I was wondering, if it is possible to show, that if I have any function $f$ over $\mathbb{R}$, such that $\int_\mathbb{R} f(x) dx = 1$, $\int_{0}^{\infty} x(f(x)-f(-x)) dx = 0$, and $\int_0^\infty x^2(f(x)+f(-x)) dx = 0$, then f cannot be non-negative.
I tried to write down the conditions on $f$ such that the individual integrals are satisfied, i.e. for the second integral we have either
(I) $f$ is even or
(II) there is some point $x$ such that $f(x)-f(-x)<0$ (and the set of points $x$ for which it holds has positive measure).
And similarly for the last integral, either
(III) $f$ is odd or
(IV) there is some point $x'$ such that $f(x')+f(-x')<0$ (and the set of points $x'$ such that this holds has positive measure).
Based on these conditions, I tried to show that f must be negative at some point in $\mathbb{R}$. And I managed for all cases except (II)+(IV).
For case (II)+(IV) I tried to assume that f was non-negative, but I couldn't reach a contradiction.
I would be very grateful, if anyone could give me a hint as how to prove it.
There is a very well known non-negative function satisfying these properties: $f(x)=\frac 1 {\sqrt {2\pi}} e^{-x^{2}/2}$.
The edited question is easy to answer: Suppose $f \geq 0$ a.e. Then the last condition implies that $f(x)+f(-x)=0$ a.e. on $(0,\infty)$ and hence $f=0$ a.e on $\mathbb R$. But then the first condition does not hold.