A strictly convex polynomial is coercive if and only if it has a positive definite Hessian 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.
 A: 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.
A: (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.
