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.
(i) Is not true. Counter example. I can't think of a better one but this works. Take $ F(x) $ is the standard normal distribution, $ F _2 (t) = \int _{-\infty} ^t F(x) dx $ is strictly convex and the $ \lim _{t \rightarrow -\infty} F _2(t) \neq + \infty $. Although this is one dimension the absolute value should be good enough. You could always have a random vector.
(ii) Is True. I'm not quite sure how the proof would go though. Positive definite is equivalent to $ < y, \nabla ^2 f(x) y> \; \; > 0, \; \forall y \in \mathbb{R}^n $. You may also have to use the equivalent definition of convexity $ f(x) > \; \; < \nabla f(y), x-y> + f(y) $. I think you would have to prove that locally in is convex and then because you let your location be any where it proves the whole system is convex.