Differentiability of $\int_{a} ^{x} F'(t) \, dt$ at $a, b$

This question is inspired from this answer where I prove that

If $$F:(a, b) \to\mathbb{R}$$ is a function with bounded and continuous derivative $$F'$$ in $$(a, b)$$ then the limits $$\lim_{x\to a^{+}} F(x)$$ and $$\lim_{x\to b^{-}}F(x)$$ exist.

The proof for the above result is based on analysis of the function $$\int_{a} ^{x} F'(t) \, dt$$.

Based on the above theorem let's extend $$F$$ continuously to $$[a, b]$$ by defining $$F(a) =\lim_{x\to a^{+}} F(x), F(b) =\lim_{x\to b^{-}} F(x)$$ and then ask

Does the left (right) hand derivative of $$F$$ at $$b$$ ($$a$$) exist?

I think the answer should be "No" and expect some kind of counter-example to demonstrate this.

Further I would like to know what happens with both these results if we change the hypotheses to "derivative $$F'$$ is bounded in $$(a, b)$$ and its extension (in any manner) to $$[a, b]$$ is Riemann integrable".

The above discussion is based on the second fundamental theorem of calculus

If $$F:[a, b] \to\mathbb {R}$$ is differentiable on $$[a, b]$$ and derivative $$F'$$ is bounded and Riemann integrable on $$[a, b]$$ then $$\int_{a} ^{b} F'(x) \, dx=F(b) - F(a)$$

Essentially I want to know if the assumption of differentiability at the end points $$a, b$$ in the above theorem can be dropped or not.

Here is a much simpler example:

$$f(0) = 0$$ and $$f(x) = x·\sin(7·\ln(1/x))$$ for every $$x \in (0,1)$$.

The idea is easy: Make it self-similar and shrinking to the endpoint. The $$7$$ is unnecessary and only to make the graph nice and easy to grasp intuitively.

• Nice example and the graph +1. – Paramanand Singh Nov 8 '18 at 7:53
• @ParamanandSingh: Thanks! Graphs are fun! All these graphs were made using some nice software called Graph, and the formulae for those fractals are here. – user21820 Nov 8 '18 at 7:59
• Thanks for the link to graphing software. – Paramanand Singh Nov 8 '18 at 8:01

There exists a continuous function $$F\colon[0,1]\to\mathbb{R}$$, continuously differentiable on $$(0,1)$$ with $$F'$$ bounded, but the one-sided derivative $$D_+F(0)$$ does not exist.

Construction

Let $$F'$$ takes value $$(-1)^n$$ on interval $$[3^{-n}(1+\epsilon),3^{-n+1}]$$, and linearly interpolate between $$\pm 1$$ on $$[3^{-n},3^{-n}(1+\epsilon)]$$. Note that $$\int_0^{3^{-n}}F'=\underbrace{\color{red}{\int_0^{3^{-n-1}(1+\epsilon)}F'}}_{\lvert\cdot\rvert\leq3^{-n-1}(1+\varepsilon)}+(-1)^n[3^{-n}-3^{-n-1}(1+\epsilon)]$$ So $$\limsup_n\frac{F(3^{-n})-F(0)}{3^{-n}}\geq 1-\frac23(1+\epsilon)$$ and $$\liminf_n\frac{F(3^{-n})-F(0)}{3^{-n}}\leq -1+\frac23(1+\epsilon)$$ If $$0<\epsilon<\frac12$$ this gives $$D_+F(0)$$ does not exist.

If $$F:(a, b) \to\mathbb{R}$$ is a differentiable function $$F'$$ such that an (hence any) extension to $$[a,b]$$ is Riemann integrable then the limits $$\lim_{x\to a^+} F(x)$$ and $$\lim_{x\to b^-}F(x)$$ exist

Proof.

Let $$G\colon[a,b]\to\mathbb{R}$$ be an extension of $$F'$$.

Let $$c\in(a,b)$$, $$d\in(c,b)$$. We can apply the second fundamental theorem of calculus to get $$F(d)-F(c)=\int_c^d F'.$$ From $$G\in\mathfrak{R}[c,b]$$ and $$d\in (c,b)$$, we have $$\int_c^d F'=\int_c^b G-\int_d^b G$$ so it suffices to show $$\lim_{d\to b^-}\int_d^b G$$ exists. Since $$G\in\mathfrak{R}[a,b]$$, $$\sup_{[a,b]}\lvert G\rvert=M<\infty$$. Hence $$\left\lvert\int_d^b G\right\rvert\leq\lvert b-d\rvert\cdot\sup_{[d,b]}\lvert G\rvert\leq M\lvert b-d\rvert\to 0$$ as $$d\to b^-$$. So the one-sided limit $$\lim_{d\to b^-}F(d)$$ exists.