The deterministic space hierarchy theorem

Define a diagonal decider $D$ on strings $z=1^{\mathrm{\mid }e\mathrm{\mid }}0e{01}^k$, where $e$ is a valid finite machine description and $k\ge 0$. Reject malformed strings. On a valid string of length $n$, compute $g(n)$ using its fixed constructor and simulate the described machine $M$ on the entire original string $z$, capping all encoded simulation and scratch at $B=\max \{1,g(n)\}$. The parser and any coded description/input copy cost $O(n)$ space. Use [L2] with the size of this fixed capped simulator's actual storage, which is $O(n+B)$, to force halting; on cap overflow or timeout reject, and on simulated halting return the opposite answer. The constructor costs $O(g(n))$, and $n=O(f(n))$ with $f=o(g)$ implies $n=O(g(n))$. Hence $D$ is a total $O(g(n))$-space decider.

Suppose a fixed machine $M$ decides the language of $D$ in at most $cf(n)$ cells for sufficiently large $n$. For its fixed description $e$, the strings in step 1.1 exist at every sufficiently large length. The space needed for faithful simulation is $O_M(n+cf(n))=o(g(n))$, so sufficiently long such strings do not trigger the encoded cap. Since $M$ halts, [L2] also guarantees that the timeout does not prevent its answer. On such a string $z$, $D(z)$ is therefore the opposite of $M(z)$, a contradiction. Thus the language of $D$ is outside $\mathrm{D}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}(f)$, while step 1.1 puts it in $\mathrm{D}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}(g)$. Finally $f=O(g)$ gives the ordinary class inclusion.