I have some difficulties in the following problem.
Thank you for all comments and helping.
Let $f:\mathbb{R}^n\rightarrow \mathbb{R} (n\in \mathbb{N})$ be a polynomial.
Suppose that $f$ is strictly convex, i.e., for all $x,y \in\mathbb{R}^n, \lambda \in (0,1)$ we have
$$
f(\lambda x+(1-\lambda)y)<\lambda f(x)+ (1-\lambda) f(y).
$$
Then the following statements are equivalent
(i) $f$ is coercive, i.e.,
$$
\lim_{\|x\|\rightarrow\infty}f(x)=+\infty;
$$
(ii) There exists $x^*\in \mathbb{R}^n$ such that $\nabla^2f(x^*)$ is positive definite. Moreover, the set of such points $x^*$ is a set of full measure.
Best Answer
This is clearly not true: $p(x1,x2)=x_1^4+x_2^4$ is strictly convex (as it is a convex and positive definite form) and coercive (as it is a positive definite form), but its Hessian at $(x_1,x_2)=(1,0)$ is not positive definite.