Proof that $(\nabla f(x) - \nabla f(y))^T(x-y) \ge 0 \ \forall \ x, y \Rightarrow f$ is convex. I've differentiated both 2 sides and got this: $\nabla^2 f(x)^T(x-y) + \nabla f(x) - \nabla f(y) \succeq 0$ Then I choose $y = 0$: $\nabla^2 f(x)^Tx + \nabla f(x) - \nabla f(0) \succeq 0$ But now I have no idea to continue. Can anyone help me in this problem ?
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1You probably want to differentiate a function of $t$ related to convexity... – Pedro Oct 19 '23 at 08:43
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@Pedro What do you mean "a function of t" ? I attempt to proof the problem at the title but i don't know if it's true or false – MY_NAME_S3M Oct 19 '23 at 09:08
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Compute the derivative of $g(t) = f(\vec{x}t + \vec{y}(1-t))$ with respect to $t$. Might give you an idea of how to proceed. – Pedro Oct 19 '23 at 10:28