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Removable Singularities

31 Aug 2024

Portrait of Bernhard Riemann

This is the 7-th post in the series with my notes on complex integration, corresponding to Chapter 4 in Ahlfors’ Complex Analysis.

In this post we’ll cover Removable Singularities in the context of holomorphic functions.

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
  6. Holomorphic Functions are Analytic

Definitions

Let $f$ be a holomorphic function in $\Omega$. A singularity is is a point $a$ where a function is not defined, that is, $a \not\in \Omega$.

A singularity is “isolated” if nothing around it is a singularity. More formally, point $a$ is an isolated singularity if $a \not\in \Omega$ and there exists a radius $r \gt 0$ such that $f$ is holomorphic in the punctured disk $0 \lt \abs{z - a} \lt r$.

A removable singularity is a special type of isolated singularity with the property that $\lim_{z \rightarrow a} (z - a) f(z) = 0$.

One nice thing about removable singularities is that it’s possible to “extend” the function $f$ such it is made holomorphic there as well (Theorem 1), so that it’s not a singularity anymore, effectively “removing” it, hence the terminology.

Theorem 1. Let $f(z)$ be a holomorphic function in $\Omega \setminus \curly{a}$. Then

\[(1) \quad \lim_{z \rightarrow a} f(z)(z - a) = 0\]

if and only if there exists an holomorphic function $g(z)$ in $\Omega$ that coincides with $f(z)$ in $\Omega \setminus \curly{a}$.

First we prove that if there exists an holomorphic function in $\Omega$ that coincides with $f(z)$ in $\Omega \setminus \curly{a}$ then $(1)$ holds. Let $g$ be such a function. By hypothesis $g$ is holomorphic in $a$, so it's also continuous in $a$ and by definition $\lim_{z \rightarrow a} g(z) = g(a)$. Since $z$ is always different from $a$ inside the limit, we can substitute $g$ for $f$ and obtain: $$(1.1) \quad \lim_{z \rightarrow a} f(z) = g(a)$$ in other words the limit of $f(z)$ as $z$ approaches $a$ is finite and unique. Consider the $\lim_{z \rightarrow a} (z - a) f(z)$. First we consider the limit of each factor: $$= \paren{\lim_{z \rightarrow a} (z - a)} \paren{\lim_{z \rightarrow a} f(z)}$$ Replacing $(1.1)$: $$= \paren{ \lim_{z \rightarrow a} (z - a) } g(a)$$ Since $g(a)$ is defined, it must be finite and thus a constant, and the limit above goes to 0 and we conclude that: $$\lim_{z \rightarrow a} (z - a) f(z) = 0$$ On the other direction, assume $(1)$ holds. Define the function $h(z)$ as follows: $$ \begin{equation} (1.B) \quad h(z)=\left\{ \begin{array}{@{}ll@{}} (z-a)^2 f(z), & z \in \Omega \setminus \curly{a} \\ 0, & z = a \end{array}\right. \end{equation} $$ Since $f(z)$ is holomorphic when $z \ne a$ and that polynomials like $(z - a)^2$ are also holomorphic, then $h(z)$ is holomorphic when $z \ne a$. Now consider $h'(a)$: $$h'(a) = \lim_{z \rightarrow a} \frac{h(z) - h(a)}{z - a}$$ Since $z \ne a$, $h(z) = (z-a)^2 f(z) $ and $h(a) = 0$ by $(1.B)$: $$= \lim_{z \rightarrow a} \frac{(z-a)^2 f(z) - 0}{z - a}$$ Which simplifies to: $$= \lim_{z \rightarrow a} (z-a)f(z)$$ Which by hypothesis is 0, so: $$(1.2) \quad h'(a) = 0$$ Thus the limit exists and hence $h'(a)$ is defined and $h$ is holomorphic in $a$ as well. As we've shown in [2], holomorphic functions are analytic, so $h(z)$ can be expressed as a power series: $$(1.3) \quad h(z) = c_0 + c_1 (z - a) + c_2 (z - a)^2 + \dots$$ We can use some identities to figure out the coefficients. By setting $z = a$, we get $c_0 = h(a)$ which by $(1.B)$ is 0. Differentiating $(1.3)$ gives us: $$h'(z) = c_1 + c_2 2 (z - a) + c_3 3 (z - a)^2 + \dots$$ Again, setting $z = a$, we get $c_1 = h'(a)$ which by $(1.2)$ is also 0. So $h(z)$ is: $$(1.2) \quad h(z) = c_2 (z - a)^2 + c_3 (z - a)^3 + \dots$$ If $z \ne a$, we can rewrite $(1.B)$ as $$f(z) = \frac{h(z)}{(z-a)^2}$$ Replacing $(1.2)$: $$= c_2 + c_3 (z - a) + c_4 (z - a)^2 + \dots$$ So this is the power series expansion of $f(z)$ when $z \ne a$. Since this is a convergent power series, by Theorem 4 in [2] there exists some corresponding holomorphic function $g(z)$. For $z \ne a$ we just saw it equals to $f(z)$. For $g(a)$ we get $c_2$. Summarizing: $$ \begin{equation} g(z)=\left\{ \begin{array}{@{}ll@{}} f(z), & z \in \Omega \setminus \curly{a} \\ c_2, & z = a \end{array}\right. \end{equation} $$ We can actually calculate $c_2$. Recall that it is the first term of the power series expansion of $f(z)$, which from Theorem 4 in [2] is: $$c_2 = \frac{1}{2\pi i} \int_{C} \frac{f(w)}{w - a}dw$$ For some circle $C$ centered in $a$ and contained in $\Omega$. QED

