Is the concatenation of a non-regular CFL and a complement of a regular upper-set always non-regular?

Question

Let $L_1$ be a non-regular CFL. Let $L_2$ be a regular language. Assume that $\left(L_1\right)^{*} \subseteq L_2$. I'm looking at $L_3 = \left( L_1 \right) ^{*} \circ \overline{L_2}$. Is $L_3$ always non-regular? I can find multiple examples of it being non-regular, yet none of it being regular.

Edit: $\overline{L_2} \neq \emptyset$.

Answer 1

Consider the language $L = \{uv\mid u, v\in\Sigma^*, |u|=|v|\wedge u\neq v\}$. I claim that $L$ is context-free, it is not regular (a simple pumping lemma proof can do the trick), but $L^*$ is regular, without being equal to $\Sigma^*$, so that means that $L^*\overline{L^*}$ is regular and satisfies your conditions.

Now let's show that if we denote $L'= (\Sigma\Sigma)^*\setminus\left(\bigcup\limits_{a\in\Sigma}a^*\right)\cup \{\varepsilon\}$ (words of even length, not composed of the same letter), then $L^*=L'$:

• the inclusion $\subseteq$ is the easiest, since words of $L$ are of even length, and are not composed of the same letter. We added $\varepsilon$ to $L'$ by definition of the Kleene star;

• let's prove by induction that $L'\subseteq L^*$:

  • consider $w\in L'$; if $w = \varepsilon$, then $w\in L^*$, if $|w| = 2$, then $w\in L$ (because it contains two different letters), so $w\in L^*$;

  • suppose the result proved for words of length up to a certain $2n\in\mathbb{N}$, and consider $w\in L'$ of length $2n+2$. If $w\in L$, then trivially, $w\in L^*$. Otherwise, since $|w|$ is even, then $w = uu$, with $u\in\Sigma^*$, $u$ containing different letters.

    • If $|u|$ is even, then it means that $u \in L'$
    • Otherwise, $u = u_1u_2…u_{2k+1}$, and we can show that $uu = xy$ verifying $x,y\in L'$ (with either $x = u_1…u_{2k}$ and $y = u_{2k+1}u$ or $x=uu_1$ and $y = u_2…u_{2k+1}$).

    In all cases, we can use the induction hypothesis to show that $w\in L^*$.