How do you show that $e^{-asqrt{1+|x|^2}}$ is in the Schwartz space
$begingroup$
I want to show that $e^{-asqrt{1+|x|^2}}$ is in $mathcal{S}(mathbb{R}^d)$. ($a>0$)
Please tell me proof.
Where,
$$|x|^2 = left(sum_{j=1}^{d} |x_j|^2right)$$
$$ f(x) in mathcal{S} overset {mathrm{def}} {Leftrightarrow} displaystyle sup_{x in mathbb{R}^d} |x^alphapartial^beta_x f(x)| < infty $$
$alpha,beta$ is multi index notations.
real-analysis distribution-theory schwartz-space
edited Dec 30 '18 at 11:47
Bernard
119k639112
asked Dec 30 '18 at 11:38
NikolaiNikolai
65
$endgroup$
add a comment |
$begingroup$
I want to show that $e^{-asqrt{1+|x|^2}}$ is in $mathcal{S}(mathbb{R}^d)$. ($a>0$)
Please tell me proof.
Where,
$$|x|^2 = left(sum_{j=1}^{d} |x_j|^2right)$$
$$ f(x) in mathcal{S} overset {mathrm{def}} {Leftrightarrow} displaystyle sup_{x in mathbb{R}^d} |x^alphapartial^beta_x f(x)| < infty $$
$alpha,beta$ is multi index notations.
real-analysis distribution-theory schwartz-space
edited Dec 30 '18 at 11:47
Bernard
119k639112
asked Dec 30 '18 at 11:38
NikolaiNikolai
65
$endgroup$
$begingroup$
Hint: Calculate $ sup_{x in mathbb{R}^d} |x^alphapartial^beta_x f(x)| $ and see if it is finite. No?
$endgroup$
– Paul
Dec 30 '18 at 11:51
$begingroup$
@Paul Thank you for showing hint, but I don't know how to calculate $partial^beta_xe^{-asqrt{1+|x|^2}}$.
$endgroup$
– Nikolai
Dec 30 '18 at 12:01
$begingroup$
@Kavi Rama Murthy Hmmm... let $d$=1. When $f(x)=e^{-asqrt{1+x^2}}$ ,$f'(x)=-ax(1+x^2)^{-1/2}e^{-asqrt{1+x^2}}$. $-ax(1+x^2)^{-1/2}$ is polynomial ?
$endgroup$
– Nikolai
Dec 30 '18 at 12:54
add a comment |
$begingroup$
I want to show that $e^{-asqrt{1+|x|^2}}$ is in $mathcal{S}(mathbb{R}^d)$. ($a>0$)
Please tell me proof.
Where,
$$|x|^2 = left(sum_{j=1}^{d} |x_j|^2right)$$
$$ f(x) in mathcal{S} overset {mathrm{def}} {Leftrightarrow} displaystyle sup_{x in mathbb{R}^d} |x^alphapartial^beta_x f(x)| < infty $$
$alpha,beta$ is multi index notations.
real-analysis distribution-theory schwartz-space
edited Dec 30 '18 at 11:47
Bernard
119k639112
asked Dec 30 '18 at 11:38
NikolaiNikolai
65
$endgroup$
I want to show that $e^{-asqrt{1+|x|^2}}$ is in $mathcal{S}(mathbb{R}^d)$. ($a>0$)
Please tell me proof.
Where,
$$|x|^2 = left(sum_{j=1}^{d} |x_j|^2right)$$
$$ f(x) in mathcal{S} overset {mathrm{def}} {Leftrightarrow} displaystyle sup_{x in mathbb{R}^d} |x^alphapartial^beta_x f(x)| < infty $$
$alpha,beta$ is multi index notations.
real-analysis distribution-theory schwartz-space
real-analysis distribution-theory schwartz-space
edited Dec 30 '18 at 11:47
Bernard
119k639112
asked Dec 30 '18 at 11:38
NikolaiNikolai
65
edited Dec 30 '18 at 11:47
Bernard
119k639112
asked Dec 30 '18 at 11:38
NikolaiNikolai
65
edited Dec 30 '18 at 11:47
Bernard
119k639112
edited Dec 30 '18 at 11:47
Bernard
119k639112
edited Dec 30 '18 at 11:47
Bernard
119k639112
119k639112
asked Dec 30 '18 at 11:38
NikolaiNikolai
65
asked Dec 30 '18 at 11:38
NikolaiNikolai
65
asked Dec 30 '18 at 11:38
NikolaiNikolai
65
65
$begingroup$
Hint: Calculate $ sup_{x in mathbb{R}^d} |x^alphapartial^beta_x f(x)| $ and see if it is finite. No?
$endgroup$
– Paul
Dec 30 '18 at 11:51
$begingroup$
@Paul Thank you for showing hint, but I don't know how to calculate $partial^beta_xe^{-asqrt{1+|x|^2}}$.
$endgroup$
– Nikolai
Dec 30 '18 at 12:01
$begingroup$
@Kavi Rama Murthy Hmmm... let $d$=1. When $f(x)=e^{-asqrt{1+x^2}}$ ,$f'(x)=-ax(1+x^2)^{-1/2}e^{-asqrt{1+x^2}}$. $-ax(1+x^2)^{-1/2}$ is polynomial ?
