Let $\bar{x}\in\mathbb{R}^n$ and $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a $C^2$ function.

We have known that if $\nabla f(\bar{x})=0$ and $\nabla^2f(\bar{x})>0$, i.e. $\nabla^2f(\bar{x})$ is positive definite, then $\bar{x}$ is a strict local minimum of $f$ and moreover the linear perturbation of $f$, the function $f_v(x):=f(x)+v^Tx$ also has a strict local minimum point for each $v$ with sufficiently small norm. Moreover the condition $\nabla^2f(\bar{x})>0$ implies $f$ is locally convex around $\bar{x}$. This fact motivates us to the following question:

Suppose that $\bar{x}$ satisfies the following properties.

There exists $r>0$ such that:

  • $\bar{x}$ is unique local minimum of $f$ on $\overline{B}(\bar{x},r)$;
  • The linear perturbation function $f_v(x):=f(x)+v^Tx$ has a unique local minimum in $\overline{B}(\bar{x},r)$ for each $v$ with sufficiently small norm.

Here $\overline{B}(\bar{x},r)$ is the closed ball with center $\bar{x}$ and radius $r$.

Could we conclude that $f$ is locally convex around $\bar{x}$.

Thank you for all answers, constructive comments and useful references.

My question is related to the following topics:

  1. Is a smooth function convex near a local minimum?
  2. Local minimum implies local convexity?
  3. Does a unique global and local minimum imply (strict) convexity?

This question has an open bounty worth +100 reputation from Blind ending in 4 days.

Looking for an answer drawing from credible and/or official sources.

The answer is incorrect since the function $f$ is not $C^2$.

  • 1
    It's not true that $\nabla ^2 f(\overline{x})>0$. Take for instance $f(x) = x^4$. – CVdeFire Sep 7 at 18:05
  • @CVdeFire Thanks for your comment. I change a little in my question. – Blind Sep 7 at 18:19

The existence of such an $r$-ball follows from the fact that positive definite matrices form an open set (see The openness of the set of positive definite square matrices).

  • The Hessian $\bar{H} := \nabla^2 f(\bar{x})\in\mathbb{R}^{n\times n}$ is positive definite (p.d.) per assumption. Since the set of p.d. matrices is open, perturbations that are small enough do not change its positive definiteness. In other words, there exists a $r_1 > 0$ such that $\forall H\in B(\bar{H},r_1): H$ is p.d.

Now we need to link this result back to our input space $\mathbb{R}^n$. To this end, we use a continuity argument.

  • Since $f\in C^2 $, the function $x\mapsto\nabla^2 f(x)$ is continuous. Then there exists a $r_2>0$ such that $\forall x\in B(\bar{x},r_2): \nabla^2 f(x)\in B(\bar{H},r_1)$. This shows that all Hessians in a ball around $\bar{x}$ are p.d.

  • Finally, we observe that positive definiteness of a smooth function over a convex set is a sufficient condition for convexity (see Relation between Positive definite matrix and strictly convex function).

This proves that $f$ is convex in the ball $B(\bar{x},r_2)$.

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  • Your solution is incorrect since we do not assume $\nabla^2f(\bar{x})>0.$ Perhaps you do not understand my question. – Blind yesterday

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