Sequences of periodic functions converging to their average value

Elysian
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Homework Statement



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)


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.
 
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Elysian said:

Homework Statement



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)
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?
 
jbunniii said:
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?

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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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