# Convergence in the sense of distributions

1. May 30, 2012

### QuArK21343

I have the following problem: prove that the sequence $e^{inx}$ tends to $0$, in the sense of distributions, when $n\to \infty$. Here it is how I approached the problem. I have to prove this:

$$\lim \int e^{inx}\phi(x)\,dx=0$$

, where $\phi$ is a test-function. I changed variable: $nx=x'$ and got:

$$\lim \frac{1}{n}\int e^{ix'}\phi(x'/n)dx'$$

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, $c|\phi|$, which is summable and the limit of the integrand exists, because $\phi$ is continuos. So,

$$\phi(0)\lim \frac{1}{n}\int e^{ix'}dx'$$

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?

2. May 30, 2012

### micromass

Staff Emeritus
The idea is right. But I don't quite understand why you use things like "summable". Can you tell us what you mean with dominated convergence exactly?

3. May 30, 2012

### mathman

What are the limits of integration? ∫eixdx = 0 when integrating over an interval of length 2π.

4. May 30, 2012

### lugita15

I think summable is an older term for integrable.

5. May 30, 2012

### micromass

Staff Emeritus
Oh, that would make sense actually.

6. May 30, 2012

### QuArK21343

Ok, by summable I mean that the integral over all space exists and it is finite. I don't know if the same terminology is used in english. Also, I didn't write the limits of integration, but they are over all space (let's say over all R or R^N). The dominated converge theorem I have studied says that if a sequence of measurable functions converges point-wise and the modulus of these functions (indexed by n) is dominated by a nonnegative summable function I can exchange limit and integral and the limit function is automatically summable:

$$\lim\int f_n dx=\int f dx$$

if $|f_n|<$ some summable function (at least almost everywhere etc). Is this correct? In my part of the world it is late in the night, so I will check for answers tomorrow morning. Thanks again for your help!

7. May 30, 2012

### micromass

Staff Emeritus
In that case, everything looks ok.

8. May 31, 2012

### QuArK21343

What worries me is that the last integral I get (the integral of the complex exponential) does not exists, does it? Shouldn't the limit function be summable?

9. May 31, 2012

### micromass

Staff Emeritus
Note that the test function $\varphi$ is nonzero on a compact domain K. So what you're actually doing is calculating

$$\lim_{n\rightarrow +\infty} \int_K e^{inx}\varphi(x)dx$$

If you work from this then you end up with

$$\int_K e^{ix}$$

which does exist.

10. May 31, 2012

### QuArK21343

I see, but my definition of a test-function is an infinitely differentiable function that goes to zero faster than any inverse power. Is this equivalent to saying that it has a compact support?

11. May 31, 2012

### micromass

Staff Emeritus
No. But it goes faster than zero than what?? Inverse powers of polynomials??

In that case, applying dominated convergence does not seem correct here. You say that it is smaller than $c|\varphi|$. But I don't see how that can be true. I get

$$\left| \int e^{ix}\varphi(x/n)dx\right| \leq \int |\varphi(x/n)dx|=n\int |\varphi(u)du|=n|\varphi|$$

But this is dependent of n and is not bounded by a summable function.

12. May 31, 2012

### QuArK21343

They must be $o(||x||^{-m})$ but not only that, also all their derivatives must satisfy the same condition (I was in a hurry this afternoon). And yes, you are right, my estimate is clearly wrong. My new attempted solution is this: just notice that it is a limit as $n\to \infty$ of the Fourier transform $\hat \phi(n)$. By a lemma of Riemann and Lebesgue, the Fourier transform of a summable function vanishes at infinity.

13. May 31, 2012

### theorem4.5.9

The only problem is that the fourier coefficients are for $L1$ functions on the circle. For the infinite domain case you need to appeal to the monotone convergence theorem, approximating from compact intervals.