Minimum on all the lines is local minimum for $C^1$ functions?

Question

Let $f(x,y)\in C^1(\mathbb R^2)$ and let $z=(x_0,y_0)$ be a stationary point where the Hessian is null; suppose that the $z$ is a local minimum (or maximum) for $(x,mx) \ \ \forall m\in \mathbb R$; in other words, the point $z$ is a local minimum (maximum) on every line. Is it true that then $z$ is a local minimum (max) for $f$? I think it's not, but I can't find a counterexample... Actually, the differentiability is a very strong condition and it could be even true. Any suggestion?

Answer 1

On $\mathbb R,$ define $g(t) = e^{1/(t^2-1)}, |t|<1,$ $g=0$ elsewhere. Then $g\in C^\infty(\mathbb R),$ and the support of $g$ is $[-1,1].$ On $\mathbb R^2,$ define

$$ f(x,y) = \begin{cases}e^{-1/x}g(2(y-x^2)/x^2)& x>0\\ = 0&x\le 0 \end{cases}$$

Then $f(x,y) > 0$ iff $x>0$ and $x^2/2 < y < 3x^2/2.$ Note that each line through the origin misses this region in an open line segment containing the origin. Thus $f(0,0)=0$ is a local maximum value on each such line. But $f(0,0)$ is not a local maximum value relative to $\mathbb R^2,$ since $f(x,x^2) = e^{-1/x}g(0) = e^{-1/x}\cdot e^{-1} >0$ for all $x>0.$

It remains to verify that $f\in C^1(\mathbb R^2).$ In fact $f\in C^\infty(\mathbb R^2).$ I'll leave that to the reader for now.