Help needed in proving non-convexity of the following two variable function

Question

I have following function $$f(x,y)=x^2\exp(y)$$ Where $x>0$ and $y>0$. The constraint $f(x,y)\leq c$ (where $c$ is some positive value) results in non-convex set. If I am able to prove that $f(x,y)$ is concave then I can understand how the constraint results in non-convex set. To prove that $f(x,y)$ is concave I get the Hessian of the $f(x,y)$ as follows $$H(x,y)=\left[\begin{matrix}2\exp(y) & 2x\exp(y) \\ 2x\exp(y) & x^2\exp(y) \end{matrix}\right]$$ Now for any real valued vector $\textbf{v}=[v_1 \quad v_2]$ if $$\textbf{v}H(x,y)\textbf{v}^T\leq 0$$ then it means that $f(x,y)$ is concave. But I get $$\textbf{v}H(x,y)\textbf{v}^T=\exp(y)[2v_1^2+4v_1v_2x+v_2^2x^2]$$ which I think is not always negative. Therefore, How to prove that $f(x,y)$ is a concave function? Any help in this regard will be much appreciated. Thanks in advance.

Answer 1

You don't need to prove that f(x,y) is concave. Indeed, it is not.

Take, for example, x = y = 1. Then the eigenvalues of the Hessian of f(x,y) are -1.526458808 and 9.681304292, which shows that the Hessian is indefinite there. Therefore, f(x,y) is neither convex nor concave.

Answer 2

For any $c>1$ consider the points $(1,\ln c)$ and $(3,\ln (c/9))$. Both these points satisfy $f(x,y) \leq c$ but the mid-point does not.