Contents

# Contents

## Functional equation characterization

The nLab page on pi characterizes the sine function as the unique continuous function $f: \mathbb{R} \to \mathbb{R}$ such that

• $1 - x^2 \leq \textstyle{\frac{f(x)}{x}} \leq 1$,

• $f(3 x) = 3f(x) - 4f(x)^3$.

In this section we will check that this really is a definite description (i.e., that only one function satisfies this). It’s clear that $f(x) = \sin(x)$ indeed satisfies these conditions.

By far the more significant condition is the second one, which we’ll call the functional equation. The first condition could be considered a tameness condition; its utility here is that it implies $f(0) = 0$ and $f'(0) = 1$, which effectively eliminates pathological solutions.

The functional equation implies that $f$ is uniquely determined from its germ at $x = 0$ (i.e., that the restriction of $f$ to any neighborhood $U$ of $0$ determines $f$ globally: this is because $\mathbb{R}$ is the union of neighborhoods of the form $3^n U$).

Also we may deduce that ${|f(x)|} \leq 1$ for all $x$. Certainly $-1 \leq f(x) \leq 1$ is true for all $x$ in some small neighborhood $U$ of $0$. One then proves that $-1 \leq f(x) \leq 1$ for all $x \in 3^n U$ by induction on $n$ and with the help of the functional equation, using the fact that the function $u$ defined by $u(x) \coloneqq 3 x - 4 x^3$ maps $[-1, 1]$ to itself.

The plan of attack will be to reconstruct from $f$ (which plays the role of the sine) a second function $g$ that plays the role of the cosine. (You can’t just define $g = \sqrt{1 - f^2}$ since we know the sign has to change at appropriate spots.) The trick is first to get the germ of $g$ at zero, and invoke a companion functional equation which can be used to propagate out to a function $g$ defined on all of $\mathbb{R}$. Then, one considers the corresponding complex exponential $g + i f$, with its own companion functional equation which is used to deliver the coup de grace.

###### Lemma

Let $a \gt 0$ be chosen small enough so that ${|f(x)|} \lt 1/2$ whenever $x \in [-a, a]$. Define $g$ on $[-a, a]$ by $g(x) = \sqrt{1 - f(x)^2}$. Then

$g(3 x) = 4g(x)^3 - 3g(x)$

for all $x \in [-a/3, a/3]$.

###### Proof

For $x \in [-a/3, a/3]$ we have

$\array{ 4g(x)^3 - 3g(x) & = & g(x)(g(x)^2 - 3(1 - g(x)^2)) \\ & = & g(x)(1 - f(x)^2 - 3f(x)^2) \\ & = & g(x)(1 - 4f(x)^2) }$

which is positive by choice of $a$. Hence both $g(3 x)$ and $4g(x)^3 - 3g(x)$ are positive for $x \in [-a/3, a/3]$, so it suffices to show their squares are equal. In other words, that

$1 - f(3 x)^2 = (4g(x)^3 - 3g(x))^2 = g(x)^2(1 - 4f(x)^2)^2.$

This boils down to checking

$1 - (3f(x) - 4f(x)^3)^2 = (1 - f(x)^2)(1 - 4f(x)^2)^2$

but this follows from the polynomial identity $1 - (3t - 4t^3)^2 = (1 - t^2)(1 - 4t^2)^2$, which is easily checked.

Now extend $g$ to a function $\mathbb{R} \to \mathbb{R}$ by recursion: put $g_0 = g: [-a, a] \to \mathbb{R}$ and define $g_n: [-3^n \cdot a, 3^n \cdot a] \to \mathbb{R}$ by $g_n(x) \coloneqq 4g_{n-1}(x/3)^3 - 3g_{n-1}(x/3)$.

###### Lemma

The equation $g_{n+1}(x) = g_n(x)$ holds for $x \in [-3^{n-1} a, 3^{n-1}a]$.

###### Proof

The case for $n=0$ is just the preceding lemma. The rest is by induction: let $v$ be the function $v(x) = 4x^3 - 4x$, and use the fact that for $n \gt 0$ and $x \in [-3^{n-1}a, 3^{n-1}a]$ we have

$g_{n+1}(x) = (v \circ g_n)(x/3) = (v \circ g_{n-1})(x/3) \coloneqq g_n(x)$

where the second equation follows from the inductive hypothesis.

