Prove that $f\in L^1(\mathbb R^d)$

Question

Assume that $f(x)\in L^2(\mathbb R^d)$ and $|x|^{\frac{d+1}{2}}f(x)\in L^2(\mathbb R^d)$. Prove that $f(x)\in L^1(\mathbb R^d)$.

I tried to use Holder's inequality to bound $\int_{\mathbb R^d} |f| $, but I had no progress. Does anyone have ideas?

Thank you for your help!

Answer 1

The idea: the first assumption tells you that $f$ is not very big around $0$ and the second assumption tells you that it's not very big around $\infty$.

So: let $$A_0 = \{ x \in \mathbb{R}^d : \| x \| \leqslant 1 \ \& \ |f(x)| \leqslant 1 \} \\[1ex] A_1 = \{ x \in \mathbb{R}^d : \| x \| \leqslant 1 \ \& \ |f(x)| > 1 \}.$$

Then

$$\int \limits_{\| x \| \leqslant 1} |f(x)| \, \mathrm{d} \lambda(x) = \int \limits_{A_0} |f(x)| \, \mathrm{d} \lambda(x) + \int \limits_{A_1} |f(x)| \, \mathrm{d} \lambda(x)$$

and

$$\begin{align*} \int \limits_{A_0} |f(x)| \, \mathrm{d} \lambda(x) & \leqslant \int \limits_{A_0} 1 \, \mathrm{d} \lambda(x) = \lambda( A_0 ) < \infty \\[1ex] \int \limits_{A_1} |f(x)| \, \mathrm{d} \lambda(x) & \leqslant \int \limits_{A_1} |f(x)|^2 \, \mathrm{d} \lambda(x) < \infty \end{align*}$$

where the finiteness of the second integral follows from the assumption $f(x) \in L^2(\mathbb{R}^d)$.

The second part goes similarly: let

$$B_0 = \left\{ x \in \mathbb{R}^d : \| x \| > 1 \ \& \ |f(x)| \leqslant \frac{1}{\|x\|^{d+1}} \right\} \\[1ex] B_1 = \left\{ x \in \mathbb{R}^d : \| x \| > 1 \ \& \ |f(x)| > \frac{1}{\|x\|^{d+1}} \right\}.$$

Then

$$\int \limits_{\| x \| > 1} |f(x)| \, \mathrm{d} \lambda(x) = \int \limits_{B_0} |f(x)| \, \mathrm{d} \lambda(x) + \int \limits_{B_1} |f(x)| \, \mathrm{d} \lambda(x)$$

and

$$\begin{align*} \int \limits_{B_0} |f(x)| \, \mathrm{d} \lambda(x) & \leqslant \int \limits_{B_0} \frac{1}{\|x\|^{d+1}} \, \mathrm{d} \lambda(x) < \infty \\[1ex] \int \limits_{B_1} |f(x)| \, \mathrm{d} \lambda(x) & \leqslant \int \limits_{B_1} |f(x)|^2 \| x \|^{d+1} \, \mathrm{d} \lambda(x) < \infty \end{align*}$$

where the finiteness of the second intergral follows from the assumption $\| x \|^{\frac{d+1}{2}} f(x) \in L^2( \mathbb{R}^d )$.

Answer 2

Hint:

First, find a bound for $\int_{B_1(0)} f\mathrm dx$, this should be easy. Then, for the rest of the domain, apply Holder to $$ \int_{\mathrm R^d\setminus B_1(0)} ( |x|^{\frac{d+1}{2}} f(x) ) |x|^{-\frac{d+1}{2}} \mathrm dx. $$

For one of the resulting factors, you have to use polar coordinates: $$ \int_{\mathrm R^d\setminus B_1(0)} |x|^{-d-1} \mathrm dx. = C_d \int_1^\infty r^{-d-1} r^{d-1} \mathrm dr, $$ where $C_d$ is the surface area of the unit sphere in $d$-dimensional space.