I have the following expression
$$F(t) \cdot \int_{0}^{1} f(x) a(x) dx - \int_{0}^{t} f(x) a(x) dx.$$
$f(x)$ is a density function with $f(x) > 0$ if $x \in [0,1]$ and $f(x)=0$ otherwise. $a(x)$ is a monotoneously increasing function with $a(0) = 0$ and $a(1)=1$. Between 0 and 1, $a(x)$ is either strictly convex, strictly concave, or linear. And $0<t<1$.
I would like to know if/when this expression becomes negative. I haven't been able to find a numerical example where it does become negative and I feel that if $a(x)$ is concave, it could happen, but I would like proof either way. I tried integration by parts, but that didn't get me anywhere.
Edit: F(x) is the CDF of f(x) only. No weight is applied here.
Edit2: The second integral goes from 0 to t, not 0 to 1 (I guess it's easy to overlook with 1 and t being fairly similar).
Edit3: a(x) is continuous and differentiable.
I leave my answer, even though what follows is correct - I think - only if $a(x)$ is continuous.
Let
$$g(t) = F(t) \int_0^1 f(x)a(x)dx - \int_0^t f(x)a(x) dx,$$
continuous and differentiable in $\Bbb R$. Since, according to your hypotheses, $F(0) = 0$ and $F(1)=1$, you have
$$g(0) = 0$$ and $$g(1) = 0.$$
Also, \begin{eqnarray} g'(t) &=& f(t) \int_0^1 f(x)a(x)dx - f(t)a(t) =\\ &=& f(t) \left[\int_0^1 f(x)a(x)dx - a(t)\right]. \end{eqnarray} Since $a(x)$ is injective, and $0<\int_0^1 f(x) a(x)dx\leq1$, there is exactly one value $0<\overline t< 1$ such that $g'(\overline t) = 0$, that is the value for which $$a(\overline t) = \int_0^1 f(x)a(x) dx.$$
Thus $g(t) \neq 0$, for $t\in (0,1)$, because otherwise we would have a contradiction to Rolle's Theorem.
Furthermore, $g'(t) >0$ when $a(t) < \int_0^1 f(x)a(x)dx$, that is for $0 <t<\overline t$.
In conclusion, $g(t)>0$ when $t\in (0,1)$.