I've been doing a math problem about traffic lights and the following interesting question came up. Not homework.
say:
$\int_0^{+\infty} f(x) dx = 0 $
$\int_0^{+\infty} f(x) g(x) dx = 0$
Does this imply:
$\forall a \in \mathbb{R^+}:\int_0^{+\infty} f(x) g(x)^a\,dx =0 $
In the problem $g$ is a probability so we can assume $g(x) \ge 0$.
In the problem functions are "nice" but my intuition says this is a general rule.
To give you some intuition, let shift the problem on $[-1,1]$ (assume functions are $0$ outside and translate if necessary).
For any odd $f$ (e.g. $f(x)=x$) we have $\displaystyle\int_{-1}^1 f(x)dx=0$ and for any $g$ even (i.e. $g(x)=x^2$ to fulfil the positivity requirement) we keep $fg$ odd therefore we also have $\int fg=0$.
But you can see on this simple example that some exponents $a$ could make $g$ odd (e.g. $a=3/2$ then $g(x)^a=x^3$) and now $\int fg^a$ is potentially not zero (it is strictly $>0$ for the example given).
This is of course just some example, but the principle is that there is always a possibility to build a specific function $g$ such that the integral of the product $fg$ compensate the positivity and negativity of $f$, as I explored in this question of mine: Given $f>0$ can we find $g$ such that $\int g>0$ but $\int fg\le 0$
This is not true. For example let $f$ be $1$ on $(0,\frac{1}{2})$ and $-1$ on $(\frac{1}{2},1)$; let $g$ be $1$ on $(0,\frac{1}{2})$, $2$ on $(\frac{1}{2},\frac{3}{4})$ and $0$ on $(\frac{3}{4},1)$. Then $\int fg^2=-\frac{1}{2}$. These are not continuous but I think they can be modified to be continuous.
The point is that $\int f=0$ and $\int fg=0$ does not give much information about $\int fg^2$. Say $f$ is as above and $g$ is $1$ on $(0,\frac{1}{2})$, then $\int fg=0$ means $\int_{1/2}^{1}g=\frac{1}{2}$, but $\int_{1/2}^{1}g^2$ can be pretty much anything (modulo Cauchy-Schwarz).
