Why do I need $\vert \vert f_{n} \vert \vert _{p} \xrightarrow{n \to \infty} \vert \vert f \vert \vert _{p}$ here

Question

Let $p \in [1,\infty[$ and $f, f_{n}\in L^{p}(\mathbb R)$ for all $n \in \mathbb N$

Prove:

if $f_{n}\xrightarrow{n \to \infty} f $ in $L^{p}[-N,N]$ for all $N \in \mathbb N$ and $\vert \vert f_{n} \vert \vert _{p} \xrightarrow{n \to \infty} \vert \vert f \vert \vert _{p}$ then:

$f_{n}\xrightarrow{n \to \infty} f $ in $L^{p}(\mathbb R)$

My ideas:

$\int_{\mathbb R}\vert f_{n}-f\vert^{p}d\lambda=\int_{\mathbb R}\lim\limits_{N\to \infty}\vert f_{n}-f\vert^{p}1_{[-N,N]}d\lambda$ and now I claim that we can take the limit out given that $\vert f_{n}-f\vert^{p}1_{[-N,N]}\in L^{1}(\mathbb R)$ and then using the Dominated Convergence Theorem:

$\int_{\mathbb R}\lim\limits_{N\to \infty}\vert f_{n}-f\vert^{p}1_{[-N,N]}d\lambda=\lim\limits_{N\to \infty}\int_{\mathbb R}\vert f_{n}-f\vert^{p}1_{[-N,N]}d\lambda \xrightarrow{n\to \infty}0$

Nowhere in my proof did I use that $\vert \vert f_{n} \vert \vert _{p} \xrightarrow{n \to \infty} \vert \vert f \vert \vert _{p}$

Any ideas where I went wrong?

Answer 1

The question has been completely answered by the kimchi lover. However, for the sake of others I would like to provide a correct proof of the assertion in the title.

Let $\epsilon >0$ and choose $N$ such that $\int_{A} |f|^{p} <\epsilon$ where $A=\mathbb R \setminus [-N,N]$. Let $B=[-N,N]$. Since $\int_A|f_n|^{p}+\int_{B} |f_n|^{p}\to \int_A|f|^{p}+\int_{B} |f|^{p}$ and $\int_{B} |f_n|^{p}\to \int_{B} |f|^{p}$ it follows that $\int_{A} |f_n|^{p}\to \int_{A} |f|^{p}$. Hence there exists $n_0$ such that $\int_{A} |f_n|^{p}<\epsilon$ for $n \geq n_0$. Hence $\int|f_n-f|^{p}=\int_A |f_n-f|^{p}+\int_B |f_n-f|^{p}\leq \int_B |f_n-f|^{p}+2^{p} (\int_{A} |f_n|^{p}+\int_{A} |f|^{p})$. Rest is clear.

Answer 2

Suppose $g\in L^p(\mathbb R)$ is not the $0$ function, and let $f_n(x)=g(x-n)$ and let $f$ be the $0$ function. Then for each fixed $N$ it looks like $f_n$ converges to $f$ on $[-N,N]$, but yet $\|f_n-f\|=\|g\|\not\to 0$. The requirement that $\|f_n\|\to\|f\|$ is to prevent "mass from escaping to $\infty$" this way.