By this lemma, the $g_n$ patch together to form a continuous global function $g: \mathbb{R} \to \mathbb{R}$ that satisfies the functional equation $g(3x) = 4g(x)^3 - g(x)$.

###### Proposition

The continuous function $h = g + i f: \mathbb{R} \to \mathbb{C}$ satisfies $h(0) = 1$, $h(3 x) = h(x)^3$, and ${|h(x)|} = 1$ for all $x$.

###### Proof

By our earlier choice of $a$, we have $g(x)^2 + f(x)^2 = 1$ for $x \in [-a, a]$. One shows by induction that the equations hold for $x \in [-3^n a, 3^n a]$, using the functional equations and $g(x)^2 + f(x)^2 = 1$ for $x \in [-3^{n-1}a, 3^{n-1}a]$:

$\array{ (g(x) + i f(x))^3 & = & g(x)^3 - 3g(x) f(x)^2 + i(3g(x)^2 f(x) - f(x)^3) \\ & = & g(x)^3 - 3g(x)(1 - g(x)^2) + i(3(1 - f(x)^2)f(x) - f(x)^3) \\ & = & 4g(x)^3 - 3g(x) + i(3f(x) - 4f(x)^3) \\ & = & g(3 x) + i f(3 x) }$

so that $h(3 x) = h(x)^3$, and in particular $h(3 x)$ is on the unit circle provided that $h(x)$ is, so that the induction goes through.

We also have of course $h'(0) = i$. Intuitively, this means that for any $t$ we have

$h(t) = h(\frac{t}{3^n})^{3^n} \sim (1 + \frac{i t}{3^n})^{3^n}$

where the right side tends to $\exp(i t)$ as $n \to \infty$. In other words, $h(t) = \exp(i t)$, whence $f(t) = \sin(t)$ by matching imaginary parts. The goal now is to make this intuitive argument rigorous.

###### Proposition

For any given $t \in \mathbb{R}$ and $\eta \gt 0$, there exists $N$ such that

${|h(t) - (1 + \frac{i t}{3^n})^{3^n}|} \lt \eta$

whenever $n \geq N$.

Since (as is well-known) $lim_{m \to \infty} (1 + \frac{i t}{m})^m = \exp(i t)$, it follows at once from this proposition that $h(t) = \exp(i t)$.

###### Proof

Since $h'(0) = i$ and $h(0) = 1$, we can for any $\epsilon \gt 0$ choose $N$ so large that

${|h(\frac{t}{3^n}) - 1 - i\frac{t}{3^n}|} \leq \epsilon\; {|\frac{t}{3^n}|}$

for all $n \geq N$. In other words, for such $n$ we have $h(\frac{t}{3^n}) = 1 + \frac{i t}{3^n} + \zeta\frac{t}{3^n}$ for some $\zeta$ with ${|\zeta|} \leq \epsilon$. There we have

$h(t) = h(\frac{t}{3^n})^{3^n} = (1 + \frac{i t}{3^n} + \zeta\frac{t}{3^n})^{3^n}$

and so we proceed to bound the difference

$(1 + \frac{i t}{3^n} + \zeta\frac{t}{3^n})^{3^n} - (1 + \frac{i t}{3^n})^{3^n}.$

Put $m = 3^n$. We have

$(1 + \frac{i t}{m} + \zeta\frac{t}{m})^m = \sum_{k=0}^m \binom{m}{k} (1 + \frac{i t}{m})^{m-k} (\frac{\zeta t}{m})^k$

so that the difference above is bounded above by

$\sum_{k=1}^m \binom{m}{k} (1 + {|\frac{t}{m}|})^{m-k} ({|\frac{\zeta t}{m}|})^k.$

Clearly we have

$(1 + {|\frac{t}{m}|})^{m-k} \leq (1 + {|\frac{t}{m}|})^m \leq \exp({|t|}).$

Hence the difference is bounded above by

$\exp({|t|}) \cdot \sum_{k=1}^m \binom{m}{k} ({|\frac{\zeta t}{m}|})^k = \exp({|t|}) \cdot [(1 + {|\frac{\zeta t}{m}|})^m - 1] \leq \exp({|t|}) \cdot [\exp({|\zeta t|}) - 1] \leq \exp({|t|}) \cdot [\exp({|\epsilon t|}) - 1].$

This can be made as small as desired, by taking $\epsilon$ small enough. This completes the proof.

