# $f(\alpha x) = f(x)^{\beta}$ under different constraints

With $\alpha > 0,\, \beta \in \Bbb R^*,\, \alpha, \beta \neq 1$ and $f : \Bbb R \to \Bbb R_+^*$, let's consider the functional equation $$f(\alpha x) = f(x)^{\beta} \tag{\Xi}$$ or equivalently $g(\alpha x) = \beta g(x)$ for $g = \ln f$.

The case where $\alpha = \sqrt2$, $\beta = 2$ and $f \in \mathcal C^2$ has already been solved here : Solving $(f(x))^2 = f(\sqrt{2}x)$ (the answer is $\exists \lambda\mid f(x) = e^{\lambda x^2}$).

What if we relax/change some of the constraints, for instance:

• Keeping $f$ regular (say $\mathcal C^{\infty}$) but setting $\alpha, \beta$ generic
• $f \in \mathcal C^0$
• $f \in L^1$
• (other ideas?)
• I'm not sur how this can help but note that a function satisfying $h(\lambda x)=\lambda^ph(x)$ for every $\lambda>0$ is said to be homogeneous of degree $p$. – Surb Mar 25 '15 at 12:55

Summary at the bottom.

$$g(\alpha x)=\beta g(x)$$

$$\gamma = \frac{\ln |\beta|}{\ln\alpha}$$

If $$\beta >0$$, $$\beta=\alpha^\gamma$$ and $$g(\alpha x)=\alpha ^\gamma g(x)$$.

We define $$h(x)$$, for $$x\neq 0$$, as $$h(x)= g(x)x^{-\gamma}$$. Then: $$h(\alpha x)=g(\alpha x){\alpha^{-\gamma} x^{-\gamma}}=g(x){x^{-\gamma}}=h(x)$$

Now let $$k_1(x)=h(\alpha^x)$$ and $$k_2(x)=h(-\alpha^x)$$.

Then $$k_1$$ and $$k_2$$ can be any periodic functions, with a period of $$1$$.

Therefore:

$$g(x)=\cases{x^\gamma k_1(\log_\alpha(x)) & if x>0 \cr x^\gamma k_2(\log_\alpha(-x)) & if x<0 }$$

If $$\beta<0$$, $$\beta=-\alpha^\gamma$$. We use the same definition for $$h$$, $$k_1$$ and $$k_2$$, but now we have: $$k_1(x+1)=-k_1(x)$$ and $$k_2(x+1)=-k_2(x)$$.

So $$k_1$$ and $$k_2$$ can be any antiperiodic functions, with a period of $$1$$.

And:

$$g(x)=\cases{x^\gamma k_1(\log_\alpha(x)) & if x>0 \cr x^\gamma k_2(\log_\alpha(-x)) & if x<0 }$$

If $$g \in \mathcal C^n$$, then $$k_1$$, $$k_2 \in \mathcal C^n$$.

If $$n\geq\gamma$$, then in a neighborhood of $$0^+$$: $$g(x)=\sum\limits_{k=0}^{n}\frac{g^{(k)}(0)}{k!}x^k + o(x^n)= x^\gamma k_1(\log_\alpha(x))$$ $$k_1(\log_\alpha(x))=\sum\limits_{k=0}^{n}\frac{g^{(k)}(0)}{k!}x^{k-\gamma} + o(x^{n-\gamma})$$

So $$k_1(\log_\alpha(x))$$ has a limit at $$0^+$$ (which could be infinite), so $$k_1$$ has a limit at $$-\infty$$. So it must be a constant (since it is periodic of period at most $$2$$). We can use the same reasoning for $$k_2$$. So $$g(x)=\cases{c_1x^\gamma & \text{if } x>0 \cr c_2x^\gamma & \text{if } x<0 }$$

Therefore, all the derivatives of $$g$$ at $$0$$ of order smaller than $$\gamma$$ are $$0$$, and if $$n>\gamma$$, then $$g$$ is not $$p=\lceil \gamma\rceil$$ times differentiable near $$0$$. So $$\gamma = n$$ and $$c_1=c_2$$.

