The read-only-workspace space hierarchy theorem

Apply Effective enumeration of clocked machines with the work-space bound $f$. On a code $z$ of length $n$ for $(M,c)$, first compare the required cap $cf(n)+O(\log n)$ with a fixed fraction of $g(n)$. If it does not fit, return a fixed answer; if it fits, simulate $M(z)$ with cap $cf(n)$, stop after its finite configuration bound, and return the opposite answer. Malformed codes also receive a fixed answer.

The preliminary comparison and the guarded simulation use $O(g(n))$ work cells on every input. For each fixed decoded pair, the simulator and its counter use $O(f(n)+\log n)$ cells; because $f=o(g)$ and the logarithmic floor is absorbed by $f$, every sufficiently long padded code for that pair passes the guard. Thus the diagonal language lies in $\mathrm{D}\mathrm{W}\mathrm{O}\mathrm{R}\mathrm{K}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}(g)$ without losing the eventual self-code needed below.

If the diagonal language were in $\mathrm{D}\mathrm{W}\mathrm{O}\mathrm{R}\mathrm{K}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}(f)$, choose a clock multiplier $c$ for its decider and then a sufficiently long padded code $z$ for that pair. On input $z$ the diagonal machine simulates that same decider on $z$ and returns the opposite answer, a contradiction. Since $f=O(g)$ also gives the ordinary inclusion, the containment is strict.