$endgroup$
– Nikolai
Dec 30 '18 at 12:54
add a comment |
$begingroup$
Hint: Calculate $ sup_{x in mathbb{R}^d} |x^alphapartial^beta_x f(x)| $ and see if it is finite. No?
$endgroup$
– Paul
Dec 30 '18 at 11:51
$begingroup$
@Paul Thank you for showing hint, but I don't know how to calculate $partial^beta_xe^{-asqrt{1+|x|^2}}$.
$endgroup$
– Nikolai
Dec 30 '18 at 12:01
$begingroup$
@Kavi Rama Murthy Hmmm... let $d$=1. When $f(x)=e^{-asqrt{1+x^2}}$ ,$f'(x)=-ax(1+x^2)^{-1/2}e^{-asqrt{1+x^2}}$. $-ax(1+x^2)^{-1/2}$ is polynomial ?
$endgroup$
– Nikolai
Dec 30 '18 at 12:54
$begingroup$
Hint: Calculate $ sup_{x in mathbb{R}^d} |x^alphapartial^beta_x f(x)| $ and see if it is finite. No?
$endgroup$
– Paul
Dec 30 '18 at 11:51
$begingroup$
Hint: Calculate $ sup_{x in mathbb{R}^d} |x^alphapartial^beta_x f(x)| $ and see if it is finite. No?
$endgroup$
– Paul
Dec 30 '18 at 11:51
$begingroup$
@Paul Thank you for showing hint, but I don't know how to calculate $partial^beta_xe^{-asqrt{1+|x|^2}}$.
$endgroup$
– Nikolai
Dec 30 '18 at 12:01
$begingroup$
@Paul Thank you for showing hint, but I don't know how to calculate $partial^beta_xe^{-asqrt{1+|x|^2}}$.
$endgroup$
– Nikolai
Dec 30 '18 at 12:01
$begingroup$
@Kavi Rama Murthy Hmmm... let $d$=1. When $f(x)=e^{-asqrt{1+x^2}}$ ,$f'(x)=-ax(1+x^2)^{-1/2}e^{-asqrt{1+x^2}}$. $-ax(1+x^2)^{-1/2}$ is polynomial ?
$endgroup$
– Nikolai
Dec 30 '18 at 12:54
$begingroup$
@Kavi Rama Murthy Hmmm... let $d$=1. When $f(x)=e^{-asqrt{1+x^2}}$ ,$f'(x)=-ax(1+x^2)^{-1/2}e^{-asqrt{1+x^2}}$. $-ax(1+x^2)^{-1/2}$ is polynomial ?
$endgroup$
– Nikolai
Dec 30 '18 at 12:54
add a comment |
1 Answer
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$begingroup$
Hint: Show by induction that every derivative of $f$ is a finite sum of functions of the form
$$p(x)(1+|x|^2)^rf(x).$$
Here $p$ is a polynomial and $rin mathbb R.$
answered Dec 30 '18 at 19:42
zhw.zhw.
71.9k43075
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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votes
$begingroup$
Hint: Show by induction that every derivative of $f$ is a finite sum of functions of the form
$$p(x)(1+|x|^2)^rf(x).$$
Here $p$ is a polynomial and $rin mathbb R.$
answered Dec 30 '18 at 19:42
zhw.zhw.
71.9k43075
$endgroup$
add a comment |
$begingroup$
Hint: Show by induction that every derivative of $f$ is a finite sum of functions of the form
$$p(x)(1+|x|^2)^rf(x).$$
Here $p$ is a polynomial and $rin mathbb R.$
answered Dec 30 '18 at 19:42
zhw.zhw.
71.9k43075
$endgroup$
add a comment |
$begingroup$
Hint: Show by induction that every derivative of $f$ is a finite sum of functions of the form
$$p(x)(1+|x|^2)^rf(x).$$
Here $p$ is a polynomial and $rin mathbb R.$
answered Dec 30 '18 at 19:42
zhw.zhw.
71.9k43075
$endgroup$
Hint: Show by induction that every derivative of $f$ is a finite sum of functions of the form
$$p(x)(1+|x|^2)^rf(x).$$
Here $p$ is a polynomial and $rin mathbb R.$
answered Dec 30 '18 at 19:42
zhw.zhw.
71.9k43075
answered Dec 30 '18 at 19:42
zhw.zhw.
71.9k43075
answered Dec 30 '18 at 19:42
zhw.zhw.
71.9k43075
answered Dec 30 '18 at 19:42
zhw.zhw.
71.9k43075
71.9k43075
add a comment |
add a comment |
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$begingroup$
Hint: Calculate $ sup_{x in mathbb{R}^d} |x^alphapartial^beta_x f(x)| $ and see if it is finite. No?
$endgroup$
– Paul
Dec 30 '18 at 11:51
$begingroup$
@Paul Thank you for showing hint, but I don't know how to calculate $partial^beta_xe^{-asqrt{1+|x|^2}}$.
$endgroup$
– Nikolai
Dec 30 '18 at 12:01
$begingroup$
@Kavi Rama Murthy Hmmm... let $d$=1. When $f(x)=e^{-asqrt{1+x^2}}$ ,$f'(x)=-ax(1+x^2)^{-1/2}e^{-asqrt{1+x^2}}$. $-ax(1+x^2)^{-1/2}$ is polynomial ?
$endgroup$
– Nikolai
Dec 30 '18 at 12:54