Let $f\in L^p$. Can we say $\|f\|_{L^{p}} \leq \epsilon$ on $|x|\geq R$ for large $R$?

Question

Let $f\in L^{p}(\mathbb R), (1\leq p <\infty)$ and $\epsilon>0.$

My Question: Can we expect to find $R>1$ (may be large) so that $\|f\|_{L^{p}(B_R)} \leq \epsilon$ on $B_{R}=\{x:|x|\geq R\}$?

Answer 1

The function $ g_n = \chi_{[-n,n]} |f|^p$ satisfies

$$0\le g_1 \le g_2\le \cdots \le |f|^p$$

and $g_n \to |f|^p$ pointwisely on $\mathbb R$. Thus the monotone convergence theorem implies that

$$\left( \int_{-n}^n |f|^p\right)^{\frac{1}{p}} \to \|f\|_p.$$

If $\int_\mathbb{R} |f|^p$ is finite, then for all $\epsilon >0$ there is $n$ so that

$$ \|f\|^p_p -\epsilon^p \le \int_{-n}^n |f|^p \le \|f\|^p_p,$$

which implies

$$\left(\int_{|x|\ge n} |f|^p\right)^{\frac 1p} <\epsilon.$$

Answer 2

Let $f$ consist of small non-overlapping rectangular bumps of height 1 centered on 1,2,3,$\cdots $,n,$\cdots $ with width $w_{n}=1/n^{k}$ with sufficiently large $k$. Then $f\in L^{p}$ but will never become smaller than $\varepsilon $ as $x$ increases.

Answer 3

$$\|f\|_p^p = \int_{\{|x|<R\} }|f|^p + \int_{\{|x|>R\}} |f|^p.$$

Both sides are finite. By monotone convergence the first summand on RHS tends to $\|f\|_p^p$ as $R\to \infty$. Therefore the second summand tends to $0$.