# Tagged Questions

Harmonic analysis is the generalisation of Fourier analysis. Use this tag for analysis on locally compact groups (e.g. Pontryagin duality), eigenvalues of the Laplacian on compact manifolds or graphs, and the abstract study of Fourier transform on Euclidean spaces (singular integrals, Littlewood-...

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### Riesz transform does not preserve continuity

I've read somewhere that the Riesz operator $R_j$ defined by $$R_j f(t) := c(n) \, \text{pv} \int_{\mathbb{R}^n} \frac{x_j}{|x|^{n+1}} f(t-x) \, dx$$doesn't preserve the continuity, but I can't ...
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### Fourier transform identity on $L^{p}(\mathbb{T})$

Let $p\in (1,2]$. I want to prove there exists $C$ such that for $f\in L^{p}(\mathbb{T})$ we have $$\sum_{n\in\mathbb{Z}}|\hat{f}(n)|^{p}|n|^{p-2}\le C\|f\|^{p}_{L^{p}(\mathbb{T})}$$ If $p=1$, it's ...
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### General derivative of composition

We know that if $G\in \mathcal{C}^1(\mathbb{R})$ such that $G(0)=0$ and if $u \in W^{1,p}(I)$, then $G(u) \in W^{1,p}(I)$ and $(G(u))' = G'(u) u'$. Do we have a similar result for $W^{s,p}(I)$ with ...
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### Proof of existence of a non-trivial character on a local field of positive characteristic

Let $\mathbb{K}$ be a local field. Definition of local field: Let $\mathbb{K}$ be a field and a topological space. Then $\mathbb{K}$ is called a locally compact field if both $\mathbb{K}^+$ ...
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### where can I find the proof of this theorem of wavelet frames?

Let $\mathbb{K}$ be a local field. Definition of local field: Let $\mathbb{K}$ be a field and a topological space. Then $\mathbb{K}$ is called a locally compact field if both $\mathbb{K}^+$ ...
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### Prove that the Pontryagin dual of $\mathbb{R}$ is $\mathbb{R}$.

From Wikipedia: the group of real numbers $\mathbb{R}$, is isomorphic to its own dual; the characters on $\mathbb{R}$ are of the form $r \to e^{i\theta r}$. How can I prove this assertion?
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### Direct sum decomposition of $L^2(\mathbb{R})$ using Fourier Transform

Let $L_+^2(\mathbb{R})=\{f\in L^2(\mathbb{R}):supp \hat{f}\subset\mathbb{R^+}\}$ and $L_-^2(\mathbb{R})=\{f\in L^2(\mathbb{R}):supp \hat{f}\subset\mathbb{R^-}\}$, where $\hat{f}$ denotes the Fourier ...
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### Proof of Howe-Moore Property for SL(n,R)

On page 210 of Howe and Tan's Non-Abelian Harmonic Analysis there is the following proposition and proof: Let $(\rho, V)$ be a unitary representation of $SL(n,\mathbb{R})$. The following are ...
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### Coadjoint representation $Ad^*$ of the Heisenberg group

The Heisenberg group $H^3$ is the set $\mathbb C\times \mathbb R$ endowed with the group law $$(z,t)\cdot(w,s) =\left (z+w, \,t+s+\tfrac{1}{2}\Im m(z \bar{w})\right).$$ For $z=x+ i y \in \mathbb C$ ...
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### The $Ad^*$ $G$-Orbits in $\mathcal{g}*$ (for $G$ being the Heisenberg group)

I did not understand why, the $Ad^*$ $G$-Orbits in $\mathcal{g}*$ (for $G$ being the Heisenberg group and $\mathcal{g}*$ is the dual space of its Lie algebra $\mathcal g$) are given by i) The ...
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### Bound a complex exponential sum when we can only estimate its argument

Suppose we have a function $f : \mathbb{N_0} \to \mathbb{R}$ for which we can give an estimate of its values, and say its values $f(n)$ are roughly uniformly distributed for $n$ in some range $[1,N]$. ...
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### references for basic course in abstract harmonic analysis

I am currently studying a basic course in abstract harmonic analysis. I am really finding difficulty in understanding the concepts of representation of algebras, groups, etc. Moreover, any of the ...
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### Example of a linear functional, but not a distribution

I'm looking for an example of a linear functional $u: C_c^\infty(\Omega) \to \mathbb C$ ($\Omega \subset \mathbb R^n$ open), which is not a distribution. I could not find anything... I thought of ...
### Decay property of oscillatory integrals in $\mathbb{R}^n$
We know that an oscillatory integral in $n$ dimensions is an integral of the form \begin{equation*} I(\lambda)=\int_{\mathbb{R}^n}e^{i\lambda\phi(x)}f(x)dx \end{equation*} where $\phi\in C^{\infty}$ ...
### Concluding that if $f\in L^{1}_{\text{loc}}(\mathbb{R}^{n})$ and $\mathcal{M}(f)\in L^{1}(\mathbb{R}^{n})$ then $f=0$ a.e.
Let $\mu$ be a Lebesgue measure, $\mathcal{M}$ be the Hardy-Littlewood Centered Maximal Function, and the rest as in the title. Then to this end I define $f_{R}(x):=f(x)\chi_{|x|\le R}$ where $\chi$ ...