In chapter 4, sub-chapter 4.5 (closed ideals) in the book there is the following Theorem:
Let $G=\mathbb{R}^{n}$ with $n\geq3$, and let $S$ be the unit sphere in $\mathbb{R}^{n}$. There is a closed ideal $\mathcal{J}$ in $L^{1}\left(\mathbb{R}^{n}\right)$ such that $\nu\left(\mathcal{J}\right)=S$, but $\mathcal{J}\neq\iota\left(S\right)=\iota\left(\nu\left(\mathcal{J}\right)\right).$
The proof go as follow:
- If $f$ and $x_{1}f$ are in $L^{1}\left(\mathbb{R}^{n}\right)$ then: $-2\pi i\cdot\widehat{\left(x_{1}f\right)}\left(\xi\right)\}{=}\int\left(-2\pi ix_{1}e^{-2\pi i\xi\cdot x}\right)f\left(x\right)dx\}{=}\int\frac{\partial e^{-2\pi i\xi\cdot x}}{\partial\xi_{1}}\cdot f\left(x\right)dx\}{=}\frac{\partial\widehat{f}}{\partial\xi_{1}}\left(\xi\right)$ Hence, $\frac{\partial f}{\partial\xi_{1}}$ exists and continous.
- Let $I$ be the set of all $f\in L^{1}\left(\mathbb{R}^{n}\right)$ such that $x_{1}f\in L^{1}\left(\mathbb{R}^{n}\right)$ and $\widehat{f}|_{S}\stackrel{\left(a\right)}{=}\left(\frac{\partial\widehat{f}}{\partial\xi_{1}}\right)|_{S}\equiv0$, and let $J$ be the closure of $I$ in $L^{1}\left(\mathbb{R}^{n}\right)$. Since $\widehat{\left(\underbrace{L_{y}f}_{L_{y}f\left(x\right)=f\left(y^{-1}x\right)}\right)}\left(\xi\right)=\overline{\left\langle y,\xi\right\rangle }\cdot\widehat{f}\left(\xi\right)=\overline{\xi\left(y\right)}\widehat{f}\left(\xi\right)=e^{-2\pi iy\xi}\cdot\widehat{f}\left(\xi\right)$, $I$ is translation- invariant. and So $J$ is closed ideal by Theorem 2.45.
- Moreover, $\left\{ \widehat{f}\,|\,f\in I\right\} $ contains all $\phi\in C_{c}^{\infty}\left(\mathbb{R}^{n}\right)$ such that $\left(supp\phi\right)\cap S=\emptyset$. Therefore $\nu\left(\mathcal{J}\right)=S$. To show that $\mathcal{J\neq\iota}\left(S\right)$ we shall exhibit a bounded linear functional on $L^{1}$ that annihilates on $J$ but not on $\iota\left(S\right)$.
Then they prove the following lemma (4.57): Let $\mu$ denote the surface measure on $S$, Then $\left|\widehat{\mu}\left(x\right)\right|\leq C\left(1+\left|x\right|\right)^{-1}$ for some $C>0$.
And then they continue with the proof:
- Let $\phi\left(x\right)=-2\pi ix_{1}\widehat{\mu}\left(x\right)$, then by lemma 4.57 $\phi\in L^{\infty}$. If $f\in L^{1}$ and $x_{1}f\in L^{1}$ then: $\int\phi f\left(x\right)dx=-2\pi i\int\left(\int x_{1}e^{-2\pi ix\cdot\xi}f\left(x\right)d\mu\left(\xi\right)\right)dx=-2\pi i\int\left(\int x_{1}e^{-2\pi ix\cdot\xi}f\left(x\right)dx\right)d\mu\left(\xi\right)=\int\frac{\partial\widehat{f}}{\partial\xi_{1}}\left(\xi\right)d\mu\left(\xi\right)$.
- From this we deduce two things. On the one hand, if $f\in I$ then $\int\phi f=0$. and hence the same if $f\in J$
- On the other hand, if $f$ is the inverse Fourier transform of a function in $C_{c}^{\infty}\left(\mathbb{R}^{n}\right)$, that equals $\xi_{1}\left(\left|\xi\right|^{2}-1\right)$ when $\left|\xi\right|\leq2$ then we have $\widehat{f}=0$ on S and $\frac{\partial\widehat{f}}{\partial\xi_{1}}=\left|\xi\right|^{2}-1+2\xi_{1}^{2}=2\xi_{1}^{2}$, so $\int\phi f\left(x\right)dx=\int\frac{\partial\widehat{f}}{\partial\xi_{1}}\left(\xi\right)d\mu\left(\xi\right)=\int2\xi_{1}^{2}d\mu\left(\xi\right)>0$ , therefore,$f\in\iota\left(S\right)$ but $f\notin J$.
This is the proof, and here are my questions:
A) Regarding line 3. Why is this true? It is pretty obvious that $\phi|_{S}\equiv0$ and $\frac{\partial\phi}{\partial\xi_{1}}\equiv0$, but why the inverse Fourier of $\phi$ (denote by g) is a function such that $x_{1}g$ is in $L^{1}\left(\mathbb{R}^{n}\right)$?
B) Still regarding line 3, why $\nu\left(\mathcal{J}\right)=S$? The only thing I know is that every function in $I$ is zero on S, I don't even sure why function on $J/I$ is zero on S.
C) Line 5. Why is it also true that if $f\in J$ then $\int\phi f=0$? If I take sequence that lies on I and convergence to f (in L1 norm), I don't have uniformly convergence.
D)line 6, Why is it important to take $\left|\xi\right|\leq2$? And more important question, I understand that in this case, $f$ can't be in $J$ becasue if it is then $\int\phi f=0$ by contradiction. But Why I need the lemma? Why is it important to say the fact that $\phi\in L^{\infty}$? I guess it also have a relation to the statment in line 3 which I don't understand.
E) More general question. What is the intuition here? why in line 1 we suddenly talking about the partial derivative and want to show it existence and continuous?
Even if you can give me partial answer it will be great. Thank you very much in advance.