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I suppose that $\Sigma = \{a\}$.

Prove or Disprove: For every polynomial $p(n)$ with coefficients in $\mathbb{N}$, $L = \{a^{p(n)} \; | \; n \in \mathbb{N}\}$ is a context sensitive language.

It seems that it is a context sensitive language. I guess making LBA or context sensitive grammar is not easy for this language. Can I prove this with closure property of CSL for example like complement? Can any one help me to prove for instance $L_1 = \{a^{n^7+n^5+n^3+n^2+1} | n\in \mathbb{N}\}$ is context sensitive. Maybe I can get an idea from this to prove my first question.

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Not true for linear polynomials. E.g let $p(n)$ be $3n+2$, then L is generated by the regular grammar 'S --> aaaS | aa'. [unless you mean 'at most context-sensitive', not 'exactly context-sensitive']

The Pumping Lemmas for both regular and context-free grammars seem to imply that any language over a 1-letter alphabet can be decomposed into a (not necessarily disjoint) union of sub-languages corresponding to linear polynomials. If the set of linear polynomials in question can be made finite (which I suspect must be true, but I can't prove or disprove off the top of my head [and I am not sure it even matters]), then any higher-degree polynomial must take on values not achieved by any polynomial in the linear set, in which case such languages must be at least context-sensitive.

Also: A strategy for building a context-sensitive grammar for a polynomial language over one alphabet is to construct a sequence of $k$ sub-grammars, each of which takes $m$ consecutive $a$'s and transforms it into $b_km+c_k$ consecutive $a$s for some constants $b_k$, $c_k$ [think "Horner's Method"] and then 'cascades' to the next sub-grammar.

So if you mean 'at most context-sensitive', then yes.

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    $\begingroup$ That's not what context-sensitive means (at least not under the standard definition). Every context-free language is also a context-sensitive language. The standard definition says that $L$ is a context-sensitive language if there exists a context-sensitive grammar for $L$; there's no requirement that $L$ must be not-context-free. Perhaps you can adapt the 3rd paragraph to provide a proof of the claimed statement? I'm not quite clear about the details of what you're suggesting, though. How do you build such a sub-grammar? $\endgroup$ – D.W.♦ Jan 30 '17 at 19:04
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    $\begingroup$ A unary language $L$ is regular iff it is context-free iff the set $\{n : a^n \in L\}$ is eventually periodic. $\endgroup$ – Yuval Filmus Dec 27 '17 at 16:41

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