According to Wikipedia this is part of Riemman’s Theorem on removable singularities [6]. The theorem proves other equivalences too.

In the proof of Theorem 1, we saw that even if $f(a)$ doesn’t exist, we can compute $g(a)$ via Cauchy’s Integral Formula:

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

However, an alternative approach is to simply compute $g(a)$ as:

\[g(a) = \lim_{z \rightarrow a} f(z)\]

Since we know $g(a)$ exists, is holomorphic and thus continuous.

Example

Wikipedia provides the example of the unormalized $\mbox{sinc}$ function:

\[\mbox{sinc}(z) = \frac{\sin z}{z}\]

which is holomorphic everywhere except (in principle) at $z = 0$. We can use Theorem 1 to show $\mbox{sinc}(0)$ can be computed as $\lim_{z \rightarrow 0} \mbox{sinc}(z)$ which from Lemma 2 equals to 1.

Lemma 2. Let $z \ne 0$ be a complex number. Then:

\[\lim_{z \rightarrow 0} \frac{\sin z}{z} = 1\] The real case can be proved using Euclidean geometry and the squeeze theorem [7]. For the complex case we can use the Taylor series expansion of $\sin z$: $$ \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots $$ Dividing by $z \ne 0$: $$ \frac{\sin z}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \frac{z^6}{7!} + \cdots $$ Taking the limit $z \rightarrow 0$: $$ \lim_{z \rightarrow 0} \frac{\sin z}{z} = 1 $$ QED.

We can also compute it directly from $(2)$:

\[\mbox{sinc}(0) = \frac{1}{2\pi i} \int_{C} \frac{\sin z}{z^2}dw\]

by using the residue theorem (which we haven’t studied yet) to arrive at the same conclusion. Thus $\mbox{sinc}(z)$ is holomorphic on the entire complex plane.

Finite Taylor Series

Consider the function:

\[(3) \quad F(z) = \frac{f(z) - f(a)}{z - a}\]

which is holomorphic everywhere $f(z)$ is except at $a$, but we have at least that $\lim_{z \rightarrow a} F(z) (z - a) = 0$, so we can apply Theorem 1 for $F(z)$ to conclude that there exists some function $G(z)$ which is holomorphic on $f$’s domain.

Since $G$ is holomorphic at $a$, it’s also continuous in $a$ and thus we can express it as

\[G(a) = \lim_{z \rightarrow a} F(z)\]

Which when replacing with $(3)$, gives us the definition of $f’(a)$, so:

\[\begin{equation} G(z)=\left\{ \begin{array}{@{}ll@{}} F(z), & z \in \Omega \setminus \curly{a} \\ f'(a), & z = a \end{array}\right. \end{equation}\]

We can write:

\[f(z) = f(a) + (z - a) G(z)\]

Which holds for $z \ne a$ from $(3)$. For $z = a$ this is trivally correct since $G(a)$ is finite, and we get $f(a) = f(a)$. Renaming $G$ to $f_1$:

\[f(z) = f(a) + (z - a) f_1(z)\]

Repeating for $f_1$ and naming the corresponding function defined in $(3)$ as $f_2$:

\[f_1(z) = f_1(a) + (z - a) f_2(z)\]

If we do it $n$ times we get the recurrence:

\[f_{n - 1}(z) = f_{n - 1}(a) + (z - a) f_n(z)\]

We can “expand” these and write $f(z)$ as a function of $f_n$:

\[(4) \quad f(z) = f(a) + (z - a) f_1(a) + (z - a)^2 f_2(a) + \cdots \\ + (z - a)^{n-1} f_{n-1}(a) + (z - a)^n f_n(z)\]

We can now use Lemma 3 to express $f_k$ as the $k$-th derivative of $f$.

