# Prove decreasing convex function has decreasing differences

Let $f: \mathbb R \rightarrow \mathbb R$ be a decreasing convex function.

For any $t>0$, I want to show that for all $x_{1} \leq x_{2}$ we have: $$f(x_{1}) - f(x_{1}+t) \geq f(x_{2} ) - f(x_{2}+t)$$

I am trying to show this from the definition of a convex function and by the fact that $f$ is decreasing,

• The "decreasing" criterion is immaterial; all it does is guarantee that both sides of the inequality are positive. Aug 14, 2018 at 19:54

The decreasing assumption is not needed, as mentioned by Connor Harris in the comments. I claim that for $a < b < c$ we have $$\frac{f(b) - f(a)}{b-a} \le \frac{f(c) - f(a)}{c-a} \le \frac{f(c) - f(b)}{c-b}.\tag{*}$$ Do you see how to obtain the answer to your original question from ($*$)?
Proof of ($*$):
By convexity, $$f(b) \le \frac{b-a}{c-a} f(c) + \frac{c-b}{c-a} f(a).$$ This yields $$f(b) - f(a) \le \left(\frac{b-a}{c-a} f(c) + \frac{c-b}{c-a} f(a)\right) - f(a) = \frac{b-a}{c-a} (f(c) - f(a))$$ and $$f(c) - f(b) \ge f(c) - \left(\frac{b-a}{c-a} f(c) + \frac{c-b}{c-a}f(a)\right) = \frac{c-b}{c-a} (f(c) - f(a)).$$
If $t \le x_2 - x_1$ then $x_1 \le x_1 + t \le x_2 \le x_2 + t$. Using ($*$) twice yields $$\frac{f(x_1 + t) - f(x_1)}{t} \le \frac{f(x_2) - f(x_1 + t)}{x_2 - x_1 - t} \le \frac{f(x_2 + t) - f(x_2)}{t}.$$ Otherwise $t \ge x_2 - x_1$ which yields $x_1 \le x_2 \le x_1 + t \le x_2 + t$. Then applying ($*$) twice yields $$\frac{f(x_1 + t) - f(x_1)}{t} \le \frac{f(x_1 + t) - f(x_2)}{x_2 - x_1 - t} \le \frac{f(x_2 + t) - f(x_2)}{t}.$$