Lemma 6. Let $a, z, w \in \mathbb{C}$ be distinct values, with $\abs{w - a} \lt \abs{z - a}$. Then:

\[\frac{1}{z - w} = \sum_{n = 0}^\infty \frac{(w - a)^n}{(z - a)^{n+1}}\] Let $v = (w - a)/(z - a)$. Since $\abs{v} \lt 1$, the geometric series $$S_n = \sum_{n = 0}^\infty v^n$$ Converges to: $$S_n = \frac{1}{1 - v}$$ Replacing $v$: $$S_n = \frac{1}{1 - \frac{w - a}{z - a}} = \frac{1}{\frac{z - a - (w - a)}{z-a}} = \frac{z - a}{z -w}$$ Thus $$\sum_{n = 0}^\infty \frac{(w - a)^n}{(z - a)^{n}} = \frac{z - a}{z -w}$$ or that $$\frac{1}{z - w} = \sum_{n = 0}^\infty \frac{(w - a)^n}{(z - a)^{n+1}}$$

Lemma 7. Let $(f_k(z))$ be a sequence of Darboux integrable functions that converges uniformily to $f(z)$. Then $\int_{\gamma} f(z)dz$ is Darboux integrable and:

\[\lim_{k \rightarrow \infty} \int_{\gamma} f_k(z)dz = \int_{\gamma} f(z)dz\] Since $(f_k(z))$ converges uniformily to $f(z)$, for any $\epsilon$, there is $N$ such that for all $k \ge N$, $$\abs{f_k(z) - f(z)} \lt \epsilon$$ from this we can say that: $$\abs{f_k(z)} - \epsilon \lt \abs{f(z)} \lt \abs{f_k(z)} + \epsilon$$ since each $f_k$ is Darboux integrable, the upper and lower Darboux integrals, $U_k$ and $L_k$, exist and are equal to $\int_{\gamma} f_k(z)dz$. We have the upper and lower Darboux sums for $f(z)$ and some partition $P$ as: $$\begin{align} U(f,P) &= \sum_{i = 1}^n M_{i} (\abs{z(t_{i-1}) - z(t_{i})}) \\ L(f,P) &= \sum_{i = 1}^n m_{i} (\abs{z(t_{i-1}) - z(t_{i})}) \end{align} $$ with $$\begin{align} M_{i} &= \sup \curly{\abs{f(z(t))} : t \in [t_{i-1}, t_{i}]} \\ m_{i} &= \inf \curly{\abs{f(z(t))} : t \in [t_{i-1}, t_{i}]} \end{align} $$ Since $f(z)$ is within $\epsilon$ of $f_k(z)$, so $\abs{M_{k,i} - M_{i}} \lt \epsilon$ and $\abs{m_{k,i} - m_{i}} \lt \epsilon$ and thus $\abs{U(f_k,P) - U(f,P)} \lt \abs{\gamma} \epsilon$ and $\abs{L(f_k,P) - L(f,P)} \lt \abs{\gamma} \epsilon$ for any partition $P$. Thus $\abs{U - U_k} \lt \abs{\gamma} \epsilon$ and $\abs{L - L_k} \lt \abs{\gamma} \epsilon$. Since we can pick arbitrarily small $\epsilon$, we have that $$U = \lim_{k \rightarrow \infty} U_k = \lim_{k \rightarrow \infty} L_k = L$$ and thus $f(z)$ is Darboux integrable and equal to $\lim_{k \rightarrow \infty} \int_{\gamma} f_k(z)dz$. QED.

Lemma 8. Let $f_k(z)$ be integrable functions such that

\[\abs{f_k(z)} \le M_k\]

Such that $\sum_{n = 0}^{\infty} M_k$ converges. Then

\[\abs{\int_{\gamma} f_k(z)} \lt N_k\]

Where $\sum_{n = 0}^{\infty} N_k$ also converges.

From Lemma 5 we have that $$\abs{\int_{\gamma} f_k(z)} \le M_k \abs{gamma}$ Thus setting $N_k = M_k \abs{gamma}$ satisfies the first constaint and that $$\sum_{n = 0}^{\infty} N_k = \abs{\gamma} \sum_{n = 0}^{\infty} M_k$$ So if $\sum_{n = 0}^{\infty} M_k$ converges to a limit $L$, $\sum_{n = 0}^{\infty} N_k$ comverges to $LM$.

Lemma 9. Let $f_k(z)$ be Darboux integrable functions such that

\[\abs{f_k(z)} \lt M_k\]

and that $\sum_{n = 0}^{\infty} M_k$ converges. Then we can exchange the order of the integral and the sum:

\[\sum_{k=0}^\infty \int_\gamma f_k(z) dz = \int_\gamma \sum_{k=0}^\infty f_k(z) dz\] Since $$\abs{f_k(z)} \lt M_k$$ and that $\sum_{n = 0}^{\infty} M_k$ converges, we can use Weierstrass M-test to conclude that the series $$\sum_{k=0}^\infty f_k(z)$$ converges uniformly. In other words, if we have the partial sum: $$(9.E) \quad s_n(z) = \sum_{k=0}^n f_k(z)$$ then the M-test says that the sequence $(s_n(z))$ converges uniformily to some function $s(z)$. Uniform convergence implies pointwise convergence and hence: $$(9.F) \quad \lim_{n \rightarrow \infty} s_n(z) = s(z) = \sum_{k=0}^\infty f_k(z)$$ Since each $f_k(z)$ is integrable, their finite sum is also integrable, so we can conclude by Lemma 7 that $s(z)$ is integrable and that: $$\lim_{n \rightarrow \infty} \int_{\gamma} s_n(z)dx = \int_{\gamma} s(z)dz$$ Replacing $s_k$ with $(9.E)$ and $s$ with $(9.F)$: $$(9.C) \quad \lim_{n \rightarrow \infty} \int_{\gamma} \sum_{k=0}^n f_k(z) dz = \int_{\gamma} \sum_{k=0}^\infty f_k(z) dz$$ For finite terms we can exchange the order of an integral and sum, so we have: $$(9.A) \quad \int_\gamma \sum_{k=0}^n f_k(z) dz = \sum_{k=0}^n \int_\gamma f_k(z) dz$$ Replacing in $(9.C)$: $$(9.D) \quad \lim_{n \rightarrow \infty} \sum_{k=0}^n \int_{\gamma} f_k(z) dz = \int_{\gamma} \sum_{k=0}^\infty f_k(z) dz$$ By Lemma 8 the series of the integrals of $f_k$ also satisfies the Weierstrass M-test and hence converges uniformily, so the limit from $(9.D)$ exists: $$\lim_{n \rightarrow \infty} \sum_{k=0}^n \int_{\gamma} f_k(z) dz = \sum_{k=0}^\infty \int_{\gamma} f_k(z) dz$$ Replacing in $(9.D)$: $$\sum_{k=0}^\infty \int_\gamma f_k(z) dz = \int_{\gamma} \sum_{k=0}^\infty f_k(z) dz$$ QED.

Lemma 9 shows that we can swap a series and an integral if the functions $f_k(z)$ satisfy the Weierstrass M-test. Tao’s notes and many sites I found suggest it’s possible to show the same if $f_k(z)$ are uniformily convergent.

Satisfying Weierstrass M-test implies uniform convergence but not the other way around. So Lemma 9 is a weaker version of the result based on uniform convergence, but I couldn’t prove or find a clear proof ([10] has an answer but I don’t understand why that proves the claim).

References

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