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A differentiable and also convex function has the following property.

$$ (\nabla f (x) - \nabla f (y))^T(x-y) \geq 0 $$

I can derive it from the first order condition for a convex function,

$$ f(y) \geq f(x)+\nabla f(x)^T(y-x) \\ f(x) \geq f(y)+\nabla f(y)^T(x-y) $$

by adding the two inequalities and rearrange it.

but I don't know meanings of the inequality in a graph.

Does it mean that a tangent line should be above the graph? but how? or is something else?

Could you explain the meaning?

user3681105
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  • I believe your expression might be wrong. For a multivariate function $f: \mathbb{R}^n\to \mathbb{R}$, the condition that for $n=1$ reads $f(x)\geq f(y) + f'(y)(x-y)$ should for general $n\in \mathbb{N}$ read $f(x)\geq f(y) + \nabla f(y)^T\cdot (x-y)$, and indeed the graph should lie above all its tangents, since otherwise the epigraph wouldn't be convex, that is, we could not connect any two points in the space $E$ "over the graph" with a segment whose points lie entirely inside $E$. – Albert Feb 15 '24 at 15:20
  • @Albert Yes, the graph should lie above all its tangents. I edited the question to correct the expression. – user3681105 Feb 15 '24 at 16:35
  • If you know that convexity implies the graph lies above its tangents, why do you explicitly ask whether the opposite is true? "Does it mean that a tanget line should be above the graph?" – Albert Feb 15 '24 at 19:58
  • For $n=1$ it means that $f'$ is monotoncally increasing. – daw Feb 16 '24 at 14:45

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