Redefine a discrete compact set

Question

I need to find twice continuously differentiable functions $g_i: \mathbb{R}^2 \rightarrow \mathbb{R}$ $i=1,\ldots,I$ such that the set $\{ 0,1\}^2=\{(0,0), (1,0), (0,1), (1,1) \}$ can be written as $\{x \in \mathbb{R}^2 \text{ s.t. } g_i(x)\leq 0 \text{ } \forall i=1,...,I \}$. One idea could be to set $I=1$ and $g_1(x):=(x_1^2+x_2^2)((1-x_1)^2+x_2^2)(x_1^2+(1-x_2)^2)((1-x_1)^2+(1-x_2^2))$. However the gradient vector and hessian matrix of this function are zero for any $x \in M$ and I would like to avoid this. Do you know any other function $g$ without this undesired characteristic?

Answer 1

Let $g_1(x) = (x_1 (1-x_1))^2$, $g_2(x) = (x_2 (1-x_2))^2$.

Old, incorrect answer:

Draw a picture and use four straight lines to get four smooth $g_i$ that do the job.

Hint:

Try $g_1(x) = x_1-1, g_2(x) = -x_1$, $g_3(x) = x_2-1, g_4(x) = -x_2$.