Decay estimate of $L^2$ integral.

Question

Let $C_{R} = [\frac{-R}{2} , \frac{R}{2}]^n$, the cube centered at origin in $\mathbb{R}^n$ of length $R > 0$ and $\|x\|_{\infty} = \max_{i = 1}^n \{x_i \}$, $x \in \mathbb{R}^n$. Suppose $\phi$ is a complex valued function on $\mathbb{R}^n$ and has the decay estimate: for some $\epsilon > 0$,

$|\phi(x)| \leq C_1 (1 + \|x\|_{\infty})^{-n - \epsilon}$, for all $x \in \mathbb{R}^n$.

Then prove that $\| \phi ( 1 - \chi_{C_R} )\|_{L^2(\mathbb{R}^n)}^2 \leq C_2 R^{-n}$.

P.S: When sup norm $\|.\|_{\infty}$ is replaced by $l^2$ norm and cube $C_{R}$ replaced by ball, then I can estimate the above integral by a well known formula of integration of radial function.

Answer 1

Since $n^{-1/2}\lVert x\rVert_2\leq\lVert x\rVert_\infty$, you also have the bound \begin{align*} \lvert\phi(x)\rvert&\leq C_1(1+\lVert x\rVert_\infty)^{-n-\epsilon}\\ &\leq C_1(1+n^{-1/2}\lVert x\rVert_2)^{-n-\epsilon}\\ &\leq \underbrace{n^{(n+\epsilon)/2}C_1}_{C'_1}(1+\lVert x\rVert_2)^{-n-\epsilon}. \end{align*} Also, $C_R$ contains the $2$-ball of radius $R/2$, so $$ \lVert\phi(1-\chi_{C_R})\rVert_{L^2}^2\leq \lVert\phi(1-\chi_{B_{R/2}})\rVert_{L^2}^2 $$ and you can use your 2-norm result.