Indicator function and weak convergence The following is a qual question that I do not know how to proceed.
Let $f_n=\chi_{[n,n+10]}$. Does $f_n\rightarrow 0$ weakly in $L^4(\mathbb{R})$? 
My attempt:
I want to prove that it is true. By definition of weak convergence and Riesz representation theorem for $L^p$ spaces, it suffices to show that for any $g\in L^{4/3}(\mathbb{R})$, we have $\int_{[n,n+10]}g\rightarrow 0$ as $n\rightarrow\infty$. Since simple functions are dense in $L^{4/3}(\mathbb{R})$, for any $\varepsilon>0$, there is a simple function $f\in L^{4/3}(\mathbb{R})$ such that $\int_{\mathbb{R}}|f-g|^{4/3}<\varepsilon.$ Then 
$$\int_{[n,n+10]}|g|\leq \int_{[n,n+10]}|f-g|+\int_{[n,n+10]}|f|.$$
The second integral is $0$ when $n$ is large since $f$ is simple. To estimate the first integral, I write it as
$$\int_{[n,n+10]\\|f-g|\geq 1}|f-g|+\int_{[n,n+10]\\|f-g|\leq 1}|f-g|.$$
Now $$\int_{[n,n+10]\\|f-g|\geq 1}|f-g|\leq \int_{[n,n+10]\\|f-g|\geq 1}|f-g|^{4/3}<\varepsilon.$$ But I do not know how to estimate
$$\int_{[n,n+10]\\|f-g|\leq 1}|f-g|.$$
Any help will be appreciated.
 A: One way to do it is to see that $ f_{n}\to f$ pointwise where $f$ is the zero function . $f _{n}$ is bounded in $L^{4}$ norm. So $f_{n}$ weakly converges to O.  The following theorem is what I am using: 
If $1<p<\infty$., and if $f_{n}\to f$ pointwise a.e, and $f_{n}$  is bounded in $L^{p}$ Then $f_{n}$ converges to $f$ weakly.
The proof of the theorem uses heavy machinery. It can be proved as follows:
(1) If $1<p< \infty$, $f_{n}$ converges to $f$ weakly in $L^{p}(E)$ iff 
$$\int _{A}f_{n} \to \int_{A} f$$ for every set of finite measure $A\subseteq E$.
This is a very well known theorem regarding weak convergence in $L^{p}$ spaces.
(2) So, given that $f_{n}$ converges to $f$ pointwise a.e , if you want to prove $f_{n}$ weakly converges to $f$, you just need to verify condition in $(1)$.So, if you can justify passage under the integral sign you are done.
(3) But that is guaranteed by Vitali convergence theorem once you prove that $f_{n}$ are uniformly integrable. But that is true by the following lemma which uses the boundedness condition:
Lemma: If $1<p< \infty$ , $f_{n}$ is a sequence in $L^{p}(E)$ , which is bounded in the norm that is, $||f_{n}||_{p} < M$ for all $n\in \mathbb{N}$ for some $M>0$, then  the sequence $(f_{n})$  is uniformly integrable.
This lemma proves the theorem.
You can find all these theorem in $Real Analysis- Royden, Fitzpatrick$(4th edition)
A: You can prove this much simpler.
Let $(f, g)$ denote 
$$ (f, g) = \int { fg \, dx }  $$
for $f \in L^p$ and $g \in L^q$ with $1/p + 1/q = 1$. ($p=4$ for your case.) 
If $(f_n, g)$ does not converge to $0$, then there exists a $\delta > 0$ and an infinite subsequence $\{f_{n_i}\}$ that satisfies $|(f_{n_i}, g)| > \delta$. From the Hölder's inequality applied to $L^p[n_i, n_i + 10]$, we have
$$|(f_{n_i}, g)| \leq \int_{n_i}^{n_i+10} {|f_{n_i}g| \, dx} \leq \left(\int_{n_i}^{n_i+10} {|f_{n_i}|^p \, dx}\right)^{1/p}\left(\int_{n_i}^{n_i+10} {|g|^q \, dx}\right)^{1/q},$$
which follows
$$\frac{\delta^q}{10^{q/p}} \leq \int_{n_i}^{n_i+10} {|g|^q \, dx}. $$
This gives $\lVert g \rVert_q = \infty$ since the intervals $\{[n_i, n_i+10]\}$ appear infinitely many times in the range of integral.
A: Let $g \in L^{4/3}.$ Then
$$|\int f_ng| = |\int_n^{n+10}g|\le \int_n^{n+10}|g| = \int_n^{n+10}|g|\cdot 1$$ $$ \leq \left ( \int_n^{n+10}|g|^{4/3}\right )^{3/4}\cdot \left ( \int_n^{n+10}1^{4} \right )^{1/4}=  \left ( \int_n^{n+10}|g|^{4/3}\right )^{3/4}\cdot 10^{1/4}.$$
Because $|g|^{4/3}\in L^1,$ the last integral $\to 0$  by the dominated convergence theorem. So the answer is yes.
