Prelim problem: bizarre integral expression

[laonious]

Hi all,
Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a bounded measurable function such that
$$\lim_{x\rightarrow\infty}x^2f(x)=1.$$
Find an integral expression for
$$\lim_{\lambda\rightarrow 0^+}\frac{\int_0^{\infty}(1-\cos(x))f(\frac{x}{\lambda})dx}{\lambda^2}.$$

This one is really bizarre to me. I'm not sure how to use the information about f's endpoint behavior other than to try somehow to approximate f by 1/x^2?

I've tried changing lambda into 1/n where n goes to infinity but that seems to get me an expression that will tend to 0, not an integral expression as requested.

Any help or ideas would be greatly appreciated; a solution isn't necessary, but maybe thoughts on how to think about the problem?

Thanks!

 

[micromass]

The theorem on dominant convergence??

 

[Stephen Tashi]

$$\lim_{x\rightarrow\infty}x^2f(x)=1.$$

For k > 0, a constant

$\lim_{x \rightarrow \infty} (kx)^2 f(kx) = 1$

$k^2 \lim_{x \rightarrow \infty} x^2 f(kx) = 1$

$\lim_{x \rightarrow \infty} x^2 f(kx) = \frac{1}{k^2}$

Change variables $x = 1/ \lambda$ and let the $k$ in the above equation be represented by $x$

$\lim_{\lambda \rightarrow 0^+} \frac{1}{\lambda^2} f(\frac{x}{\lambda}) = \frac{1}{x^2}$

Then we need an excuse (dominated convergence?) to take the limit inside the integral sign, in order to do the manipulations:

$$\lim_{\lambda\rightarrow 0^+}\frac{\int_0^{\infty}(1-\cos(x))f(\frac{x}{\lambda})dx}{\lambda^2}$$

=

$$\int_{0}^{\infty} (1 - cos(x)) \lim_{\lambda \rightarrow 0^+} \frac{1}{\lambda^2} f(\frac{x}{\lambda}) dx$$

=

$$\int_{0}^{\infty} \frac{ (1 - cos(x))}{x^2} dx$$

 

[laonious]

Thank you for the insight, it was very helpful!
To apply dominated convergence I think we have to be careful because f isn't integrable, but I think we can say something like
$$\int_a^{\infty} f \leq \sum_i \int_{a_i}^{a_{i+1}}\frac{1}{x^2}+\frac{\epsilon}{2^i}\,dx<\infty.$$