Let $f_n=\chi_{[n,n+1]}$. Does $f_n$ converge weakly to $0$ in $L^1(\mathbb{R})$?

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Given a Banach space $X$ we say a sequence in $X$, $\{x_n\}$, converges weakly to $x\in X$ if $$\lim_n \varphi(x_n)=\varphi(x) \quad \forall \varphi\in X^*.$$ Clearly, if $x_n\rightarrow x$ in $X$ then $x_n$ converges weakly to $x$ since $|\varphi(x_n)-\varphi(x)|\leq \|\varphi\|\|x_n-x\|_X$.

Let $f_n=\chi_{[n,n+1]}$. Show that $f_n$ converges weakly to $0$ in $L^p(\mathbb{R})$ when $1<p<\infty$. Does $f_n \rightarrow 0$ in $L^p(\mathbb{R})$? Does $f_n$ converge weakly to $0$ in $L^1(\mathbb{R})$?

My attempt: I'm confused. I feel like $(\int_{\mathbb{R}} f_n^{p})^{\frac{1}{p}} = 1$ for all $n$, and thus it should also converge weakly to one... Can somebody wakl me through this? I'd really appreciate it.... been stuck on this one for at least a week >.<

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Hint: What is $L^1(\mathbb{R})^*$ (Use duality of $L_p$-spaces)?

We have

$$\Vert f_n \Vert_p = \left(\int_\mathbb{R} \chi_{[n,n+1]}^p\right)^{1/p} = 1$$

so indeed, $f_n \not\to 0$ in $L_p$.However, this does not mean that $f_n \to 1$ in $L_p$, since

$$\Vert f_n - 1\Vert_p^p = \int|\chi_{[n,n+1]}-1|^p = +\infty$$

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1.- Does $f_n\rightharpoonup 0$ in $L^p$ for $p\in(1,+\infty)$? Yes. In fact, let $g\in L^q$ with $\tfrac{1}{p}+\tfrac{1}{q}=1$. Then, $$ \vert \langle g,f_n\rangle\vert \leq \int_{\mathbb{R}} f_n\vert g\vert=\int f_n^2\vert g\vert\leq \left(\int_n^{n+1} \vert g\vert^q\right)^{1/q} \to 0 \quad \hbox{as} \quad n\to+\infty, $$ where the last inequality was an application of Hölder's inequality and the limit holds due to the fact that $g\in L^q$. (Notice that the equality in the middle of the procedure it's a little bit artificial, just to make explicit that I will let one $f_n$ with $g$ in the Hölder's inequality)

2.- Does $f_n\rightharpoonup 0$ in $L^1$? No. Take $g\equiv 1\in L^\infty$. Then, $$ \langle g,f_n\rangle =\int_n^{n+1}dx=1\not\rightarrow0. $$ 3.- Does $f_n\rightarrow 0$ in $L^p$ for $p\in(1,+\infty)$? No. It is enough to see that $$ \Vert f_n-0\Vert_{L^p}=\Vert f_n\Vert_{L^p}=1, $$ and hence it cannot converge to $0$.