Proof that a continuous monotone function is a.e differentiable

Question

In S&S there is the proof that a monotone function is a.e differentiable. It says it is enough to prove 2 properties

I understand why we need to prove both properties but I am confused as to why we can consider -F(-x) and how they get the final inequality.

Isn't $D_- < D_+$ trivially since the function is monotonic?

If this is true then why do we need to consider -F(-x) to get that inequality?

Answer 1

We want to show that:

$$D^{+}F(x) \leq D_{-}F(x) \implies D^{-}F(x) \leq D_{+}F(x)$$

By definition, $$D^{-}F(x)=\lim\sup_{h \to 0 \\ h<0} \Delta_h(F)(x)$$

We know that $\lim\sup \fbox{ something }=-\lim\inf \fbox{ - something }$

Therefore,

$$D^{-}F(x)=\lim\sup_{h \to 0 \\ h<0} \Delta_h(F)(x)=-\lim\inf_{h \to 0 \\ h<0}\Delta_h(-F)(x)$$

Now, note that $x \mapsto -x$, means $h<0 \mapsto -h>0$ and it doesn't affect $h \to 0$. Therefore,

$$D^{-}F(x)=\lim\sup_{h \to 0 \\ h<0} \Delta_h(F)(x)=-\lim\inf_{-h \to 0 \\ -h>0}\Delta_h(-F)(-x)=-D_{+}(-F)(-x)=D_{+}F(-x)$$

Similarly,

$$D_{+}F(x)=\lim\inf_{h \to 0 \\ h>0} \Delta_h(F)(x)=-\lim\sup_{-h \to 0 \\ -h>0}\Delta_h(-F)(-x)=-D^{-}(-F)(-x)=D^{-}F(-x)$$

$$D^{-}F(-x) = D_{+}F(x)\leq D^{+}F(x) \leq D_{-}F(x) \leq D^{-}F(x) = D_{+}F(-x)$$

Hence, $$D^{-}F(-x) \leq D_{+}F(-x)$$

Since the inequality we began with is true for almost every $x$, this inequality is also true for almost every $x$ and therefore, $-x$ can be changed to $x$ to finish the proof that:

$$D^{-}F(x) \leq D_{+}F(x)$$

Now the chain of inequalities mentioned in the book shows you that they all have to be equal to each other.

Typesetting all this with all of these subscripts and superscripts and plus and minus signs was a nightmare. So, do check that I haven't made any typos.

Answer 2

Regarding your question Isn't $D_- < D_+$ trivially since the function is monotonic?, even for a strictly increasing and continuous function, we can have $D_- > D_+$ at some points.

Consider the function defined by $f(x) = 4x$ if $x < 0$ and $f(0) = 0$ and $f(x) = 2x$ if $x > 0.$ Then $f$ is continuous and strictly increasing on the reals, and at $x=0$ the lower left derivate is $4$ and the lower right derivate is $2.$ In fact, at $x=0$ we have $D_{-} = D^{-} = 4$ and $D_{+} = D^{+} = 2.$