Any machine that fully reads every input of length n needs at least linear time

For $n=0$ the claimed lower bound is immediate. For $n\ge 1$, the input head starts at cell $0$ and moves at most one cell per transition. Reaching cell $n-1$ therefore takes at least $n-1$ transitions. The hypothesis says the machine scans that cell before halting, so it must make a further transition from a nonhalting configuration scanning the cell before it can enter an accepting or rejecting state. Thus every such halting branch takes at least $n$ steps.

The hypothesis says this lower bound applies on every input of length $n$, and the input set is nonempty. By [L1], its worst-case maximum is at least the time on any one such input, so ${\mathrm{Time}}_M(n)\ge n$. Without the nonemptiness premise the claim fails for an empty input alphabet and $n>0$, because the revised worst-case convention gives value zero.