Density of a set of functions in Schwartz space I have a difficulty doing the following problem:
Let $S(\mathbb{R}^n)$ be the Schwartz space. I need to determine whether the following set of functions $A$:
$$A= \{f\in S(\mathbb{R}^n): \text{supp}(\hat{f}) \text{ is compact and } 0\not\in \text{supp}(\hat{f})\}$$
is dense in $L^p(\mathbb{R}^n)$, where $\hat{f}$ is the Fourier transform of $f$.
I think that the above space is dense in $L^p$ when $p\in (1,\infty)$, but not when $p=1$ or $\infty$, but I don't know how to prove it. Any help is appreciated.
 A: I will denote the vector space of Schwartz functions with compactly supported Fourier transforms not containing the origin by $\mathcal{S}_{0}(\mathbb{R}^{n})$.
$\mathcal{S}_{0}(\mathbb{R}^{n})$ is obviously not dense in $L^{\infty}(\mathbb{R}^{n})$ since $\mathcal{S}(\mathbb{R}^{n})$ is not dense in $L^{\infty}(\mathbb{R}^{n})$. Decay at $\infty$ means that $\left\|1-f\right\|_{L^{\infty}}\geq\epsilon$, for any given $f\in\mathcal{S}(\mathbb{R}^{n})$ and $0<\epsilon<1$. $\mathcal{S}_{0}(\mathbb{R}^{n})$ is also not dense in $L^{1}(\mathbb{R}^{n})$, since every $f\in S_{0}$ has mean zero. 
However, $\mathcal{S}_{0}$ is dense in $L^{p}(\mathbb{R}^{n})$, for $1< p<\infty$. We will use Littlewood-Paley theory to show this.

Given a Schwartz function $\phi$, we denote the $k^{th}$ Littlewood-Paley projection associated to $\phi$, where $k\in\mathbb{Z}$, by
$$\widehat{\Delta_{k}^{\phi}f}=\widehat{\phi}(\cdot/2^{k})\widehat{f},\quad f\in\mathcal{S}'(\mathbb{R}^{n})$$
If there is no ambiguity, I may omit the $\phi$ and just write $\Delta_{k}$.

Lemma 1. Let $\eta$ be a Schwartz function such that $\widehat{\eta}$ is compactly supported away from the origin. Then for $1<p<\infty$,
  $$\left\|\sum_{j\in\mathbb{Z}}\Delta_{j}^{\eta}f\right\|_{L^{p}}\lesssim_{\eta,p}\left\|f\right\|_{L^{p}},\qquad f\in\mathcal{S}(\mathbb{R}^{n})$$

For a proof, see this post of mine.

Lemma 2. Let $\eta$ and $\zeta$ be Schwartz functions as above. Then for any $j,k\in\mathbb{Z}$,
  $$\sup_{j,k}\left|\Delta_{j}^{\eta}\Delta_{k}^{\zeta}f\right|\lesssim_{\eta,\zeta}\mathcal{M}(\mathcal{M}(f)) \quad \text{a.e.}$$
  where $\mathcal{M}$ denotes the Hardy-Littlewood maximal operator.

Proof. Observe first that by the invariance properties of the Fourier transform,
$$\Delta_{j}^{\eta}\Delta_{k}^{\zeta}f=2^{jn}\eta(2^{j}\cdot)\ast 2^{kn}\zeta(2^{k}\cdot)\ast f$$
Since $\eta$ and $\zeta$ have $L^{1}$ radially decreasing majorants, being Schwartz functions, it follows (see Stein, Singular Integrals and Differentiability Properties of Functions, Theorem 2, pg. 63) ) that
$$\sup_{k} \left|2^{kn}\zeta(2^{k}\cdot)\ast f\right|\lesssim_{\zeta} \mathcal{M}(f) \quad\text{a.e.}$$
and by the same argument,
$$\sup_{j}\left|2^{jn}\eta(2^{j}\cdot)\ast\mathcal{M}(f)\right|\lesssim_{\eta}\mathcal{M}(\mathcal{M}(f))\quad\text{a.e.}$$
$\Box$

Proposition 3. Let $\eta$ be as in the statement of the preceding lemma. Then the operator
  $$f\mapsto \sup_{N\in\mathbb{Z}}\left|\sum_{j<N}\Delta_{j}^{\eta}(f)\right|$$
  is bounded on $L^{p}(\mathbb{R}^{n})$, for $1<p<\infty$.

