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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.

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    $\begingroup$ Do you want to truncate your domain of integration to $[z,2z] \cap [a,b]$? $\endgroup$
    – user101142
    Commented Jan 4, 2019 at 13:53
  • $\begingroup$ Good point, thanks! I updated the question slightly. $\endgroup$
    – TomH
    Commented Jan 4, 2019 at 15:00
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    $\begingroup$ 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? $\endgroup$
    – David E Speyer
    Commented Jan 4, 2019 at 15:20
  • $\begingroup$ 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. $\endgroup$
    – TomH
    Commented Jan 4, 2019 at 15:38

1 Answer 1

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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 $\text{Re}(p) > 0$.

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  • $\begingroup$ "... $p$ be a zero of $2^{p+1} - p - 2$ ...": should there be an $x$ somewhere? $\endgroup$
    – pinaki
    Commented Jan 4, 2019 at 16:28
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    $\begingroup$ No, there shouldn't. I mean a solution of the equation $2^{p+1} - p - 2 = 0$. $\endgroup$
    – Robert Israel
    Commented Jan 4, 2019 at 16:43
  • $\begingroup$ 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 $x$. $\endgroup$
    – TomH
    Commented Jan 4, 2019 at 17:11
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    $\begingroup$ 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 $0 < x < 1$. $\endgroup$
    – Robert Israel
    Commented Jan 4, 2019 at 19:03
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    $\begingroup$ $\text{Re}(x^p)$ is the derivative of $\text{Re}(x^{p+1}/(p+1))$. $\endgroup$
    – Robert Israel
    Commented Jan 5, 2019 at 23:50

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