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I recently found out that $f \in L^{p}(\mathbb R)$ does not necessarily imply that $\lim\limits_{\vert x \vert \to \infty} f(x)=0$. Take for example, $f(x)=\begin{cases} n, \text{if } x \in [n,n+\frac{1}{n^{3}}]\\ 0, \text{if } x \in ]n+\frac{1}{n^{3}},n+1[\end{cases}$, $n \in \mathbb N$ it is clear that $\lim\limits_{\vert x \vert \to \infty} f(x)$ does not exist and further $\vert\vert f\vert \vert_{p}=\sum\limits_{n \in \mathbb N}\frac{1}{n^{2}}<\infty$.

But in my proof to show that $f_{n}\xrightarrow{L^{p}([-N,N])} f$ on every $N \in \mathbb N$ and $\vert \vert f_{n} \vert \vert_{p}\xrightarrow{n \to \infty}\vert\vert f\vert \vert$ implies that $f_{n}\xrightarrow{L^{p}(\mathbb R)} f$

In this proof I used the fact that since $\vert \vert f_{n} \vert \vert_{L^{p}([-N,N])}\xrightarrow{n \to \infty}\vert \vert f \vert \vert_{L^{p}([-N,N])}$ and $\vert \vert f_{n} \vert \vert_{p}\xrightarrow{n \to \infty}\vert\vert f\vert \vert$ then for $A_{N}:=\mathbb R\setminus[-N,N]$: $\vert \vert f_{n} \vert \vert_{L^{p}(A_{N})}\xrightarrow{n \to \infty}\vert \vert f \vert \vert_{L^{p}(A_{N})}$

This is all logical, but now I see a flaw in my following argument:

Let $\epsilon > 0$. We can thus choose an $N \in \mathbb N$, so that $\int_{A_{N}}\vert f_{n}\vert^{p}d\mu<\epsilon$. I used ths path in my solution, and it was marked as correct, but surely this cannot be true by the paragraph I highlighted above? Am I missing something in this proof?

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  • $\begingroup$ Is the inequality supposed to hold for all $n \in \mathbb N$ simultaneously or for some fixed $n$? $\endgroup$ – user159517 Jul 9 at 21:43
  • $\begingroup$ I think you are confusing $f_n$ and $f$. There is no $n$ is $f$ $\endgroup$ – MPW Jul 9 at 22:03
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Fix $p\in ]1,\infty]$. Then $f\in L^p(\Bbb R)$ doesn't imply $\lim_{|x|\to\infty} f(x)=0$, but it does imply that $\lim_{N\to\infty} \|f_n |_{A_N}\|_p=0$. But the $L^p$ norms of "tail of the function" does converge to $0$:

Define the sequence $a_N:=\int_0^N |f_n|^p \,\mathrm d\mu$. By definition, this sequence converges and $$\lim_{N\to\infty} a_N = \underbrace{\int_0^\infty |f_n|^p \,\mathrm d\mu}_{\displaystyle=:l}.$$

So (see remark) $$\lim_{N\to\infty} \int_{N}^\infty |f_n|^p \,\mathrm d\mu = \lim_{N\to\infty} \int_{N+1}^\infty |f_n|^p \,\mathrm d\mu = \lim_{N\to\infty} l -a_N = l -\lim_{N\to\infty} a_N = \color{orange}0.$$

By the same argumentation, we have $$\lim_{N\to-\infty} \int_{-\infty}^N |f_n|^p \,\mathrm d\mu= \color{blue}0.$$

It follows that $$\lim_{N\to\infty} \int_{A_N} |f_n|^p \,\mathrm d\mu = \color{orange}0+\color{blue}0=0.$$

Remark. In order to use $\int_N^\infty |f_n|^p \,\mathrm d\mu = \lim_{N\to\infty} l -a_N$, I would first have to prove that $\int_N^\infty |f_n|^p \,\mathrm d\mu$ converges and then use the fact that $a_N+\int_{N+1}^\infty |f_n|^p \,\mathrm d\mu = l$. The proof of convergence is very similar to the proof for series.

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For $f \in L^p$, define $(g_N)_{N\geq 1}$ as $$g_N (x) := |f(x)| 1_{\{|f(x)| \leq N\}}.$$ Clearly, it holds that $$\lim_{N\to\infty} g_N(x) = |f(x)|, \quad x \in \mathbb R, $$ so the monotone convergence theorem implies $$\lim_{N\to\infty} \|g_N\|_p = \|f\|_p,$$ which in turn gives $$\lim_{N\to\infty} (\|f\|_p^p - \|g_N\|_p^p) = \lim_{N\to\infty} \int_{A_N} |f(x)|^p dx = 0.$$

To show the "fact" you want to use more directly, consider that $$ \|f_n\|_{L^p (A_N)}^p = \|f_n\|_{L^p (\mathbb R)}^p -\|f_n\|_{L^p (-N,N)}^p \to \|f\|_{L^p (\mathbb R)}^p -\|f\|_{L^p (-N,N)}^p = \|f\|_{L^p (A_N)}^p$$

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