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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.

Rodrigo de Azevedo
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blindman
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  • 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 00:46
  • 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 01:31
  • But how is $\nabla^2 f(x^*)$ a matrix? – user782220 Apr 12 '13 at 01:39
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    $\nabla^2f(x^)=\left(\frac{\partial^2f}{\partial x_i\partial x_j}(x^)\right)_{i=\overline{1, n}, j=\overline{1, n}}$ – blindman Apr 12 '13 at 02:12
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    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 02:36
  • @user782220 Maybe $f(x_1,x_2) = x_1 - x_2$? If we consider our "movement" to infinity over the line $x_1 = x_2$, the limit will be zero, even though $|x| = \sqrt{{x_1}^2 + {x_2}^2}$ goes to infinity. – Daniel Cunha Aug 28 '23 at 03:31

2 Answers2

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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.

Rodrigo de Azevedo
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Amir Ali Ahmadi
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  • But the statement is that there exist a set of full measure in which the Hessian is positive definite. A single-point counter-example is not enough to disprove it... – Daniel Cunha Aug 28 '23 at 03:11
  • The Hessian for this function is $12,\begin{bmatrix}{x_1}^2 & 0 \0 & {x_2}^2\end{bmatrix}$, which is positive definite for any point that is not on the lines $(x_1,0)$ and $(0,x_2)$, in which it is only positive semi-definite. This means it is a set of full measure, right? – Daniel Cunha Aug 28 '23 at 03:20
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(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.

DiegoNolan
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    $F_2(t)$ 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