Does $\{f_n\}$ converges to $f$ in $L_2$? Definition We say that a sequence $\{f_n\}$ converges quadratically to $f$ in $L_2$ when $$\lim_{n \rightarrow \infty} \int_X[f_n(x)-f(x)]^2d\mu(x)=0$$
Let $f_n:[0,1] \rightarrow \mathbb{R}$ ($n=1,2,...$) defined by $$\begin{align*}
f(x) = 
\begin{cases}
n^{\frac{4}{7}}\sin(\frac{1}{x^2}),  & \text{if $x \in [0,\frac{1}{n})$} \\
0, & \text{if $ x \in (\frac{1}{n},1)$}\\
1, & \text{if $x=0$}
\end{cases}
\end{align*}$$
Does $\{f_n\}$ converges converges quadratically in $L_2$?
Attempt We can see that $\{f_n\}$ converges to $\begin{align*}
f(x) = 
\begin{cases}
0,  & \text{if $x \in [0,1)$} \\
1, & \text{if $ x=1$}
\end{cases}
\end{align*}$
So I can try to prove $\{f_n\}$ converges converges quadratically. But, $$\int_{[0,1]}[f_n(x)-f(x)]^2dx$= \int_0^{\frac{1}{n}}[n^{\frac{4}{7}}\sin(\frac{1}{x^2})-0]^2dx+\int_{\frac{1}{n}}^1[0-0]^2dx$$
But I can not calculate $\int_{0}^{\frac{1}{n}}n^\frac{8}{7}\sin^2(\frac{1}{x^2})dx$, so I suppose that $\{f_n\}$ doesn't converge quadratically but what I couldn't, but what I had is
$$\int_{0}^{\frac{1}{n}}n^\frac{8}{7}\sin^2(\frac{1}{x^2})dx \geq -\int_0^{\frac{1}{n}}n^{\frac{8}{7}}dx=-n^{\frac{1}{7}}$$ but it isn't enough, so could you help me?, please.
 A: Functions $f_n$ converge pointwise to zero. Thus if the limit in $L^2$ existed it would have to be equal to zero almost everywhere. We shall prove this is not the case.
As Oliver pointed out, the change of variables reduces the problem to proving thath the expression
$$
n^{8/7}\int^\infty_{n^2}u^{-3/2}\sin^2(u)\,du
$$
diverges to infinity (then of course it doesn't converge to zero).
Note that 
$$
\sin^2(u)>\frac{1}{2}, \qquad \textrm{for}\qquad u\in \bigcup_{k\in\mathbb{Z}}\Big((k+1/4)\pi,(k+3/4)\pi)\Big).
$$
Therefore
\begin{align*}
\int^\infty_{n^2}u^{-3/2}\sin^2(u)\,du&>\frac{1}{2}\sum_{k=\lfloor\frac{n^2}{\pi}\rfloor}^\infty \int_{(k+1/4)\pi}^{(k+3/4)\pi}u^{-3/2}du 
>\frac{1}{2}\sum_{k=\lfloor\frac{n^2}{\pi}\rfloor}^\infty \frac{\pi}{2}((k+3/4)\pi)^{-3/2}du\\
&\simeq \sum_{k=n^2}^\infty k^{-3/2}\simeq \int_{k=n^2}^\infty x^{-3/2}\,dx\simeq \frac{1}{n},
\end{align*}
and finally
$$
n^{8/7}\int^\infty_{n^2}u^{-3/2}\sin^2(u)\,du\gtrsim n^{1/7}\rightarrow \infty, \qquad\textrm{as}\quad n\rightarrow \infty.
$$
A: We have
$\displaystyle \int\limits_0^1f_n^2(x)\,dx =n^{8/7}\int\limits_0^{1/n}\sin^2(x^{-2})\,dx$
Substituting, as in Oliver Diaz answer, $x=y^{-1/2}$ gives $2\,dx =-y^{-3/2}\,dy$ and
$$\displaylines{2\int\limits_0^{1/n}\sin^2(x^{-2})\,dx =\int\limits_{n^2}^\infty y^{-3/2}\sin^2(y)\,dy
\ge \int\limits_{n^2\pi}^\infty y^{-3/2}\sin^2(y)\,dy\\
=\sum_{k=n^2}^\infty \int\limits_{k\pi}^{(k+1)\pi}y^{-3/2}\sin^2y\,dy\ge \pi^{-3/2}\sum_{k=n^2}^\infty (k+1)^{-3/2}\int\limits_{k\pi}^{(k+1)\pi}\sin^2y\,dy \\
=\pi^{-3/2}\,{\pi\over 2}\,\sum_{k=n^2}^\infty (k+1)^{-3/2}\ge 2^{-1}\pi^{-1/2}\int\limits_{n^2+1}^\infty y^{-3/2}\,dy\\
=\pi^{-1/2}(n^2+1)^{-1/2}\ge \pi^{-1/2}2^{-1}n^{-1}}$$
On the way we have used $\int\limits_0^\pi \sin^2y\,dy ={\pi/2}, $ the formula which follows from $\sin^2y+\cos^2y=1.$ Finally
$$\int\limits_0^1f_n^2(x)\,dx \ge 4^{-1}\pi^{-1/2}n^{1/7} $$
Remark I have posted the answer, as it avoids searching for intervals, where $\sin^2x$ is bounded away from $0.$ I make use of the fact that the integral of $\sin^2x$ is equal $\pi/2$ on every interval of length $\pi.$
A: The answer to the OP's question is no, $(f_n:n\in\mathbb{N})$ does not converge in $L_2([0,\infty))$. Here is a sketch of a solution:

*

*Making the change of variables $u=\frac{1}{x^2}$ yields
$$
I_n:=n^{8/7}\int^{1/n}_0 \sin^2(x^{-2})\,dx = \frac{1}{2}n^{8/7}\int^\infty_{n^2}u^{-3/2}\sin^2(u)\,du
$$


*Let $k_n$ be the smallest integer such that $n^2\leq (6k_n+1)\frac{\pi}{6}$. Then
$$(n^2,\infty)\supset\bigcup_{k\geq k_n}\Big((6k+1)\frac{\pi}{6},(6k+5)\frac{\pi}{6}\Big)=(n^2,\infty)\cap\Big(\pi\mathbb{Z}+\big(\pi/6,5\pi/6\big)\Big)$$
On each $k\pi$ translate ($k\in\mathbb{Z}$) of the interval $(\frac{\pi}{6},\frac{5\pi}{6})$  we have that  $\sin^2u\geq \frac14$. Thus
$$\begin{align}
\int^\infty_{n^2}u^{-3/2}\sin^2u\,du&\geq\sum_{k\geq k_n}\int_{k+(\tfrac{\pi}{6},\tfrac{5\pi}{6})}u^{-3/2}\sin^2(u)\,du\\
&\geq\frac14\sum_{k\geq k_n}\int^{(6k+5)\pi/6}_{(6k+1)\pi/6}u^{-3/2}\,du\\
&>\frac{\sqrt{6}}{\sqrt{\pi}}\sum_{k\geq k_n}\frac{1}{\big(6k+1\big)^{3/2}}\\
&>\frac{1}{6\sqrt{\pi}}\sum_{k\geq k_n}\frac{1}{(k+1)^{3/2}}\\
&>\frac{1}{6\sqrt{\pi}}\int^\infty_{k_n}\frac{dx}{(x+1)^{3/2}}=\frac{1}{3\sqrt{\pi}}\frac{1}{\sqrt{k_n+1}}
\end{align}$$
Hence
$$n^{8/7}\int^\infty_{n^2}u^{-3/2}\sin^2u\,du>\frac{1}{3\sqrt{2\pi}}\frac{n^{8/7}}{(k_n)^{1/2}}=\frac{1}{3\sqrt{2\pi}}n^{1/7}\frac{n}{(k_n)^{1/2}}$$

*

*Notice that $(6k_n-5)\frac{\pi}{6}<n^2\leq (6k_n+1)\frac{\pi}{6}$; whence we obtain that $\lim_n\frac{n^2}{k_n}=\pi$. Consequently

$$I_n\xrightarrow{n\rightarrow\infty}\infty$$
All this shows that although $f_n(x)\xrightarrow{n\rightarrow\infty}0$ a.s. in $[0,\infty)$, $f_n$ fails to converge to $0$ in $L_2([0,\infty))$. In fact, $\{f_n:n\in\mathbb{N})$ is unbounded in $L_2([0,\infty))$.