## Delange’s characterization

A very pretty characterization of the sine is due to Hubert Delange:

###### Theorem

Suppose $f: \mathbb{R} \to \mathbb{R}$ is a $C^\infty$ function such that all of its $n^{th}$ derivatives satisfy ${|f^{(n)}(x)|} \leq 1$ for all $x$, and $f'(0) = 1$. Then $f(x) = \sin(x)$ for all $x$.

A proof may be found in this online text by Jean Saint Raymond, Théorème 14.4.3 (page 156). I will follow that proof here.

First, a few preliminary observations. Put

$F(z) = \sum_{k \geq 0} f^{(k)}(0) \frac{z^k}{k!}$

which is an entire analytic function. By the Taylor remainder formula for $C^n$ functions on $\mathbb{R}$, we have

${|f(x) - \sum_{k = 0}^{n-1} f^{(k)}(0) \frac{x^k}{k!}|} \leq \sup_z {|f^{(n)}(z)|}\; \frac{{|x|}^n}{n!} \leq \frac{{|x|}^n}{n!}$

so that taking $n \to \infty$, we infer $f(x) = F(x)$ for all $x \in \mathbb{R}$. Thus the real-valued $f$ analytically continues to an entire analytic function $F$, and we get to invoke the theory of entire analytic functions.

The order of $F$ as an entire function is $1$ because we have the growth rate

${|F(x + i y)|} = {|\sum_{n \geq 0} f^{(n)}(x) \frac{(i y)^n}{n!}|} \leq \sum_{n \geq 0} \frac{{|i y|}^n}{n!} = \exp({|y|}).$

Now let $a$ be an arbitrary point in the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$; we will prove $f(a) = \sin(a)$ (whence $f(x) = \sin(x)$ by uniqueness of analytic continuation).

###### Lemma

$D_{x = a} \frac{f(x)}{\cos(x)} = \sum_{k \in \mathbb{Z}} (-1)^k \frac{f(k\pi + \frac{\pi}{2})}{(k\pi + \frac{\pi}{2} - a)^2}$.

###### Proof

Let $g$ be the meromorphic function defined by

$g(z) \coloneqq \frac{F(z)}{(z - a)^2 \cos(z)}$

where $\cos(z) = \frac{e^{i z} + e^{-i z}}{2}$. Consider the square contour $C_n$ with vertices at the points $n\pi(\pm 1 \pm i)$, $n \geq 1$. We have

• $\cos(x + i y) = \cos(x)\cosh(y) + i \sin(x)\sinh(y)$,

• ${|\cos(x + i y)|}^2 = \cos^2(x)\cosh^2(y) + \sin^2(x)\sinh^2(y) = \cos^2(x)(1 + \sinh^2(y)) + \sin^2(x)\sinh^2(y) = \cos^2(x) + \sinh^2(y)$

where $\cos^2(x) + \sinh^2(y) = 1 + \sinh^2(y) = \cosh^2(y)$ along the vertical line segments of $C_n$, and $\cos^2(x) + \sinh^2(n\pi) \geq \sinh^2(n\pi)$ along the horizontal line segments of $C_n$ (where $y = \pm n\pi$). Either way we have the easy estimate

${|cos(x + i y)|} \geq \frac{e^{{|y|}}}{3}$

over the contour $C_n$. Moreover $\frac1{{|z-a|}^2} \leq \frac1{(n\pi - {|a|})^2}$ over the contour, and

$\frac{{|F(z)|}}{{|cos(z)|}} \leq e^{{|y|}} \cdot \frac{3}{e^{{|y|}}} \leq 3$

over the contour according to the growth rate of $F(z)$ observed above, and so

${|\int_{C_n} g(z) d z|} \leq (\sup {|g(z)|}) \cdot length(C_n) \leq \frac{3}{(n\pi - {|a|})^2} \cdot (8 n \pi)$

which tends to $0$ as $n \to \infty$. On the other hand, by a simple residue calculation, we calculate

