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
$$
f(\lambda x+(1-\lambda)y)&lt;\lambda f(x)+ (1-\lambda) f(y).
$$ 
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.
This is clearly not true: p(x1,x2)=x1^4+x2^4 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 (x1,x2)=(1,0) is not positive definite.
(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 $ &lt; y, \nabla ^2 f(x) y&gt; \; \; &gt; 0, \; \forall y \in \mathbb{R}^n $.  You may also have to use the equivalent definition of convexity $ f(x) &gt; \; \;  &lt; \nabla f(y), x-y&gt; + 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.