Lower bounds and class separation Consider the language $A=\{0^{k}1^{k}|k\geq0\}$ . On Sipser's book "Introduction to the Theory of Computation" an algorithm with running time $O(n\log n)$ is given, on single-tape TM. We also know that the set of languages that run in time $o(n\log n)$ is exactly the set of regular languages. Since $A$ is not a regular language, we can be sure that this is a strict lower bound, on a single-tape TM.

Therefore, if $f(x) = n\log n$ and $g(x) = o(f(x))$ , the existence of this language allows us to say that 
$DTIME(g(x)) \subset DTIME(f(x))$ . Therefore:

By using the padding argument, couldn't we argue that for any time function $h(x) = \Omega(n\log n) $ the same result applies, i.e. that if $g(x) = o(h(x))$ then $DTIME(g(x)) \subset DTIME(h(x))$
I will try to answer the question more generally so does not depend on the specific problem. Let $L$ be a language in $DTime(O(t(n))-DTime(o(t(n))$. Can we use padding to get a language in  $DTime(O(s(n))-DTime(o(s(n))$ where $s(n) \in \Omega(t(n))$ is an arbitrary (time constructible) function?

The answer is no:


  the complexity class separations using padding translate downwards but not upwards, see Why do equalities between complexity classes translate upwards and not downwards?.