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](https://doi.org/10.1016/S0304-4149(98)00049-0)." Stochastic Processes and their Applications 77.2 (1998): 253-281.



Baffling proof using function convexity
As [suggested](https://mathoverflow.net/questions/365504/baffling-proof-using-function-convexity#comment922614_365504), 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.


