Given a Finsler-Minkowski norm $F$ on $R^n$, let $\mathfrak{F}=\{\Sigma_r\}_{r\geq0}$, where $\Sigma_r=\{y\in R^n: F(y)=r\}$, be a partition of $R^n$. There exists any (infinitely) smooth Morse function $f:R^n\to R$ such that $f^{-1}(c)=\Sigma_r$ (that is gave the same partition)?
FYI:
A function $F:R^n\to R$ that is (infinitely) smooth on $R^n\backslash 0$ is called a Finsler-Minkowski norm if $F(y)>0$, for $y\neq 0$, $F(\lambda y)=\lambda F(y)$, for $\lambda\in R>0$, and if $E:=\frac12 F^2$, then $\forall y\in R^n\backslash 0$, $g_y:=E''(y):R^n\times R^n\to R$ is non-degenerate.
$f:R^n\to R$ is called a Morse function if all of the critical points are non-degenerate, that is the Hessian of $f$ at each critical point is non-degenerate.
As $g_y$ is non-degenerate for nonzero values of $y$, one candidate could be the composition of $g_y$ with some other functions like Bump or ....
Any suggestion?