Lemma 3. Let $f$ be a holomorphic function in $\Omega$ and $f_k$ be defined as

\[f_k(z) = \frac{f_{k-1}(z) - f_{k-1}(a)}{z - a}\]

Then

\[(5) \quad f_n(a) = \frac{f^{(n)}(a)}{n!}\] We already have the identity $(4)$. If we differentiate it we get: $$ f'(z) = f_1(a) + 2(z - a) f_2(a) + \cdots + (n - 1) (z - a)^{n-2} f_{n-1}(a) \\ + n(z - a)^{n - 1} f_n(z) + (z - a)^n f_n'(z) $$ Differentiating again: $$ f''(z) = 2 f_2(a) + \cdots + (n - 1)(n - 2) (z - a)^{n-3} f_{n-1}(a) \\ + n(n - 1)(z - a)^{n - 2} f_n(z) + (z - a)^{n-1} f_n'(z) + n (z - a)^{n - 1} f_n'' + (z - a)^n f_n''(z) $$ It's tedious work, but differentiating it $n$ times, we'll see that the terms with $f_k$ for $k \lt n$ will vanish and the other terms besides $f_n(z)$ will have a factor $(z - a)$, so when we set $z = a$ we obtain: $$ f^{(n)}(a) = n!f_n(a) $$ QED

Using $(5)$ we can now rewrite $(4)$ using only the derivarites of $f$ and $f_n$:

\[f(z) = f(a) + \frac{f'(a)}{1!} (z - a) + \frac{f''(a)}{2!} (z - a)^2 + \cdots \\ + \frac{f^{n-1}(a)}{(n - 1)!} (z - a)^{n - 1} + f_n(z) (z - a)^n\]

or more concisely:

\[(6) \quad f(z) = f_n(z) (z - a)^n + \sum_{k = 0}^{n - 1} \frac{f^{(k)}(a)}{k!} (z - a)^k\]

Noting that if $f(z)$ is holomorphic in $\Omega$, so is $f_n(z)$. We can use Lemma 4 to express $f_n$ as a function of $f$ as well.

Lemma 4. Let $f_n$ be a holomorphic function defined as:

\[(7) \quad f_n(z) = \frac{f(z)}{(z - a)^n} - \sum_{k = 0}^{n - 1} \frac{f^{(k)}(a)}{k!(z - a)^{n-k}}\]

For $n \ge 1$. Then

\[(8) \quad f_n(z) = \frac{1}{2\pi i} \int_{C} \frac{f(w)dw}{(w - a)^n(w - z)}\] Because $f_n$ is holomorphic, we can use Cauchy's Integral Formula (Lemma 1 in [3]): $$ (4.1) \quad f_n(z) = \frac{1}{2\pi i} \int_{C} \frac{f_n(w)dz}{w - z} $$ We can then express $f_n(w)$ via $(7)$: $$f_n(w) = \frac{f(w)}{(w - a)^n} - \sum_{k = 0}^{n - 1} \frac{f^{(k)}(a)}{k!(w - a)^{n-k}}$$ Plugging this into $(4.1)$ we get: $$ f_n(z) = \frac{1}{2\pi i} \paren{ \int_{C} \frac{f(w)dw}{(w - a)^n(w - z)} - \int_C \sum_{k = 0}^{n - 1} \frac{f^{(k)}(a)}{k!(w - a)^{n-k}(w - z)} dw} $$ Exchanging the integral and the finite sum, and moving the terms that do not depend on $w$ out of the integral: $$ = \frac{1}{2\pi i} \paren{ \int_{C} \frac{f(w)dw}{(w - a)^n(w - z)} - \sum_{k = 0}^{n - 1} \frac{f^{(k)}(a)}{k!} \int_C \frac{dw}{(w - a)^{n-k}(w - z)}} $$ We'll now show that the integral inside the sum is actually 0. Let's define: $$ (4.2) \quad F_v(a) = \int_C \frac{dw}{(w - a)^{v}(w - z)} $$ For $v = 1, \dots, n$ (noting this is the same index range $n-k$ has for $k = 0, n-1$). For $v = 1$ we have: $$ F_1(a) = \int_C \frac{dw}{(w - a)(w - z)} $$ We can use the identity: $$\frac{1}{(w - a)(w - z)} = \frac{1}{z - a} \paren{\frac{1}{w - a} - \frac{1}{w - z}}$$ to obtain: $$ F_1(a) = \frac{1}{z - a} \int_C \frac{1}{w - a} - \frac{1}{w - z} dw $$ Since points $a$ and $z$ are inside the circle and $w$ is on the circumference, $w \ne a$, $w \ne z$ and thus the integrand is a holomorphic function and we can use *Cauchy Integral Theorem* [4] to conclude the integral is 0 and thus: $$F_1(a) = 0$$ is a constant function. Hence all its $n$-th derivatives are also zero, $F_1^{(n)}(a) = 0$. Now let $g(w) = \frac{1}{w - z}$. We can write $(4.2)$ as: $$ F_v(a) = \int_C \frac{g(w) dw}{(w - a)^{v}} $$ Since $g(w)$ is continuous in $C$ (again, $z \ne w$), we can use *Lemma 2* from [3] to obtain $F_v'(a) = v F_{v+1}(a)$, or expanding it $v$ times: $$ F_{v + 1}(a) = \frac{F^{(n)}_1(a)}{v!} $$ Which implies that $F_v(a) = 0$ for all $v$. Going back to $f_n(z)$, we can now conclude: $$ f_n(z) = \frac{1}{2\pi i} \int_{C} \frac{f(w)dw}{(w - a)^n(w - z)} $$ QED

