Lower bound for co-NP
Yes.  If you can decide all languages in NP on a DTM in polynomial time, then P = NP = co-NP, i.e., you can also decide all languages in co-NP on a DTM in polynomial time, basically by negating the output of the DTM.  Taking the converse... if there exists a language $L$ in co-NP that can't be solved in polynomial time on a DTM, then there exists a language $L'$ in NP that can't be solved in polynomial time on a DTM (for instance, you can take $L'$ to be the complement of $L$).