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).