Spring 2012 - Problem 4

Solution.

1. Let $\set{f_n}_n \subseteq S$ be a sequence such that $f_n \to f$ in $L^1$. Then

   $$\abs{\int f \,\diff{x}} = \abs{\int f - f_n \,\diff{x}} \leq \norm{f - f_n}_{L^1} \xrightarrow{n\to\infty} 0,$$

   so $f \in S$.

2. Compactly supported functions $L^2$ functions are dense in $L^2\p{\R^3}$, so it suffices to prove that $S \cap L^2\p{\R^3}$ is dense for such functions.

   Let $f \in L^2\p{\R^3}$ have compact support. Set $I = \int f \,\diff{x}$ and $R > 0$ be such that the support of $f$ is contained in $B\p{0, R}$. Let $\sigma$ denote the surface measure on the unit ball $S^2$ so that $\int_{S^2} \diff\sigma = 4\pi$. Thus, by polar coordinates,

   $$m\p{B\p{0, a}} = \int_0^a \int_{S^2} r^2 \,\diff\sigma \,\diff{r} = \frac{4\pi r^3}{3}.$$

   Fix $\epsilon > 0$. Set $g = f$ on $B\p{0, R}$ and for $R \leq \abs{x} \leq M$ (where $M > R$ will be chosen later), let $g = -I\epsilon$ (it's only important that $g$ has the opposite sign of $I$ on this interval). Then

   $$\int g \,\diff{x} = \int f \,\diff{x} - \int_{B\p{0, M} \setminus B\p{0, R}} I\epsilon \,\diff{x} = I - \frac{4\pi I\epsilon}{3}\p{M^3 - R^3}.$$

   Pick $M = \p{R^3 + \frac{3}{4\pi\epsilon}}^{1/3}$, i.e., so that $\int g \,\diff{x} = 0$. Thus, $g \in S$ and because $f \in L^2\p{\R^3}$ and $g$ is bounded on $R \leq \abs{x} \leq M$, $g \in L^2\p{\R^3}$ as well. Finally,

   $$\begin{aligned} \norm{f - g}_{L^2}^2 &= \int_{B\p{0, M} \setminus B\p{0, R}} I^2\epsilon^2 \,\diff{x} \\ &= m\p{B\p{0, M} \setminus B\p{0, R}} I^2\epsilon^2 \\ &= \frac{4\pi\p{M^3 - R^3}}{3} I^2\epsilon^2 \\ &= I^2\epsilon. \end{aligned}$$

   Since $I$ is independent of $\epsilon$, we may send $\epsilon \to 0$, and that gives the claim.