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Let $L_{0}^{1}(\mathbb{R}^{n})$ denote the the closed subspace of $L^{1}$ functions whose Fourier transform vanishes at the origin (equivalently, $\int f=0$). At the top of pg. 231 in E.M. Stein, Singular Integrals and Differentiability Properties of Functions, the author claims that a dense subspace of $L_{0}^{1}(\mathbb{R}^{n})$ is the subspace

$$\left\{f\in L^{1}(\mathbb{R}^{n}) : \text{supp}(\widehat{f}) \text{ compact }, \text{supp}(\widehat{f})\subset\mathbb{R}^{n}\setminus\left\{0\right\}\right\}$$

and that density

"...can be proved directly by elementary computation, or one can appeal to Wiener's theorem characterizing the maximal ideals of $L^{1}(\mathbb{R}^{n})$."

I am not familiar with Wiener's theorem, so I tried the direct route. I know that $L^{1}$ functions with compactly supported Fourier transforms are dense in $L^{1}$, but I don't see how to extend this result where the support is disjoint from the origin in frequency space. I tried taking a cutoff function $\varphi$ which is $\equiv 1$ on the unit ball $B_{1}(0)$ and supported in $B_{2}(0)$ and then considering the sequence of functions defined by

$$f_{\delta}(x):=(\widehat{f}[1-\varphi(\cdot/\delta)])^{\vee}(x)$$

But it's not clear to me at the moment how to estimate the $L^{1}$-norm of $f-f_{\delta}$. Any suggestions?

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Let $S_0\subset L^1_0(\mathbb{R}^n)$ be the set of Schwartz functions with mean $0$. $S_0$ is clearly dense (Edit: see the OP's comment below).

Thus it suffices to approximate $f\in S_0$.

Now let us try to finish your argument.

Compute $$f-f_\delta=\mathcal{F}^{-1} ( \hat{f}\cdot \varphi(\xi/\delta) )=f*\psi_\delta,$$

where $\psi(x)=\hat{\varphi}(-x)$ and $\psi_\delta(x)=\delta \psi(\delta x)$. $\psi$ is not compactly supported, but it is a Schwartz function. We have,

$$ \int_{\mathbb{R}^n} |f(x)-f_\delta(x)| dx = \int_{\mathbb{R}^n}\Big|\int_{\mathbb{R}^n} f(y)\psi_\delta(x-y) dy\Big| dx.$$

Using that $\int_{\mathbb{R}^n} f(y) dy=0$, this equals

$$\int_{\mathbb{R}^n}\Big|\int_{\mathbb{R}^n} f(y)(\psi_\delta(x-y)-\psi_\delta(x)) dy\Big| dx=\int_{\mathbb{R}^n}\Big|\int_{\mathbb{R}^n} f(y)(\psi(x-\delta y)-\psi(x)) dy\Big| dx$$

By the fundamental theorem of calculus,

$$\psi(x-\delta y)-\psi(x)=\int_0^\delta \nabla\psi(x-ty)\cdot y\,dt $$

(if $h(t)=\psi(x-t y)$, then $h'(t)=\nabla\psi(x-ty)\cdot y$ )

Let us plug this in and use the triangle ineqality, Cauchy-Schwarz and Fubini to estimate

$$\|f-f_\delta\|_1\le \int_0^\delta \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |\nabla\psi(x-ty)| dx |y| |f(y)| dy dt=\|\nabla\psi\|_1 \int_0^\delta \int_{\mathbb{R}^n} |y| |f(y)| dy dt= C\delta,$$

where $C=\|\nabla\psi\|_1 \int_{\mathbb{R}^n} |y| |f(y)| dy<\infty$ since $f$ is Schwartz.

Thus, $\|f_\delta-f\|_1\to 0$ as $\delta\to 0+$.

But since $f_\delta$ does not necessarily have compact Fourier support, this proves only that $\{f\in L^1(\mathbb{R}^n)\,:\,0\not\in\mathrm{supp}(\hat{f})\}$ is dense in $L^1_0(\mathbb{R}^n)$.

