Can a discontinuous function be increasing or decreasing

Question

How would we decide increasing or decreasing function if the function is not differentiable? Can a discontinuous function be increasing or decreasing? And can it be monotonic?

Answer 1

Let $f: \mathbb R \to \mathbb R, x \in \mathbb Q \implies f(x)=\tan x. x\in \mathbb Q^c \implies f(x)=0.$

It's only continuous at $x=0$. Not differentiable anywhere. If $x,y \in \mathbb Q \cap(-\pi/2,\pi/2), x<y\implies f(x)<f(y)$. Since $\tan$ is periodic, the function is understood by analyzing only this interval.

Since the rationals are dense in $\mathbb R$ ,$\forall p \in \mathbb Q, \delta \in \mathbb R, \delta>0, \exists q\in \mathbb Q, |p-q|<\delta$. Something similar can be said about two elements from $\mathbb Q^c$.

So the limit of $\frac{\tan p-\tan q}{p-q}, q\to p$ exists and is equal to $\sec^2 p$ if $p$ and $q$ are both rational. If they are both irrational, that limit is 0. Since the limits aren't equal, the function isn't differentiable.

So there's a clear sense in which a non-trivial part of the function is increasing, yet it isn't differentiable.

Answer 2

Here's an example of a function $f$ which is discontinuous in every neighbourhood and strictly increasing everywhere. Let $r_0,r_1,r_2,...$ be an enumeration of the set $D$ of finite decimals between $r_0=0$ and $1$ such that the $0$-place decimals are counted first, the $1$-place decimals next, and so on.* Let $f(r_0)=0$ and, for each $k\in\Bbb N$, define $f(r_{k+1})=f(r_k)+2^{-k}$. For $x\in[0\,\pmb,\,1)\setminus D$, define $$f(x):=\sup\{f(r):r\in D\,;\,r<x\}.$$ All other values of $f$ are determined by the identity $f(x+1)=f(x)+2\,$ ($x\in\Bbb R$).

* For example, $(r_0,r_1,...)=(0; 0.1, ...,0.9;0.01,...,0.99;\,...)$.