All Questions
7 questions
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Convolution of $\mathscr{F}\{ \log \}(x) * \mu$ with compactly supported measure $\mu$
As I read in this post the Fourier transform of $\psi(\lambda) = \log{|\lambda|}$ must be interpreted in distributional sense and it is given by:
$$\mathscr{F}\{\psi\}(x)=-2\pi \gamma \delta(x)-\pi \...
5
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0
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194
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When does the Fourier transform of a measure decay?
Let $\mu$ be a Borel measure on $\Bbb R^d$.
It is well known that $\mu= |f|dx$ with $f\in L^1(\Bbb R^d)$ then its Fourier transform satisfies
$$\widehat{\mu}(\xi)\to0,\qquad \xi\to\infty.$$
However if ...
3
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2
answers
590
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On the Fourier inversion formula
For a given function $f\in L^1(\mathbb{R})$, suppose that the
$$\check{f}(x)=\int_\mathbb{R} \hat{f}(\zeta)e^{2\pi i\zeta x}d\zeta$$
almost every where converges in $\mathbb{R}$. Then, can we say that
...
3
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1
answer
304
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Existence of probability measure on the circle with given Fourier coefficients
We say that a Hermitian symmetric (i.e., $f_{-n} = f_n^*$ for any $n \in \mathbb{Z})$ sequence $(f_n)_{n\in \mathbb{Z}}$ is positive-definite if, for any $N \geq 0$ and any $z_0 , \ldots, z_N \in \...
4
votes
1
answer
295
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Fourier coeffients of Cantor measure
For $0<\theta<\frac{1}{2}$, denote by $\mu_\theta$ the uniform Cantor measure with dissection ratio $\theta$. It is not hard to show that the Fourier–Stieltjes transform of $\mu_\theta$ is
$$
\...
5
votes
0
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286
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$f, \hat{f} \in L^{p}\cap L^{\infty} \implies f\in B(\mathbb R)$ (algebra of Fourier- Stieltjes transforms )?
For a bounded complex Borel measure $\mu$ on $\mathbb R$, we define, its Fourier-Stieltjes transform, $\hat{\mu}(y)= \int_{\mathbb R} e^{-2\pi ix\cdot y} d\mu(x); (y\in \mathbb R).$
Let $1\leq p \leq ...
1
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1
answer
367
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How to get Fourier–Stieltjes transform on $\mathbb R$ from the nice function on $\mathbb T$ (periodic on $\mathbb R$)? [closed]
We put, $M(\mathbb R)= $The set of bounded complex Borel measure $\mu$ on $\mathbb R$ and for $\mu \in M(\mathbb R)$, we define $||\mu||:= |\mu| (\mathbb R) = \text {total variation of } \ \mu $; and
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