I would like find (if it exists) a function $F:\mathbb R^2\to\mathbb R$ such that the following conditions hold true:
- $F\in C^\infty(\mathbb R^2)$
- $F$ is radial, that is $F(x,y)=f(\sqrt{x^2+y^2}\,)$
- $f$ is strongly convex on $[0,\infty)$ (i.e., $\min_{r\geq0} f''(r)>0)$
- $\min_{r\geq0} f(r)=f(1)=0$
- $\lim_{r\to\infty}f(r)/r^4\in(0,\infty)$
For example $f(r)\equiv(r^2-1)^2$ gives $F(x,y)=(x^2+y^2-1)^2$ and conditions 1., 2., 4., 5. are satisfied but convexity is not, since $f$ has an inflection point at $r_0\equiv\frac{1}{\sqrt{3}}$. Is it possible to modify $f$ in order to satisfy all the conditions 1.-5. ?
Notice that if $f$ is polynomial with odd power terms, condition 1. is not satisfied.
Such an $F$ doesn't exist:
Conditions 1 and 2 imply $x \mapsto f(\vert x\vert)$ is differentiable at $0$, and then
$$\lim_{x \uparrow0}\frac{f(\vert x\vert)-f(0)}{x} = \lim_{x \uparrow0}\frac{f(-x)-f(0)}{x} = \lim_{h \downarrow0}\frac{f(h)-f(0)}{-h} = -\lim_{h \downarrow0}\frac{f(\vert h \vert)-f(0)}{h}$$
shows that the derivative must be $0$, which also implies $f'(0) = 0$.
But condition 4 implies $f'(1) = 0$, so Rolle's theorem gives us a point in $(0, 1)$ where the second derivative vanishes, contradicting condition 3.