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Let $f$ be a strictly positive function such that $\int_{0}^{\infty}f(x)dx=\int_{0}^{\infty}xf(x)=1$ (i.e., a probability density function with expectation one). Let also $g$ be a nonnegative nonconstant function which satisfies $\int_{0}^{\infty}g(ax)f(x)dx=a$, $\forall a>0$. Does this imply that $g(x)=x$ a.e.?

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The substitution $x=e^t$ will turn this integral into a convolution. Then use Fourier transform. –  Michael Renardy Feb 5 at 0:12
    
@MichaelRenardy: The problem is, $g(e^t)$ is not quite in the domain where Fourier transform is defined. –  Alexander Shamov Feb 5 at 0:19
    
@StephanSturm: Or better, consider functions up to equality almost everywhere. –  Alexander Shamov Feb 5 at 3:10
    
@StephanSturm: You are right. I should have asked: Does this imply that $g(x)=x$ a.e.? (That's my actual question.) –  Gili Feb 5 at 7:44

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The answer is no. Let $h$ be a positive function on $R$ with the following properties:

a) Fourier transform has a pair of non-real zeros $\lambda, -\overline{\lambda}$, symmetric with respect to the imaginary axis.

b) $\int_{-\infty}^\infty h(t)dt=1,$

c) $\int_{-\infty}^\infty e^t h(t)dt=1.$

Condition a) means $$\int_{-\infty}^\infty e^{-i\lambda t}h(t)dt=0.$$ Multiplying this by $a^{-i\lambda}$, $a>0$ we obtain $$\int_{-\infty}^\infty e^{-i\lambda(t+\log a)}h(t)dt=0,$$ for every $a>0$. Making change of the variable $t=\log x$, we obtain $$\int_0^\infty e^{-i\lambda\log (ax)}h(\log x)\frac{dx}{x}=0.$$ Putting $f(x)=h(\log x)/x$, we obtain a function with the properties you stated; they follow from b), c). Thus $$\int_0^\infty (ax)^{-i\lambda} f(x)dx=0.$$ Now $g(x)=x+x^{-i\lambda}$ is a function $g$ which satisfies your identity. If a positive function is required, we need $\lambda=\sigma-i$, and $$g(x)=x+k(x^{i\lambda}+x^{-i\overline{\lambda}})=x(1+k\cos(\sigma\log x)),$$ which is positive if $0<k<1$.

Existence of such $h$ is pretty evident. Take any probability density whose Fourier transform is analytic in some strip $|Im z|<c$ and has some zeros with negative imaginary part. Scaling will give you a zero whose imaginary part is $-1$. To achieve c), shift $h$, replacing it with $h(t+c)$. This does not affect the zeros of Fourier transform. Taking $h$ with Fourier transform having infinitely many zeros on a horizontal line, you obtain an infinite dimensional space of $g$ with fixed $f$.

Edit. Sorry, I did not notice that you also need $g$ positive. I modified the example to achieve this additional property.

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As far as I understand this makes $f$ to have bounded support, right? But $f$ is strictly positive, so the support is the whole $(0,\infty)$. –  Gili Feb 5 at 7:37
    
No problem. All we need is that Fourier transform of $h$ is analytic in some strip of complex plane and has some non-real zeros. –  Alexandre Eremenko Feb 5 at 14:25
    
I'm accepting your answer, thanks. But I'll return with additional restrictions on $f$. –  Gili Feb 6 at 5:03
    
OK, try to impose more restrictions on $f$. –  Alexandre Eremenko Feb 6 at 13:43

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