Sobolev Space of Order $s$
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For $fin L^2(mathbb{R})$, define $K(x,t)=dfrac{1}{sqrt{t}}e^{-pi frac{x^2}{t}}$, and let $u(cdot,t)= K(cdot,t)ast f$. Show that for $t>0$, $u(cdot,t)in H^s(mathbb{R})$, $forall sgeq 0$ and conclude that $u(cdot,t)in C^infty$.
By taking the Fourier transform, we have
$widehat{u}(xi, t)=e^{-pi xi^2 t}widehat{f}(xi).$
I know that the definition of the Sobolev space of order $s$ is that $s^{th}$ weak derivative exists but I am having trouble deducing this from the Fourier transform.
real-analysis distribution-theory
asked Dec 16 '18 at 19:05
user519208user519208
665
$endgroup$
add a comment |
$begingroup$
For $fin L^2(mathbb{R})$, define $K(x,t)=dfrac{1}{sqrt{t}}e^{-pi frac{x^2}{t}}$, and let $u(cdot,t)= K(cdot,t)ast f$. Show that for $t>0$, $u(cdot,t)in H^s(mathbb{R})$, $forall sgeq 0$ and conclude that $u(cdot,t)in C^infty$.
By taking the Fourier transform, we have
$widehat{u}(xi, t)=e^{-pi xi^2 t}widehat{f}(xi).$
I know that the definition of the Sobolev space of order $s$ is that $s^{th}$ weak derivative exists but I am having trouble deducing this from the Fourier transform.
real-analysis distribution-theory
asked Dec 16 '18 at 19:05
user519208user519208
665
$endgroup$
2
$begingroup$
$H^s$ is defined in terms of the Fourier transform. The result follows immediately from the fact that exponential decay outweighs polynomial growth.
$endgroup$
– Dunham
Dec 16 '18 at 23:44
add a comment |
$begingroup$
For $fin L^2(mathbb{R})$, define $K(x,t)=dfrac{1}{sqrt{t}}e^{-pi frac{x^2}{t}}$, and let $u(cdot,t)= K(cdot,t)ast f$. Show that for $t>0$, $u(cdot,t)in H^s(mathbb{R})$, $forall sgeq 0$ and conclude that $u(cdot,t)in C^infty$.
By taking the Fourier transform, we have
$widehat{u}(xi, t)=e^{-pi xi^2 t}widehat{f}(xi).$
I know that the definition of the Sobolev space of order $s$ is that $s^{th}$ weak derivative exists but I am having trouble deducing this from the Fourier transform.
real-analysis distribution-theory
asked Dec 16 '18 at 19:05
user519208user519208
665
$endgroup$
For $fin L^2(mathbb{R})$, define $K(x,t)=dfrac{1}{sqrt{t}}e^{-pi frac{x^2}{t}}$, and let $u(cdot,t)= K(cdot,t)ast f$. Show that for $t>0$, $u(cdot,t)in H^s(mathbb{R})$, $forall sgeq 0$ and conclude that $u(cdot,t)in C^infty$.
By taking the Fourier transform, we have
$widehat{u}(xi, t)=e^{-pi xi^2 t}widehat{f}(xi).$
I know that the definition of the Sobolev space of order $s$ is that $s^{th}$ weak derivative exists but I am having trouble deducing this from the Fourier transform.
real-analysis distribution-theory
real-analysis distribution-theory
asked Dec 16 '18 at 19:05
user519208user519208
665
asked Dec 16 '18 at 19:05
user519208user519208
665
asked Dec 16 '18 at 19:05
user519208user519208
665
asked Dec 16 '18 at 19:05
user519208user519208
665
asked Dec 16 '18 at 19:05
user519208user519208
665
665
2
$begingroup$
$H^s$ is defined in terms of the Fourier transform. The result follows immediately from the fact that exponential decay outweighs polynomial growth.
$endgroup$
– Dunham
Dec 16 '18 at 23:44
add a comment |
2
$begingroup$
$H^s$ is defined in terms of the Fourier transform. The result follows immediately from the fact that exponential decay outweighs polynomial growth.
$endgroup$
– Dunham
Dec 16 '18 at 23:44
2
2
$begingroup$
$H^s$ is defined in terms of the Fourier transform. The result follows immediately from the fact that exponential decay outweighs polynomial growth.
$endgroup$
– Dunham
Dec 16 '18 at 23:44
$begingroup$
$H^s$ is defined in terms of the Fourier transform. The result follows immediately from the fact that exponential decay outweighs polynomial growth.
$endgroup$
– Dunham
Dec 16 '18 at 23:44
add a comment |
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$begingroup$
$H^s$ is defined in terms of the Fourier transform. The result follows immediately from the fact that exponential decay outweighs polynomial growth.
$endgroup$
– Dunham
Dec 16 '18 at 23:44