What assumptions must be made about f for this limit and integral to be correct?

[jostpuur]

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

<br /> \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)<br />

If we first fix the integration domain like this

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

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.

 

[jostpuur]

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

<br /> |\int dx\; g(x)| \leq \int dx\;|g(x)|<br />

is not suitable here. This instead,

<br /> |\int dx\; f(x)g(x)| \leq \|f\|_{\textrm{sup}} |\int dx\; g(x)|<br />

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.

 

[jostpuur]

<br /> |\int dx\; f(x)g(x)| \leq \|f\|_{\textrm{sup}} |\int dx\; g(x)|<br />

Argh! Not even correct!

<br /> g(x) = \cos(x^2)<br />

<br /> \int\limits_0^{\infty} \cos(x^2)dx = \frac{1}{2}\sqrt{\frac{\pi}{2}}<br />

<br /> f(x)=\left\{\begin{array}{ll}<br /> 1, \quad &amp; \cos(x^2)\geq 0\\<br /> 0,\quad &amp;\cos(x^2) &lt; 0 \\<br /> \end{array}\right.<br />

<br /> |\int dx\; f(x)g(x)| = \infty &gt; \frac{1}{2}\sqrt{\frac{\pi}{2}} = \|f\|_{\textrm{sup}} |\int dx\; g(x)|<br />

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

 

[jostpuur]

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

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

<br /> |\int dx\; (f\chi_{[0,M]})(x) g(x)| &gt; \|f\chi_{[0,M]}\|_{\textrm{sup}} |\int dx\; g(x)|<br />

I'm back in the starting point