The deterministic time hierarchy theorem

Apply [L1] to obtain its language $D\in \mathrm{D}\mathrm{T}\mathrm{I}\mathrm{M}\mathrm{E}(g)$.

Since $f(n)$ is eventually at least $n$ and $f(n)\log f(n)=o(g(n))$, we have $f(n)=O(g(n))$. Any $O(f)$ decider is therefore also an $O(g)$ decider, proving $\mathrm{D}\mathrm{T}\mathrm{I}\mathrm{M}\mathrm{E}(f)\subseteq \mathrm{D}\mathrm{T}\mathrm{I}\mathrm{M}\mathrm{E}(g)$.

Suppose a machine $M$ running in $O(f)$ decided $D$. Fix an integer $c$ large enough to bound its running time by $cf(n)$ at all sufficiently large $n$; enlarge $c$ to cover the finitely many shorter lengths as well. By [L1], the clocked pair $(M,c)$ has arbitrarily long valid padded codes $w$, and on each sufficiently long one the diagonal decider returns the complement of $M(w)$, since the inner clock does not expire. But $M$ decides $D$ on $w$, a contradiction. Thus $D\mathrm{\notin }\mathrm{D}\mathrm{T}\mathrm{I}\mathrm{M}\mathrm{E}(f)$; combine this with steps 1.1–1.2 for strict containment.