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Given $a>0$, $frac1x^2+a^2$ is not a Schwarz function.
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Given $a>0$, $f(x) = frac1x^2+a^2$ is not a Schwarz function.
Please verify if this is correct:
Although Poisson Sumation formula is working for this function $f$, I think it is not Schwarz, since $f’(x)=-2x(x^2+a^2)^-2$, and $lim_x to infty |-2x^5(x^2+a^2)^-2| = infty$, so for $alpha = 4$ and $beta = 1$ there cannot exist $C_alpha,beta$ such that $|x^alphaf^(beta)(x)|leq C_alpha,beta$.
Also, in Schwarz space can it be taken $alpha, beta = 0$?
fourier-series fourier-transform schwartz-space
asked Jul 25 at 0:03
dude3221
11510
 |Â
show 1 more comment
up vote
1
down vote
favorite
Given $a>0$, $f(x) = frac1x^2+a^2$ is not a Schwarz function.
Please verify if this is correct:
Although Poisson Sumation formula is working for this function $f$, I think it is not Schwarz, since $f’(x)=-2x(x^2+a^2)^-2$, and $lim_x to infty |-2x^5(x^2+a^2)^-2| = infty$, so for $alpha = 4$ and $beta = 1$ there cannot exist $C_alpha,beta$ such that $|x^alphaf^(beta)(x)|leq C_alpha,beta$.
Also, in Schwarz space can it be taken $alpha, beta = 0$?
fourier-series fourier-transform schwartz-space
asked Jul 25 at 0:03
dude3221
11510
(my derivative was wrong. fixed.)
– dude3221
Jul 25 at 0:10
1
One (necessary but not sufficient) condition for a function to be a Schwarz function is that $forall n in mathbb N, lim_x rightarrow infty x^n f(x) = 0$.
– Matrefeytontias
Jul 25 at 0:16
this answers both of my questions, right? it is not schwarz and I can take $alpha,beta=0$ in schwarz definition.
– dude3221
Jul 25 at 0:18
1
It is indeed not Schwarz, but the Schwarz definition must hold for all $alpha, beta$, not just one couple.
– Matrefeytontias
Jul 25 at 0:20
1
Yeah, both being 0 is also included in the set of all $alpha, beta$ indeed.
– Matrefeytontias
Jul 25 at 0:22
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given $a>0$, $f(x) = frac1x^2+a^2$ is not a Schwarz function.
Please verify if this is correct:
Although Poisson Sumation formula is working for this function $f$, I think it is not Schwarz, since $f’(x)=-2x(x^2+a^2)^-2$, and $lim_x to infty |-2x^5(x^2+a^2)^-2| = infty$, so for $alpha = 4$ and $beta = 1$ there cannot exist $C_alpha,beta$ such that $|x^alphaf^(beta)(x)|leq C_alpha,beta$.
Also, in Schwarz space can it be taken $alpha, beta = 0$?
fourier-series fourier-transform schwartz-space
asked Jul 25 at 0:03
dude3221
11510
Given $a>0$, $f(x) = frac1x^2+a^2$ is not a Schwarz function.
Please verify if this is correct:
Although Poisson Sumation formula is working for this function $f$, I think it is not Schwarz, since $f’(x)=-2x(x^2+a^2)^-2$, and $lim_x to infty |-2x^5(x^2+a^2)^-2| = infty$, so for $alpha = 4$ and $beta = 1$ there cannot exist $C_alpha,beta$ such that $|x^alphaf^(beta)(x)|leq C_alpha,beta$.
Also, in Schwarz space can it be taken $alpha, beta = 0$?
fourier-series fourier-transform schwartz-space
asked Jul 25 at 0:03
dude3221
11510
edited Jul 25 at 0:09
asked Jul 25 at 0:03
dude3221
11510
asked Jul 25 at 0:03
dude3221
11510
asked Jul 25 at 0:03
dude3221
11510
11510
(my derivative was wrong. fixed.)