We can replace $(8)$ in $(6)$ and obtain the final form of the finite Taylor series:

Corollary 5. Let $f(z)$ be a holomorphic function in $\Omega$, with $a \in \Omega$. Then

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

For a circle $C$ centered in $a$ and contained within $\Omega$ and $n \ge 1$.

I’m not sure how this result is useful though. It seems to be that computing $f_n$ via $(8)$ is strictly harder than just computing $f(z)$ directly:

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

Alternative Characterizations

It’s useful to consider other definitions of removable singularity, because they might be used to contrast with other types of isolated singularities.

Corollary 6. A point $a$ is a removable singularity if and only if $\lim_{z \rightarrow a} f(z)$ exists.

One direction follows from Theorem 1: we know that $g(a)$ is holomorphic and hence continuous and that $g(a) = \lim_{z \rightarrow a} f(z)$, so the limit exists.

If $\lim_{z \rightarrow a} f(z)$ exists, then $f(z)$ is bounded in a neighborhood of $a$ by some $L$, so $$ \lim_{z \rightarrow a} \abs{z - a}\abs{f(z)} \lt \lim_{z \rightarrow a} \abs{z - a} L = 0 $$ and thus $\lim_{z \rightarrow a} (z - a) f(z) = 0$ which is the definition of removable singularity.

Conclusion

In this post we learned that if a function is holomorphic in a domain $\Omega$ except at a point $a$, we can extend it to be holomorphic in $a$, as long as $\lim_{z \rightarrow a}f(z) (z - a) = 0$.

We’ve seen a similar property before with Cauchy Integral Theorem in a disk (Theorem 4 in [4]) states that:

Let $f(z)$ be holomorphic in the region $\Delta’$ obtained by omitting a finite number of points $\xi_j$ from the open disk $\Delta$. If $f(z)$ is such that

\[\lim_{z \rightarrow \xi_j} (z - \xi_j) f(z) = 0\]

then

\[\int_{\gamma} f(z) dz = 0\]

for any closed curve $\gamma$ in $\Delta’$.

So these $\xi_j$ are removable singularities. With Theorem 1, we wouldn’t need to prove Theorem 4 in [4], we could just show that if $f(z)$ is holomorphic in $\Delta’$, it’s also holomorphic in $\Delta$ and use Theorem 3 in [4] which is the version without removable singularities.

We must however be careful with circular arguments: Theorem 1 relies on the fact that holomorphic functions are analytic, whose proof depends on Cauchy’s Integral Formula, which in turn relies on Theorem 4.

That’s one concern I have in using multiple sources to complement Alfhors’ textbook: By relying on results out of order I might risk running into circular arguments. Alfhors proves Theorem 1 by claiming that Cauchy’s Integral Formula allows holomorphic functions to have missing points as long as $\lim_{z \rightarrow a}f(z) (z - a) = 0$, but it doesn’t provide an explicit proof of that and I couldn’t prove it myself.

I relied on Tao’s blog [4] and Wikipedia [6] that prove Theorem 1 by using that holomorphic functions are analytic.

References

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