Show that $\int_{|x|<R}\frac{|f(x)|}{|x|}\mathrm dx=o(R),\quad R\to\infty.$

Question

Suppose $f\in L^3(\mathbb R^3)$. Show that $$\int_{|x|<R}\frac{|f(x)|}{|x|}\mathrm dx=o(R),\quad R\to\infty.$$

First, I try to show that for a fixed $R_0$, $\int_{|x|<R_0}\frac{|f(x)|}{|x|}\mathrm dx<\infty$. This follows from Hölder's inequality: $$\int_{|x|<R_0}\frac{|f(x)|}{|x|}\mathrm dx\le\left(\int_{|x|<R_0}|f(x)|^3\mathrm dx\right)^{1/3}\left(\int_{|x|<R_0}\frac{\mathrm dx}{|x|^{3/2}}\right)^{2/3}.$$ But it can only yield $$\int_{|x|<R}\frac{|f(x)|}{|x|}\mathrm dx=O(R),\quad R\to\infty.$$

How can I improve my result?

Answer 1

Let $\epsilon>0$, and let $R_{\epsilon}$ satisfy $\int\limits_{|x|>R_{\epsilon}}{|f(x)|^3\text{ d}x}<\epsilon^3$. For $R>R_{\epsilon}$, have \begin{align} \int\limits_{|x|<R}{\frac{|f(x)|}{|x|}\text{ d}x}&=\int\limits_{|x|<R_{\epsilon}}{\frac{|f(x)|}{|x|}\text{ d}x} + \int\limits_{R_{\epsilon}\le |x|<R}{\frac{|f(x)|}{|x|}\text{ d}x} \\ &\le C_{\epsilon} + \left(\,\int\limits_{R_{\epsilon}\le|x|<R}{|f(x)|^3\text{ d}x}\,\right)^{\frac{1}{3}}\left(\,\int\limits_{R_{\epsilon}\le|x|<R}{\frac{1}{|x|^{\frac{3}{2}}}}\,\right)^{\frac{2}{3}} \\ &\le C_{\epsilon} + \left(\,\int\limits_{|x|>R_{\epsilon}}{|f(x)|^3\text{ d}x}\,\right)^{\frac{1}{3}}\left(\,\int\limits_{|x|<R}{\frac{1}{|x|^{\frac{3}{2}}}}\,\right)^{\frac{2}{3}} \\ &\le C_{\epsilon} + C\epsilon R \end{align} and hence $$ \limsup\limits_{R\rightarrow\infty}{\frac{1}{R}\int\limits_{|x|<R}{\frac{|f(x)|}{|x|}\text{ d}x}}\le C\epsilon. $$ Letting $\epsilon\rightarrow 0$ gives the desired result.

Answer 2

Edit: I've claimed that the arguments in the two answers are in some sense really the same. People have disputed this. I think it's a Good Thing for a person to see how it is that they are really the same. A Good Thing because if you like one better than the other, or find one easier to follow, then in the future when you see another argument presented in the style you don't prefer you'll be better able to understand it by seeing how it's the same as an argument you do prefer.

So I'm going to give a second version of my proof; I think it will be clear that my second version is really the same as my first version, while it may be more clear how my second version is really the same as Joey Zou's presentation.

Original:

Let $$\phi(f,R)=\frac1R\int_{|x|\le R}\frac{|f(x)|}{|x|}\,dx.$$You've shown that $$\phi(f,R)\le c||f||_3.$$

If $f$ has compact support then certainly $$\lim_{R\to\infty}\phi(f,R)=0.$$ In general, choose $f_n$ with compact support such that $||f_n-f||_3\to0$.Then $$\phi(f_n,R)\to\phi(f,R)\quad(n\to\infty)$$_uniformly_ in $R$; hence $\lim_{R\to\infty}\phi(f,R)=0$.

Revised: This is the same as above, except instead of using the fact about uniform convergence that's used above I don't mention uniform convergence, instead inserting and using the proof of said fact about uniform convergence:

Let $$\phi(f,R)=\frac1R\int_{|x|\le R}\frac{|f(x)|}{|x|}\,dx.$$You've shown that $$\phi(f,R)\le c||f||_3.$$

If $f$ has compact support then certainly $$\lim_{R\to\infty}\phi(f,R)=0.$$

In general, suppose that $f\in L^3$ and $\epsilon>0$. Choose $g$ with compact support so that $$||f-g||_3<\epsilon.$$ Now $$|\phi(f,R)-\phi(g,R)|<C\epsilon$$for every $R$; since $\phi(g,R)\to0$ this shows that $$\limsup_{R\to\infty}\phi(f,R)\le C\epsilon.$$Since $\epsilon>0$ was arbitrary, $\phi(f,R)\to0$.