Is a kind of reverse Kleene star of a context-free language context-free?

Question

Recently I had a question on one of my assignments asking to prove or disprove the following:

Let $L$ be a language. If $L^*$ is context-free then $L$ is context-free.

Now obviously this is false since we can take some non-context free language $L_1$ and the alphabet $\Sigma$, then make $(L_1 \cup \Sigma)^* = \Sigma^*$ is context-free and clearly $L_1\cup \Sigma$ is not.

Now I was thinking about a similar problem but now with respect to the words in the language, and was wondering if it is true or not. If $C$ is a context-free language then $C'$ is a context-free language where $w\in C'$ if and only if $\{w\}^*$ is contained in $C$.

My suspicion is that this is also false, but I can't come up with a counterexample. Somehow one needs to construct a CFL $C$ such that the subset of all the "periodic" words of $C$ together are not context-free.

Answer 1

Let me solve a simpler question first. Suppose we'd like to know

If $C$ is context-free, must $F(C)=\{w: ww \in C\}$ be context-free?

The answer is no: $F(\{a^i b^i c^j a^j b^k c^k\})=\{a^i b^i c^i\}$.

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As for the original problem, consider the context-free language that contains words of the form $\{a^i b^i c^j d a^j b^k c^k d\}$ or those which don't contain exactly two $d$'s.

Define $D=\{w: w^\ast \subseteq C\}$. Suppose $D$ is context-free. Then $E=D\cap a^\ast b^\ast c^\ast d$ is context-free.

I claim $E=\{a^i b^i c^i d\}$, which is not context-free. Suppose $w = a^i b^j c^k d$. We have $w \in E$ exactly if $w^n \in C$ for every $n$. This condition is clearly satisfied when $n \neq 2$, because $w^n$ will not have exactly two $d$'s. For $n=2$, $w w=a^i b^j c^k d a^i b^j c^k d\in C$ which is equivalent to $i=j=k$.

Answer 2

Take the language $L = \{a^{n^2} \colon n \ge 1\}$, it is easy to prove non-regular (and non-context-free). However, $L^*$ is the language denoted by $a^*$, which is regular (and thus context free).