The time-diagonal language respects its budget

Define one fixed multitape procedure $R$ on $w$: decode $(M,c)$, reject malformed codes, construct $cf(n)$ for $n=\mathrm{\mid }w\mathrm{\mid }$, and run the clocked universal simulator on $M$ with simulated input equal to the entire word $w$. If the inner simulation returns, output the opposite of its result. This is a fixed interpreter; its decoding, clock construction, and universal-simulator transitions are all actual transitions of $R$.

The diagonal decider first computes $g(n)$ and constructs a unary timer of that length. The constructor costs $O(n+g(n))$. Converting its binary output to unary also costs $O(g(n)+\log (g(n)+2))$: repeatedly decrement a least-significant-bit-first counter and emit a mark, charging the traversed carry/borrow runs over the whole countdown. Their total is linear in the initial counter value. Rewind the timer, then execute $R$ with its fixed tapes and one additional timer tape. Each simulated transition of this fixed $R$ advances the timer once; at its end output a default rejection unless $R$ has already returned. The fixed transition table is incorporated directly, so this outer clock adds constant overhead per transition, not another universal-simulation factor.

On every input, including malformed descriptions and arbitrarily large encoded multipliers, the outer clock permits at most $g(n)$ transitions of $R$. All input-dependent decoding and inner-clock work are inside that clock. Thus total time is $O(n+g(n))=O(g(n))$, after treating finitely many lengths below the eventual floor separately.

Fix $(M,c)$. On its valid padded codes, decoding and copying the full simulated input cost $O_{M,c}(n)$; constructing the inner clock costs $O_c(n+f(n))$. By An efficient universal simulation with a clock, simulation costs $O_M(n+cf(n)\log (cf(n)+2))$. For this fixed pair the combined runtime of $R$ is $O_{M,c}(n+f(n)\log (f(n)+2))=o(g(n))$. It therefore finishes before the outer timer on every sufficiently long code of that pair. Arbitrarily long such codes exist by the enumeration, and the returned answer is the required complement.