## New answers tagged fourier-analysis

3

I think the legend is that an official introduction to $BMO$ was given in the paper "Of Functions of Bounded Mean Oscillation" by F. John and L. Nirenberg , but the initial mention was in the paper by F. John "Rotation and Strain". F. John was looking at mappings and rotations. For example, the class of mappings $f(x)$ satisfying the following:
...

1

The following papers may help:
"Besov Spaces on Domains in $\mathbb{R}^d$" by R. Devore and R. Sharpley
This paper is for Besov spaces $B^{\alpha}_{q}(L_p(\Omega)),~p,q,\alpha \in (0,\infty )$ on domains $\Omega\subset \mathbb{R}^d$
http://www.ams.org/journals/tran/1993-335-02/S0002-9947-1993-1152321-6/S0002-9947-1993-1152321-6.pdf
"Elliptic and ...

0

I got the wrong conlcusion. There must be some mistakes. I will edit it correctly if I find the error.
By fourier inversion formula and Fubini theorem,
$$f(x)=\int \widehat{f}(\xi)e^{2\pi ix \cdot \xi}d\xi = \int_{|t| \leq \frac{1}{R}}\int_{S+te_{n}} \widehat{f}(\xi)e^{2\pi ix \cdot \xi}d\sigma(\xi)dt=\int_{|t| \leq ...

0

I'm answering your second question.
If $f(x)=\int_{-M}^M\hat f(w) e^{2\pi i wx}dw$ in $L^2(R)$, with $f\in L^2(R),$ we have
$\vert f'(x)\vert=\vert\int_{-M}^{M} 2\pi iw\hat f(w) e^{2\pi i wx}dw\vert\leq 2\pi \Vert f\Vert_2 \sqrt{2M^3/3}$
If you take now
$$f_n(w)={{\sqrt 3}\over {\sqrt 2 n^{3/2}}}({{ -n \cos(2 n \pi x)}\over{\pi x}} + {{\sin(2 n \pi ...

1

Let $\mathbb{H}$ denote the field of quaternions and $\mathbb{S}$ denote the unit sphere of imaginary quaternions. We have a Riemann mapping theorem for axially symmetric sets inside the field of quaternions $\mathbb{H}$. The set $\Omega$ is axially symmetric if the set $\{ x+Jy~|~J\in\mathbb{S} \},~J\in\mathbb{S}$ are contained in $\Omega$ (we have rotation ...

5

The answer is "yes". We have
$$f(x)=\int_{-\omega}^\omega g(t)e^{2\pi itx}dt,$$
so
$$|f'(x)|^2\leq\left(2\pi\int_{-\omega}^\omega |t||g(t)|dt\right)^2\leq\frac{8\pi^2\omega^3}{3}\| g\|_2,$$
by Cauchy-Bounyakovski-Schwarz inequality. It remains to notice that $\| g\|_2=\| f\|_2$ according to Parseval . Equality when $g(t)=t$.

0

I think the sequence does not converge in $L^2.$
We have
$$\vert \hat f_{4m-1}(x)\vert= e^{i2xm}\hat f_{4m-1}(x)= {1\over{(2x\sin(x/2))^{4m}\sum_{l\in Z} {1\over{(x+2\pi l)^{8m}}}}}$$
Let us suppose $ \lim \hat f_m = f ,\ \lim \vert \hat f_m\vert = \vert f \vert$ in $L^2$.
If we define $g_m(x)=e^{2imx}f(x),$
we have
$$\lim\Vert g_m-\vert \hat ...

0

Hint at a partial answer to the revised question, extended to $p\in (0,\infty)$: for $\frac12<s<1$, $p\geq 1$ is necessary and sufficient.
The "if" part is straightforward using the double integral definition of $H^s$. The "only if" part will better be dealt with using Fourier series $i\ \sum \pm|n|^{-\alpha} e^{int}$ ($\pm :=$ sign of $n$) or $\sum ...

1

The real Hardy spaces are equivalent to $L^p(\mathbb{R}^n)$ space when $p>1$, and they are much easier to to use than $L^p(\mathbb{R}^n)$ when $p\leq 1$. Since $H^p(\mathbb{R}^n)$ has a maximal function and singular integral generalisation, $H^p(\mathbb{R}^n)$ gives us an extension of maximal function/singular integral results to $p\leq 1$, when they were ...

2

You ask for a Fourier reconstruction mechanism that avoids the Gibbs phenomenon.
One strategy in this direction was developed by David Gottlieb and collaborators, in a series of papers entitled ￼"On the Gibbs phenomenon: Recovering exponential accuracy from the Fourier partial sum of a nonperiodic analytic function" (part I, part II, part III, part IV, part ...

2

assuming $c>0$, this limit is dominated by the upper end of the integration interval, so $f(\omega)=a/2$ independent of $\omega$; as a check, try $\gamma=3$, when a closed form expression exists,
$$f(\omega) = \lim_\limits{R \to \infty} \frac{a}{R^2}\left( 1 - (1+bR)e^{-cR}\right) \int_{r=0}^{R}{re^{i \omega r^{-3}}} \mathrm{d}r$$
$$=\lim_\limits{R \to ...

