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For a differentiable function $f:\mathbb R^d\rightarrow \mathbb R$, show that if $\langle x-y,\nabla f(x)-f(y)\rangle\ge0$, then f is convex.

Can someone explain the steps in the Link in baby language and baby symbols. That is, step 2) in the linked answer.

What does $A=\{x|f(x)\le a\}$ represent and why are are there $x,y\in \partial A$ (what is $\partial A)$ such that $\nabla f(x)\cdot (y-x)>0$ if A is not convex? Is A a convex set, and what's its relationship with a convex function?

Edit: So I see that it represents a "sublevel set" and has to be convex for f to be convex. I still don't know why A being not convex implies that $\nabla f(x)\cdot (y-x)>0,\nabla f(y)\cdot(x-y)>0$ for some "$x,y\in\partial A"$.

Vons
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  • You have to be more specific with what you're asking. I don't think your question is difficult to answer, but I'm not sure exactly what you're looking for. Instead of referring to another thread, you should explicitly write the topic you're trying to understand and exactly what you don't understand – iarbel84 Feb 06 '21 at 10:21
  • Shouldn’t it be $\nabla f(y)$ rather than $f(y)$? – littleO Oct 13 '21 at 02:05

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