Concave function applied to equally distant points

Question

Is the next statement true?

Let $f$ be concave and $a \leq b \in dom(f)$. For any $c \geq 0$ such that $a+c, b+c \in dom(f)$ then $$ f(b+c) - f(b) \leq f(a+c) - f(a) $$

If it is, how would you prove it?

Answer 1

There's an easy way to show it directly from the definition of concavity. Let $\theta = c/(c + b - a)$. We need $c\ge0$ so that $\theta \in [0,1]$. Then we have: $$ \begin{aligned} b &= \theta a + (1-\theta)(b+c)\\ a+c &= (1-\theta) a + \theta (b+c) \end{aligned} $$ Then concavity of $f$ implies: $$ \begin{aligned} f(b) &\ge \theta f(a) + (1-\theta)f(b+c)\\ f(a+c) &\ge (1-\theta) f(a) + \theta f(b+c) \end{aligned} $$ Adding the inequalities gives: $$ f(b) + f(a+c) \ge f(a) + f(b+c) \quad \square $$

On the other hand, if $c<0$, we can transform $a' = a+c$, $b' = b+c$, $c' = -c$ and apply the above inequality to the transformed variables. $$ f(b') + f(a'+c') \ge f(a') + f(b'+c')\\ f(b+c) + f(a) \ge f(a+c) + f(b) $$ So if $c<0$ the inequality is reversed.

Answer 2

Yes, it is true. Note that (if necessary relabelling the points, and after dealing with some special cases) we can assume wlog that $a < a+c < b < b+c$. Concavity of $f$ implies that the slopes of the segments $[(a,f(a)), (a+c, f(a+c))]$, $[(a+c,f(a+c)), (b,f(b))]$ and $[(b,f(b)), (b+c,f(b+c))]$ are sorted in decreasing order.