– dude3221
Jul 25 at 0:10
1
One (necessary but not sufficient) condition for a function to be a Schwarz function is that $forall n in mathbb N, lim_x rightarrow infty x^n f(x) = 0$.
– Matrefeytontias
Jul 25 at 0:16
this answers both of my questions, right? it is not schwarz and I can take $alpha,beta=0$ in schwarz definition.
– dude3221
Jul 25 at 0:18
1
It is indeed not Schwarz, but the Schwarz definition must hold for all $alpha, beta$, not just one couple.
– Matrefeytontias
Jul 25 at 0:20
1
Yeah, both being 0 is also included in the set of all $alpha, beta$ indeed.
– Matrefeytontias
Jul 25 at 0:22
 |Â
show 1 more comment
(my derivative was wrong. fixed.)
– dude3221
Jul 25 at 0:10
1
One (necessary but not sufficient) condition for a function to be a Schwarz function is that $forall n in mathbb N, lim_x rightarrow infty x^n f(x) = 0$.
– Matrefeytontias
Jul 25 at 0:16
this answers both of my questions, right? it is not schwarz and I can take $alpha,beta=0$ in schwarz definition.
– dude3221
Jul 25 at 0:18
1
It is indeed not Schwarz, but the Schwarz definition must hold for all $alpha, beta$, not just one couple.
– Matrefeytontias
Jul 25 at 0:20
1
Yeah, both being 0 is also included in the set of all $alpha, beta$ indeed.
– Matrefeytontias
Jul 25 at 0:22
(my derivative was wrong. fixed.)
– dude3221
Jul 25 at 0:10
(my derivative was wrong. fixed.)
– dude3221
Jul 25 at 0:10
1
1
One (necessary but not sufficient) condition for a function to be a Schwarz function is that $forall n in mathbb N, lim_x rightarrow infty x^n f(x) = 0$.
– Matrefeytontias
Jul 25 at 0:16
One (necessary but not sufficient) condition for a function to be a Schwarz function is that $forall n in mathbb N, lim_x rightarrow infty x^n f(x) = 0$.
– Matrefeytontias
Jul 25 at 0:16
this answers both of my questions, right? it is not schwarz and I can take $alpha,beta=0$ in schwarz definition.
– dude3221
Jul 25 at 0:18
this answers both of my questions, right? it is not schwarz and I can take $alpha,beta=0$ in schwarz definition.
– dude3221
Jul 25 at 0:18
1
1
It is indeed not Schwarz, but the Schwarz definition must hold for all $alpha, beta$, not just one couple.
– Matrefeytontias
Jul 25 at 0:20
It is indeed not Schwarz, but the Schwarz definition must hold for all $alpha, beta$, not just one couple.
– Matrefeytontias
Jul 25 at 0:20
1
1
Yeah, both being 0 is also included in the set of all $alpha, beta$ indeed.
– Matrefeytontias
Jul 25 at 0:22
Yeah, both being 0 is also included in the set of all $alpha, beta$ indeed.
– Matrefeytontias
Jul 25 at 0:22
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The Schwartz space on $mathbbR$ is the set of all smooth functions $f:mathbbRrightarrowmathbbR$ such that for all nonnegative integers $alpha$ and $beta$, the function $xmapsto x^alphaf^(beta)(x)$ is bounded.
Hint: In your case, take $alpha=3$ and $beta=0$ to get
$$
x^3f^(0)(x)=fracx^3x^2+a^2.
$$
What can you say about the above as $|x|rightarrow infty$?
answered Jul 25 at 0:19
parsiad
16k32253
goes to $infty$ for $xto infty$ and to $-infty$ for $x to -infty$, then it´s not bounded.
– dude3221
Jul 25 at 0:24
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The Schwartz space on $mathbbR$ is the set of all smooth functions $f:mathbbRrightarrowmathbbR$ such that for all nonnegative integers $alpha$ and $beta$, the function $xmapsto x^alphaf^(beta)(x)$ is bounded.
