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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 $f$)

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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 $f(y) \le f(x^*)$.

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