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.

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

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.

  • $\begingroup$ $F_2(t)$ is not convex $\endgroup$ – user782220 Apr 13 '13 at 22:24
  • $\begingroup$ @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. $\endgroup$ – DiegoNolan Apr 14 '13 at 10:21
  • $\begingroup$ 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. $\endgroup$ – user782220 Apr 14 '13 at 11:18
  • $\begingroup$ @user782220 Yes, the CDF. Ahh, I missed the polynomial part. $\endgroup$ – DiegoNolan Apr 14 '13 at 21:14

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