Let \(f : \mathbb{R}^3 \to \mathbb{R}\) be a continuous function of compact support. Show that \(u(x) = \int_{\mathbb{R}^3} \frac{f(y)}{|x - y|} \, dy\) is well defined and that \(\lim_{|x| \to \infty} u(x) |x| = \int_{\mathbb{R}^3} f(y) \, dy\).
Solution
Choose \(R>0\) such that \[ \operatorname{supp}f\subseteq\{y\in\RR^3:\abs y\leq R\}, \] and set \[ M\coloneqq\sup_{y\in\RR^3}\abs{f(y)}<\infty. \]
1. For every fixed \(x\in\RR^3\), the integral defining \(u(x)\) isabsolutely convergent.
Since \(f\) vanishes outside the ball of radius \(R\), \[ \int_{\RR^3} \frac{\abs{f(y)}}{\abs{x-y}}\,dy \leq M\int_{\abs y\leq R}\frac{dy}{\abs{x-y}}. \] Put \(z=y-x\). If \(\abs y\leq R\), then \[ \abs z = \abs{y-x} \leq R+\abs x. \] Therefore \[ \begin{aligned} \int_{\abs y\leq R}\frac{dy}{\abs{x-y}} &\leq \int_{\abs z\leq R+\abs x}\frac{dz}{\abs z}\\ &= 4\pi\int_0^{R+\abs x}r\,dr <\infty. \end{aligned} \] Thus \(u(x)\) is well defined.
2. If \(L>R\), \(\abs x\geq L\), and \(y\in\operatorname{supp}f\), then\[ \left| \frac{\abs x}{\abs{x-y}}-1 \right| \leq \frac{R}{L-R}. \]
The reverse triangle inequality gives \[ \bigl|\abs x-\abs{x-y}\bigr| \leq \abs y \leq R, \] while \[ \abs{x-y} \geq \abs x-\abs y \geq L-R. \] Consequently \[ \left| \frac{\abs x}{\abs{x-y}}-1 \right| = \frac{\bigl|\abs x-\abs{x-y}\bigr|}{\abs{x-y}} \leq \frac{R}{L-R}. \]
3. If \(L>R\) and \(\abs x\geq L\), then\[ \left| \abs x\,u(x)-\int_{\RR^3}f(y)\,dy \right| \leq \frac{R}{L-R} \int_{\RR^3}\abs{f(y)}\,dy. \]
For such \(x\), the denominator \(\abs{x-y}\) is nonzero on \(\operatorname{supp}f\). Hence \[ \abs x\,u(x)-\int f(y)\,dy = \int f(y) \left( \frac{\abs x}{\abs{x-y}}-1 \right)dy. \] Taking absolute values and applying step 2 on the support of \(f\) gives the stated estimate.
4. One has\[ \boxed{ \lim_{\abs x\to\infty}\abs x\,u(x) = \int_{\RR^3}f(y)\,dy }. \]
The integral \[ \int_{\RR^3}\abs{f(y)}\,dy \] is finite because \(f\) is continuous with compact support. The right-hand side of the estimate in step 3 tends to \(0\) as \(L\to\infty\). Therefore, given \(\varepsilon>0\), choosing \(L\) sufficiently large makes the difference in step 3 smaller than \(\varepsilon\) for every \(\abs x\geq L\). This is exactly the asserted limit.
5. Q.E.D.
Step 1 proves that \(u\) is well defined, and step 4 proves the required asymptotic formula.