Finding a sequence that converges weakly but not strongly in $L^2(-1,1)$ Suppose $\Omega=(-1,1)$, and $f_n$ is a sequence of functions with $n=1,2,3,..$ 
I am asked to find such sequence if $f_n\rightharpoonup 0$ in $L^2(\Omega)$, $f_n\rightarrow0$ in $L^{3/2}(\Omega)$ but $f_n\nrightarrow 0 $ in $L^2(\Omega)$.
In other words I neet to find sequence such that
$$\|f_n\|_{3/2}\to0,\,\,\,\lim_{n\to\infty}\int_{\Omega} f_n (x)g(x)\,d\mu(x)=0,\,\,\,\,\|f_n\|_2\nrightarrow0$$
for all $g\in L^{2}\left(\Omega\right).$
I was thinking of choosing 
$f_n(x):=1/\sqrt{n|x|}$, and strong convergence in $L^3$ can be easily verified as follows,
$$\|f_n-0\|_{3/2}^{3/2}=\int_{-1}^{1}\left(\frac{1}{\sqrt{nx}}\right)^{3/2}\,dx\to0, \quad n\to\infty.$$
On the other hand, to disprove strong convergence in $L^2$, 
$$\|f_n-0\|_2^2=\int_{-1}^{1}\left(\frac{1}{\sqrt{n|x|}}\right)^{2}\,dx=\int_{-1}^1\frac{1}{n|x|}\,dx=\infty.$$
Now, I only need to prove weak convergence in $L^2(\Omega)$,
$$\lim_{n\to\infty}\int_{-1}^{1}\frac{g(x)}{\sqrt{n|x|}}\,dx=\lim_{n\to\infty}\frac1{\sqrt{n}}\int_{-1}^{1}\frac{g(x)}{\sqrt{|x|}}\,dx$$
However, I tried using Holder but I do not think $f_n$ converges weakly because of the asymptote at $0$ (which I initially added to make it not strongly convergent in $L^2$). Is there a way I can tweak my function so that it converges weakly in $L^2$?
 A: In your previous post, some useful indication for answer is ready.
First of all, let us prove that, for $\alpha>0$, 
\begin{align*}
\int_{0}^{1/2}\dfrac{1}{|\log x|^{\alpha}}dx<\infty.
\end{align*}
This can be seen that, $\log x<0$ for $x\in(0,1/3]$, then
\begin{align*}
\int_{0}^{1/3}\dfrac{1}{|\log x|^{\alpha}}dx&=\int_{0}^{1/3}\dfrac{1}{(-\log x)^{\alpha}}dx\\
&=\int_{0}^{1/3}\dfrac{1}{(\log(1/x))^{\alpha}}dx\\
&=\int_{3}^{\infty}\dfrac{1}{u^{2}(\log u)^{\alpha}}du\\
&\leq\int_{3}^{\infty}\dfrac{1}{u^{2}}du\\
&<\infty.
\end{align*}
Let $f(x)=\dfrac{1}{|\log x|^{\alpha}}\chi_{(0,1/3]}(x)$ and $f_{n}(x)=n^{1/2}f(nx)$, just what we have just shown, $f\in L^{2}$, so $f_{n}\rightarrow 0$ weakly but not in norm.
We compute that
\begin{align*}
\int_{-1}^{1}f_{n}(x)^{3/2}dx&=\int_{-1}^{1}n^{3/4}f(nx)^{3/2}dx\\
&=n^{3/4}\int_{-1}^{1}\dfrac{1}{|\log(nx)|^{3\alpha/2}}\chi_{(0,1/3]}(nx)dx\\
&=n^{3/4}\int_{0}^{1/(3n)}\dfrac{1}{|\log(nx)|^{3\alpha/2}}dx\\
&=n^{3/4}\int_{0}^{1/3}\dfrac{1}{|\log x|^{3\alpha/2}}\dfrac{1}{n}dx\\
&=\dfrac{1}{n^{1/4}}\int_{0}^{1/3}\dfrac{1}{|\log x|^{3\alpha/2}}dx\\
&\rightarrow 0,
\end{align*}
as $n\rightarrow\infty$.
