# Does existence of only 1 critical point and local convexity imply global minimum?

For a multivariate smooth function $$f: [0,1]^m\rightarrow \mathbb{R}$$, does the existence of only one point where the Jacobian $$=0$$, plus the fact that the Hessian is PSD at that point (local convexity) imply that it is a global minimum? Suppose the minimum is not attained at boundaries (In other words, f evaluated at the boundaries will be larger than the local minimum).

My thoughts: I saw some counterexamples in multivariate case but not sure this would fail in general with the additional boundary assumption. The proof in the bounded univariate case seems well extendable in this case, as long as we take care of the "boundaries". Intuitively, what could possibly go wrong if boundary is not the issue?

• Yes it is correct. May 18 at 18:46

The answer is Yes! In fact, if $$\Omega$$ is an open in $$\mathbb R^m$$ and $$f: \Omega \to \mathbb R$$ a $$C^2$$ function, then $$x$$ is a local minimum of $$f$$ implies the two conditions: $$\nabla f(x) = 0$$ and $$\nabla^2 f(x)\succeq 0$$.