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.