Suppose $NP\subseteq DTIME[n^{f(n)}]$ where $f(n)$ is any function satisfying $\omega(1)$ then is it true $P=NP$ holds? Ladner's theorem states infinite time hierarchy between $P$ and $NP$. That is there are $NP$ problems in $DTIME[n^{g(n)}]$ for any $g(n)$ such that $n^c\leq n^{g(n)}\leq 2^{n^{1/c}}$ at some fixed $c\geq1$. If $NP\subseteq DTIME[n^{f(n)}]$ where $f(n)$ is any function satisfying $\omega(1)$ then for every such $g(n)$ we can find a $f(n)\ll g(n)$ and so Ladner's theorem is violated and so $P=NP$ has to hold.And so if $P\neq NP$ there is $g(n)$ with $n^c\leq n^{g(n)}\leq 2^{n^{1/c}}$ such that $NP\subsetneq DTIME[n^{g(n)}]$ and $NP\subseteq DTIME[n^{f(n)}]$ where $f(n)$ is any function satisfying $\omega(1)$ implies $P=NP$. Are these observations correct?What if I replace $DTIME$ by $RTIME$? Would this yield $RP=NP$ by some randomized version Ladner's theorem?
Is following observation on Ladner's theorem correct?
complexity theory;time complexity;randomized algorithms;p vs np;np intermediate
I'm annoyed at both [any being used as a quantifier] and [the placement of quantifiers making things even more ambiguous], so I'll start with the two interpretations of your initial block-quote.Suppose there is a function $\hspace{.04 in}f$ in $\omega(1)$ such that  NP $\subseteq$ DTIME$\left[\hspace{-0.02 in}n^{\hspace{.04 in}f(n)}\hspace{-0.03 in}\right]$ .    Does it follow that  P = NP ?It's not known to, since    ($n\mapsto 2^{\hspace{.02 in}n}\hspace{-0.03 in}$)  $\in$  $\omega(1)$    and  NP $\subseteq$ EXP $\subseteq$ DTIME$\left[\hspace{-0.02 in}n^{\hspace{.04 in}(2^{\hspace{.02 in}n})}\hspace{-0.04 in}\right]$  are both known.Suppose that for all functions $\hspace{.04 in}f$ in $\omega(1)$,  NP $\subseteq$ DTIME$\left[\hspace{-0.02 in}n^{\hspace{.04 in}f(n)}\hspace{-0.03 in}\right]$ .    Does it follow that  P = NP ?Yes, since as described in this answer, there existsa function $\hspace{.04 in}f$ in $\omega(1)$ such that  DTIME$\left[\hspace{-0.02 in}n^{\hspace{.04 in}f(n)}\hspace{-0.03 in}\right]$ = P .The truth of your That is ... fixed $c\geq 1$. sentence has nothing to do with Ladner's Theorem- The empty language and full language, for example, are such NP problems.For the sentence after that, the g given by  g(n) = c  is such a g, and functions f cannot simultaneously be in $\omega(1)$ and satisfy  f(n) << c ,  so where's the violation of Ladner's theorem?By the function g I just mentioned, the part of yourAnd so if ... sentence before that sentence's and is true.For the rest of that sentence, see this answer's two block-quotes and its responses to those.This answer's response to its there is block-quote would still apply.I believe this answer's response to its for all block-quotewould also still apply, by a similar argument.