Functional Equation Problem?

Question

Assume that f: R -> R is a continously differentiable and periodic function of period 2*pi which satisfies the inequality: |f(x+y)| <= |f(x)| + |f(y)| for all x, y in RWhat can you say about f ?Prove your statement.

Answer 1

m not good at math sorry

Answer 2

Let x such that f(x)=0. Suppose x is the smallest possible such that x>0. Then it has to be of the form 2pi/n for some n. Otherwise by looking at the sequence kx, one would get a contradiction. If the smallest x possible does not exist, either the function never vanishes, or it vanishes infinitely often near 0, and therefore vanishes identically.

If it never vanishes, a sufficient condition is that max |f|<= 2 min |f| and I don't know that you can say much more.

If it vanishes on a set of the form { 2kpi/n, k in ZZ} for some n then it can be f(x)= sin (nx/2). But I don't know if there are other examples... But you can see that |f(x+2pi/n)|<=|f(x)|. From which you deduce that 2pi/n is a period of |f|. Also it is clear that f and f' cannot vanish simultaneously, otherwise the funtion would be identically zero. So f changes sign each time it vanishes and by periodicity it vanishes an even number of times in a given period. So n has to be even. It also implies that 4pi/n is a period of f. Letting n=2N, we have that f vanishes at the points kpi/N and is 2pi/N periodic. So it looks like sin Nx, but it could be a variation of it. For instance it doesn't have to be odd

Answer 3

f does not seem to a trigo function.
Right now that's all I can say.

Answer 4

(x,y)=(0,0) (?)