The standard one-tape simulation of a fixed multitape machine has quadratic overhead

After $j$ steps of $M$, each simulated head has moved by at most $j$ cells from its initial position, so tape $i$ can have visited only the cells in an interval of length at most $\mathrm{\mid }w\mathrm{\mid }+j+1$. Therefore the explicit block for tape $i$ in the encoding from [L1] has length $O(\mathrm{\mid }w\mathrm{\mid }+j+1)$, with an implied constant depending only on the fixed machine $M$.

Summing over the fixed number $k$ of tapes, and absorbing the fixed delimiter overhead, the whole one-tape encoding after $j$ simulated steps has length at most $c_M^{\mathrm{\prime }}(\mathrm{\mid }w\mathrm{\mid }+j+1)$ for some constant $c_M^{\mathrm{\prime }}>0$.

One simulation cycle of $S_M$ at stage $j$ performs a constant number of full sweeps across that length-$O(\mathrm{\mid }w\mathrm{\mid }+j+1)$ encoding and, when needed, one local shift of a suffix of the same order. Hence the time for the $j$th simulated step is at most $c_M^{\mathrm{\prime }\mathrm{\prime }}(\mathrm{\mid }w\mathrm{\mid }+j+1)$ for some constant $c_M^{\mathrm{\prime }\mathrm{\prime }}>0$.

The initial encoding from [L1] takes $O_M(\mathrm{\mid }w\mathrm{\mid }+1)$ time and space: the simulator makes a fixed number of sweeps over the input and writes a fixed number of blocks and delimiters. Adding this setup cost to the estimate from step 3.1 for $j=0,1,\dots ,t-1$ gives $O_M(\mathrm{\mid }w\mathrm{\mid }+1)+{\sum }_{j=0}^{t-1}{c}_M^{\mathrm{\prime }\mathrm{\prime }}(\mathrm{\mid }w\mathrm{\mid }+j+1)\le O_M(\mathrm{\mid }w\mathrm{\mid }+1)+c_M^{\mathrm{\prime }\mathrm{\prime }}t(\mathrm{\mid }w\mathrm{\mid }+1)+c_M^{\mathrm{\prime }\mathrm{\prime }}\frac{t(t-1)}{2}\le c_M(\mathrm{\mid }w\mathrm{\mid }+t+1{)}^2$ for a suitable constant $c_M$, including when $t=0$. Together with step 2.1, this proves the claimed linear space bound and quadratic time bound.