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.