# Homework Help: Sequences of periodic functions converging to their average value

1. Nov 11, 2012

### Elysian

1. The problem statement, all variables and given/known data

Let f be a 2π-periodic function (can be any periodic really, not only 2π), and let g be a smooth function. Then

lim$_{n\rightarrow∞}$$\int^{B}_{A} f(nx)g(x)$ converges to $\frac{1}{2π}$$\int^{2π}_{0}f(x)$

3. The attempt at a solution

So far, I've come up with somewhat of an intuitive idea but I can't really put it into math.

For any interval 0 to t where t is the period, the integral of that function from 0 to t will be equal to the integral of the function with n times the period from 0 to t. In this case, the function g because it is smooth, can be broken down into little sections in which f will hit the function g many times so there becomes a maximum M which starts to resemble a straight line, and a minimum m which also starts to resemble a straight line. For f(nx) the period will go to infinity and it will start to hit the graph of g and pretty much fill it in from the interval [A,B]. I'm not exactly sure where the average value of f(x) comes in now, but I'd expect it to be from little intervals of the g and having f hit it so many times you could find that it has an average value. Not sure about that part.

I'm also not sure about the part in which if g(x) = 0, this really doesn't hold for all periodic functions. But it would hold for something like sin(x). Apparently the problem could be wrong so if it is wrong I'm supposed to present a counterexample to show why it's wrong.

2. Nov 11, 2012

### jbunniii

This is not true. Suppose $g(x) = 0$ (identically zero), and $f(x)$ is any $2\pi$-periodic function with nonzero mean. Are there any hypotheses which you have not mentioned?

3. Nov 11, 2012

### Elysian

Thanks, and yeah I get you, It's pretty much what I said in the last part of the question. What happens then if it is a zero mean? It holds but how would we go about proving that for zero means it is true but for nonzero means it isn't? I've just given you what was given to me so I don't believe there's anything missing. Sorry