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which function could satisfy the following, for a certain $c\ne1$

$f'(x) = f(cx)$

...beyond the trivial $f=0$

i've been thinking about it for a while.

for a simpler case:

$f'(x)= f(x+c)$

i've found $e^{xe^v}$ where $v$ is the solution of $c=\frac{v}{e^v}$

Zoltán Haindrich
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    Use Taylor at the origin. The equation allows you to compute the value of all derivatives at the origin in terms of $f(0)$ and $c$. – OR. Jul 17 '13 at 17:29

2 Answers2

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We get that $$f^{(k)}(x)=c^{\frac{k(k-1)}{2}}f(c^kx).$$

From this we get that $$f(x)=\sum_{k=0}^{\infty}\frac{c^{\frac{k(k-1)}{2}}}{k!}f(0)x^k.$$

OR.
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For $c>1$, let $f_0\colon [1,c]\to\mathbb R$ be smooth with $f_0'(1)=f_0(c)$. Then extend this to the right and left using your functional equation. That is, if we already have some smooth $f_n\colon[a,b]\to \mathbb R$, define $$ f_{n+1}(x)=\begin{cases}f_n(x)&\text{if }a\le x\le b,\\ f_n'(x/c)&\text{if }b\le x\le bc,\\ f_n(a)-\int_x^af_n(ct)\,\mathrm dt&\text{if }\frac ac\le x\le a. \end{cases}$$ This is a smooth function and taking the union over all $n$, we get a smooth solution $f\colon(0,\infty)\to\mathbb R$ of your functional equation.

Hagen von Eitzen
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    Why is this smooth at powers of $c$? It seems that $f'(c+\epsilon)$ is determined by $f''(1+\epsilon/c)$, which I could choose completely independently of $f'(c-\epsilon)$. I agree that there are smooth solutions constructed as you have suggested, but you have to insist on this funny version of periodic boundary conditions for $f_0$ to all orders. – Theo Johnson-Freyd Jul 17 '13 at 18:05
  • I guess it is clear that $f$ is continuous. But then also $x\mapsto f'(x)=f(cx)$, $x\mapsto f''(x)=f(c^2x)$ and so on are continuous. – Hagen von Eitzen Jul 17 '13 at 18:21
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    Well, if $f'(x) = f(cx)$, then $f''(x) = cf'(cx) = cf(c^2x)$, but I don't see when $f$ automatically is continuous at $c^2$. Indeed, let's take $f_0(x) = x+1-c$ for $x\in [1,c]$. Then $f'_0(x) = 1 = f_0(c)$. Your formula would extend this to $f_1(x) = 1$ for $x\in [c,c^2]$, and $f_2(x) = 0$ for $x>c^2$, and there is a discontinuity at $c^2$. – Theo Johnson-Freyd Jul 17 '13 at 19:44