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If $f$ is differentiable on $(a,b)$ continuous at a, and $f$ has bounded derivative, must $f$ be right differentiable at $a$?

In case answer to previous question is true, is the statement still true if we replace bounded derivative with finite vertical distance travelled by $f$?

To clarify what I mean by finite vertical distance travelled, is there a constant $K$ such that for all $e>0$ if we let $g$ be the derivative of $f$ on $(a+e,b-e)$ then the integral of $|g|$ on $[a+e,b-e]$ is less than $K$?

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2 Answers 2

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If we don't require continuity at $a$, the answer is an easy "no". The answer is still "no" if we require continuity, but not quite as easily.

If we look at the function $g\colon (0,1) \to \mathbb{R}$,

$$g(x) = (-1)^n\quad\text{ if } \frac{1}{(n+1)!} \leqslant x < \frac{1}{n!},$$

and consider

$$f(x) = \int_0^x g(t)\,dt,$$

we have an almost-example: For $a_k = \frac{1}{k!}$ we have

\begin{align} \frac{f(a_{2k})-f(0)}{a_{2k}} &= (2k)!\int_0^{1/(2k)!} g(t)\,dt\\ &= (2k)!\biggl(\frac{1}{(2k)!} - \frac{1}{(2k+1)!}\biggr) + (2k)!\int_0^{1/(2k+1)!} g(t)\,dt\\ &= \frac{2k}{2k+1} + O\biggl(\frac{1}{2k+1}\biggr) \end{align}

and

\begin{align} \frac{f(a_{2k+1}) - f(0)}{a_{2k+1}} &= (2k+1)!\int_0^{1/(2k+1)!}g(t)\,dt\\ &= -(2k+1)!\biggl(\frac{1}{(2k+1)!}-\frac{1}{(2k+2)!}\biggr) + (2k+1)!\int_0^{1/(2k+2)!} g(t)\,dt\\ &= -\frac{2k+1}{2k+2} + O\biggl(\frac{1}{2k+2}\biggr), \end{align}

so the difference quotients $\frac{f(x)-f(0)}{x}$ vary from very close to $1$ to very close to $-1$ and back when $x$ traverses the interval $[a_{2k+2},a_{2k}]$, and the set of all limit points of the difference quotients for sequences $x_k \searrow 0$ is the interval $[-1,1]$.

The $f$ considered above is only an almost-example, since $f$ is not differentiable on the whole interval $(0,1)$, there are countably many points where $f$ isn't differentiable.

We can convert it to a true example if we alter $g$ to linearly interpolate instead of jump at the points $a_k$ if we choose a steep enough slope for the linear interpolation. Then the modification $\tilde{g}$ is continuous, and its integral $\tilde{f}$ is differentiable on $(0,1)$ but not right-differentiable at $0$.

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No. There are functions which are not even continuous on $[a,b]$ even though they are differentiable on $(a,b)$. One such example is

$$f(x) = \left\{ \begin{array}{ll} 0 & x=0, \\ 1 & 0<x<1, \\ 0 & x=1. \end{array} \right. $$

It is easy to verify that $f'(x)=0$ whenever $x\in (0,1)$ but not right differentiable at $x=0$.

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