Hint: In your case, take $alpha=3$ and $beta=0$ to get
$$
x^3f^(0)(x)=fracx^3x^2+a^2.
$$
What can you say about the above as $|x|rightarrow infty$?
answered Jul 25 at 0:19
parsiad
16k32253
goes to $infty$ for $xto infty$ and to $-infty$ for $x to -infty$, then it´s not bounded.
– dude3221
Jul 25 at 0:24
add a comment |Â
up vote
2
down vote
accepted
The Schwartz space on $mathbbR$ is the set of all smooth functions $f:mathbbRrightarrowmathbbR$ such that for all nonnegative integers $alpha$ and $beta$, the function $xmapsto x^alphaf^(beta)(x)$ is bounded.
Hint: In your case, take $alpha=3$ and $beta=0$ to get
$$
x^3f^(0)(x)=fracx^3x^2+a^2.
$$
What can you say about the above as $|x|rightarrow infty$?
answered Jul 25 at 0:19
parsiad
16k32253
goes to $infty$ for $xto infty$ and to $-infty$ for $x to -infty$, then it´s not bounded.
– dude3221
Jul 25 at 0:24
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The Schwartz space on $mathbbR$ is the set of all smooth functions $f:mathbbRrightarrowmathbbR$ such that for all nonnegative integers $alpha$ and $beta$, the function $xmapsto x^alphaf^(beta)(x)$ is bounded.
Hint: In your case, take $alpha=3$ and $beta=0$ to get
$$
x^3f^(0)(x)=fracx^3x^2+a^2.
$$
What can you say about the above as $|x|rightarrow infty$?
answered Jul 25 at 0:19
parsiad
16k32253
The Schwartz space on $mathbbR$ is the set of all smooth functions $f:mathbbRrightarrowmathbbR$ such that for all nonnegative integers $alpha$ and $beta$, the function $xmapsto x^alphaf^(beta)(x)$ is bounded.
Hint: In your case, take $alpha=3$ and $beta=0$ to get
$$
x^3f^(0)(x)=fracx^3x^2+a^2.
$$
What can you say about the above as $|x|rightarrow infty$?
answered Jul 25 at 0:19
parsiad
16k32253
answered Jul 25 at 0:19
parsiad
16k32253
answered Jul 25 at 0:19
parsiad
16k32253
answered Jul 25 at 0:19
parsiad
16k32253
16k32253
goes to $infty$ for $xto infty$ and to $-infty$ for $x to -infty$, then it´s not bounded.
– dude3221
Jul 25 at 0:24
add a comment |Â
goes to $infty$ for $xto infty$ and to $-infty$ for $x to -infty$, then it´s not bounded.
– dude3221
Jul 25 at 0:24
goes to $infty$ for $xto infty$ and to $-infty$ for $x to -infty$, then it´s not bounded.
– dude3221
Jul 25 at 0:24
goes to $infty$ for $xto infty$ and to $-infty$ for $x to -infty$, then it´s not bounded.
– dude3221
Jul 25 at 0:24
add a comment |Â
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(my derivative was wrong. fixed.)
– dude3221
Jul 25 at 0:10
1
One (necessary but not sufficient) condition for a function to be a Schwarz function is that $forall n in mathbb N, lim_x rightarrow infty x^n f(x) = 0$.
– Matrefeytontias
Jul 25 at 0:16
this answers both of my questions, right? it is not schwarz and I can take $alpha,beta=0$ in schwarz definition.
– dude3221
Jul 25 at 0:18
1
It is indeed not Schwarz, but the Schwarz definition must hold for all $alpha, beta$, not just one couple.
– Matrefeytontias
Jul 25 at 0:20
1
Yeah, both being 0 is also included in the set of all $alpha, beta$ indeed.
– Matrefeytontias
Jul 25 at 0:22