Sequence of functions converges weakly but not strongly This is my problem:

Let $f\in L^2(\mathbb{R}^n)\backslash \{0\}$. For $j\in\mathbb{N}$ we set $f_j(x):=j^{\frac{n}{2}}f(jx)$. Show that the sequence $(f_j)_{j\in\mathbb{N}}$ converges weakly in $L^2(\mathbb{R}^n)$ to $0$, but not strongly.

I know that $(L^2(\mathbb{R}^n)\backslash\{0\})$'=$L^2(\mathbb{R}^n)\backslash\{0\}$ so for the weak convergence, I have to show that for all $g(x)\in L^2(\mathbb{R}^n)\backslash\{0\}$ we get $$\int_{\mathbb{R}^n/\{0\}}j^{\frac{n}{2}}f(jx)g(x)dx\rightarrow0$$ for $j\rightarrow \infty$. But I don't know how to show that this is true. Could you show me how?
For the strong convergence, I tried the following:
$$||f_j(x)-0||\not\xrightarrow{\text{$j\rightarrow\infty$}}0$$ 
$$\Leftrightarrow||f_j(x)||\not\xrightarrow{\text{$j\rightarrow\infty$}}0$$ 
$$\Leftrightarrow\sqrt{\int_{\mathbb{R}^n/\{0\}}(f_j(x))^2dx}\not\xrightarrow{\text{$j\rightarrow\infty$}}0$$
$$\Leftrightarrow\sqrt{\int_{\mathbb{R}^n/\{0\}}j^n(f(jx))^2dx}\not\xrightarrow{\text{$j\rightarrow\infty$}}0$$
The last statement is true since $j^n\xrightarrow{\text{$j\rightarrow\infty$}}\infty$ for $n\in\mathbb{N}$ and $(f(jx))^2$ is bounded since $f(jx)\in L^2(\mathbb{R}^n)\backslash\{0\}$, so $f$ does not converge strongly. Is that proof correct?
Thanks in advance.
 A: For the weak convergence, after making a change of variables $t=jx\Rightarrow dt=j^ndx$,  you can use Cauchy-Schwarz inequality :
$$\bigg|\int_{\mathbb{R}^n/\{0\}}f_j(x)g(x)dx\bigg|=\bigg|\int_{\mathbb{R}^n/\{0\}}j^{\frac{n}{2}}f(jx)g(x)dx\bigg|=\bigg|\int_{\mathbb{R}^n/\{0\}}j^{-\frac{n}{2}}f(t)g(\frac{t}{j})dx\bigg| $$ $$\leq j^{-\frac{n}{2}}\|f\|\|g\|\rightarrow 0,\,\text{as}\,j\rightarrow \infty,$$
for all $g$ in $L^2(\mathbb{R}^n/\{0\})$, where $\|.\|$ denotes the norm on $L^2(\mathbb{R}^n/\{0\})$. For (not) strong convergence we must show that $\|f_j\|$ does not converge to zero as $j\rightarrow \infty$, but this does not hold. Since $f_j\in L^2$
for each $j$, in particular you can take $g=f_j$ and use the above integral to have $$\int_{\mathbb{R}^n/\{0\}}f_j(x)g(x)dx=\int_{\mathbb{R}^n/\{0\}}|f_j(x)|^2dx=\int_{\mathbb{R}^n/\{0\}}|f(t)|^2dt,$$
where we have used the substitution $t=jx$ once more. Thus we see that $\|f_j\|=\|f\|$, implying $\|f_j\|\rightarrow\|f\|\neq 0$ as $j\rightarrow \infty$
A: Your argument for strong convergence is not quite correct because actually, the $\mathbb L^2$-norm of $g_j\colon x\mapsto f\left(jx\right)$ is $j^{-n/2}$ and therefore, there exists an increasing sequence of integers $j_l$ such that $g_{j_l}(x)\to 0$ for any $x\neq 0$.
However, the beginning of your computation combined with the substitution $t_i:=j x_i$, $1\leqslant i\leqslant n$ show that the $\mathbb L^2$-norm of $f_j$ is equal to that of $f$.
For the weak convergence, using boundedness in $\mathbb L^2$ of $(f_j)$ and a density argument,  it suffices to establish that $\int_{\mathbb R^n}f_j(x)g(x)\mathrm dx\to 0$ for each function $g$ which is continuous with compact support, or each function $g$ which can be expressed as a product of indicator functions of intervals.
