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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.

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    $\begingroup$ 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 $f(b) - f(a) + f(c) - f(d) \le f'(c)(b - d + c - a)$. $\endgroup$
    – LSpice
    Jul 12, 2020 at 22:21
  • $\begingroup$ 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 $\endgroup$
    – icewater
    Jul 12, 2020 at 22:25
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    $\begingroup$ 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. $\endgroup$
    – LSpice
    Jul 12, 2020 at 22:26

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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.

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  • $\begingroup$ Thank you LSpice, you are 100% right here $\endgroup$
    – icewater
    Jul 13, 2020 at 0:40

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