Stable strict local minimum implies local convexity 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:


*

*Is a smooth function convex near a local minimum?

*Local minimum implies local convexity?

*Does a unique global and local minimum imply (strict) convexity?

I was thinking this has a chance in $\mathbb R$ but once you step in $\mathbb R^2$ there should be counterexamples. Let us have a look at the polynomial function: $f\colon (x,y)\in \mathbb R^2 \longmapsto x^4 + 12 x^2 y^2 + y^4 + (x+2y)^2$. You can factor it as $f(x,y) = (x^2+y^2)^2 + 10x^2y^2 + (x+2y)^2$, this establishes that $f(0,0) = 0$ is the unique minimum.
We should note that $f$ is not locally convex around $(0,0)$. For example you can check that,
\begin{equation}
\begin{bmatrix} 2 \\ -1 \end{bmatrix}^\top  \nabla^2 f(t,t) \begin{bmatrix} 2 \\ -1 \end{bmatrix} = -12t^2 < 0,
\end{equation}
for all $t \neq 0$, no matter how small.
It seemed that there existed a unique minimum for a small enough linear perturbation, however upon closer inspection the gradient map is not injective. I have plotted (rotated and resized) $\nabla f(r\cos \theta, r\sin\theta)$ with $r$ fixed smaller and smaller, and with $\theta$ varying from $0$ to $2 \pi$. It seems to describe a deformed circle around $(0,0)$ (it actually becomes close to a segment) but those circles actually overlap (see the attached plot). In any case the equation is,
\begin{equation}
\begin{bmatrix}
4 x^3 + 24 x y^2 + 2 (x + 2 y) \\ 24 x^2 y + 4 y^3 + 4 (x + 2 y)
\end{bmatrix}
=
\begin{bmatrix}
-u \\ -v
\end{bmatrix}.
\end{equation}
Arcs of the gradient are drawn, $\nabla f(r \cos t, r \sin t)$, for $r=10^{-7}(1+n/4)$ with $n=1,\dots,10$ and $t = 0 \dots 2\pi$. The picture has been rotated and stretched to show clearly the overlap.

