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Suppose $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is once (and only once) continuously differentiable.

Are there any characterizations of convexity that rely only on the gradient $\nabla f$?

In the one-dimensional case this would be that $f'$ is non-decreasing.

Thilo
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1 Answers1

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For $a\neq b$ consider $g_{a,b}(t) = f((1-t)a + tb)$. You have

$$g_{a,b}'(t) = \langle\nabla f((1-t)a+tb),b-a\rangle.$$

If $f$ is convex, so is $g_{a,b}$ (and $f$ is convex if $g_{a,b}$ is for all $a,b$), and from the monotonicity of $g_{a,b}'$ we obtain

$$\langle \nabla f(b) - \nabla f(a), b-a\rangle \geqslant 0.\tag{1}$$

$(1)$ holds for all $a,b$, and that means that $\nabla f$ is monotonic. Convince yourself that this is a necessary and sufficient condition for the convexity of $f$.

Daniel Fischer
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  • Thank you for your answer. At a first glance, your reasoning looks reasonable. Tomorrow I will have time to verify everything and hopefully accept the answer afterwards. – Thilo Nov 25 '13 at 15:09