A time bound always 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.

Applying step 1.1 to the worst halting run on an input of length $n$ gives ${\mathrm{Space}}_M(n)\le k({\mathrm{Time}}_M(n)+1)$. The additive constant is harmless asymptotically, and [L2] converts this inequality together with the hypothesis ${\mathrm{Time}}_M(n)=O(T(n))$ into ${\mathrm{Space}}_M(n)=O(T(n))$. When an argument also requires the budget to dominate input reading, [L3] supplies the usual linear lower bound.