# If $x$ is a strict local minimal point, is the Hessian at $x$ a positive definite matrix?

Given a function $$f$$ and its local minima $$x$$. We have known that if $$H(x)$$ is positive definite, then $$x$$ is a strict local minimum for $$f$$. https://planetmath.org/relationsbetweenhessianmatrixandlocalextrema

But does the inverse also hold? Namely, if $$x$$ is a strict local minimal point, is the Hessian at $$x$$ a positive definite matrix?

• I don't why this is downvoted. If there is any problem, just leave a comment here and I will modify it. Commented Sep 3, 2022 at 21:05

The Hessian must be weakly positive definite (that is, $$v\cdot H(x)v\ge0$$ for all $$v$$), but need not be strictly positive definite. For example, if $$f:\Bbb R\to\Bbb R$$ the Hessian is just the $$1\times1$$ matrix $$[f'']$$; the function $$f(x)=x^4$$ has a strict global minimum at $$x=0$$ but $$f''(0)=0$$.
No, that is not true. Consider for example $$f:x \to \|x\|^4$$. Clearly this has a strict local minimum of in $$x=0$$. But $$d_v^2 f(0) = 0$$ for all $$v$$. So in fact the hessian vanishes in $$0$$.