Smaller, yes: if a TM uses n bits of space, its execution time can exceed any polynomial of n in steps. But any such space-bounded deterministic computation th

Smaller, yes: if a TM uses n bits of space, its execution time can exceed any polynomial of n in steps. But any such space-bounded deterministic computation that takes 2^n steps will not terminate, by exhaustion of possible state transitions, and nontermination would mean it does not compute a function over all inputs. This constraint is small in terms of the kind of growth functions this kind of maths deals with.