Time Hierarchies in DSPACE(O(s(n))) The time hierarchy theorem states that turing machines can solve more problems if they have (enough) more time.
Does it hold in some way if the space is limited asymptotically?
How does $\textrm{DTISP}(g(n), O(s(n)))$ relate to $\textrm{DTISP}(f(n), O(s(n)))$ if $\frac{f}{g}$ grows fast enough?

I am especially interested in the case that $s(n) = n$, $g(n) = n^3$ and $f(n) = 2^n$.

In particular, I considered the following language:
$ L_k := \{ (\langle M \rangle, w) \; : \; \text{M rejects } (\langle M \rangle, w)  \text{ using at most } |\langle M \rangle, w|^3 \text{ time steps}, $ $ k * |\langle M \rangle, w| \text{ cells and four different tape symbols} \} $

However, $L_k$ could be decided in $n^3$ steps by using $(k+1)n \in O(n)$ space.

Without limiting $M$ to four tape symbols and thus allowing to compress $O(n)$ cells into $n$ cells, we get space issues when simulating an $M$ with too many tape symbols. In this case, the language is not in $\text{DSPACE}(O(n))$ anymore. The same happens when setting $k = h(|w|)$ for some $h$ that can be computed fast enough.

This question is basically a rephrase of my question here.

Edit Summary:
Changed $\textrm{DSPACE}(s(n)) \cap \textrm{DTIME}(f(n))$ to $\textrm{DTISP}(f(n), s(n))$, however, I think the intersection is also worth to think about.
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.