# No minimum at the origin but a minimum along all lines

We look here at an example, from the Italian mathematician Giuseppe Peano of a real function $$f$$ defined on $$\mathbb{R}^2$$. $$f$$ is having a local minimum at the origin along all lines passing through the origin, however $$f$$ does not have a local minimum at the origin as a function of two variables.

The function $$f$$ is defined as follows
$\begin{array}{l|rcl} f : & \mathbb{R}^2 & \longrightarrow & \mathbb{R} \\ & (x,y) & \longmapsto & f(x,y)=3x^4-4x^2y+y^2 \end{array}$ One can notice that $$f(x, y) = (y-3x^2)(y-x^2)$$. In particular, $$f$$ is strictly negative on the open set $$U=\{(x,y) \in \mathbb{R}^2 \ : \ x^2 < y < 3x^2\}$$, vanishes on the parabolas $$y=x^2$$ and $$y=3 x^2$$ and is strictly positive elsewhere. Consider a line $$D$$ passing through the origin. If $$D$$ is different from the coordinate axes, the equation of $$D$$ is $$y = \lambda x$$ with $$\lambda > 0$$. We have $f(x, \lambda x)= x^2(\lambda-3x)(\lambda -x).$ For $$x \in (-\infty,\frac{\lambda}{3}) \setminus \{0\}$$, $$f(x, \lambda x) > 0$$ while $$f(0,0)=0$$ which proves that $$f$$ has a local minimum at the origin along the line $$D \equiv y – \lambda x=0$$. Along the $$x$$-axis, we have $$f(x,0)=3 x^ 4$$ which has a minimum at the origin. And finally, $$f$$ also has a minimum at the origin along the $$y$$-axis as $$f(0,y)=y^2$$.

However, along the parabola $$\mathcal{P} \equiv y = 2 x^2$$ we have $$f(x,2 x^2)=-x^4$$ which is strictly negative for $$x \neq 0$$. As $$\mathcal{P}$$ is passing through the origin, $$f$$ assumes both positive and negative values in all neighborhood of the origin.

This proves that $$f$$ does not have a minimum at $$(0,0)$$.