## New answers tagged fourier-analysis

1

No, it does not. It implies convergence in the $L^2$-norm.
If you assume $C^\infty$ functions, it converges in the $C^\infty$-sense. Check Harish-Chandra Discretes Series Representations II. HC does actually consider non--compact groups $G$ there and decomposes $C^\infty(G)$ as $K-K$-bimdule for $K$ compact.
Again you want to work with smooth functions, ...

25

A class of maps including both continuous and $H^{1/2}$, where an extension is available, is Vanishing Mean Oscillation, VMO. This has been treated by several authors starting I think with Haïm Brezis. You can find quite a lot googling "degree theory for VMO maps".

1

It is intuitively clear that for the simplest case $G = \delta_\theta$ and $G' = \delta_{\theta'}$ in $\mathbb{R}^n$ we have $W_2(G,G') = W_2(f_G,f_{G'})$ for all $f$.
However, in general it is harder to give useful bounds:
Example
Let $X = \Theta = \mathbb{R}$ and $f(x) = \frac12\chi_{[-1,1]}(x)$. Take
$n \in \mathbb{N}$ and define
$$
G = \sum_{i=1}^n ...

3

The answer is no as the following example shows: Let
$$
F(t)= \begin{cases} a_nt^{-\frac 12+b_n},& t\in [t_{n+1},t_n),\quad n\geq 1,\\
0,& \text{otherwise,}
\end{cases}
$$
where $0< t_{n+1}< t_n<1$, $n\geq 1$. Then $F\in L^2(0,1)$ provided
$$
\sum_{n=1}^\infty ...

12

Your first two relations follow quickly from the asymptotic behavior of the functions $s\mapsto\zeta(s)$ and $s\mapsto d^{1-s}$ as $s\to 1$. Let me demonstrate this for the second relation for $n>1$ (note that it fails for $n=1$).
For $|s-1|<1$ we have
$$\zeta(s)=1/(s-1)+O(1)\quad\text{and}\quad d^{1-s}=1-(s-1)\log d+O_d((s-1)^2),$$ hence
$$ ...

4

(1) The space is obviously complete since you defined the norm in such a way that $\mathcal{F}$ is an isometry $L^1 \cong \mathcal{F}L^1$ (maybe an isometry up to constants depending on your definition of the fourier transform). Consequently the answer to (3) is that the closed sets of $\mathcal{F}L^1$ are exactly the fourier transforms of closed sets of ...

1

The answer in the second case is even easier. In order to avoid circularity, define the Stieltjes transform of a measure only on the complex plane minus the real axis . Then the measure has support in a compact interval if and only if this transform has an analytic continuous over the real axis outside of this interval. This has suitable generalisations ...

1

The question is whether the trigonometric functions form an unconditional basis for $L^p$. Unfortunately, I have no access to the literature at the moment but if memory serves me well the answer to the basis problem is negative (and so positive for your problem) and can be found in "Classical Banach spaces" by Lindenstrauss and Tzafriri (probably vol. 2).
...

7

There are two recent and very comprehensive monographs on this subject:
Groemer: Geometric applications of Fourier series
and
Koldobsky: Fourier analysis in convex geometry
which might interest you.

9

see: Harmonic representation of closed curves (1992)
The centroid is given by the zeroth order Fourier coefficient, the two first harmonics give the orientation and scale of an elliptic approximation to the curve.

1

I found a related paper myself:
http://link.springer.com/article/10.1007%2Fs13324-012-0025-6

1

In general, clearly no.
Let $f$ be a fixed function of your class with $f(0) = 1$. Let $f_K(x) = \frac{1}{K}f(Kx)$. Then $f_K$ is an admissible sequence by your hypothesis. However
$$\int_0^\infty \cos(Kx) f_K(x) \mathrm{d}x = \int_0^\infty \frac{1}{K^2} \cos(y) f(y) \mathrm{d}y $$
and similarly
$$\int_0^\infty \sin(Kx) f_K(x) \mathrm{d}x = ...

0

I suggest you read Charlie Epstein's analysis.

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