I want to prove that $\textbf{NP}\neq \textbf{DTIME}(2^{\sqrt{n}}).$
My thoughts is:
if I try to prove $\textbf{NP}\not\subseteq \textbf{DTIME}(2^{\sqrt{n}})$ would imply $\textbf{NP $\neq$ P}$.
if I try to prove $\textbf{DTIME}(2^{\sqrt{n}})\not\subseteq \textbf{NP}$ would imply $\textbf{NP $\neq$ EXP}$.
I think both are open questions, is there any other way to prove it?
One way to prove that those classes are different is to prove that, like the class $\mathsf{E}$, $\mathsf{DTIME}(2^{\sqrt{n}})$ is not closed under polynomial-time many-one reductions, whereas $\mathsf{NP}$ is closed under such reductions.
To prove that, consider any problem $A\in \mathsf{DTIME}(2^n)$. Let $f : x\mapsto \underbrace{(x, x, …, x)}_{|x|\text{ times}}$ be a reduction, and $B$ be the (artificial) decision problem:
Input: $y = (x_1, …, x_k)$ with $|x_1| = |x_2| = … = |x_k| = k$
Output: $x_1\in A$?
Then for any input $y=(x_1, …, x_k)$, since $A\in \mathsf{DTIME}(2^n)$, $B$ can be solved in time $\mathcal{O}(2^{|x_1|}) = \mathcal{O}(2^{k}) = \mathcal{O}(2^{\sqrt{|y|}})$.
That means that $B\in \mathsf{DTIME}(2^{\sqrt{n}})$. Since $f$ is computable in polynomial time, that means that $A\leqslant_m^p B$.
If $\mathsf{DTIME}(2^{\sqrt{n}})$ was closed under polytime many-one reductions, that would mean that $\mathsf{DTIME}(2^{n}) \subseteq \mathsf{DTIME}(2^{\sqrt{n}})$, which is false.
