- #1
QuArK21343
- 47
- 0
I have the following problem: prove that the sequence [itex]e^{inx}[/itex] tends to [itex]0[/itex], in the sense of distributions, when [itex]n\to \infty[/itex]. Here it is how I approached the problem. I have to prove this:
[tex]\lim \int e^{inx}\phi(x)\,dx=0[/tex]
, where [itex]\phi[/itex] is a test-function. I changed variable: [itex]nx=x'[/itex] and got:
[tex]\lim \frac{1}{n}\int e^{ix'}\phi(x'/n)dx'[/tex]
Now, can I exchange limit and integral? I would say yes, because of dominated convergence: the absolute value of the integrand is less than, say, [itex]c|\phi|[/itex], which is summable and the limit of the integrand exists, because [itex]\phi[/itex] is continuos. So,
[tex]\phi(0)\lim \frac{1}{n}\int e^{ix'}dx'[/tex]
But can I say that this last limit is zero? I mean, shouldn't the limit function be summable, again by dominated convergence? I suspect that I have a wrong understanding either of convergence in the sense of distribution or of dominated convergence. Can you clear up my doubts?
[tex]\lim \int e^{inx}\phi(x)\,dx=0[/tex]
, where [itex]\phi[/itex] is a test-function. I changed variable: [itex]nx=x'[/itex] and got:
[tex]\lim \frac{1}{n}\int e^{ix'}\phi(x'/n)dx'[/tex]
Now, can I exchange limit and integral? I would say yes, because of dominated convergence: the absolute value of the integrand is less than, say, [itex]c|\phi|[/itex], which is summable and the limit of the integrand exists, because [itex]\phi[/itex] is continuos. So,
[tex]\phi(0)\lim \frac{1}{n}\int e^{ix'}dx'[/tex]
But can I say that this last limit is zero? I mean, shouldn't the limit function be summable, again by dominated convergence? I suspect that I have a wrong understanding either of convergence in the sense of distribution or of dominated convergence. Can you clear up my doubts?