Integrability of the Fourier transform in Sobolev space
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I believe that the statement below is a standard fact but I haven't figured out yet:
Suppose $fin L^{1}(mathbb{R}^{n})$ has integrable partial derivatives of order $n+1$ and $D^{alpha}fin L^{1}(mathbb{R}^{n})$ for all multiindex with $|alpha|le n+1$ (here $D^{alpha}$ is the mixed partial derivative). Prove that the Fourier transform $hat{f}$ is in $L^{1}(mathbb{R}^{n})$.
Does anyone know the proof or a reference for this statement?
sobolev-spaces fourier-transform harmonic-analysis
edited Dec 11 '18 at 17:16
Bernard
123k741117
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
$endgroup$
add a comment |
$begingroup$
I believe that the statement below is a standard fact but I haven't figured out yet:
Suppose $fin L^{1}(mathbb{R}^{n})$ has integrable partial derivatives of order $n+1$ and $D^{alpha}fin L^{1}(mathbb{R}^{n})$ for all multiindex with $|alpha|le n+1$ (here $D^{alpha}$ is the mixed partial derivative). Prove that the Fourier transform $hat{f}$ is in $L^{1}(mathbb{R}^{n})$.
Does anyone know the proof or a reference for this statement?
sobolev-spaces fourier-transform harmonic-analysis
edited Dec 11 '18 at 17:16
Bernard
123k741117
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
$endgroup$
2
$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57
add a comment |
$begingroup$
I believe that the statement below is a standard fact but I haven't figured out yet:
Suppose $fin L^{1}(mathbb{R}^{n})$ has integrable partial derivatives of order $n+1$ and $D^{alpha}fin L^{1}(mathbb{R}^{n})$ for all multiindex with $|alpha|le n+1$ (here $D^{alpha}$ is the mixed partial derivative). Prove that the Fourier transform $hat{f}$ is in $L^{1}(mathbb{R}^{n})$.
Does anyone know the proof or a reference for this statement?
sobolev-spaces fourier-transform harmonic-analysis
edited Dec 11 '18 at 17:16
Bernard
123k741117
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
$endgroup$
I believe that the statement below is a standard fact but I haven't figured out yet:
Suppose $fin L^{1}(mathbb{R}^{n})$ has integrable partial derivatives of order $n+1$ and $D^{alpha}fin L^{1}(mathbb{R}^{n})$ for all multiindex with $|alpha|le n+1$ (here $D^{alpha}$ is the mixed partial derivative). Prove that the Fourier transform $hat{f}$ is in $L^{1}(mathbb{R}^{n})$.
Does anyone know the proof or a reference for this statement?
sobolev-spaces fourier-transform harmonic-analysis
sobolev-spaces fourier-transform harmonic-analysis
edited Dec 11 '18 at 17:16
Bernard
123k741117
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
edited Dec 11 '18 at 17:16
Bernard
123k741117
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
edited Dec 11 '18 at 17:16
Bernard
123k741117
edited Dec 11 '18 at 17:16
Bernard
123k741117
edited Dec 11 '18 at 17:16
Bernard
123k741117
123k741117
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
568
2
$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57
add a comment |
2
$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57
2
2
$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57
add a comment |
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$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57