# What is it that makes holomorphic functions so rigid?

This semester I took a complex analysis class and, as far as I've seen, holomorphic functions on the complex plane have very "powerful" properties; For example, the identifying theorem, or the Casorati-Weierstrass theorem, Morera's Theorem, or even Gauss' mean value theorem.

Thing is, most of the theorems have very "light" conditions but lead to "heavy" results. One comes to the conclusion that the holomorphy of a function is itself a very strong property.

But what is it that makes it that way? In real analysis, one can say some things about differentiable functions, but not that much as one can say about complex-differentiable functions.

EDIT: As stated in the answers, complex-differentiability implies complex-analyticity and that is a big thing indeed. Actually, the proof of this theorem uses Cauchy's integral formula and the uniform convergence of a series on a circle. I'm already familiar with this things though and I was kinda hoping for an answer orientated around the fact that the field $\mathbb{C}$ is algebraicly closed. Does this algebraic property of the plane play a crucial role in the implicitation "differentiability $\rightarrow$ analyticity" ?

• There are many ways to think about it: One very mechanical way is that for a $C^1$ function on (an open subset of) the plane, the Cauchy-Riemann equations impose two differential conditions on the function at every point and that are not generically satisfied. So, from this point of view, the condition of being holomorphic is itself very strong---this is of course obscured some by the fact that most of the functions we work with satisfy it already. – Travis Willse May 20 '17 at 14:50
• by the way, I wouldn't say that holomorphic functions are "versatile". One word that is more commonly used to express your observations is "rigid". Think of the identity theorem: you can't paste and glue bits of holomorphic functions like you can for say, continuous functions. – Glougloubarbaki May 20 '17 at 19:08
• I'm sorry, english is not my native language and the lessons I take are in greek, so I'm doing all the translations myself. I'll edit my title, thanks – JustDroppedIn May 20 '17 at 19:15

## 3 Answers

Most of those striking theorems (identity theorem, Liouville theorem, Cauchy inequality...) are rather direct consequences of the fact that holomorphic functions are analytic.

An analytic function is one that can be locally written on the form $f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n$ near every $z_0$ in its domain of definition. The true magic is that being holomorphic (i.e. being complex differentiable) implies being analytic (something far from obvious, that usually takes a good chunk of a first course on complex analysis).

But if you admit that non-obvious fact, it should not be very surprising that analytic functions possess all kinds of strong properties (for one thing, all those coefficients $a_n$, hence all of the function, can be read with just local information near $z_0$).

Note that real analytic functions also satisfy those very strong properties, but of course as we know real differentiability does not imply real analyticity.

EDIT

I have no deep insight as to why complex differentiability implies analyticity whereas real differentiability does not. However, I can say that the fact that $\mathbb C$ is algebraically closed is not a sufficient reason. Indeed, there are algebraically closed and complete fields denoted by $\mathbb{C}_p$ (called $p$-adic numbers), on which this is not true. In fact, $\mathbb{C}_p$ is isomorphic as a field to $\mathbb{C}$ (though they are not homeomorphic). So this is more topological than algebraic in nature.

• Thank you for this answer, It's close to what I'm more or less expecting. I would appreciate it if you could take a look at the edit I just did :) – JustDroppedIn May 20 '17 at 18:48
• A key argument is the Cauchy representation theorem. A topological argument could be the following : $\mathbb{R} \backslash \{x\}$ is disconnected but $\mathbb{C}\backslash \{z\}$ remains connected. Hence, one can find continuous paths which enclose a singularity. – C. Dubussy Jan 16 '18 at 20:45

We get the feeling of surprise in the first encounter with complex variable theory because we are trained for years in the very "loose" world of real functions. However, even though the general theorems in calculus (e.g. mean value theorem) apply to all differentiable functions, all of the actual examples that we work with (differentiating and integrating) are analytic. Even when we define a function to be one expression in one interval, and a different one in another, it still is perfectly analytic (hence rigid) in each domain separately.

So, instead of thinking of functions of a complex variable as surprisingly rigid, it may be better to think of real non-analytic functions as being "too loose" or too general to be of much use in applications. That perhaps was the more dominant view up to (and including) Weierstrass's time.

Here's another twist. Even though a general continuous function on $\mathbb R$ appears to be too loose, one often(?) can extend it to a holomorphic function in a neighborhood just above the real line, so it is "as rigid as holomorphic functions". (You can't extend to both sides, unless the function was analytic to begin with.) In fact, Weierstrass's famous example of continuous but nowhere differentiable function was first discovered as the boundary value of a holomorphic function on the open unit disk.

(BTW, by taking the boundary value of a holomorphic function, one may also obtain generalized functions such as Schwartz's distributions.)

For another piece of "evidence" that holomorphic functions (of a complex variable) are "as many as" functions of a real variable, there is Paley-Wiener type theorem: smooth compactly-supported functions on $\mathbb R$ are in one to one correspondence, under Fourier transform, with entire functions with special growth condition.

