Exercise · ZW-CONVEXITY-AND-THE-HESSIAN-009 · ★★☆ Intermediate

Let \(f(x,y)=x^3+y^2\). What can be concluded about the convexity of \(f\) on \(\{x\geq 0\}\)?

Choose one option.

Show answer
  • D. It is convex, because \(\nabla^2 f=\operatorname{diag}(6x,2)\succeq 0\) on that set

Explanation

With \(x\geq 0\), the entry \(6x\) is nonnegative, so the Hessian is positive semidefinite. Having a zero in the Hessian does not prevent convexity, and a zero gradient does not determine convexity.

Study Continuous optimization in ZWMath →

Review the concept: Convexity and the Hessian

Criterion characterizing convexity of a twice differentiable function through positive semidefiniteness of its Hessian matrix \(\nabla^2 f(x)\) at every point of the domain.

\[f \text{ convexa} \iff \nabla^2 f(x) \succeq 0 \ \forall x \in \operatorname{dom} f\]

More practice problems: convexity and the hessian

← ZW-CONVEXITY-AND-THE-HESSIAN-008 ZW-CONVEXITY-AND-THE-HESSIAN-010 →