Let $\Omega$ be an open set in $\mathbb R^n$ and let $F : \Omega \times \mathbb R\to \mathbb R$ be a smooth function. Let $G$ be the "average of $F$" $$ G(x) = \int_0^1 F(x, t) dt.$$
By the intermediate value theorem, for each $x\in \Omega$, there is $t(x) \in [0,1]$ so that $$ G(x) = F(x, t(x)).$$
Question: Can I choose this $t$ smoothly?
Some observation: If $t$ is smooth and $F_t (x, t) \neq 0$ for all $(x, t)$, $t$ satisfies the PDE
$$ \nabla t = \frac{1}{F_t (x, t(x))} \big(\nabla G (x)- \nabla F(x, t(x))\big).$$
so when e.g. $n=1$, this becomes an ODE and is always solvable and $t$ is smooth. What happens for general $n$ and when $F_t$ is possibly zero somewhere? I believe that $t(x)$ can be chosen to be continuous, but how about smoothness?