I checked the Hessian which is $\begin{bmatrix}-2&0\\0&0\end{bmatrix}$ which is negative semidefinite but according to the sketch of the function it is convex. What am I missing?

  • 4
    $\begingroup$ The function on the LHS is concave so the set specified by the inequality is convex. $\endgroup$ Apr 7, 2020 at 12:20

1 Answer 1


I think you are trying to use the following property:

  • $\{x:g(x) \le 0\}$ is convex if $g$ is convex.

Note the direction of the inequality.

Notice that

\begin{align}\{(x,y): -x^2+y-1 \ge 0\}&=\{(x,y): -(-x^2+y-1 ) \le 0\} \\ &=\{(x,y): x^2-y+1 \le 0\} \end{align}

If you compute the Hessian of $x^2-y+1$, you will obtain $\begin{bmatrix} 2 & 0 \\ 0 & 0\end{bmatrix} \succeq 0$, hence the corresponding region is a convex set.


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