If you put the same topology on the domain and the codomain, a classical example of this is $f(x) = -x$, which is continuous as a map $f:(\mathbb{R}, \mathbb{R}_{\text{usual}}) \rightarrow (\mathbb{R}, \mathbb{R}_{\text{usual}})$ but not as a map $f: (\mathbb{R}, \mathbb{R}_{\text{lower}}) \rightarrow (\mathbb{R}, \mathbb{R}_{\text{lower}})$, the latter because $f^{-1}[[0,1)] = (-1,0]$ which is not open as $0$ is not an interior point.
This also shows that $\mathbb{R}$ is not a topological group under addition in the lower limit topology.
If you have the same topology on the codomain, then as $(\mathbb{R}, \mathbb{R}_{\text{usual}})$ is coarser than $(\mathbb{R}, \mathbb{R}_{\text{lower}})$, every function continuous on usual reals would be continuous in the lower limit topology. So it's essential we change the topology on the codomain as well.
I define the continuous from the right at $a \in \Bbb R$ condition for $f:\Bbb R \to \Bbb R$ as follows:
$$\forall \varepsilon>0: \exists \delta > 0: \forall x \in \Bbb R: (x>a \land |x-a| < \delta) \to |f(x)-f(a)| < \varepsilon\tag{1}$$
which is I think pretty standard. Let the smallest topology that makes all the functions that are continuous from the right at every point, continuous, be called $\mathcal{T}_w$
Then if $f: (\Bbb R, \mathcal{T}_l) \to (\Bbb R, \mathcal{T}_e)$ (from lower limit topology to Euclidean topology) is continuous, then $f$ fulfills $(1)$ at all points:
Having $a$ and $\varepsilon >0$ arbitrary, $f^{-1}[(f(a)-\varepsilon, f(a)+\varepsilon)]$ is open in $\mathcal{T}_l$ (by continuity of $f$, and the fact that open intervals are open in $\mathcal{T}_e$) and contains $a$, so there is some $\delta>0$ such that $$[a,a+\delta) \subseteq f^{-1}[(f(a)-\varepsilon, f(a)+\varepsilon)]$$ which immediately implies $(1)$ by definition.
On the other hand if $f:\Bbb R \to \Bbb R$ is continuous on the right at every point, we have that $f$ seen as a function from $(\Bbb R, \mathcal{T}_l)$ to $(\Bbb R, \mathcal{T}_e)$ is continuous: let $O$ be open in the Euclidean topology and let $a \in f^{-1}[O]$; we need to show that $a$ is an interior point of $f^{-1}[O]$. First note that $f(a) \in O$ and $O$ is Euclidean open so there is some $\varepsilon>0$ such that $(f(a)-\varepsilon, f(a)+\varepsilon) \subseteq O$. Apply $(1)$ to $a$ and this $\varepsilon$ and the condition gives that $x \in [a,a+\delta)$ implies that $f(x) \in (f(a)-\varepsilon, f(a)+\varepsilon)$, or equivalently $$[a,a+\delta) \subseteq f^{-1}[(f(a)-\varepsilon, f(a)+\varepsilon)] \subseteq f^{-1}[O]$$ showing that indeed $a$ is an interior point of $f^{-1}[O]$, as required. So $f^{-1}[O]$ is lower-limit open and $f$ is continuous.
Now $\mathcal{T}_l$ is one topology on $\Bbb R$ that makes all right-continuous $f$ continuous, and $\mathcal{T}_w$ is the minimal such, so $$\mathcal{T}_w \subseteq \mathcal{T}_l$$ And if $\mathcal{T}$ is any topology such that all right continuous $f$ are continuous when seen as $f(\Bbb R,\mathcal{T}) \to (\Bbb R, \mathcal{T}_e)$ then $\mathcal{T}_l \subseteq \mathcal{T}$: let $[a,b)$ be a basic open set of the lower limit topology, then the function $\chi: \Bbb R \to \Bbb R$ defined by $\chi(x)=1$ if $x \in [a,b)$ and $\chi(x)=0$ otherwise, is continuous from the right at any point. This is easy to check by the definition $(1)$. Now, as $\chi$ thus is continuous with domain $(\Bbb R, \mathcal{T})$ we know that $[a,b) = \chi^{-1}(\frac12, \rightarrow)] \in \mathcal{T}$ and so $\mathcal{T}$ contains all basic open sets of $\mathcal{T}_l$, so $\mathcal{T}_l \subseteq \mathcal{T}$. This shows directly that $\mathcal{T}_l$ is the minimal topology that makes all right continuous functions continuous.
Best Answer
There is no continuous map from the usual topology to the lower-limit topology, except constant maps.
If such a map $f$ were to attain distinct values $a$ and $b$, where $a<b$, then $f^{-1}[b,\infty)$ would be closed and also open in $\tau_u$, which is impossible because the only clopen subsets of $\tau_u$ are the whole set and the empty set.