Theorem 3 (Rainbow)

Let $\phi(x)$ be a proper frequency distribution such that there exist a positive integer k >0 and a real finite constant $\alpha \gt 0$ for which $\phi^k (\alpha x) = \phi(x)$:

1.

If $\alpha <1$, then $\phi(x)$ is of the same type as

$$\begin{array} {r@{~}c@{~}l} \psi_a(x) & = & \begin{cases} e^... ... a & = & \displaystyle - \frac{\log k}{\log \alpha}.\end{array}$$

2.

If $\alpha \gt 1$, then $\phi(x)$ is of the same type as

$$\begin{array} {r@{~}c@{~}l} \phi_a (x) & = & \begin{cases} 0... ...e } a & = & \displaystyle\frac{\log k}{\log \alpha}.\end{array}$$

Proof: The possibility that $\alpha = 1$ is excluded, since if $\alpha = 1$, $\phi^k(x) = \phi(x)$ and $\phi (x)\equiv 1$ or $\phi(x) \equiv 0$. This contradicts the hypothesis that $\phi(x)$ is a proper frequency distribution.

If x = 0, then $\phi^k (0) = \phi(0)$ and so $\phi(0) = 0$ or $\phi (0) = 1$. In the former case, $\phi(x) = 0$ for x < 0. In the latter, $\phi (x) = 1$ for x > 0.

Let x₀ be any value of x such that $0 < \phi (x_0) < 1$. Such values of x exist, since $\phi(x)$ is proper. By hypothesis,

$$\phi^k (\alpha x_0) = \phi (x_0).$$

It follows that

$$\phi (\alpha x_0) \gt \phi (x_0).$$

But, since $\phi(x)$ is monotonic increasing and $\alpha \neq 1$,

$$\alpha x_0 \gt x_0 \quad \text{and} \quad (\alpha - 1) x_0 \gt 0.$$

As a consequence,

$$\begin{array} {rl} \text{if } \alpha < 1, \quad & \text{then... ...t{if } \alpha \gt 1, \quad & \text{then } x_0 \gt 0.\end{array}$$

Because of this, together with the previous result,

$$\begin{array} {rl} \text{if } \alpha < 1, \quad & \text{then... ... \text{then } \phi (0) = 0, \quad \text{for } x < 0.\end{array}$$

It will be demonstrated that

$$\begin{array} {rl} \text{if } \alpha < 1, \quad & \text{then... ...xt{then } \phi (x) \gt 0, \quad \text{for } x \gt 0.\end{array}$$

Consider first $\alpha <1$ and suppose that there exists x < 0 such that $\phi (x) = 1$. Let x' be the greatest lower bound of such x values. If $x' = -\infty$, then $\phi (x)\equiv 1$ for all x, and $\phi(x)$ is not a proper frequency distribution. Hence, x' is finite. Choose p such that $\alpha < p < 1$. Then, since x' < 0, it follows that px' > x' and $\phi (px') = 1$. Now, set $\alpha z = px'$. By hypothesis,

$$\phi (z) = \phi^k (\alpha z) = \phi^k (px') = 1.$$

But, $z/x' = p/\alpha \gt 1$ and z < x'. This contradicts the definition of x' and establishes that if $\alpha <1$, then $\phi(x) < 1$ for x < 0.

Now, consider $\alpha \gt 1$ and suppose that there exists x > 0 such that $\phi(x) = 0$. Let x' be the least upper bound of such x values. If $x' = \infty$, then $\phi(x) \equiv 0$ for all x, and $\phi(x)$ is not a proper frequency distribution. Hence, x' must be finite. Choose p so that $1/\alpha < p < 1$ and set $z/\alpha = px'$. By hypothesis,

$$\phi^k (z) = \phi \left( \frac{z}{\alpha} \right) = \phi (px') = 0.$$

Hence, $\phi (z) = 0$. But, $z/x' = \alpha p \gt 1$ and z > x'. This contradicts the definition of x' and establishes that if $\alpha \gt 1$, then $\phi (x) \gt 0$ for x > 0. In summary,

$$\begin{array} {r@{~}c@{~}l} \text{(a)} \quad \text{if } \alp... ...\leq 0, \\ \gt 0, & \text{for } x \gt 0.\end{cases}\end{array}$$

The definite type of $\phi(x)$ for (a) and (b) cases will now be determined. Cases (a) and (b) will each be considered in turn.

