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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!!

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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 $t=0$.

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