# Schwartz functions dense in weighted $L^p$ space?

Upon learning about Schwartz functions, one result that is usually presented to students is that not only are Schwartz functions dense in $L^2(\mathbb{R}^n)$ but in $L^p(\mathbb{R}^n)$ for all $1 \le p < \infty$. Here, $L^p(\mathbb{R}^n)$ is the standard $L^p$ space with respect to Lebesgue measure.

With how nicely Schwartz functions behave (with their smooth rapid decay), does this result extend to weighted $L^p$ spaces? For instance, suppose we have a weight $w \in A_p$ ($1 < p < \infty$) where $A_p$ is the class of Muckenhoupt weights of parameter $p$. Let $L^p(w)$ be the space of Lebesgue measurable functions over $\mathbb{R}^n$ such that

$$\| f\|_{L^p(w)} = \left( \int_{\mathbb{R}^n} |f(x)|^p w(x) \,dx \right)^{1/p} < \infty$$

(so by taking $w=1$, we get back to standard $L^p$ space.)

Can we say Schwartz functions are still dense in $L^p(w)$?

First, we should replace $$w$$ by an suitable modifcation such that $$1 \in L^1(w)$$.

If $$\omega$$ is locally integrable (that's the case for Muckenhoupt weights), we may take smooth partition of unity $$(u_m)_{m \in \mathbb{Z}^n}$$ with $$\mathrm{supp} \ u_m \subset [-1,1]^n + m$$. Define $$\tag{1} u(x) := \sum_{n \in \mathbb{Z}^d} \frac{1}{2^{\|m\|_2^2} (1+w_m)} u_m(x),$$ where $$w_m = \int_{[-1,1]^n+m} w(x) \, dx$$. The function (1) is smooth and non-negative. Moreover let $$\widetilde{w}(x) = w(x) u(x)^p$$.

Then $$f \in L^p(w)$$ if and only if $$f/u \in L^1(\widetilde{w})$$.

If for any $$g \in L^p(\widetilde{w})$$ there exists at least one $$h \in C_c^\infty(\mathbb{R}^n)$$ with $$\|f-h\|_{L^p(\widetilde{w})} < \varepsilon$$, then we can take $$g= f/u \in L^p(\widetilde{w})$$ if $$f \in L^p(w)$$ to get an $$h \in C_c^\infty(\mathbb{R}^n)$$ with $$\|f-hu\|_{L^p(w)} =\|g- h\|_{L^p(\widetilde{w})} < \varepsilon.$$ Note that $$hu \in C_c(\mathbb{R}^n)$$. Since $$\int u(x)^p w(x) dx \le C \sum_{m \in \mathbb{Z}^n} \frac{1}{2^{\|m\|_2^2 p/2}} \frac{w_m}{1+w_m} <\infty$$ we also have $$1 \in L^1(w)$$.

We are left to prove that the statement is true for $$\widetilde{w}$$. We write $$w$$ instead of $$\widetilde{w}$$.

We only need that $$1 \in L^1(w)$$. By the dominated convergence theorem, we can chose $$R>0$$ such that $$\int |f(x)|^p w(x) (1_{\{|f|>R\}}+1_{\{|x|>R\}}) \, dx < \varepsilon^p.$$ Set $$g(x) = f(x) 1_{\{|x| \le R, |f| \le R\}}$$. Then $$\|f-g\|_{L^p(\omega)} \le \varepsilon$$. So we may work with $$g$$ instead of $$f$$.

Let $$h \in C_c^\infty(\mathbb{R}^n)$$ be non-negative with $$\int h(y) dy = 1$$ and $$\mathrm{supp} \ h \subset [-1,1]^n$$. Set $$h_\delta(y):= \delta^{-n} h(y/\delta)$$ (Approximation to the identity) and define $$g_\delta(x) := \int h_\delta(x-y) g(y) dy.$$ Then $$g_\delta$$ is smooth and has compact support. ($$g$$ is bounded and has bounded suport.) Now we have $$g_\delta(x) \rightarrow g(x)$$ for almost all $$x \in \mathbb{R}^n$$ by Lebesgue's differentiation theorem. Since $$|g_\delta| \le R$$ and $$1 \in L^1(w)$$ we can apply the dominated convergence theorem (in $$L^p(w)$$) again to see that there exists $$\delta>0$$ with $$\|g-g_\delta\|_{L^p(w)} < \varepsilon$$.