# 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.

• What exactly do you mean by positive definite? Do you mean for example that $\sum_{j,k=1}^m z_j \bar{z_k} \phi(\mathbf{x_j}-\mathbf{x_k})\ge 0$. – user782220 Apr 12 '13 at 0:46
• A matrix $A\in \mathbb{R}^{n\times n}$ is said to be positive semidefinite if $$\langle Ax, x\rangle>0$$ for all $x\in \mathbb{R}^n\setminus\{0\}$. – blindman Apr 12 '13 at 1:31
• But how is $\nabla^2 f(x^*)$ a matrix? – user782220 Apr 12 '13 at 1:39
• $\nabla^2f(x^*)=\left(\frac{\partial^2f}{\partial x_i\partial x_j}(x^*)\right)_{i=\overline{1, n}, j=\overline{1, n}}$ – blindman Apr 12 '13 at 2:12
• Another question how is it even possible for $f$ to be a strictly convex polynomial but not coercive? Do you have an example? – user782220 Apr 12 '13 at 2:36

## 2 Answers

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 $< y, \nabla ^2 f(x) y> \; \; > 0, \; \forall y \in \mathbb{R}^n$. 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.

• $F_2(t)$ is not convex – user782220 Apr 13 '13 at 22:24
• @user782220 Yes, it is. Distribution functions are non decreasing so their integrals are convex. Or the derivative of a distribution function is non-negative thus convex. – DiegoNolan Apr 14 '13 at 10:21
• By distribution function you must mean the cumulative distribution function and not the probability distribution function. Also the problem is asking about polynomials which confuses me since I don't see how a strictly convex polynomial can be not coercive. – user782220 Apr 14 '13 at 11:18
• @user782220 Yes, the CDF. Ahh, I missed the polynomial part. – DiegoNolan Apr 14 '13 at 21:14