Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The problem is: Let $\vec{x}\in\mathbb{R}^d$ be the variable and $f(\vec{x})$ be a scalar function that is globally strictly convex in $\mathbb{R}^d$. We assume the unique optimum of $f$ to be finite(in all dimensions). Denote the optimum as $\hat{x}$. Suppose $f$ is smooth enough, can we know the sign of each dimension in $\hat{x}$ by looking at the sign of each dimension in $\nabla f(\vec{x})|_{\vec{x}=\vec{0}}$?

If yes, any proof or justification? If no, could you provide a counterexample?

Thanks much!!

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Try $d = 2$, $f(x_1,x_2) = (x_1+1)^2 + (x_1+1)(x_2 + t) + (x_2 + t)^2$ where $t$ is a parameter. The minimum is at $(-1,-t)$. The sign of the second component of the gradient changes at $t = -1/2$, not at $t=0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.