Exercise · ZW-CONVEXITY-AND-THE-HESSIAN-008 · ★★☆ Intermediate

Let \(f(x,y)=e^{x}+y^2\). Is \(f\) convex on \(\mathbb{R}^2\)?

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  • A. Yes, because the Hessian \(\operatorname{diag}(e^{x},2)\) is positive semidefinite

Explanation

The terms \(e^{x}\) and \(2\) are positive, so the diagonal Hessian is positive semidefinite. The Hessian has no negative entries, and the function is smooth.

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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

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