Let be continuous and for all Prove that is constant.
Source: Berkely Preliminary Exam
If T is periodic with period and also periodic with period then must be an integral multiple of which is impossible.
This means that is periodic but its period is undefined.
The situation is similar to
where is constant.Here is periodic but its period is not defined.
Thus may be a constant function.
But my solution lacks rigour.
Or going from the first idea...if is bounded, then for some and for all . So suppose for contradiction that was not constant. Then it is monotonic increasing or decreasing. But this contradicts the fact that is bounded. Hence is constant.
I think I have it.
,where is the fundamental period or the smallest period and is any positive integer.
Similarly, results in
,where is a positive integer and is the fundamental period.
which is impossible since m and n are integers.Therefore is undefined.But since the function repeats itself it can only be constant
Consider the set . This set is dense in , and must be equal to everywhere in . Since is continuous and is dense, for all real and is constant.
To see that is dense in , notice that it is a subgroup of , hence it suffices to prove that it is dense on some interval; we choose the interval . Consider ; this is a subset of . Moreover if then because is irrational. Hence there are infinitely many points of in , hence two of them are arbitrarily close; but then since is closed under addition, is dense in , Q.E.D.
The fact that contains infinitely many points does indeed imply that is dense in : it implies that (and hence ) has an accumulation point, and any additive subgroup of having an accumulation point is dense in .what do you mean by "them"? whatever it means, obviously having infinitely many points doesn't imply that is dense in
To see this rigorously consider the map
This map, restricted to the subset of consisting of elements of the form , is an injection; moreover when , we have ; hence there are infinitely many points of in the bounded set . Immediately from the Bolzano-Weierstrass theorem we know that has an accumulation point. It is then an easy matter to show that any additive subgroup of having an accumulation point is dense in . To see this, suppose you have an open interval of length in . Then by the above there is an element of with 0<|s|<L; taking all its multiples (which are all elements of ) we see that one of them must fall in .