Proof. Fix $\eta$ and let $\phi$ be as in the statement of the lemma below. I leave it to you to check (it's not hard) that there is a constant $c_{0}>0$, which depends on $\eta$ and $\phi$, such that
$$\left|j-k\right|>c_{0}\Rightarrow\Delta_{j}^{\phi}\Delta_{k}^{\eta}=0$$
So we may write
$$\sum_{k\leq N}\Delta_{k}^{\eta}=\sum_{j\leq N+c_{0}}\Delta_{j}^{\phi}\left(\sum_{k\leq N}\Delta_{k}^{\eta}\right)=\sum_{j\leq N+c_{0}}\Delta_{j}^{\phi}\left(\sum_{k\in\mathbb{Z}}\Delta_{k}^{\eta}\right)-\sum_{j\leq N+c_{0}}\Delta_{j}^{\phi}\left(\sum_{k>N}\Delta_{k}^{\eta}\right)$$
Observe that the second term is a finite sum, for any $N$, with $\lesssim c_{0}^{2}$ terms. So by the preceding lemma,
$$\sup_{j,k}\left|\sum_{j\leq N+c_{0}}\Delta_{j}^{\phi}\left(\sum_{k>N}\Delta_{k}^{\eta}\right)\right|\lesssim_{\phi,\eta}c_{0}^{2}\mathcal{M}(\mathcal{M}(f))\quad\text{a.e.}$$
For the first term, we need the following little lemma. We can define a Schwartz function $\psi$ by
$$\widehat{\psi}(\xi)=1-\sum_{j\geq 1}\widehat{\phi}(2^{-j}\cdot)$$
Observe that $\widehat{\psi}$ is compactly supported in the ball $\left|\xi\right|\leq 2$ and $\equiv 1$ on the ball $\left|\xi\right|\leq 1$. Moreover,
$$\widehat{\psi}(\xi)=\sum_{j\leq 0}\widehat{\phi}(2^{-j}\xi), \forall\xi\neq 0$$
and therefore
$$2^{Nn}\psi(2^{N}\cdot)\ast g=\sum_{j\leq N}\Delta_{j}^{\phi}g,\quad\forall g\in\mathcal{S}(\mathbb{R}^{n})$$
By density and continuity of the convolution, this equality holds for all tempered distributions, in particular $L^{p}$ functions. Since $\psi$ has an $L^{1}$ radially decreasing majorant, the same argument as above yields
$$\sup_{N}\left|\sum_{j\leq N+c_{0}}\Delta_{j}^{\phi}\left(\sum_{k\in\mathbb{Z}}\Delta_{k}^{\eta}f\right)\right|=\sup_{M}\left|\sum_{j\leq M}2^{nM}\psi(2^{M}\cdot)\ast\left(\sum_{k\in\mathbb{Z}}\Delta_{k}^{\eta}f\right)\right|\lesssim_{\phi}\mathcal{M}\left(\sum_{k\in\mathbb{Z}}\Delta_{k}^{\eta}f\right)$$
Since $f\in L^{p}$, $\sum_{k}\Delta_{k}^{\eta}f\in L^{p}$, and the $\mathcal{M}$ is bounded $L^{p}\rightarrow L^{p}$, when $1<p\leq\infty$, we conclude that
\begin{align*}
\left\|\sup_{N}\left|\sum_{k\leq N}\Delta_{k}^{\eta}f\right|\right\|_{L^{p}}&\lesssim_{\phi,\eta}\left\|\mathcal{M}(\mathcal{M}(f))\right\|_{L^{p}}+\left\|\mathcal{M}\left(\sum_{k\in\mathbb{Z}}\Delta_{k}^{\eta}f\right)\right\|_{L^{p}}\\
&\lesssim_{n,p}\left\|f\right\|_{L^{p}}+\left\|\sum_{k}\Delta_{k}^{\eta}f\right\|_{L^{p}}\\
&\lesssim_{\eta,p}\left\|f\right\|_{L^{p}},
\end{align*}
where we use the first lemma to obtain the ultimate inequality.
$\Box$

Lemma 4. Let $\phi\in\mathcal{S}(\mathbb{R}^{n})$ be such that $\text{supp}(\widehat{\phi})\subset\left\{1/2\leq\left|\xi\right|\leq 2\right\}$ and
  $$\sum_{j\in\mathbb{Z}}\widehat{\phi}(\xi/2^{j})=1,\qquad\forall \xi\neq 0$$
  Then for any $f\in\mathcal{S}(\mathbb{R}^{n})$,
  $$\left\|f-\sum_{\left|j\right|\leq N}\Delta_{j}f\right\|_{L^{\infty}}\rightarrow 0, \quad N\rightarrow\infty$$

Proof. By Hausdorff-Young, it suffices to show that
$$\left\|\widehat{f}-\sum_{\left|j\right|\leq N}\widehat{\Delta_{j}f}\right\|_{L^{1}}\rightarrow 0, \quad N\rightarrow\infty$$
But this is trivial, as
\begin{align*}
\widehat{f}(\xi)\left(1-\sum_{\left|j\right|\leq N}\widehat{\phi}(2^{-j}\xi)\right)\rightarrow 0, \quad N\rightarrow \infty
\end{align*}
a.e., and the conclusion follows by dominated convergence.$\Box$
Observe that for any $N\in\mathbb{N}$,
$$\sum_{\left|j\right|\leq N}\Delta_{j}=\sum_{j\leq N}\Delta_{j}-\sum_{j<-N}\Delta_{j}$$
and therefore for any $f\in L^{p}$,
$$\left|\sum_{\left|j\right|\leq N}\Delta_{j}f\right|\leq 2\sup_{N\in\mathbb{Z}}\left|\sum_{j\leq N}\Delta_{j}f\right|$$
which is in $L^{p}$, if $1<p<\infty$. Since the partial sums converge to $f$ uniformly, dominated convergence implies that
$$\left\|f-\sum_{\left|j\right|\leq N}\Delta_{j}f\right\|_{L^{p}}\rightarrow 0,\quad N\rightarrow\infty$$
If $f\in\mathcal{S}(\mathbb{R}^{n})$, then for any $N$, $\sum_{\left|j\right|\leq N}\Delta_{j}f$ is a Schwartz function whose Fourier transform is compactly supported away from the origin. Since Schwartz functions are dense in $L^{p}$, we conclude that $\mathcal{S}_{0}(\mathbb{R}^{n})$ is dense in $L^{p}$, if $1<p<\infty$.
