The deterministic space hierarchy theorem

Define a total diagonalizer $D$ on inputs of the self-delimiting form $x=⟨M,1^k⟩$. On such an input of length $n$, universally simulate $M(x)$ inside a fixed reserved fraction of the constructible $g(n)$-space budget, and reverse its answer; reject malformed inputs and use the forced-halting configuration cutoff if the simulation does not halt within that cap. The parser, retained input, simulated configuration, and cutoff counter together use $O(n+g(n))=O(g(n))$ space.

Suppose that a machine $M$ decides $D$ in at most $cf(n)$ space for all sufficiently large $n$. Since $n=O(f(n))$ and $f=o(g)$, sufficiently long paddings $x=⟨M,1^k⟩$ make the simulator's $O(n+cf(n))$ space fit strictly inside the reserved $g(n)$ cap. On any such $x$, the cutoff does not fire and $D(x)$ is the opposite of $M(x)$, a contradiction. Hence $D\mathrm{\notin }\mathrm{D}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}(f)$, while step 1.1 puts $D\in \mathrm{D}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}(g)$. The reverse class inclusion is immediate from $f=o(g)$.