Is the constraint $xy\leqslant 0.001$ convex?

Question

I would like to ask whether the constraint $xy\leqslant 0.001$ ($x,y\geqslant0$) is convex.

Since its Hessian matrix is positive semi-definite for $x,y\geqslant0$, the constraint $xy\leqslant 0.001$ ($x,y\geqslant0$) is convex. Is it correct?

Answer 1

Notice that a twice differentiable function $f:U\to\mathbb R$ on a convex and open set $U\subseteq\mathbb R^2$ is convex if and only if the hessian $\nabla^2 f(x,y)$ is positive semidefinite for every $(x,y)\in U$. Concerning your function $f(x,y) = xy$, we have $$ \nabla^2 f(x,y) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, $$ which is indefinite, as its eigenvalues are $-1$ and $1$.

However, the convexity of $f$ is sufficient but not necessary for the convexity of the sublevel set $S = \{ (x,y)\in\mathbb R^2 \mid f(x,y) \le 10^{-3}, x,y\ge 0 \}$. To show that $S$ is not convex, consider $(x_1, y_1) = (1,0)$ and $(x_2, y_2) = (0,1)$. Both are elements of $S$. However, $$\frac12((x_1,y_1) + (x_2,y_2)) = \left(\frac12, \frac12\right)$$ is not an element of $S$, as $$ f\left(\frac12, \frac12\right) = \frac14 > 10^{-3}. $$

Answer 2

Draw a picture: the upper boundary is a hyperbola, with the same shape as $$x=1/y.$$ If you choose two points far out in the legs, the line segment between them will cut above the curve, so it's not convex.

(The solutions by way of the Hessian of the defining function essentially exhibit exactly this counterexample in a less visual, more general way.)