Mixing
Holder continuity of Fourier transform of measure
I am trying to prove the following, say $mu$ is a non-negative Borel measure on $mathbb{R}$ with $mu(mathbb{R}) = 1$ and
$$ f(t) = widehat{mu}(t) = int_{mathbb{R}} e^{itx};dmu(x).$$
Assuming that
$$int_{mathbb{R}} |x|^{2+delta};dmu(x) < infty$$
Can we show that $fin mathrm{C}^{2,delta}(mathbb{R})$?
real-analysis measure-theory fourier-analysis fourier-transform
asked Nov 20 '18 at 0:09
Sean
527513
add a comment |
I am trying to prove the following, say $mu$ is a non-negative Borel measure on $mathbb{R}$ with $mu(mathbb{R}) = 1$ and
$$ f(t) = widehat{mu}(t) = int_{mathbb{R}} e^{itx};dmu(x).$$
Assuming that
$$int_{mathbb{R}} |x|^{2+delta};dmu(x) < infty$$
Can we show that $fin mathrm{C}^{2,delta}(mathbb{R})$?
real-analysis measure-theory fourier-analysis fourier-transform
asked Nov 20 '18 at 0:09
Sean
527513
2
I would start with $f''(t)-f''(t+c) = int_{|x| < c^{-r}}+int_{|x| > c^{-r}} x^2e^{itx} (e^{icx}-1) dmu(x)$ $= O( int |x|^2 c^{1-r} dmu(x))+O(int_{|x| > c^{-r}} |x|^2 dmu(x))$
– reuns
Nov 20 '18 at 0:42
add a comment |
I am trying to prove the following, say $mu$ is a non-negative Borel measure on $mathbb{R}$ with $mu(mathbb{R}) = 1$ and
$$ f(t) = widehat{mu}(t) = int_{mathbb{R}} e^{itx};dmu(x).$$
Assuming that
$$int_{mathbb{R}} |x|^{2+delta};dmu(x) < infty$$
Can we show that $fin mathrm{C}^{2,delta}(mathbb{R})$?
real-analysis measure-theory fourier-analysis fourier-transform
asked Nov 20 '18 at 0:09
Sean
527513
I am trying to prove the following, say $mu$ is a non-negative Borel measure on $mathbb{R}$ with $mu(mathbb{R}) = 1$ and
$$ f(t) = widehat{mu}(t) = int_{mathbb{R}} e^{itx};dmu(x).$$
Assuming that
$$int_{mathbb{R}} |x|^{2+delta};dmu(x) < infty$$
Can we show that $fin mathrm{C}^{2,delta}(mathbb{R})$?
real-analysis measure-theory fourier-analysis fourier-transform
real-analysis measure-theory fourier-analysis fourier-transform
asked Nov 20 '18 at 0:09
Sean
527513
asked Nov 20 '18 at 0:09
Sean
527513
asked Nov 20 '18 at 0:09
Sean
527513
asked Nov 20 '18 at 0:09
Sean
527513
asked Nov 20 '18 at 0:09
Sean
527513
527513
2
I would start with $f''(t)-f''(t+c) = int_{|x| < c^{-r}}+int_{|x| > c^{-r}} x^2e^{itx} (e^{icx}-1) dmu(x)$ $= O( int |x|^2 c^{1-r} dmu(x))+O(int_{|x| > c^{-r}} |x|^2 dmu(x))$
– reuns
Nov 20 '18 at 0:42
add a comment |
2
I would start with $f''(t)-f''(t+c) = int_{|x| < c^{-r}}+int_{|x| > c^{-r}} x^2e^{itx} (e^{icx}-1) dmu(x)$ $= O( int |x|^2 c^{1-r} dmu(x))+O(int_{|x| > c^{-r}} |x|^2 dmu(x))$
– reuns
Nov 20 '18 at 0:42
2
2
I would start with $f''(t)-f''(t+c) = int_{|x| < c^{-r}}+int_{|x| > c^{-r}} x^2e^{itx} (e^{icx}-1) dmu(x)$ $= O( int |x|^2 c^{1-r} dmu(x))+O(int_{|x| > c^{-r}} |x|^2 dmu(x))$
– reuns
Nov 20 '18 at 0:42
I would start with $f''(t)-f''(t+c) = int_{|x| < c^{-r}}+int_{|x| > c^{-r}} x^2e^{itx} (e^{icx}-1) dmu(x)$ $= O( int |x|^2 c^{1-r} dmu(x))+O(int_{|x| > c^{-r}} |x|^2 dmu(x))$
– reuns
Nov 20 '18 at 0:42
add a comment |
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2
I would start with $f''(t)-f''(t+c) = int_{|x| < c^{-r}}+int_{|x| > c^{-r}} x^2e^{itx} (e^{icx}-1) dmu(x)$ $= O( int |x|^2 c^{1-r} dmu(x))+O(int_{|x| > c^{-r}} |x|^2 dmu(x))$
– reuns
Nov 20 '18 at 0:42