TITLE: Are all continuous one one functions differentiable?
QUESTION [14 upvotes]: I was reading about one one functions and found out that they cannot have maxima or minima except at endpoints of domain. So their derivative , if it exists,  must not change it sign , i.e. , the function should be either strictly increasing or strictly decreasing. From this I've a feeling that all continuous one one functions must be differentiable . Is this true?

REPLY [4 votes]: $\newcommand{\Reals}{\mathbf{R}}$There exists a strictly increasing continuous function $F:\Reals \to \Reals$ that fails to be differentiable at each rational number.
Let $(a_{k})_{k=1}^{\infty}$ be an enumeration of your favorite dense countable set $A$, such as the set of rational numbers, and $H:\Reals \to \Reals$ the unit step function
$$
H(x) = \begin{cases}
  0 & \text{if $x \leq 0$,} \\
  1 & \text{if $0 < x$.}
\end{cases}
$$
Form the sum of scaled translates
$$
f(x) = \sum_{k=1}^{\infty} 2^{-k} H(x - a_{k}),
$$
and its definite integral
$$
F(x) = \int_{0}^{x} f(t)\, dt.
$$
The following are easy "honors calculus"/elementary analysis exercises:

The function $f$ is strictly positive (in fact, $0 < f(x) < 1$ for all real $x$), strictly increasing (hence Riemann integrable over an arbitrary compact interval), has a jump discontinuity at each point of $A$, and is continuous elsewhere.
The function $F$ is continuous (as a definite integral), strictly increasing (positive integrand), strictly convex (increasing integrand), and differentiable at $x$ if and only if $f$ is continuous at $x$ (fundamental theorem of calculus, since $f$ has only jump discontinuities), hence non-differentiable at each point of $A$.