$\mathsf{DTime}(O(n^k)) \subseteq \mathsf{NTime}(g)$ for some $g \in o(n^k)$? Can this statement be confirmed or disproved:


  $\mathsf{DTime}(O(n^k)) \subseteq \mathsf{NTime}(g)$ for some $g \in o(n^k)$


[Question changed to use Kaveh's brilliant formulation.]

Here the NDTM must "outrun" the DTM.

This seems similar to the PvNP question, but I'm not sure...

Thanks!

EDIT: This question seeks an inequality between the run time of polytime DTM deciders and their corresponding NP verifiers. If k=1, the proposition fails (e.g.: determining parity requires n steps on a TM and therefore a verifier cannot take any shortcuts). But if k>=2 ...?

I wonder if the statement can be disproved without leading to any major or unexpected complexity class separations... Is there a diagonalization argument that could work here.
It seems to me that you are not familiar with the basic complexity theory and therefore your question is probably not suitable for cstheory.

You probably mean to ask if the following statement is true:


  $\mathsf{DTime}(O(n^k)) \subseteq \mathsf{NTime}(g)$ for some $g \in o(n^k)$.


Note that the $\mathsf{NTime}$ hierarchy theorem is tight. If the answer to your question was positive then it would separate $\mathsf{NTime}$ from $\mathsf{DTime}$ which is not known as explained in the answers to the question linked by Robin.