Roughly speaking, then,

{holomorphic functions on a complex domain} $\approx$ {continuous (or generalized) functions on a real interval}.

With this dictum in mind, there must be a lot more holomorphic functions out there, more than one might expect from a first course in complex analysis. This partially accounts for the ubiquity of holomorphic functions in many disparate areas of mathematics, perhaps most prominently in number theory (L-functions and modular forms come to mind).

If you really want to compare the situation of real differentiable functions vs holomorphic functions, the "light" or "mild" condition (of being holomorphic) is actually very strong because in two dimensions there's a lot more "room" (or directions) for the condition to hold, instead of just the two directions in the real world.

What is actually "more loose" is the general (say, continuous) functions on the complex domain, or even $C^1$ functions in the multi-variable sense (i.e., as $\mathbb R^2\to \mathbb R^2$). The derivative at a point has four degrees of freedom (being a $2\times 2$ matrix), whereas a holomorphic function has two (rotation and dilation only).

• I enjoyed reading this, thank you, but your statement "Even though a general continuous function on $\mathbb R$ appears to be too loose, one can extend it to a holomorphic function in a neighborhood just above the real line" is false. Continuous boundary functions for holomorphic functions must satisfy properties well beyond those of general continuous functions. For example they cannot vanish on an interval (or even a set of positive measure) without being identically $0.$ – zhw. Jan 16 '18 at 20:38
• Also "In fact, Weierstrass's famous example of continuous but nowhere differentiable function was constructed as the boundary value of a holomorphic function on the open disk" Do you have a reference for this? I never heard that before. Thanks again for the post. – zhw. Jan 16 '18 at 20:39
• @zhw, I thought of being more precise there. Being holomorphic is only meaningful on an open subset of $\mathbb C$, and what I meant there is that it has an extension to a slim neighborhood $U$ bordering on the interval $I$ such that it is holomorphic on $U$, has a limit as $z$ approaches $I$, and the limit is equal to the original function there. (The identity theorem states that a holomorphic function can not vanish on a set with an accumulation point inside the domain.) – liuyao Jan 16 '18 at 21:55
• This does sound bizarre and hard to believe. However, if we wrap the real line to a closed loop (say with a Möbius transformation), it can be constructed by the Cauchy integral formula: $f(z)=\frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta \qquad z\in U$ (The moral being that the value of the function on the boundary completely determines the holomorphic function inside.) One caveat is that the original function probably needs to be integrable; I'll try to find a reference. – liuyao Jan 16 '18 at 21:55
• I'm not sure if you understood me. Let's take something simple like the open unit disc $D.$ Suppose $f$ is continuous on $\partial D,$ equals $0$ on the arc from $1$ to $i,$ and equals $1$ on the arc from $-1$ to $-i.$ Then there is no $F$ continuous on $\bar D$ such that $F|_D$ is holomorphic. Continuous functions on the boundary that arise from holomorphic function are very special creatures, well studied. Perhaps you are confusing the Cauchy integral with the Poisson integral. The latter will do what you say in the context of harmonic functions. – zhw. Jan 16 '18 at 23:41

Let $U$ be an open subset of $\mathbb{C}$ and $f\colon U\rightarrow\mathbb{C}$ be a map. Let us identify $\mathbb{R}^2$ and $\mathbb{C}$ through the following isometry $(x,y)\mapsto x+iy$, then $f$ is holomorphic at $z_0\in U$ if and only if $f$ is differentiable and $\mathrm{d}_{z_0}f$ is either zero or a direct similarity, namely there exists $(a,b)\in\mathbb{R}^2\setminus\{(0,0)\}$ such that: $$\mathrm{d}_{z_0}f=\begin{pmatrix}a&-b\\b&a\end{pmatrix}.$$ In fact, $f'(z_0)=a+ib$. From there, on can derives two facts:

• If $f$ is holomorphic at $z_0$, then $f$ satisfies the Cauchy-Riemann equation: $$\frac{\partial f}{\partial x}(z_0)=-i\frac{\partial f}{\partial y}(z_0).$$ This is a pretty strong condition that implies that $\textrm{Re}(f)$ and $\textrm{Im}(f)$ are harmonic maps.

• If $f$ is holomorphic at $z_0$ and its derivative is nonzero, then $f$ is a conformal mapping around $z_0$, namely if $\gamma_1$ and $\gamma_2$ are two smooth curves crossing at $z_0$ with angle $\theta$, then $f\circ\gamma_1$ and $f\circ\gamma_2$ cross at $f(z_0)$ with angle $\theta$. One can write this relation as: $$\frac{\langle{\gamma_1}'(0),{\gamma_2}'(0)\rangle}{\|{\gamma_1}'(0)\|\|{\gamma_2}'(0)\|}=\frac{\langle(f\circ\gamma_1)'(0),(f\circ\gamma_2)'(0)\rangle}{\|(f\circ\gamma_1)'(0)\|\|(f\circ\gamma_2)'(0)\|}.$$

I believe that this two consequences can give some insight on how strong being holomorphic is. The first point is actually equivalent to the holomorphy.