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I have redone this question:

On $\mathbb R^n$ the Carleson Operator if defined by $$Cf(x) = \sup_{R>0} \left \vert \int_{B_R(0)} e^{2\pi i x\cdot \xi} \widehat{f}(\xi) d \xi \right \vert. $$ (In the previous version I had an incorrect version of this written down which lead to some stupid conclusions).

• For $n=1$ Carleson proved that this operator is strong (p,p) which is equivalent $L^p(\mathbb R)$ convergence of $S_Rf(x)=\int_{B_R(0)} e^{2\pi i x\cdot \xi}\widehat{f}(\xi)d\xi\to f(x)$ as $R\to \infty$. The modern proof uses boundedness of the Hilbert-Transform and the ``translation trick'' $$\chi_{[-r,r]}(\xi)\widehat{f}(\xi) = \widehat{[\frac{i}{2}(m_{-r}H m_r - m_r H m_{-r})f] }(\xi).$$
• For $n>1$ Fefferman proved (using a Becosivitch set) that the transform $T$ defined by $$ \widehat{Tf}(\xi) = \chi_{B_1(0)}(\xi) \widehat{f}(\xi)$$ is unbounded in $L^p$ for $p\neq 2$. (There is no Hilbert Transform in higher dimensions, these become the Reisz Transform and the ``translation tricks'' to express $S_R$ in terms of the Hilbert Transform don't work).
• The Bochner-Reisz conjecture is about boundedness of smoothened characteristic functions and an MO question about it can be found here Recent progress on Bochner-Riesz conjecture

I was originally thinking that $Cf(x)$ could be defined easily for locally compact abelian groups and compact groups and that perhaps a ``translation trick'' existed in some generality. This appears to not be the case for two reasons

1. The second bullet shows that it is not true in $\mathbb{R}^n$ for $n>1$.
2. The third bullet shows you that multiplier operator in $\mathbb{R}^n$ are delicate which makes the definition of a Carleson operator for general abelian groups difficult.

Is the operator $$Cf(x) = \sup_{n\geq 0} \left \vert \int_{\frac{1}{p^n}\mathbb{Z}_p} e^{2\pi i \lbrace x \xi\rbrace } \widehat{f}(\xi) d \mu(\xi) \right \vert. $$ bounded as an operator $L^r(\mathbb Q_p) \to L^r(\mathbb Q_p)$. (I think one can define a similar operator on the Adeles no?). How does smoothness of multipliers effect the problem in more general groups?

I'll make it a community wiki so people can fix it if it has errors.

I have redone this question:

On $\mathbb R^n$ the Carleson Operator if defined by $$Cf(x) = \sup_{R>0} \left \vert \int_{B_R(0)} e^{2\pi i x\cdot \xi} \widehat{f}(\xi) d \xi \right \vert. $$ (In the previous version I had an incorrect version of this written down which lead to some stupid conclusions).

• For $n=1$ Carleson proved that this operator is strong (p,p) which is equivalent $L^p(\mathbb R)$ convergence of $S_Rf(x)=\int_{B_R(0)} e^{2\pi i x\cdot \xi}\widehat{f}(\xi)d\xi\to f(x)$ as $R\to \infty$. The modern proof uses boundedness of the Hilbert-Transform and the ``translation trick'' $$\chi_{[-r,r]}(\xi)\widehat{f}(\xi) = \widehat{[\frac{i}{2}(m_{-r}H m_r - m_r H m_{-r})f] }(\xi).$$
• For $n>1$ Fefferman proved (using a Becosivitch set) that the transform $T$ defined by $$ \widehat{Tf}(\xi) = \chi_{B_1(0)}(\xi) \widehat{f}(\xi)$$ is unbounded in $L^p$ for $p\neq 2$. (There is no Hilbert Transform in higher dimensions, these become the Reisz Transform and the ``translation tricks'' to express $S_R$ in terms of the Hilbert Transform don't work).
• The Bochner-Reisz conjecture is about boundedness of smoothened characteristic functions and an MO question about it can be found here Recent progress on Bochner-Riesz conjecture

I was originally thinking that $Cf(x)$ could be defined easily for locally compact abelian groups and compact groups and that perhaps a ``translation trick'' existed in some generality. This appears to not be the case for two reasons

1. The second bullet shows that it is not true in $\mathbb{R}^n$ for $n>1$.
2. The third bullet shows you that multiplier operator in $\mathbb{R}^n$ are delicate which makes the definition of a Carleson operator for general abelian groups difficult.

