6
$\begingroup$

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 $L^p(w)$?

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $\|g-g_\delta\|_{L^p(w)} < \varepsilon$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.