This question is from the book The Integrals of Lebesgue, Denjoy, Perron, and Henstock.
I'm currently reading part of the book The Integrals of Lebesgue, Denjoy, Perron, and Henstock. Slight confusion arise on the following definition of upper and lower derivative.
Im kinda stuck on understanding why $$\overline{D} F(x) = \max \{D^+F(x), D^-F(x)\},$$
On my a perspective, for $\delta>0$, $x<y<x+\delta$ and $x-\delta<y<x$, implies $$0<|x-y|<\delta.$$
The following link $\lim_{n\to\infty}\sup x_n= \max(x,y)$ and $\lim_{n\to\infty}\inf x_n= \min(x,y)$
Makes up the same conclusion, but sequence taken is not of the form as supremum in this case runs over $y$ with $x<y<x+\delta$.
Any insight is highly appreciated.
All you're doing is merging two one-sided $\limsup$s into one two-sided one.
Throwing out all the symbols en masse: $$\begin{align*}\overline{D}F(x)&=\lim_{\delta\rightarrow0^+}\sup_{0<|y-x|<\delta}\frac{F(y)-F(x)}{y-x}\\ &=\lim_{\delta\rightarrow0^+}\sup_{\substack{0<y-x<\delta\\\text{or}\\0<x-y<\delta}}(...)\\ &=\lim_{\delta\rightarrow0^+}\max\left(\sup_{x<y<x+\delta}(...),\sup_{x-\delta<y<x}(...)\right)\\ &=\max\left(\lim_{\delta\rightarrow0^+}\sup_{x<y<x+\delta}(...), \lim_{\delta\rightarrow0^+}\sup_{x-\delta<y<x}(...)\right)\\ &=\max(D^+F(x),D^-F(x)) \end{align*}$$ where of course, you switch the $\lim$ and $\max$ by requiring everything be defined.