Is the operator $$Cf(x) = \sup_{n\geq 0} \left \vert \int_{\frac{1}{p^n}\mathbb{Z}_p} e^{2\pi i \lbrace x \xi\rbrace } \widehat{f}(\xi) d \mu(\xi) \right \vert. $$ bounded as an operator $L^r(\mathbb Q_p) \to L^r(\mathbb Q_p)$. (I think one can define a similar operator on the Adeles no?). How does smoothness of multipliers effect the problem in more general groups?

I'll make it a community wiki so people can fix it if it has errors.

I have redone this question:

On $\mathbb R^n$ the Carleson Operator if defined by $$Cf(x) = \sup_{R>0} \left \vert \int_{B_R(0)} e^{2\pi i x\cdot \xi} \widehat{f}(\xi) d \xi \right \vert. $$ (In the previous version I had an incorrect version of this written down which lead to some stupid conclusions).

• For $n=1$ Carleson proved that this operator is strong (p,p) which is equivalent $L^p(\mathbb R)$ convergence of $S_Rf(x)=\int_{B_R(0)} e^{2\pi i x\cdot \xi}\widehat{f}(\xi)d\xi\to f(x)$ as $R\to \infty$. The modern proof uses boundedness of the Hilbert-Transform and the ``translation trick'' $$\chi_{[-r,r]}(\xi)\widehat{f}(\xi) = \widehat{[\frac{i}{2}(m_{-r}H m_r - m_r H m_{-r})f] }(\xi).$$
• For $n>1$ Fefferman proved (using a Becosivitch set) that the transform $T$ defined by $$ \widehat{Tf}(\xi) = \chi_{B_1(0)}(\xi) \widehat{f}(\xi)$$ is unbounded in $L^p$. (There is no Hilbert Transform in higher dimensions, these become the Reisz Transform and the ``translation tricks'' to express $S_R$ in terms of the Hilbert Transform don't work).
• The Bochner-Reisz conjecture is about boundedness of smoothened characteristic functions and an MO question about it can be found here Recent progress on Bochner-Riesz conjecture

I was originally thinking that $Cf(x)$ could be defined easily for locally compact abelian groups and compact groups and that perhaps a ``translation trick'' existed in some generality. This appears to not be the case for two reasons

1. The second bullet shows that it is not true in $\mathbb{R}^n$ for $n>1$.
2. The third bullet shows you that multiplier operator in $\mathbb{R}^n$ are delicate which makes the definition of a Carleson operator for general abelian groups difficult.

Is the operator $$Cf(x) = \sup_{n\geq 0} \left \vert \int_{\frac{1}{p^n}\mathbb{Z}_p} e^{2\pi i \lbrace x \xi\rbrace } \widehat{f}(\xi) d \mu(\xi) \right \vert. $$ bounded as an operator $L^r(\mathbb Q_p) \to L^r(\mathbb Q_p)$. (I think one can define a similar operator on the Adeles no?). How does smoothness of multipliers effect the problem in more general groups?

I'll make it a community wiki so people can fix it if it has errors.

I have redone this question:

On $\mathbb R^n$ the Carleson Operator if defined by $$Cf(x) = \sup_{R>0} \left \vert \int_{B_R(0)} e^{2\pi i x\cdot \xi} \widehat{f}(\xi) d \xi \right \vert. $$ (In the previous version I had an incorrect version of this written down which lead to some stupid conclusions).

• For $n=1$ Carleson proved that this operator is strong (p,p) which is equivalent $L^p(\mathbb R)$ convergence of $S_Rf(x)=\int_{B_R(0)} e^{2\pi i x\cdot \xi}\widehat{f}(\xi)d\xi\to f(x)$ as $R\to \infty$. The modern proof uses boundedness of the Hilbert-Transform and the ``translation trick'' $$\chi_{[-r,r]}(\xi)\widehat{f}(\xi) = \widehat{[\frac{i}{2}(m_{-r}H m_r - m_r H m_{-r})f] }(\xi).$$
• For $n>1$ Fefferman proved (using a Becosivitch set) that the transform $T$ defined by $$ \widehat{Tf}(\xi) = \chi_{B_1(0)}(\xi) \widehat{f}(\xi)$$ is unbounded in $L^p$ for $p\neq 2$. (There is no Hilbert Transform in higher dimensions, these become the Reisz Transform and the ``translation tricks'' to express $S_R$ in terms of the Hilbert Transform don't work).
• The Bochner-Reisz conjecture is about boundedness of smoothened characteristic functions and an MO question about it can be found here Recent progress on Bochner-Riesz conjecture

