Continuous functions $\mathbb R_\ell\to\mathbb R_\ell$ This question has been asked at least twice, but the first has an incorrect answer (as I point out in the comments) and the second only gives a "starter" instead of a solution.
I want to characterize the continuous functions $\mathbb R_\ell\to\mathbb R_\ell$, where $\mathbb R_\ell$ is the lower limit topology. Let's call these "lower limit continuous". They must of course also be continuous as functions $\mathbb R_\ell\to\mathbb R$, and these functions are known to be exactly the right-continuous functions. However, not all right-continuous functions are lower limit continuous. A counterexample is given by $f(x)=-x$, where the preimage of $[0,1)$ is the not open $(-1,0]$.
An initial conjecture I had is that lower limit continuous functions might be the "piecewise increasing" right-continuous functions. However, it turns out that right-continuous functions are not necessarily piecewise continuous, so the obvious definition of "piecewise increasing" doesn't work. A second guess was that these might be right-continuous functions where the domain can be partitioned into intervals $[a,b)$ where the function is increasing on each interval. This also turns out to be false, with a counterexample given by
$$f(x)=\begin{cases}0,&x\le 0\\x-e^{\lfloor\ln x\rfloor},&x>0.\end{cases}$$
This prompts the main question and a bonus question:

*

*Is there any "nice" characterization of lower limit continuous functions?

*Are there any "nowhere increasing" lower limit continuous functions, as in functions that are not increasing in any open interval $(a,b)$?


Edit: take $q:\mathbb N\to\mathbb Q$ an arbitrary bijection and define $$g(x)=\sum_{k=0}^\infty 2^{-k}f(x+q(k)),$$ where $f$ is the previous counterexample. I think $g$ might be lower limit continuous and nowhere increasing, but I have no idea how to show this.
 A: I belive this could be a characterization of lower limit continuous functions:
Let $f:\mathbb R_\ell\to\mathbb R_\ell$ thus $f$ is continous iff $f$ is a right-continuous function and for all $a \in \mathbb R_\ell$ there exsist $b \in \mathbb R_\ell$ such that for all $x \in [a,b)$, $f(x) \ge f(a)$
$\Rightarrow$
Let $f:\mathbb R_\ell\to\mathbb R_\ell$ be a continous function
first we will prove that $f$ is a right-continuous function, let $x_0 \in \mathbb R_\ell$ and let $\epsilon > 0$ we want to find $\delta > 0$ such that for all $x \in [x_0,x_0+\delta)$, $|f(x)-f(x_0)| <\epsilon$.
indeed, the set $[f(x_0),f(x_0)+\epsilon)$ is open in $\mathbb R_\ell$ and thus from the continouty of $f$ we get that $f^{-1}([f(x_0),f(x_0)+\epsilon))$ is open in $\mathbb R_\ell$, now $x_0 \in f^{-1}([f(x_0),f(x_0)+\epsilon))$ and thus there exist $\delta > 0$ such that $[x_0,x_0 + \delta) \subseteq f^{-1}([f(x_0),f(x_0)+\epsilon))$ and thus for every $x \in [x_0,x_0 + \delta)$ $f(x) \in [f(x_0),f(x_0)+\epsilon)$ and thus $|f(x)-f(x_0)| <\epsilon$ as needed.
in addition we get that for every $x \in [x_0,x_0 + \delta)$, $f(x) \in [f(x_0),f(x_0)+\epsilon)$ and thus $f(x)\ge f(x_0)$. so the second condition is also met.
$\Rightarrow$
Let's assume that $f$ is a right-continuous function and for all $a \in \mathbb R_\ell$ there exsist $b \in \mathbb R_\ell$ such that for all $x \in [a,b)$, $f(x) \ge f(a)$.
we shall prove that for every $x_0 \in \mathbb R_\ell$, $f$ is continous at $x_0$. Let $U \subseteq \mathbb R_\ell$ be an open neighberhood of $f(x_0)$ thus there exist $\epsilon$ such that $[f(x_0),f(x_0) + \epsilon) \subseteq U$.
now $f$ is a right-continuous function thus there exist $\delta_1>0$ shuch that $f([x_0,x_0 + \delta_1)) \subseteq (f(x0) - \epsilon,f(x0) + \epsilon)$. from the assumption there exist $\delta_2>0$ such that $f([x_0,x_0 + \delta_2)) \subseteq [f(x_0),\infty)$. define $\delta = \min (\delta_1, \delta_2)$ and thus
$f([x_0,x_0 + \delta)) \subseteq f([x_0,x_0 + \delta_1) \cap f([x_0,x_0 + \delta_2) \subseteq (f(x0) - \epsilon,f(x0) + \epsilon) \cap [f(x_0),\infty) =[f(x_0),f(x_0) + \epsilon) \subseteq U$. and thus $[x_0,x_0 + \min (\delta_1, \delta_2))$ is an open neighberhood of $x_0$ such that $[x_0,x_0 + \min (\delta_1, \delta_2)) \subseteq f^{-1}(U)$. we get that $f$ is continous at $x_0$. and thus $f$ is continous in $\mathbb R_\ell$.
