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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.

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Here is a much simpler example:

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

continuous derivative without limit at endpoint

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.

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

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