To sum it up:

In general:

$$g(x)=\cases{x^\gamma k_1(\log_\alpha(x)) & if x>0 \cr x^\gamma k_2(\log_\alpha(-x)) & if x<0 }$$

Where $$k_1$$ and $$k_2$$ are two periodic (if $$\beta>0$$) or antiperiodic (if $$\beta<0$$) functions, of period $$1$$.

If $$g\in\mathcal C^{n}$$, so are $$k_1$$ and $$k_2$$. And if $$n\geq\gamma$$, $$n = \gamma$$, and $$g(x)=\lambda x^n$$.

If there are any mistakes, please let me know.

• Very nice, much better approach than mine. Comment #1 : if $\beta \in (0,1)$ and $\alpha > 1$, $\gamma < 0$ and if $g$ is continuous in $0$, that implies $k_1 = k_2 = g = 0$ (see my answer). – Alexandre Halm Apr 2 '15 at 6:24

Assume $$f$$ (and $$g$$) is $$\mathcal C^{\infty}$$ (we will see that $$\mathcal C^p$$ with $$p$$ large enough is sufficient to get the same result).

Derivating the equality $$g(\alpha x) = \beta g(x)$$ $$k$$ times gives $$\alpha^k g^{(k)}(\alpha x) = \beta g^{(k)}(x)$$.

Let's assume first that $$\alpha, \beta > 1$$. Replacing $$x$$ by $$(x/\alpha)$$ $$n$$ times in the last equality yields

$$g^{(k)}(x) = \frac{\beta}{\alpha^k}g^{(k)}\left(\frac x{\alpha}\right) = \ldots = \left(\frac{\beta}{\alpha^k}\right)^n g^{(k)}\left(\frac x{\alpha^n}\right) \tag{\phi}$$

Now let's pick $$k = \left\lceil \frac{\ln \beta}{\ln \alpha} \right\rceil$$ so that $$\left|\frac{\beta}{\alpha^k}\right| \le 1$$ and $$\frac{\beta}{\alpha^{k-1}} > 1$$. Since $$g^{(k)}$$ is continuous in $$0$$, taking the limit in $$(\phi)$$ when $$n \to \infty$$ commands $$g^{(k)}(x) = 0$$ for all $$x$$. So $$g$$ must be a polynomial of degree $$\le k$$.

If we replace $$k$$ with $$p \le k-1$$ in $$(\phi)$$, the limit when $$n \to \infty$$ now commands $$g^{(p)}(0) = 0$$ since $$\frac{\beta}{\alpha^{p}} \ge \frac{\beta}{\alpha^{k-1}} > 1$$. That implies that $$g(x)$$ is of the form $$g(x) = \lambda x^k$$. But $$g(\alpha x) = \lambda \alpha^k x^k= \beta g(x) = \beta \lambda x^k$$ forces $$\beta = \alpha^k$$.

Now if we assume $$\alpha, \beta < 1$$, it is obvious that $$(\Xi)$$ is equivalent to $$g(\tfrac1{\alpha}x) = \tfrac1{\beta}g(x)$$ and since $$\tfrac1{\alpha}, \tfrac1{\beta} >1$$, we are back in the former case.

Wrap-up: if $$\alpha, \beta>1$$ or $$\alpha, \beta<1$$, noting $$k = \frac{\ln \beta}{\ln \alpha}$$, $$(\Xi)$$ only has $$\mathcal C^{\lceil k \rceil}$$ solutions if $$k \in \Bbb N$$ and in this case the solutions are of the form $$f(x) = e^{\lambda x^k}$$.

Let's now assume $$\alpha >1$$ and $$0 < \beta <1$$, with $$f$$ (and $$g$$) only $$\mathcal C^0$$ (continuous). Rewriting $$(\Xi)$$ as $$g(x) = \beta g(\tfrac x{\alpha})$$, we get by iterating $$\forall n, g(x) = \beta^n g(\tfrac x{\alpha^n})$$ and, by continuity of $$g$$ in $$0$$, $$g(x) = 0$$. As earlier, the case $$\alpha < 1$$ and $$\beta >1$$ has the same solution.

Wrap-up 2: if $$\alpha >1, 0<\beta<1$$ or $$\alpha <1, \beta>1$$, the only $$\mathcal C^0$$ (even bounded) solution to $$(\Xi)$$ is $$f = constant$$.

(source: mit.edu)