Let $f_{n}(x):=\frac{n\sqrt{x}\sin{x}}{1+nx^2}$ and $f(x)=x^{-\frac{3}{2}}\sin{(x)}$. Let $p \geq 2$.
Show that $ f \notin L^{p}(]0,\infty[)$ and that $f_{n} \nrightarrow f$ in $L^{p}(]0,\infty[)$
On $ f \notin L^{p}(]0,\infty[)$:
$$\int_{0}^{\infty}\left\vert x^{-\frac{3}{2}}\sin{(x)}\right\vert^{p}dx=\int_{0}^{\infty} x^{-\frac{3}{2}p}\left\vert\sin{(x)}\right\vert^{p}dx=\int_{0}^{\infty} x^{-\frac{p}{2}}\frac{\left\vert\sin{(x)}\right\vert^{p}}{x^{p}}dx=\int_{0}^{\infty} x^{-\frac{p}{2}}\left(\frac{\left\vert\sin{(x)}\right\vert}{x}\right)^{p}dx$$
Note that there exists $ c > 0$ so that $\frac{\vert\sin{(x)}\vert}{x}\geq c$ when $x \in ]0,1[$
Therefore, $\int_{0}^{1} x^{-\frac{p}{2}}\left(\frac{\vert\sin{(x)}\vert}{x}\right)^{p}dx\geq \int_{0}^{1} x^{-\frac{p}{2}}c^{p}dx=\infty$ since $p \geq 2$. That solves the first part of the problem. Is there any theorem that I can use to show that $f_{n} \nrightarrow f$ in $L^{p}$.
My attempts:
$$\vert\vert f_{n} -f \vert\vert_{p}^{p}= \int_{0}^{\infty}\vert f_{n}(x)-f(x)\vert^{p}dx= \int_{0}^{\infty}\left\vert \frac{n\sqrt{x}\sin{x}}{1+nx^2}-x^{-\frac{3}{2}}\sin{(x)}\right\vert^{p}dx$$
and then note that $\frac{n\sqrt{x}\sin{x}}{1+nx^2}=\frac{\sqrt{x}\sin{(x)}}{\frac{1}{n}+x^{2}}$ and $\frac{\sqrt{x}\sin{(x)}}{\frac{1}{n}+x^{2}}\nearrow \frac{\sin{(x)}}{x^{\frac{3}{2}}}$ for all $x \in ]0,\infty[$ so the sequence is monotone increasing. I do not know whether this helps or not but it is the best I can come up with. Any ideas are greatly appreciated.
Can you show that the $f_n$ are elements of $L^p$? If so suppose that $\lVert f_n - f \rVert_{L^p} \to 0$ and $f \not \in L^p$. Then by the triangle inequality, for all $n$ we have $\lVert f \rVert_{L^p} \leq \lVert f_n \rVert_{L^p} + \lVert f - f_n \rVert_{L^p}$. For $n$ large enough, $\lVert f_n - f \rVert_{L^p}$ must be finite (since we've got the much stronger condition that it tends to 0). Therefore $\lVert f \rVert_{L^p} < \infty$, hence $f \in L^p$. Contradiction.