Let $X$ be a compact smooth manifold without boundary and let $Y \subset X$ be a compact submanifold. (In my situation they are compact complex manifolds, if that changes anything.) Let $f : X \times \mathbb{R} \to \mathbb{R}$ be a smooth function such that:
- $f(x,t) > 0$ for all $x \in Y$ and all $t$.
- The function $g(t) := \inf_{x \in Y} f(x, t)$ is smooth, positive, strictly decreasing and tends to $0$ as $t \to \infty$.
Does there exist a neighborhood $U$ of $Y$ and $t_0 \in \mathbb{R}$ such that $f(x,t) > 0$ for all $x$ in $U$ and all $t \geq t_0$?
If $g(t) \geq c > 0$ this is true by compactness and continuity. I feel like this shouldn't be true as written, but can't come up with a counterexample.
No there isn't. For example, the function
$$ f_0(x, t) = e^{-t} ( 1- tx^2)$$
on $[-1, 1]\times \mathbb R$ is smooth, positive in $\{0\} \times \mathbb R$, and $g_0(t) = f_0(0,t)=e^{-t}$ is smooth positive and is decreasing to $0$. But for all $t_0\in \mathbb R$, there is no $\epsilon >0$ so that $f$ is positive in $[-\epsilon, \epsilon] \times [t_0 , \infty)$.
In general for any compact manifold $Y$, take $X = Y\times [-1, 1]$ and consider $$f((p, x), t) = f_0(x , t),\ \ \ \forall (p, x)\in Y\times[-1, 1].$$
Since every neighborhood of $Y$ in $X$ must contains $Y\times (-\epsilon, \epsilon)$ for some $\epsilon >0$, one cannot find $U$ and $t_0\in \mathbb R$ so that $f>0$ for all $y\in U$ and $t\ge t_0$.