This theorem was inspired to me by this thread. I have found a proof of it but I'll post it later; I want to see if somebody can find another one!
Let be continuous real-valued functions having periods . Show that is periodic if and only if are linearly dependent over .
How can you be sure that the period of the sum is a multiple of both ? You need to show that this is true; it's the hard part of the problem.
If are linearly independent over , then whenever with we must have . If they are linearly dependent then they are not linearly independent. In your post, is a linear dependence relation over the rationals.
But you have to show that we must have for some integers .
For a,b integers:
m(x + ap) + n(x + bq) = m(x) + n(x) (1)
Therefore ap=bq defines all periods of the sum function.
If the sum is periodic, p = b/a * q, where b/a obviously belongs to Q.
Alas c1*p + c2*q = (c1 * b/a + c2) * q, for c1=-a/b and c2=1 (both € Q) it is 0, that is if the sum is periodic p,q are linearly dependent over Q (2)
If p,q are linearly dependent over Q, p = -c2/c1 * q for c1,c2 rational numbers, then expressing c1,c2 as c1N/c1D and c2N/c2D, where c1N,c1D,c2N,c2D are integers, we get c1N*c2D*p = -c2N*c1D*q.
Taking (1) and making a=c1N*c2D and b=-c2N*c1D, ab=pq, so If p,q are linearly independent over Q, the sum function is periodic (3).
m(x) + n(x) is periodic If(3) and only if(2) p,q are linearly independent over Q.
(Sorry for the ugly proof)
Ok so far.For a,b integers:
m(x + ap) + n(x + bq) = m(x) + n(x) (1)
How so? If is the period of , then ; but that does NOT imply and . You are making a huge leap here.Therefore ap=bq defines all periods of the sum function.
If you do not use the continuity of your proof is certainly flawed because we can construct non-continuous whose sum is periodic, but whose periods are linearly independent over .