2

Here is an example where $T(f,g)\notin l^2$ while $m\in L^2$ and $f=g\in l^2$:
Let $a(n)=|n|^{-1+s}$ with $\frac38<s<\frac12$ (so that $a\in l^2$), $m(\xi,\eta)=\hat{a}(\xi)\hat{a}(\eta)$. Take also $f=g=a$.
One has $\hat{a}(\xi)\propto |\xi|^{-s}$ near $0$. Then $\hat{a}(\xi)\propto |\xi|^{-2s}$, so that $(a * a)(n)\propto n^{-1+2s}$ for large $n$. ...

0

1') As stated before, $\epsilon$-thin sets need not have finite measure, for instance, if $E=F = \bigcup\limits_{n \in \mathbb{Z}}[n-\frac{1}{200|n|},n+\frac{1}{200|n|}]$, then $(E,F)$ is strongly annihilating.
2) Take $E = F = \bigcup\limits_{n \in \mathbb{Z}} [n-\frac{1}{3},n+\frac{1}{3}]$. Their complements are clearly of infinite measure, yet notice ...

1

Let $\frac12<s<1$, let $u\in H^s(\mathbb R)$, then also $u\in C_0$, and let $f$ be Lipschitz on $[-||u||_\infty \ ,+||u||_\infty]$ with $|f(x)-f(y)|\leq M|x-y|$. Then $$\int\int \frac{|f(u(t))-f(u(t'))|^2}{|t-t'|^{1+2s}}dt\ dt'\leq M^2\int\int\frac{|u(t)-u(t')|^2}{|t-t'|^{1+2s}}$$With $f(0)=0$ you also have $|f(u)|\leq M|u|$ so that $f(u)\in L^2$.
...

3

I think all your questions are answered by the following calculation (assume $m\geq 1$ and $\Re(s)>1$):
$$ \sum_{c=1}^\infty\frac{r_m(c)}{c^{2s}} = \sum_{c=1}^\infty\frac{1}{c^{2s}}\sum_{\substack{\text{$d$ mod $c$}\\{(d,c)=1}}}e\left(m\frac{d}{c}\right) = \sum_{c=1}^\infty\frac{1}{c^{2s}}\sum_{\text{$d$ mod ...

0

Let me expand my comment into an answer.
Let $s>\frac12$ and pick any $r\in(\frac12,s)$.
Interpolating between Sobolev spaces gives $\|u\|_{H^r_P}\lesssim\|u\|_{H^0_P}^\delta\|u\|_{H^s_P}^{1-\delta}$ for some $\delta>0$.
As was shown to you in the MSE answer to the question, $\|u\|_{L^\infty}\lesssim\|u\|_{H^r_P}$.
Putting these estimates together ...

0

This may be too crude for what you need, but you could expand y in a geometric series:
$$y=\sum_{n=1}^\infty (-1)^{n+1}({u\over b})^n.$$
Clearly every finite truncation of the series has finite bandwidth. Then you can combine this with an estimate of the error of the series.

5

Prof. Tao's answer is excellent. I also found two research papers answering the question so I list them below as complementary reference:
G.I.Arkhipov and K.I.Oskolkov, On a special trigonometric series and its applications, 1989 Math. USSR Sb. 62 145
Link to the article:
http://iopscience.iop.org/0025-5734/62/1/A10
See Theorem 1.
E.S.Stein and S.Wainger, ...

13

The answer is yes. Fix $x,y$, and write $e(\alpha) := e^{2\pi i \alpha}$.
Using a Littlewood-Paley partition of unity and the triangle inequality, we may bound
$$ |f(x,y)| \leq \sum_N a_N$$
where $N$ ranges over powers of two,
$$ a_N := \left|\sum_{n \in {\bf Z} \backslash 0} \psi( \frac{n}{N}) \frac{1}{n} e(x n + yn^2)\right|, $$
and $\psi$ is a suitable ...

-4

As of now this answer stands incorrect, as $I_{\omega}(t)$ is a bounded function for any given $\omega$, but the expression I derived is not a bounded function of $t$ and blows up at jump discontinuties. I guess my calculations are incorrect and will update soon (possibly missing a negative sign .
Fixed (finally after thorough check and fixing errors)
...

5

Observe that if $n$ and $m$ are relatively prime, $\gamma_{nm}\leq \gamma_n \gamma_m$ because you can combine functions on $\mathbb Z/n$ and $\mathbb Z/m$ using the Chinese remainder theorem.
So if we show $\gamma_n \leq 1- \epsilon$ for some $\epsilon$ and all sufficiently large $n$, then we can make $\gamma_n$ arbitrarily small.
Probably $x(t) = \sin( ...

3

I believe no such constant exists.
$\newcommand{\cF}{{\mathcal F}}\newcommand{\FA}{{\rm A}}$
Let $G$ be a LCA group with Pontryagin dual $\Gamma$, let $\cF_\Gamma: L^1(\Gamma)\to C_0(G)$ be the Fourier transform for $\Gamma$, and let $\FA(G)$ denote the image $\cF_\Gamma(L^1(\Gamma))$ equipped with the norm pushed forward from $L^1(\Gamma)$. Then if I ...

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