Let $f(\boldsymbol{x})=f(x_1,x_2,\cdots,x_N)$ with $N>2$ be a real and continuous function and $f(\boldsymbol{x})\ge f_0$ for any $\boldsymbol{x}\in\mathbb{R}^N$. Now let $x_1,x_2,\cdots,x_N$ be the i.i.d random variables drawn from the normal distribution. I want to estimate the value of the probability density function $P(f)$ at $f_0$, or specifically, I want to know whethere $P(f_0)$ is zero, finite or divergent. The definition of $P(f_0)$ is $$ P(f_0)=\int\delta(f(\boldsymbol{x})-f_0)\rho(\boldsymbol{x})dx_1\cdots dx_N, $$ where $\rho(\boldsymbol{x})$ is the $N$-dimensional Gaussian distribution. Let us consider a concrete example. The function $f$ is given by $$ f^{(n)}(\boldsymbol{x})=-\sum_{m=n+1}^{N}x^2_m+\left(\sum_{m=n+1}^{N}x^2_m\right)^2+\sum_{m=1}^{n}x^2_m. $$ The minimal value is $f_0=-1/4$, which is taken as long as the condition $$ \sum_{m=n+1}^{N}x^2_m=\frac{1}{2},\ x_{1}=x_{2}=\cdots x_n=0$$ is satisfied. The analytical expression can be obtained by calculating $P(f)$ and letting $f\to f_0$. For $f-f_0\ll1$, the analytical expression of $P(f)$ can be approximately obtained by transforming $(x_1,\cdots,x_{N-n})$ into the $N-n-1$ sphere and transforming $(x_{N-n+1},\cdots,x_{N})$ into the $n-1$ sphere coordinates. After some calculations, I found that $$ \begin{equation} \left\{ \begin{aligned} & P(f_0)\to \infty,\ n=0, \\ & P(f_0)\to {\rm constant},\ n=1, \\ & P(f_0)\to 0,\ n>1. \end{aligned} \right. \end{equation} $$ That is, whethere $P(f_0)$ is zero, finite or divergent seems to depend on the dimension of the solution that satisfies $f(\boldsymbol{x})=f_0$. If the dimension of solution set is $N-1$, $P(f_0)$ is divergent; if the dimension of solution set is $N-2$ then $P(f_0)$ is finite; otherwise $P(f_0)$ is zero.
Q: For a generic continuous function $f$ as we specified in the begining, does the relation between the dimension of solution set and the behavior of $P(f_0)$ still hold?
