Weak convergence of a sequence $(f_n) \subset L^2(\mathbb{R})$ and convergence of $\chi_n f_n$

Question

I am asked to solve the following problem.

Let $(f_n)_n \subset L^2(\mathbb{R})$ be such that $f_n \rightharpoonup f$ in $L^2(\mathbb{R})$. Let $I_n = (-n, n)$ and denote with $\chi_n$ the characteristic function of $I_n$.

1. Prove that for all $\psi \in L^1(\mathbb{R})$, $\int_{\mathbb{R} \setminus I_n} |\psi(x)| dx \rightarrow 0$.

This follows from the absolute continuity of the integral. In fact, $I_n$ is Lebesgue measurable thus $\mathbb{R} \setminus I_n$, being the complement of $I_n$, is Lebesgue measurable, that is $\mu (\mathbb{R} \setminus I_n) = \mu ( (-\infty, -n] \cup [n, \infty)) = \mu ( (-\infty, -n]) + \mu ( [n, \infty]) \rightarrow 0$ as $n$ approaches $\infty$ (I believe that an alternative way to see it is from the fact that $\mu(I_n) \rightarrow \infty$). Hence the thesis.

2. Prove that for all $\phi \in L^2(\mathbb{R})$, $\int_{\mathbb{R} \setminus I_n} |f_n(x) \phi(x)| dx \rightarrow 0$

My idea was to use both the result of point 1. and the weak convergence of $(f_n)_n$. However, I am struggling to conclude since here the set on which I am integrating depends on $n$ as well. By rewriting the integral in this way $\int_{\mathbb{R}\setminus I_n} |f_n(x) \phi(x)| dx = \int_{\mathbb{R}} f_n(x) \chi_n(x) \phi(x) dx$ my attempt was to prove that also $\chi_n f_n \rightharpoonup f$. Then, I could conclude from the weak convergence of $\chi_n f_n$. Nevertheless, I cannot formalize this reasoning.

3. Exhibit an example of a sequence $h_n = f_n - \chi_n f_n$ which converges weakly to $0$ but not strongly.

I was thinking about a sequence with norm equal to $1$ for all $n$ which converges weakly to zero, but I have few ideas.

Any hint on how to proceed would be greatly appreciated.

Answer 1

For the first part, your argument works if you define $\mu(B)=\int_B \lvert \psi\rvert d\lambda$, where $\lambda$ denotes the Lebesgue measure. Alternatively, you can approximate in $\mathbb L^1$ by a linear combination of indicator functions of Borel sets of finite measure to get the result.

2. Denote $$\left\langle u,v \right\rangle := \int_{\mathbb R}u(x)v(x) dx$$ and $g_n(x):=\operatorname{sgn}\left(f_n(x)\right)\mathbf{1}_{\mathbb R\setminus I_n}(x)\lvert \psi(x)\rvert $, where $\operatorname{sgn}(y)=1$ if $y>0$, $\operatorname{sgn}(y)=-1$ if $y<0$ and $\operatorname{sgn}(0)=0$. Then $$\int_{\mathbb R \setminus I_n} |f_n(x) \phi(x)| dx=\left\langle f_n,g_n \right\rangle $$ and since $g_n\to 0$ strongly in $\mathbb L^2$ (by the first question), we get the wanted result.