Problem: Let $f: \mathbb{R}^n \to \overline{\mathbb{R}}$ be the convex function. Let $x_1,x_3
\in E$ (Euclidean space in $\mathbb{R}^n$) and $x_2 \in (x_1,x_3)$.
Prove that $$\dfrac{f(x_3) -f(x_2)}{\Vert x_3-x_2 \Vert} \ge\dfrac{f(x_2)-f(x_1)}{\Vert x_2-x_1\Vert}.$$
My attempt: Since $x_2 \in (x_1,x_3)$ then there exists $t \in (0,1)$ such that $x_2 = tx_1 + (1-t)x_3$. Thus we have
$$f(x_3) – f(x_2) \ge f(x_3)-tf(x_1)-(1-t)f(x_3) = tf(x_3)-tf(x_1).$$
Therefore
$$\dfrac{f(x_3)-f(x_2)}{\Vert x_3-x_2\Vert} \ge \dfrac{t}{\vert t \vert}\dfrac{f(x_3)-f(x_1)}{\Vert x_3-x_1\Vert} \ge -\dfrac{f(x_3)-f(x_1)}{\Vert x_3-x_1\Vert} = \dfrac{f(x_1)-f(x_3)}{\Vert x_1-x_3\Vert}.$$
I have tried many different ways to have a result like the solution above but I failed. I wonder that the problem is right or not?
Best Answer
Since $x_2 \in (x_1,x_3)$, there exists $t \in (0,1)$ such that $x_2 = tx_1 + (1-t)x_3$. Thus we have $$f(x_2) = f(tx_1+(1-t)x_3) \le tf(x_1) + (1-t)f(x_3).\ (1)$$ Firstly, from (1) we have \begin{align*} & \dfrac{f(x_2)-f(x_1)}{\Vert x_2-x_1 \Vert} \le \dfrac{(1-t)\left(f(x_3)-f(x_1)\right)}{(1-t)\Vert x_3-x_1\Vert} = \dfrac{f(x_3)-f(x_1)}{\Vert x_3 - x_1\Vert}\\ \text{and }& \dfrac{f(x_3) - f(x_2)}{\Vert x_3-x_2 \Vert} \ge \dfrac{t(f(x_3)-f(x_1)}{t\Vert x_3-x_1\Vert} = \dfrac{f(x_3)-f(x_1)}{\Vert x_3-x_1\Vert}. \end{align*} This yields $$\dfrac{f(x_3) -f(x_2)}{\Vert x_3-x_2 \Vert} \ge \dfrac{f(x_2)-f(x_1)}{\Vert x_2-x_1\Vert}. (\text{qed}).$$