Kac-Rice formula and Borell-TIS inequalities for gradient-flow of centered gaussian random field Let $x\mapsto g(x)$  be a centered gaussian random field on $\mathbb R^m$. Let $x_0 \in \mathbb R^n$, and (assuming regularity conditions) consider the gradient-flow
$$
\dot{x}(t) = -\nabla g(x(t)), \;x(0) = x_0.
$$
Integrating the above system gives
$$
g(x(t)) = g(x_0) + \int_0^t \langle \nabla g(x(t)),\dot{x}(t)\rangle dt = g(x_0)-\int_0^t\|\nabla g(x(t))\|^2 dt.
$$

Question 1. Is there a (Kac-)Rice formula for the zero-crossings of $g(x(t)$, namely $\#\{t \in [0, T] \mid g(x(t)) = 0\}$ ?

One of the things I'm interested in are upper-bounds for $g(x(t))$ as a function of $t$.

Question 2. Can one obtain a Borell-TIS bound for the $\sup_{0 \le t \le T} g(x(t))$.

Note. I'm not necessarily looking for a clear-cut answer (though this would be really cool), but general guidelines on how to go about this.
Examples
As a working examples, one could consider the following "simple" fields

*

*Linear gaussian random field wherein $g(x) = w^Th(x)$, with $w \sim N(0,I_k)$ and deterministic $h \in \mathcal C^1(\mathbb R^n \to \mathbb R^k)$.

*Stationary gaussian random fields
 A: Here's a toy model  that is truly linear
$$g(x) =\frac{1}{2}\sum_{i=1}^n \Lambda_i x_i^2, $$
where $\Lambda_i$ are i.i.d. $N(0,1)$   then
$$x(t)= \Big(e^{-t\Lambda_1} x_1(0),\dotsc,  e^{-t\Lambda_n} x_n(0)\Big)$$
$$U(t):= g(x(t))=\frac{1}{2}\sum_{i=1}^n \Lambda_ie^{-2t\Lambda_i} x_i(0)^2. $$
Denote by $N_T(U)$ the number of zeros of $U(t)$ on  the interval $[0,T]$. Denote by $p_{U(t)}(u)$ the probability density of $U(t)$. Then the Kac-Rice formula state that $\newcommand{\bE}{\mathbb{E}}$
$$\bE\big[\; N_T(U)\;\big]=\int_0^T\bE\big[ \; |U'(t)|\;|\; U(t)=0\;\big] p_{U(t)}(0) dt, $$
where $\bE[-|-]$ denotes the conditional expectation.
Alternatively we have $\newcommand{\bP}{\mathbb{P}}$
$$
\bE\big[\; N_T(U)\;\big]=\underbrace{\bE\big[\; N_T(U)\;|\;g(U(T))>0\;\big]}_{=0}\bP\big[ g(U(T))>0\big]+\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big]
$$
$$
=\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big]
$$
$$ =\underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))>0,g(U(T))<0\;\big]}_{=1}\bP\big[ g(U(0))>0,g(U(T))<0\big]+ \underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))<0,g(U(T))
<0\;\big]}_{=0}\bP\big[ g(U(0))<0,g(U(T))>0\big]
$$
$$
=\bP\big[ g(U(0))>0,g(U(T))<0\big].
$$
