Existence of function $f:R^2 \rightarrow R$ s.t. f is convex in x- and y- directions and f has multiple minima.

Question

Does there exist a function $f\colon \mathbb{R}^2 \rightarrow \mathbb{R}$ such that

(1) for all $(x,y) \in \mathbb{R}^2$, f is convex in the x-direction and y-direction

(2) $f$ has multiple non-degenerate minima? Non-degenerate: for any path P between minima $m_1$ and $m_2$, there exists $p \in P$ s.t. $f(p) > f(m_1)$ and $f(p) > f(m_2)$.

A function satisfying (1) is for example $f(x,y) = (xy)^2$ and a function satisfying (2) is for example $$f(x,y) = -\left(e^{-\left((x-1)^2 + (y-1)^2\right)} + e^{-\left((x+1)^2 + (y+1)^2\right)}\right).$$

If not (as I suspect) prove so, and how does this generalise for $f\colon \mathbb{R}^n \rightarrow \mathbb{R}$?

With $f(x,y) = (x-y)^2$ we come close, but minima are degenerate in the sense described in (2).

Answer 1

This function should satisfy your requirements $$ f(x,y) =((x-y)^2-1)^2 +2(x+y)^2. $$ It has global minima at $(-\frac12,\frac12)$, $(\frac12,-\frac12)$ - according to wolframalpha.

The second derivatives are non-negative: $$ \frac{\partial^2}{\partial x^2} f(x,y) = \frac{\partial^2}{\partial y^2} f(x,y)= 12(x-y)^2. $$

Answer 2

I guess that with the current writing of your constraints, (1) and (2) are compatible. Indeed, we consider the function

$$ f(x,y) \mapsto (x - y)^{2} $$

One immediately has $\frac{\partial^{2} f}{\partial x^{2}} = \frac{\partial^{2} f}{\partial y^{2}} = 1 > 0$, so that $f$ is always strictly convex in the $x$ and $y$ direction. However, the minimum value of $f$ is $0$ reached for all the points $f (t , - t) = 0$, so that $f$ reaches its minimal value along the entire axis $y = -x$, having an infinite number of minima.