Case (a) $\alpha <1$: Let $G (\log (-x))$ be the function for which

$$\begin{array} {r@{~}c@{~}l} \phi (x) & \equiv & e^{(-x)^a} G... ... & \displaystyle - \frac{\log k}{\log \alpha} \gt 0.\end{array}$$

$G (\log (-x))$ may have discontinuities for any other features required to maintain the identity. Any other functional representation of $\phi(x)$ would necessarily be identical with this representation for x < 0. (It has already been determined, of course, that $\phi (x) = 1$ for $x \geq 0$.)

It will be demonstrated that $G (\log (-x))$ is a monotonic function of $\log (-x)$ for x < 0. Assume that $\log (-x_1) < \log (-x_2)$.Then, -x₁ < -x₂ and x₁ > x₂. Since $\phi(x)$ is monotonic increasing, $\phi (x_1) \geq \phi (x_2)$. As a consequence of the properties of $\log x$ it follows that

$$\begin{array} {r@{~}c@{~}l} 0 \geq \log \phi (x_2) - \log \p... ...)^a \left[ G (\log (-x_2)) - G (\log (-x_1))\right].\end{array}$$

Since (-x₂)^a > 0, it results that

$$G (\log (-x_2)) \leq G (\log (-x_1)),$$

and $G (\log (-x))$ is a monotonic function of $\log (-x)$. If the $G (\log (-x))$ representation of $\phi(x)$ is exact, then by hypothesis,

$$\begin{array} {r@{~}c@{~}l} \left[ e^{(-\alpha x)^a G(\log (... ...x)^a G(\log (-\alpha x)) & = & (-x)^a G (\log (-x)).\end{array}$$

But, by definition of a, $k \alpha^a = 1$. So

$$G (\log \alpha + \log (-x)) = G (\log (-x)).$$

Hence, $G (\log (-x))$ is a periodic function of $\log (-x)$ with period $\log \alpha$. But, it was shown previously that $G (\log (-x))$ was monotonic. Hence, the only possibility for $G (\log (-x))$is

$$G (\log (-x)) = c, \qquad c \text{ constant}.$$

Thus,

$$\phi (x) = e^{c(-x)^a}, \qquad \text{for } x < 0.$$

Now, c must be negative if $0 = \lim_{x \to-\infty} \phi(x)$. In conclusion, $\phi(x)$ is of the type

$$\psi_a (x) = \begin{cases} e^{-(-x)^a}, & \text{for } x \leq... ... & \text{for } x \gt 0, \quad \text{if } \alpha < 1.\end{cases}$$

Case (b) $\alpha \gt 1$: Let $G(\log x)$ be the function for which

$$\begin{array} {r@{~}c@{~}l} \phi(x) & \equiv & e^{x^{-a}G(\l... ...& = & \displaystyle\frac{\log k}{\log \alpha} \gt 0.\end{array}$$

As before, $G(\log x)$ may have discontinuities or any other properties needed to maintain the identity. It will be demonstrated that $G(\log x)$ is a monotonic function of $\log x$. Assume that $\log x_1 < \log x_2$. Then, x₁ < x₂ and $\phi (x_1) \leq \phi (x_2)$. It follows that

$$\begin{array} {r@{~}c@{~}l} 0 & \leq & \log \phi (x_2) - \lo... ...{x^a_2} \left[ G (\log x_2) - G (\log x_1) \right]. \end{array}$$

Hence, $G(\log x_2) \geq G (\log x_1)$. Now, by hypothesis, if the $G(\log x)$ representation of $\phi(x)$ is exact,

$$\begin{array} {r@{~}c@{~}l} \left[ e^{(\alpha x)^{-a}G(\log ... ...a} G (\log \alpha + \log x) & = & x^{-a} G (\log x).\end{array}$$

But, a was defined so that $k \alpha^{-1} = 1$, and as a result,

$$G (\log x + \log \alpha) = G (\log x).$$

Thus, $G(\log x)$ is both periodic with period $\log \alpha$ and monotonic. It follows that

$$G(\log x) = c, \qquad c \text{ constant}.$$

Thus,

$$\phi (x) = e^{cx^{-a}}, \qquad \text{for } x \gt 0.$$

By inspection, c must be negative if $\lim_{x \to 0} \phi (x) = 0$.

In summary then, $\phi(x)$ must be of the same type as

$$\phi_a (x) = \begin{cases} 0, & \text{for } x < 0, \\ e^{-x^{-a}}, & \text{for } x \geq 0. \end{cases}$$