Time Hierarchies in DSPACE(O(s(n)))
This is an open problem: It is open whether $\mathrm{DTISP}(O(n \log n),O(n)) = \mathrm{DSPACE}(O(n))$ (or even $\mathrm{NSPACE}(O(n))$). We only know that $\mathrm{DTIME}(O(n))⊆\mathrm{DSPACE}(O(n/\log n))$.

However, under plausible computational complexity conjectures, there is a proper hierarchy.  For example, if for every $ε&gt;0$, CIRCUIT-SAT ∉ i.o.-$O(2^{n-ε})$, then
$\mathrm{DTISP}(O(f),O(s(n))) ⊊ \mathrm{DTISP}(O(f^{1+ε}),O(s(n)))$
where $f(n)≥n$, $f(n)$ is $2^{o(\min(n,s(n)))}$, and $f$ is time-space constructible.

In particular (under the hypothesis), existence of a satisfying assignment for circuits with $⌊\mathrm{lg}(f^{1+ε/2})⌋$ inputs and size $(\log f)^{O(1)}$ serves as a counterexample to the equality of the classes.

Notes:


CIRCUIT-SAT is at least as hard as $k$-SAT (which is used in the strong exponential time hypothesis).
Per convention, in CIRCUIT-SAT, $n$ is the number of input wires; circuit size is $n^{O(1)}$.
If the assumption used CIRCUIT-SAT for quasilinear circuit sizes, then the bound on $f(n)$ can be relaxed to $O((2-ε)^{\min(n,s(n))})$.  Also, weaker/stronger assumptions on hardness of CIRCUIT-SAT give weaker/stronger hierarchies (that we can currently prove).
i.o. means infinitely often, and can be dropped for $f$ that are in a certain sense continuous (including $f(n)=n^a$).
It appears likely that the DTISP hierarchy is sharp enough to distinguish $O(f)$ from $o(f/\log f)$ (and perhaps $o(f)$) (when $f$ is not too large relative to the permitted space).
To distinguish $n^a$ from $2^n$, we only need the weaker assumption P≠PSPACE.