Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $D$ be a convex set in $\mathbb{R}^n$ and $f: D \to \mathbb{R}$ a concave and $C^1$ function. How do I show that $x^*$ is a global maximum for $f$ if and only if $f^{(1)}(x^*)y \leq 0$ for all $y$ pointing into $D$ at $x^*$ (Here $f^{(1)}$ denotes the first derivative of $f$)

share|improve this question
add comment

1 Answer

If we had $f^{(1)} (x^*)y > 0$ then by the definition of the derivative we could find a point close by in that direction for which the value $f$ is higher. That shows the only if part.

For the other direction, take any other point $y \in D$. Consider a straight path between the two points $v(t) : [0,1] \to \mathbb{R}^d$, $v(t) = x^* + t(y-x^*)$. $f(v(t))$ is concave and continuously differentiable with respect to $t$, so must have nonincreasing derivative. Combining this observation with the fact that $f^{(1)} (x^*)y \le 0$ and the fundamental theorem of calculus, we see that $f(y) \le f(x^*)$.

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.