$\frac1{2\pi i} \int_{C_n} g(z) d z = D_{z = a} \frac{F(z)}{\cos(z)} + \sum_{k = -n}^{n-1} (-1)^{k+1} \frac{F(\frac{\pi}{2} + k\pi)}{(\frac{\pi}{2} + k\pi - a)^2}$

where the first term on the right comes from the double pole of $g(z)$ at $z = a$ and the rest come from the zeroes of $\cos(z)$. Thus in the limit we have

$0 = D_{x = a} \frac{f(x)}{\cos(x)} + \sum_{k = -\infty}^{\infty} (-1)^{k+1} \frac{f(\frac{\pi}{2} + k\pi)}{(\frac{\pi}{2} + k\pi - a)^2}$

which completes the proof.

In particular, for the case $f(x) = \sin(x)$, we have

###### Corollary

$D_{x = a} \frac{\sin(x)}{\cos(x)} = \sum_{k \in \mathbb{Z}} \frac1{(\frac{\pi}{2} +k\pi - a)^2}$.

In particular, taking $a = 0$, we have

$1 = \sum_{k \in \mathbb{Z}} \frac1{(\frac{\pi}{2} +k\pi)^2}$

and so

$0 = 1 - f'(0) = \sum_{k \in \mathbb{Z}} \frac{1 - (-1)^k f(\frac{\pi}{2} + k\pi)}{(\frac{\pi}{2} + k\pi)^2}.$

But since ${|f(x)|} \leq 1$ for all $x$, we have $1 - (-1)^k f(\frac{\pi}{2} + k\pi) \geq 0$. Thus the last equation forces

$(-1)^k f(\frac{\pi}{2} + k\pi) = 1$

for all $k \in \mathbb{Z}$. Plugging this back into the lemma, we deduce

$D_{x = a} \frac{f(x)}{\cos(x)} = \sum_{k \in \mathbb{Z}} \frac1{(k\pi + \frac{\pi}{2} - a)^2} = D_{x = a} \frac{\sin(x)}{\cos(x)}.$

As this is true for every $a \in (-\frac{\pi}{2}, \frac{\pi}{2})$, we see $f(a) - \sin(a) = \lambda \cos(a)$ for some constant $\lambda$, for all $a$ in this range. Thus $f'(a) = \cos(a) - \lambda \sin(a)$.

Finally, $f(\frac{\pi}{2}) = 1$ and also $f'(\frac{\pi}{2}) = -\lambda$. But as $1$ is a maximum value, this forces $-\lambda = 0$, and thus we have shown $f(a) = \sin(a)$. The proof of Delange’s theorem is now complete.

## Roe’s characterization

A slightly different characterization was given by J. Roe (first name is?):

###### Theorem

Suppose there is a collection of functions $\ldots, f_{-2}, f_{-1}, f_0, f_1, f_2, \ldots$ such that $f_{n+1} = f_{n}'$ and $\sup {|f_n|} \leq M$ for all $n$, for some uniform bound $M \geq 0$. Then $f = f_0$ is of the form $f(x) = a\sin(x + b)$ for some constants $a, b$.

At first this might appear to be a weakening of Delange’s theorem, in that one considers a sequence $\{f_n\}_{n \in \mathbb{Z}}$ that is infinite in both positive and negative directions, rather than the positive direction only. Notice however that Delange assumes that ${|f'|}$ actually attains this bound $M$ (by shifting the sequence, he could ask merely that one of the ${|f^{(n)}|}$ attains this $M$, for some $n \geq 1$). For example, as pointed out by Roe, the function $f(x) = \sin(x) + \sin(x/\pi)$ has all of its derivatives bounded – but observe that no derivative actually hits a uniform upper bound $M$.

It seems odd to me that Roe makes no reference whatever to Delange’s earlier work, as their statements are so close in spirit. His method of proof is different to the complex-analytic one given above; Roe uses Fourier analysis and distribution theory instead.

## References

• nLab, pi
• H. Delange, Caractérisations des fonctions circulaires, Bull. Sc. Math. 91 (1967), 65-73.

• J. Roe, A characterization of the sine function, Math. Proc. Cam. Phil. Soc., Vol. 87 Issue 1 (January 1980), pp 69-73.

Revised on August 3, 2019 at 09:39:59 by Todd Trimble