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*Function f is convex if and only if for all x,y*

[tex](*)\quad f(x)-f(y)\ge\nabla f(y)^T(x-y)[/tex]

It's proven in this way: From definition of convexity

[tex]f(\lambda x+(1-\lambda)x)\le \lambda f(x)+(1-\lambda)f(y)[/tex]

we have

[tex]\frac{f(y+\lambda(x-y))-f(y)}{\lambda}\le f(x)-f(y)[/tex]

Setting [tex]\lambda\to0[/tex] we have (*).

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My problem is in last sentece. I understand formula on left-hand side as directional derivative. But in definition of directional derivative is needed to (x-y) be an unit vector. It is not. So it is not a directional derivative. If I define

[tex]\lambda:=\frac{\mu}{\Vert x-y\Vert}[/tex]

I have directional derivative on left hand side

[tex]\frac{f(y+\mu\frac{(x-y)}{\Vert x-y\Vert})-f(y)}{\mu}\le\frac{f(x)-f(y)}{\Vert x-y\Vert}[/tex]

But in this way I don't obtain result (*) but I obtain this

[tex]\nabla f(y)^T(x-y)\le\frac{f(x)-f(y)}{\Vert x-y\Vert}[/tex]