Does there exist a function $F : C^\infty(\mathbb{R}, [0, \infty)) \to \mathbb{R}$ with the following properties:


* $F(f) = 0$ if and only if there exists an $x \in [0,1]$ such that $f(x) = 0$.
* $F$ is *smooth* in the following sense: if $f(x,t) \in C^\infty(\mathbb{R} \times \mathbb{R}, [0,\infty))$ and $F$ is applied to $f$ in the $x$ variable, the function of $t$ that results is smooth ($C^\infty$).


The second question is whether the following candidate satisfies the smoothness property. For $f \in C^\infty(\mathbb{R}, [0, \infty))$, define $G(f)$ to be
$$
 \left( \int\_0^1 \frac{1}{f(x)} \, dx \right)^{-2}
$$
if $f$ has no zeros in $[0,1]$, and $0$ otherwise.
Then $G$ satisfies the first condition by definition,
and one can show that when $f$ is a function of $x$ and $t$ as above, $G(f)$ is (continuous and) differentiable as a function of $t$. Is it smooth?


**Edit:** Willie Wong answered the second question in the negative. So let's instead define $G(f)$ to be
$$
 \exp \left( -\int\_0^1 \frac{1}{f(x)} \, dx \right)
$$
if $f$ has no zeros in $[0,1]$, and $0$ otherwise.
Is this $G$ smooth?



smooth functional to detect whether a function has a zero
A functional that works is
$$
 F(f)=\begin{cases} \exp(-\exp(\int\_0^1 \frac{1}{f(x)}dx)), & \textrm{if $f(x)>0$ for all $x \in [0,1]$} \\
 0, & \textrm{otherwise}.
 \end{cases}
$$
The idea to use this formula and a sketch of the proof that it works are both due to Chengjie Yu. The proof is in the appendix of a paper I wrote with Enxin Wu, Smooth classifying spaces, <http://arxiv.org/abs/1709.10517>


I don't know whether the functional described in the question, $G(f) = \exp(-\int\_0^1 \frac{1}{f(x)} dx)$, works.


