I'm doing this exercise, and I wish for corrections on my solution:
a) Find the index set $I:=\{ \alpha \in R\} : \{f_n\} \in L^2(R^3), > f_n(x) = n^\alpha X_{B_{1/n}}$ converge weakly in $L^2(R^3)$ and for those $\alpha$ find the weak limit.
b) Find the index set $J:=\{ \alpha \in R\} : f_n$ converge strongly in $L^2(R^3)$.
c) Find the index set $K:=\{ \alpha \in R\} : f_n$ converge a.e. in $L^2(R^3)$.
For point a), I know that given $g \in L^2(R^3)$ we have $||f_ng||_{L^1} =\int f_ng \rightarrow \int fg$. But thanks to Holder we have $||f_ng||_{L^1} \leq ||f_n||_{L^2}||g||_{L^2}$. For definition, $g$ converge in $L^2$. Now I check the other one (I square now the norm to slim the notation).
$||f_n||_{L^2} ^2= \int_R |n^\alpha X_{B_{1/n}}|^2 dx = \int_{B_{1/n}} |n^{2\alpha}|^2 dx =$ I use now the spherical coordinates and I have $=c \int_0^{1/n} \rho^2n^{2 \alpha} d \rho =$ where $c$ is the "radial stuff"$= {c\over 3}n^{2 \alpha -3}$. For $n \rightarrow +\infty$ if $\alpha \in [- \infty, {3 \over 2}]$ there is convergence. This should be the set $I$.
The weak limit is $0$ for $ \alpha < {3 \over 2}$ and ${c\over 3}$ for $ \alpha = {3 \over 2}$ (some constant).
For point b), I have to satisfy the condition $lim_n \rightarrow +\infty$ of $||f_n - f||_{L^2}^2 = 0$, but with that kind of sequence, how am I suppose to compute this? I can compute $||f_n - f||_{L^2}^2 =||f_n||_{L^2}^2 + ||f||_{L^2}^2 -2<f_n,f>$, using for the distributional term the convergence to $||f||_{L^2}^2$ according to index in a), but then...?
For point c), isn't it the set $I$ from point a)?
For a) you got the interval right. Why is the weak limit $0$ for $\alpha<3/2$? What is the weak limit for $\alpha =3/2$?
b) a strong limit is a weak limit, so you should look among the weak limits for the strong limits...
c) try to pass to the limit $\lim_{n\to\infty} f_n(x)$ for fixed $x$ to see what are candidates for a.e. limits. Also the pointwise limit coincides with weak/strong limit if they exists, so it will give you an idea what to do in a) and b).