# One limit and integral

1. Feb 22, 2008

### jostpuur

What do we have to assume of the function f so that following limit is correct,

$$\lim_{L\to\infty} \int\limits_{-L}^L f(\frac{x}{L}) \frac{\sin(x)}{x} dx = f(0)\int\limits_{-\infty}^{\infty} \frac{\sin(x)}{x} dx = \pi f(0)$$

If we first fix the integration domain like this

$$\lim_{L\to\infty} \int\limits_{-\infty}^{\infty} f(\frac{x}{L})\frac{\sin(x)}{x} \chi_{[-L,L]}(x)dx$$

the problem is that the limit of the integrand is not Lebesgue integrable over $\mathbb{R}$, so the standard convergence results do not settle this immediately.

2. Mar 4, 2008

### jostpuur

I figured out one way to do this, but I had to use one inequality thats correctness is not clear to me. The triangle inequality

$$|\int dx\; g(x)| \leq \int dx\;|g(x)|$$

is not suitable here. This instead,

$$|\int dx\; f(x)g(x)| \leq \|f\|_{\textrm{sup}} |\int dx\; g(x)|$$

becomes useful. But is this correct? I don't know how to prove this. I wouldn't want to do much assumptions about g, but f can be assumed to be as nice as necessary.

3. Mar 4, 2008

### jostpuur

Argh! Not even correct!

$$g(x) = \cos(x^2)$$

$$\int\limits_0^{\infty} \cos(x^2)dx = \frac{1}{2}\sqrt{\frac{\pi}{2}}$$

$$f(x)=\left\{\begin{array}{ll} 1, \quad & \cos(x^2)\geq 0\\ 0,\quad &\cos(x^2) < 0 \\ \end{array}\right.$$

$$|\int dx\; f(x)g(x)| = \infty > \frac{1}{2}\sqrt{\frac{\pi}{2}} = \|f\|_{\textrm{sup}} |\int dx\; g(x)|$$

hmhmh.... but here f is not integrable itself. It could be that the inequality is true if f's integral exists...

Last edited: Mar 4, 2008
4. Mar 4, 2008

### jostpuur

Even this is not true. By fixing sufficiently large M, we obtain an integrable $f\chi_{[0,M]}$ such that

$$|\int dx\; (f\chi_{[0,M]})(x) g(x)| > \|f\chi_{[0,M]}\|_{\textrm{sup}} |\int dx\; g(x)|$$

I'm back in the starting point