1

I'm having trouble with the following.

Let $k(t)=e^{-at}(t-\frac{1}{2}t^2)$, so k(t) has the property that $\int k(t)=0$, and let $k_{\nu}(t)=\nu^2k(\nu t)$.

Show that $\int k_\nu(t-u)f(u)d u \rightarrow f'(t)\int |k(t)|d t $ as $\nu\rightarrow\infty$, where $f$ is an arbitrary function.

Thanks for the help in advance.

adamG
  • 665
  • 1
  • 10
  • 23

1 Answers1

1

Note that $$ \begin{align} \int k_\nu(t-u)f(u)\,\mathrm{d}u &=\int k_\nu(u)f(t-u)\,\mathrm{d}u\\ &=\int k_\nu(u)\left(f(t)-f'(t)u+O\!\left(u^2\right)\right)\mathrm{d}u\\ &=\int k(u)\left(\nu f(t)-f'(t)u+O\!\left(u^2/\nu\right)\right)\mathrm{d}u\\ &=-\int k(u)\,f'(t)\,u\,\mathrm{d}u+O\left(\frac1\nu\int k(u)\,u^2\,\mathrm{d}u\right)\\ \end{align} $$ Therefore, $$ \lim_{\nu\to\infty}\int k_\nu(t-u)f(u)\,\mathrm{d}u=-\,f'(t)\int k(u)\,u\,\mathrm{d}u $$

robjohn
  • 345,667
  • Note that the change of variables also changes the order of the limits of the integration. This is why the absolute value of the Jacobian is used in the change of variables formula. See this answer for a bit more detail. – robjohn May 01 '18 at 22:45