# Baffling proof using function convexity

Let the function $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be convex, differentiable with derivative $$f_x$$ and Lipschitz continuous with constant $$L$$. Then, for $$a,b,c,d \in \mathbb{R}$$ such that $$a \ge b\ge d$$ and $$a \ge c\ge d$$, $$\begin{equation*} \begin{split} & f(\max\{ b,c\}) - f(a) + f(\min\{ b,c\}) - f(d)\\ & \le f_x(\min\{ b,c\})(b -d + c - a). \\ \end{split} \end{equation*}$$ Apparently this can be proven easily using $$\max\{ b,c\} - a \le 0$$ and the convexity of $$f$$, but I am stumped about how exactly that is done … almost seems like a mistake! Would really appreciate any pointers or tips.

For reference, this is from the proof of Lemma 3.2 in the paper: Boetius, Frederik, and Michael Kohlmann. "Connections between optimal stopping and singular stochastic control." Stochastic Processes and their Applications 77.2 (1998): 253-281.

• Since the statement is symmetric in $b$ and $c$, it's probably easier just to assume that $a \ge b \ge c \ge d$ and re-write it as $f(b) - f(a) + f(c) - f(d) \le f'(c)(b - d + c - a)$. Jul 12, 2020 at 22:21
• Hi LSpice - thanks for answering - the problem is that this is an argument used in a proof that $b \ge c$, so can't really make that assumption here. I'm seriously thinking this entire argument is a mistake Jul 12, 2020 at 22:25
• It's just notation. If you can prove the result in the form in which I've stated it, then you can switch the labels $b$ and $c$ in case they're in the other order. Jul 12, 2020 at 22:26

As suggested, assume that $$a \ge b \ge c \ge d$$ and re-write the desired inequality as $$f(b) - f(a) + f(c) - f(d) \le f'(c)(b - d + c - a).$$ We have $$f(a) - f(b) \ge f'(b)(a - b)$$, $$f(c) - f(d) \le f'(c)(c - d)$$, and $$f'(b) \ge f'(c)$$ by convexity, so the result follows.