Proving Convexity of a Function with Directional Derivative

[stanley.st]

I'm reading book and there's proposition with convex function

Function f is convex if and only if for all x,y
$$(*)\quad f(x)-f(y)\ge\nabla f(y)^T(x-y)$$

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

$$f(\lambda x+(1-\lambda)x)\le \lambda f(x)+(1-\lambda)f(y)$$

we have

$$\frac{f(y+\lambda(x-y))-f(y)}{\lambda}\le f(x)-f(y)$$

Setting $$\lambda\to0$$ 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

$$\lambda:=\frac{\mu}{\Vert x-y\Vert}$$

I have directional derivative on left hand side

$$\frac{f(y+\mu\frac{(x-y)}{\Vert x-y\Vert})-f(y)}{\mu}\le\frac{f(x)-f(y)}{\Vert x-y\Vert}$$

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

$$\nabla f(y)^T(x-y)\le\frac{f(x)-f(y)}{\Vert x-y\Vert}$$

 

[fresh_42]

There is no need for a unit vector, and if, we can simply assume $\|x-y\|=1$ as the crucial point is $\mu \to 0$. See https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/