# Functional equation $\int_z^{2z} [f(x)-f(z)] dx = 0$

Suppose a continuous function $$f:[0,1] \to \mathbb{R}$$ satisfies the following equation for all $$z \in \left(0,\frac{1}{2}\right)$$, $$\int_z^{2z} [f(x)-f(z)] dx = 0.$$ It is clear that a constant function $$f(x)=c$$ satisfies it. I would like to prove that there are no other such continous functions.

Note: this is a missing part of a larger proof I'm working on. I've already verified that if $$f$$ is a polynomial then it must be constant. Any hints on how to prove it for arbitrary continuous functions would be appreciated.

• Do you want to truncate your domain of integration to $[z,2z] \cap [a,b]$? – user101142 Jan 4 '19 at 13:53
• Good point, thanks! I updated the question slightly. – TomH Jan 4 '19 at 15:00
• With the update, the statement is no longer true: If $a<b<2a$ then the condition is vacuously true, but $f$ needn't be constant. And I suspect it is also false if $[a,b] \cap [2a,2b]$ is small but nonempty. The most natural formulation of the problem seems to be for functions on $[0, \infty)$, but is this what is relevant to your application? – David E Speyer Jan 4 '19 at 15:20
• Again, good points! In my application $a=0$ and $0<b<\infty$. But it would be enough to have it for $[0,1]$ interval. I'll update the text once more. – TomH Jan 4 '19 at 15:38

Let $$p$$ be a zero of $$2^{p+1}-p-2$$ other than $$-1$$ and $$0$$ (e.g. one is approximately $$2.54536493037426+10.7539751752688 i$$). Then the real and imaginary parts of $$f(x) = x^p$$ satisfy the equation. Note that (with $$f(0)=0$$) $$f$$ is continuous on $$[0,\infty)$$ if $$\text{Re}(p) > 0$$.
• "... $p$ be a zero of $2^{p+1} - p - 2$ ...": should there be an $x$ somewhere? – auniket Jan 4 '19 at 16:28
• No, there shouldn't. I mean a solution of the equation $2^{p+1} - p - 2 = 0$. – Robert Israel Jan 4 '19 at 16:43
• This is more complex than I expected. Do you mean $f(x) = Re(x^p)$? Any hints on how to prove this? In my application, I also have that $f(x)<0$ for all $x$. – TomH Jan 4 '19 at 17:11
• Since the equation is linear in $f$, a linear combination of solutions is a solution. So take $f(x) = -1 + \text{Re}(x^p)$, which will be negative for $0 < x < 1$. – Robert Israel Jan 4 '19 at 19:03
• $\text{Re}(x^p)$ is the derivative of $\text{Re}(x^{p+1}/(p+1))$. – Robert Israel Jan 5 '19 at 23:50