We know that Linear context-free languages are not closed under complement, so I encountered a challenge in finding an example to show the above theorem. I think the complement of $L={a^nb^n}$ is not linear, but I can't prove it.
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The complement of the language $L = \{a^nb^n : n \in \mathbb{N}\}$ should actually be linear, if I'm not mistaken. Take some string $x \notin L$ over the alphabet $\Sigma = \{a, b\}$, since $L \subseteq a^*b^*$ one of three cases must be true:
- $x \notin a^*b^*$, or
- $x = a^nb^m \in a^*b^*$ and $n > m$, or
- $x = a^nb^m \in a^*b^*$ and $n < m$.
So the complement of $L$ (relative to $\Sigma$) can be expressed as $$\overline{L} = \overline{a^*b^*} \cup \{a^nb^m : n > m\} \cup \{a^nb^m : n < m\}.$$ Now, $\overline{a^*b^*} \in \texttt{REG}$ so $\overline{a^*b^*}$ must be linear. Using an induction argument over the rules of the linear grammar $G$ with rules $$S \to aS | aA$$ $$A \to aAb | \varepsilon$$ and start variable $S$, it can be shown that $L(G) = \{a^nb^m : n > m\}$ is linear. Similarly, $\{a^nb^m : n < m\}$ can be generated by a linear grammar with rules $$S \to Sb | Bb$$ $$B \to aBb | \varepsilon.$$ Since $\overline{L}$ can be written as the union of linear languages, and the linear languages are closed under union, it follows that $\overline{L}$ must be linear as well.
Counterexample
I think I might've found a valid counterexample. The language $L = \{a^nb^mc^k : n \neq m \land m \neq k\}$ is well known to be context-sensitive.
Observe that $$\overline{L} = \overline{a^*b^*c^*} \cup \{a^nb^mc^k : n = m\} \cup \{a^nb^mc^k : m = k\}.$$ It can be verified that the linear Grammars $$S \to Sc | A$$ $$A \to aAb | \varepsilon$$ and $$S \to aS | A$$ $$A \to bAc | \varepsilon$$ with start variable $S$ generate $\{a^nb^mc^k : n = m\}$ and $\{a^nb^mc^k : m = k\}$ respectively, and that $\overline{a^*b^*c^*} \in \texttt{REG}$. So $\overline{L}$ is linear whilst $\overline{\overline{L}} = L$ is not, counterexample ↯
