Prove or disprove the concavity of the function

Question

Prove or disprove the concavity of $f$ over the following two domains.

$$f(x_1,x_2)=10-2(x_2-x^{2}_{1})^{2}$$

defined either over

$$S_1=\{(x_1,x_2) : -1\leq x_1 \leq 1, -1 \leq x_2 \leq 1\}$$

or

$$S_2=\{ (x_1,x_2) : x^{2}_{1} >= x_2 \}$$

Answer 1

Looking only at $\textit{local}$ concavity, we can restrain ourself to looking only at the Hessian of the function

$$ f(x,y) = 10 - 2 (y - x^{2})^{2} $$

Its Hessian is given by

$$ H (x,y) = \{\{-16 x^2 + 8 (-x^2 + y), 8 x\}, \{8 x, -4\}\} $$

To prove concavity, we need to study the sign of the eigenvalues of $H$, labelled as $\lambda_{1}$ and $\lambda_{2}$. We have

$$ det (H) = 32 (x^{2} - y) = \lambda_{1} \, \lambda_{2}$$

Looking at the trace of $H$, we have

$$ Tr (H) = -4 - 16 x^{2} - 8 (x^{2} - y) = \lambda_{1} + \lambda_{2}$$

Let's now focus on the case where the function is defined on $S_{1}$. If we place ourself in $(x,y) = (\frac{1}{2},\frac{1}{2})$, we have $det (H) < 0$. As a consequence, at least one the two eigenvalues $\lambda_{i}$ is strictly positive, so that the function is not concave on $S_{1}$

If we suppose that the domain of definition of f is $S_{2}$, one can see that we have $det (H) \geq 0 $. (The limit case $det (H) = 0$ is at the boundary of the definition ensemble $S_{2}$, so that there is no issue with inflexion points.). As $det (H) \geq 0$, either both eigenvalues are positive or both negative. One can see that for $(x,y) \in S_{2}$, we also have $Tr(H) < 0$, so that we necessarily have $\lambda_{i} \leq 0$. Hence, the function $f$ is $\textit{locally}$ concave, when defined on $S_{2}$.