Are compactly supported Lipschitz functions dense in $L^2$ for some measures?

Question

I have been searching for such a statement in the book,https://www.math.ucdavis.edu/~hunter/book/pdfbook.html but I can't find any! (Any reference for such a statement in that book would be most helpful!)

The closest thing I could find here is Theorem 12.50 on page 354 saying that w.r.t the Lebesgue measure the compactly supported continuous functions are dense in $L^p$. Does the corresponding statement about compactly supported Lipschitz functions with arbitrary measures somehow follow from this?

Answer 1

Even stronger, $C^{\infty}$-smooth functions with compact support are dense in $L^p(\mathbb R^n)$ with the Lebesgue measure (where $1\le p < \infty$). Of course, the class of compactly supported Lipschitz funtions is a larger class than that of smooth, compactly supported functions.

For a proof of the above result, see for example Theorem 3 here: http://texas.math.ttu.edu/~gilliam/f06/m5340_f06/mollifiers_approx.pdf

Answer 2

Such a result should follow pretty easily from standard results. First, if $\mu$ is a positive regular Borel measure on $\mathbb R^n$ such that $\mu(K) < \infty$ for each compact $K$, then $C_c$ is dense in $L^p(\mu)$ for $1\le p < \infty$. Whatever proof you know for Lebesgue measure should work here as well.

Now take any $f\in C_c.$ Choose $R$ such that $B(0,2R)$ contains the support of $f$. By Stone-Weierstrass, there is a sequence of polynomials $P_k$ that converges uniformly to $f$ on $\overline B (0,2R)$. Take any $g\in C^\infty_c$ that equals $1$ on $B(0,R)$ with support in $B(0,2R)$. Then $gP_k \to gf = f$ uniformly on $\overline B (0,2R)$. We thus obtain a sequence $gP_k$ in $C^\infty_c$ such that $gP_k \to f$ in $L^p(\mu), 1\le p < \infty$.