So the last step is to approximate by Schwartz functions with compact Fourier support. Allow me to omit this, because it will be very similar. We define $f_N=\mathcal{F}^{-1}(\hat{f} \hat{\varphi}(\xi/N) )$ with $\varphi$ a suitable bump function and show that $\|f_N-f\|_1\to 0$ as $N\to\infty$.


Addendum: To make this answer more complete, let me add the "alternative" approach via maximal ideals and Wiener's theorem.

We look at $L^1(\mathbb{R}^n)$ as an algebra wrt. convolution. For a subset $\mathcal{I}\subset L^1(\mathbb{R}^n)$ let us define $$\nu(\mathcal{I})=\{\xi\in\mathbb{R}^n\,:\,\hat{f}(\xi)=0\,\forall f\in\mathcal{I}\}.$$

Let us give Wiener's theorem in the following form.

Theorem. A proper closed ideal $\mathcal{I}\subset L^1(\mathbb{R}^n)$ is maximal if and only if $\nu(\mathcal{I})=\{\xi_0\}.$

In particular, all the maximal ideals are given by $\{f\in L^1(\mathbb{R}^n)\,:\,\hat{f}(\xi_0)=0\}$ for some $\xi_0\in\mathbb{R}^n$ (it is easy to see that these are indeed maximal ideals).

Now take $\mathcal{I}$ to be the $L^1$ closure of $\{f\in L^1_0(\mathbb{R}^n)\,:\,0\not\in\text{supp}(\hat{f})\,\text{compact}\}$. It is clear that $\nu(\mathcal{I})=\{0\}$, so by Wiener's theorem $\mathcal{I}=L^1_0(\mathbb{R}^n)$.

For the proof of that theorem see Corollary 4.67 and 4.69 in Folland - A course on abstract harmonic analysis. You will also find much more precise information on the structure of other closed ideals.

Finally note that this is not an honest alternative proof, because if you dig into the proof given in Folland's book, you will see that he hinges on a technical lemma that shows exactly the difficult part of our problem (see Lemma 4.57 and 4.58).

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  • $\begingroup$ At first read this looks, except for your argument about convolution with a mean zero mollifier. If $\int \phi=0$ and we set $\phi_{\delta}:=\delta^{-n}\phi(\cdot/\delta)$, then $\widehat{f\ast\phi_{\delta}}(\xi)=\widehat{f}(\xi)\widehat{\phi}(\delta\xi) \rightarrow \widehat{f}(\xi)\widehat{\phi}(0)=0$ as $\delta\rightarrow 0$. Thus, $f\ast\phi_{\delta}\not\rightarrow f$ in $L^{1}$ if $f\neq 0$. Instead, let $g\in\mathcal{S}$ be such that $\left\|f-g\right\|_{L^{1}}<\epsilon$, where $\int f=0$. Then $\left|\int g\right|<\epsilon$. If $\eta\in C_{c}^{\infty}$ with $\int\eta=1$, then... $\endgroup$ – Matt Rosenzweig Nov 16 '15 at 2:54
  • $\begingroup$ $h:=g-c\eta\in\mathcal{S}$, $\int h=0$, and $\left\|h-f\right\|_{L^{1}}\leq 2\epsilon$. $\endgroup$ – Matt Rosenzweig Nov 16 '15 at 2:57
  • $\begingroup$ Yes, thanks for catching that! I was using the word mollifier too carelessly. Your correction is very clever. $\endgroup$ – J.R. Nov 16 '15 at 10:06
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Hint: you can Apply stone weirstrass, differentiable functions in $L^1_0$ with compact support are dense in continuous functions in $L^1_0$ which are dense in $L^1_0$.

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  • $\begingroup$ I'm sorry, but would you please elaborate a bit more--it's late and my brain is very tired. I need $\text{supp}(\widehat{f})\subset\mathbb{R}^{n}\setminus\left\{0\right\}$ to be compact, and I can't have both $\text{supp}(f)$ and $\text{supp}(\widehat{f})$ compact (by the Paley-Wiener theorem), which is what I'm reading your hint to suggest I do. $\endgroup$ – Matt Rosenzweig Nov 12 '15 at 3:57

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