Every multitape time-T computation has a single-tape simulation in O((n+T(n))^2) time

By [L1], choose the one-tape simulator $S_M$ that reproduces the accept, reject, and divergence behavior of $M$ on every input.

For an input $w$ of length $n$, let $t={\mathrm{time}}_M(w)$. The quadratic-overhead lemma [L2] gives ${\mathrm{time}}_{S_M}(w)\le c_M(\mathrm{\mid }w\mathrm{\mid }+t+1{)}^2$ for a constant $c_M$ depending only on $M$. Because $t\le {\mathrm{Time}}_M(n)$ and $\mathrm{\mid }w\mathrm{\mid }=n$, the hypothesis and [L3] imply ${\mathrm{time}}_{S_M}(w)=O((n+T(n){)}^2)$. Taking the maximum over all $w$ of length $n$ yields ${\mathrm{Time}}_{S_M}(n)=O((n+T(n){)}^2)$.

The simulator from step 1.1 decides the same language as $M$, and step 2.1 gives the quantitative bound. Therefore the language decided by $M$ also lies in the one-tape deterministic time class described by [L4].