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Consider a convex function $f(x)$ on interval $(a,b) \subseteq \mathbb{R}$. According to "A user's guide to measure theoretic probability" by Pollard (see Appendix C), its right-hand $D_{+}(x)$ and left-hand $D_{-}(x)$ derivatives

$D_{-}(x_0) = \lim_{x \rightarrow x_0^{-}} \frac{f(x) - f(x_0)}{x - x_0} , \quad D_{+}(x_0) = \lim_{x \rightarrow x_0^{+}} \frac{f(x) - f(x_0)}{x - x_0}$

exist at any $x_0 \in (a,b)$. Further, we know that $D_{+}(x)$ is increasing and right-continuous, and $D_{-}(x)$ is increasing and left-continuous, w.r.t. domain $(a,b)$.

My question is following: given $D_{+}(x)$, can we recover $D_{-}(x)$? If no, then under what additional conditions it is possible?

My current guess is to do the recovery as:

$D_{-}(x_0) = \lim_{x \rightarrow x_0^{-}} D_{+}(x)$

Yet, I'm not sure if the left-hand limit exist at $D_{+}(x_0)$. Is the above statement correct? How can we prove it?

Dimka Kopitkov
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1 Answers1

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The answer is yes.

First you can check that $$\limsup_{x\to x_0^-}D_+(x)\leq D_-(x_0)\,.$$ To show it fix $x<x_0$. Since a convex function is differentiable almost everywhere, there must be a point $x'\in(x,x_0)$ such that $f$ is differentiable in $x'$. So $D_+(x')=D_-(x')$. Using the monotonicity of $D_+$ and $D_-$ you get $$D_+(x)\leq D_+(x')=D_-(x')\leq D_-(x_0)\,.$$ So in particular $D_+(x)\leq D_-(x_0)$ for all $x<x_0$ and the inequality for the $\limsup$ follows immediately.

Then you need to show that $$\liminf_{x\to x_0^-}D_+(x)\geq D_-(x_0)\,.$$ Here you can use the fact that the convexity of $f$ implies $D_+(x)\geq D_-(x)$ for all $x$. Hence you have $$\liminf_{x\to x_0^-}D_+(x)\geq \liminf_{x\to x_0^-}D_-(x) = \lim_{x\to x_0^-}D_-(x)=D_-(x_0)\,,$$ where you use the fact that $D_-$ is left continuous.

So eventually $$\lim_{x\to x_0^-} D_+(x) = D_-(x_0)\,.$$

ECL
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