Does the sequence $(\sqrt{n} \cdot 1_{[0, 1/n]})_n$ converge weakly in $L^2$?

Question

Let $f_n (x) = \sqrt n 1_{[0,1/n ] } (x) \in L^2 (\mathbb R)$. Does $f_n$ converges weakly to $0$ in $L^2$?

I tried to prove it by using Hölder's inequality or the Lebesgue differentiation theorem, but neither seems to work. I also can't find a counter example. Any help is appreciated.

Answer 1

Often, the best way to establish weak convergence is a two step procedure:

1) Show that the sequence in question is bounded. In this case, this is straightforward (do it!).

2) Show that there is a dense subset $X \subset L^2$ for which $\langle x_n, y\rangle \to \langle x, y\rangle$ holds for all $y \in X$, where $x$ is the (supposed) weak limit.

This will suffice, since for an arbitrary $z \in L^2$, we have $$ |\langle x_n - x, z\rangle| \leq |\langle x_n - x, y\rangle| + |\langle x_n - x, y - z\rangle| \leq |\langle x_n - x, y\rangle| + C \cdot \Vert y - z \Vert \to C \cdot \Vert y - z \Vert $$ for arbitrary $y \in X$. By choosing $y$ close enough to $z$, this shows the desired convergence.

In your case, one possible choice for $X$ is $X = L^2 \cap L^\infty$. Indeed, for $g \in L^\infty \cap L^2$, we get $$ |\langle f_n, g\rangle| \leq \Vert g \Vert_\infty \cdot \sqrt{n} \int_0^{1/n} \, dx = \Vert g \Vert_\infty / \sqrt{n} \to 0, $$ as desired.