Let be continuous and for all Prove that is constant.

Source: Berkely Preliminary Exam

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- June 25th 2009, 01:49 PMNonCommAlgContinuous & Periodic
Let be continuous and for all Prove that is constant.

__Source__: Berkely Preliminary Exam - June 25th 2009, 07:01 PMpankaj

Also,

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. - June 25th 2009, 07:21 PMSampras
We can say that is bounded and uniformly continuous. Maybe we can say that is a doubly periodic function (e.g. ). And somehow use Liouville's theorem to show that it is constant? The imaginary parts would be .

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. - June 27th 2009, 06:17 PMpankaj
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 - June 28th 2009, 05:51 AMNonCommAlg
- June 29th 2009, 08:58 PMBruno J.
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. - June 30th 2009, 04:14 AMNonCommAlg
- June 30th 2009, 08:01 AMpankaj
__'S' is dense in R__

What does this statement mean - June 30th 2009, 11:12 AMMoo
- June 30th 2009, 03:22 PMBruno J.
Oops. What I was thinking is that if and then ; since there are infinitely many choices of , there are infinitely many points in .

Quote:

what do you mean by "them"? whatever it means, obviously having infinitely many points doesn't imply that is dense in

- July 1st 2009, 04:43 PMJose27
- July 1st 2009, 06:34 PMBruno J.
Where do you see that I said is compact or closed? It is neither. I said 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 . - July 1st 2009, 06:53 PMJose27
- July 1st 2009, 07:34 PMBruno J.
You should read carefully what I wrote.

and the sequence

is a sequence of distinct elements of , all in . - July 3rd 2009, 08:43 AMpankaj
I think it is time that solution be revealed