This is another problem I am not sure about:
Determine if the following sequence of functions on R :
$$f_n(x) =\frac{\sin [x \sqrt n + \log(n)]}{(1 + nx^2)^\frac13}$$
- converges in quadratic mean?
- converges in the sense of distributions?
I know that convergence in quadratic mean means that
$$ \lim_{n \to \infty} |f_n-f_{\infty}|^2=0$$
Where $f_{\infty}=\lim_{n\to\infty}f_n$.
But I have troubles computing $f_\infty$, I get that:
if $x \ne 0$ then $f_{\infty}=0$
if $x=0$ then $f_{\infty}=$ not defined (since it must jump on the line between -1 and 1)
So my conclusion is that it does not converge in quadratic mean,(?)
What about in the distributional sense? help?
For the estimate of $f_n$: Consider $|x|>1$ and for $n\geq 1$. Since $|\sin (anything)|\leq 1$ you find $$|f_n|^2\leq (1/(1+nx^2)^{1/3})^2\leq (1/(nx^2)^{1/3})^2\leq (1/x^{2/3})^2=x^{-4/3} $$. For $|x|\leq 1$ we find $|f_n|\leq 1$. Now by Lebesgue dominated convergence theorem the integral converges in quadratic mean to the a.e pointwise limit $f=0$. Showing that $L^2$ convergence implies ditributional convergence should yield the rest.
This is roughly done in the following way. Associate with the distributions $T_n$ and $T$ the integral $T_n(\varphi)=\int f_n \varphi$ and $T(\varphi)=\int 0 \varphi$. Then $$|(T_n-T )(\varphi)|^2= |\int f_n \varphi|^2\leq \int|\varphi|^2 \int |f_n|^2 \to 0$$
I hope this is clearer and that there are no mistakes in my reasoning