An eventually positive time bound yields the same-order space bound

During one step, each of the $k$ heads can enter at most one new cell on its own tape. Therefore after $t$ steps, the total number of visited cells across all tapes is at most the initial $k$ cells plus $kt$. By [L1], every halting computation using $t$ steps uses at most $k(t+1)$ space.

If there are no length-$n$ inputs, the revised convention in [L1] makes both worst-case values zero. At an inhabited length, applying step 1.1 to the worst halting run gives ${\mathrm{Space}}_M(n)\le k({\mathrm{Time}}_M(n)+1)$. For sufficiently large inhabited lengths the hypothesis $T(n)\ge 1$ absorbs the additive constant, and [L2] converts the time bound into ${\mathrm{Space}}_M(n)=O(T(n))$.