I am trying do this exercise:
Exercise: Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$ a convex funciton. Show that: $$\langle y^1 - y^2 , x^1 - x^2 \rangle \geq 0, \qquad \forall x^i \in \mathbb{R}^n, \forall y^i \in \partial f(x^i), i= 1,2.$$
Notation: $\partial f(\star)$ means subdifferential of $f$ in $\star$.
And we know the following defenition:
Let $f : X \in \mathbb{R}^n \rightarrow \mathbb{R}$ a convex function, then
$$f(x) - f(y) \leq \langle \partial f(y) , x - y \rangle$$.
My attemp:
Since $f$ is convex, by defenition above:
$$f(x^1) - f(x^2) \leq \langle \partial f(x^2) , x^1 - x^2 \rangle, \qquad \forall x^1,x^2 \in \mathbb{R^n}$$
Let $ y^2 \in \partial f(x^2)$ and some inequality properties, we obtain:
$$f(x^1) - f(x^2) - \langle y^2 , x^1 - x^2 \rangle \leq 0 $$
I believe wich by iteration or anything similar we can "transform" $f(x^1)- f(x^2)$ in derivate of one point in $D$ (f dominian)and developing the inner product we got what we wanted. But I cannot undertsand one way to do that. Anyone can help me?
Your inequality is incorrect, it should be $$ f(x) - f(y) \geq \langle \partial f(y), x-y\rangle. $$ that is, a convex function always lies above its supporting hyperplane.
Let $y_1\in \partial f(x_1)$ and $y_2\in \partial f(x_2)$. Then, $$ f(x_2) - f(x_1) \geq \langle y_1, x_2 - x_1\rangle $$ and $$ f(x_1) - f(x_2) \geq \langle y_2, x_1 - x_2\rangle $$ Summing these inequalities yields, $$ 0 \geq \langle y_1, x_2 - x_1\rangle + \langle y_2, x_1 - x_2\rangle = \langle y_1-y_2, x_2 - x_1\rangle $$ and the result follows.