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 $$ \exp -\left( \int_0^1 \frac{1}{f(x)} \, dx \right). $$ Is this $G$ smooth?

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?

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?

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 $$ \exp -\left( \int_0^1 \frac{1}{f(x)} \, dx \right). $$ Is this $G$ smooth?