I was originally thinking that $Cf(x)$ could be defined easily for locally compact abelian groups and compact groups and that perhaps a ``translation trick'' existed in some generality. This appears to not be the case for two reasons

1. The second bullet shows that it is not true in $\mathbb{R}^n$ for $n>1$.
2. The third bullet shows you that multiplier operator in $\mathbb{R}^n$ are delicate which makes the definition of a Carleson operator for general abelian groups difficult.

Is the operator $$Cf(x) = \sup_{n\geq 0} \left \vert \int_{\frac{1}{p^n}\mathbb{Z}_p} e^{2\pi i \lbrace x \xi\rbrace } \widehat{f}(\xi) d \mu(\xi) \right \vert. $$ bounded as an operator $L^r(\mathbb Q_p) \to L^r(\mathbb Q_p)$. (I think one can define a similar operator on the Adeles no?). How does smoothness of multipliers effect the problem in more general groups?

I'll make it a community wiki so people can fix it if it has errors.

On RR^n It is well known that the fourier transform is strong (p,p) for p in (1,2): There exists some $C>0$ such that for all $f\in L^p(\mathbb{R}^n)$ we have $$\Vert f \Vert_p \leq C \Vert \widehat{f} \Vert_p$$

-Where can I find a proof of this theorem over for nonabelian compact groups? Is it even true?

-Where can i find a proof of this theorem in the case of locally compact abelian groups? Is it even true?

Here is a special case: Let $\mathbb{A}$ be the group of Adeles, if $f\in L^p(\mathbb{A})$ is it true that $\widehat{f}\in L^p(\widehat{\mathbb{A}}).$

I have redone this question:

On $\mathbb R^n$ the Carleson Operator if defined by $$Cf(x) = \sup_{R>0} \left \vert \int_{B_R(0)} e^{2\pi i x\cdot \xi} \widehat{f}(\xi) d \xi \right \vert. $$ (In the previous version I had an incorrect version of this written down which lead to some stupid conclusions).

• For $n=1$ Carleson proved that this operator is strong (p,p) which is equivalent $L^p(\mathbb R)$ convergence of $S_Rf(x)=\int_{B_R(0)} e^{2\pi i x\cdot \xi}\widehat{f}(\xi)d\xi\to f(x)$ as $R\to \infty$. The modern proof uses boundedness of the Hilbert-Transform and the ``translation trick'' $$\chi_{[-r,r]}(\xi)\widehat{f}(\xi) = \widehat{[\frac{i}{2}(m_{-r}H m_r - m_r H m_{-r})f] }(\xi).$$
• For $n>1$ Fefferman proved (using a Becosivitch set) that the transform $T$ defined by $$ \widehat{Tf}(\xi) = \chi_{B_1(0)}(\xi) \widehat{f}(\xi)$$ is unbounded in $L^p$. (There is no Hilbert Transform in higher dimensions, these become the Reisz Transform and the ``translation tricks'' to express $S_R$ in terms of the Hilbert Transform don't work).
• The Bochner-Reisz conjecture is about boundedness of smoothened characteristic functions and an MO question about it can be found here Recent progress on Bochner-Riesz conjecture

I was originally thinking that $Cf(x)$ could be defined easily for locally compact abelian groups and compact groups and that perhaps a ``translation trick'' existed in some generality. This appears to not be the case for two reasons

1. The second bullet shows that it is not true in $\mathbb{R}^n$ for $n>1$.
2. The third bullet shows you that multiplier operator in $\mathbb{R}^n$ are delicate which makes the definition of a Carleson operator for general abelian groups difficult.

Is the operator $$Cf(x) = \sup_{n\geq 0} \left \vert \int_{\frac{1}{p^n}\mathbb{Z}_p} e^{2\pi i \lbrace x \xi\rbrace } \widehat{f}(\xi) d \mu(\xi) \right \vert. $$ bounded as an operator $L^r(\mathbb Q_p) \to L^r(\mathbb Q_p)$. (I think one can define a similar operator on the Adeles no?). How does smoothness of multipliers effect the problem in more general groups?

I'll make it a community wiki so people can fix it if it has errors.