About Moderately Decreasing functions
0
A function $g: mathbb{R} to mathbb{R} $ is said of moderate decrease if there is a $M in mathbb{R}$ such that $|g(x)| leq frac{M}{1 + |x|^{alpha}}$, for some $alpha > 1$, for all $x in mathbb{R}$. Supose $f$ is continuous and such that $f(x) = O(frac{1}{x^alpha})$ when $|x| to + infty$. Is $f$ of moderate decrease?
real-analysis fourier-analysis fourier-transform
asked Nov 27 at 6:57
Nuntractatuses Amável
61812
add a comment |
0
A function $g: mathbb{R} to mathbb{R} $ is said of moderate decrease if there is a $M in mathbb{R}$ such that $|g(x)| leq frac{M}{1 + |x|^{alpha}}$, for some $alpha > 1$, for all $x in mathbb{R}$. Supose $f$ is continuous and such that $f(x) = O(frac{1}{x^alpha})$ when $|x| to + infty$. Is $f$ of moderate decrease?
real-analysis fourier-analysis fourier-transform
asked Nov 27 at 6:57
Nuntractatuses Amável
61812
No, because there is no bound near 0.
– T. Bongers
Nov 27 at 6:59
0 is not included, because it is when $|x| to + infty$. So there is a $x_0 > 0 $ such that if $|x| geq x_0$, $f(x) = O(frac{1}{x^alpha})$. So, inside $[-x_0, x_0]$, $f$ is limited, and decreases slower than $frac{1}{x^alpha}$ outside the interval.
– Nuntractatuses Amável
Nov 27 at 7:14
1
If you assume continuity or bounds on f, sure. But that's not a consequence of the big O estimate you wrote down.
– T. Bongers
Nov 27 at 7:16
Yes, forgot to add continuity
– Nuntractatuses Amável
Nov 27 at 7:17
add a comment |
0
0
0
A function $g: mathbb{R} to mathbb{R} $ is said of moderate decrease if there is a $M in mathbb{R}$ such that $|g(x)| leq frac{M}{1 + |x|^{alpha}}$, for some $alpha > 1$, for all $x in mathbb{R}$. Supose $f$ is continuous and such that $f(x) = O(frac{1}{x^alpha})$ when $|x| to + infty$. Is $f$ of moderate decrease?
real-analysis fourier-analysis fourier-transform
asked Nov 27 at 6:57
Nuntractatuses Amável
61812
A function $g: mathbb{R} to mathbb{R} $ is said of moderate decrease if there is a $M in mathbb{R}$ such that $|g(x)| leq frac{M}{1 + |x|^{alpha}}$, for some $alpha > 1$, for all $x in mathbb{R}$. Supose $f$ is continuous and such that $f(x) = O(frac{1}{x^alpha})$ when $|x| to + infty$. Is $f$ of moderate decrease?
real-analysis fourier-analysis fourier-transform
real-analysis fourier-analysis fourier-transform
asked Nov 27 at 6:57
Nuntractatuses Amável
61812
asked Nov 27 at 6:57
Nuntractatuses Amável
61812
edited Nov 27 at 7:44
asked Nov 27 at 6:57
Nuntractatuses Amável
61812
asked Nov 27 at 6:57
Nuntractatuses Amável
61812
asked Nov 27 at 6:57
Nuntractatuses Amável
61812
61812
No, because there is no bound near 0.
– T. Bongers
Nov 27 at 6:59
0 is not included, because it is when $|x| to + infty$. So there is a $x_0 > 0 $ such that if $|x| geq x_0$, $f(x) = O(frac{1}{x^alpha})$. So, inside $[-x_0, x_0]$, $f$ is limited, and decreases slower than $frac{1}{x^alpha}$ outside the interval.
– Nuntractatuses Amável
Nov 27 at 7:14
1
If you assume continuity or bounds on f, sure. But that's not a consequence of the big O estimate you wrote down.
– T. Bongers
Nov 27 at 7:16
Yes, forgot to add continuity
– Nuntractatuses Amável
Nov 27 at 7:17
add a comment |
No, because there is no bound near 0.
– T. Bongers
Nov 27 at 6:59
0 is not included, because it is when $|x| to + infty$. So there is a $x_0 > 0 $ such that if $|x| geq x_0$, $f(x) = O(frac{1}{x^alpha})$. So, inside $[-x_0, x_0]$, $f$ is limited, and decreases slower than $frac{1}{x^alpha}$ outside the interval.
– Nuntractatuses Amável
Nov 27 at 7:14
1
If you assume continuity or bounds on f, sure. But that's not a consequence of the big O estimate you wrote down.
– T. Bongers
Nov 27 at 7:16
Yes, forgot to add continuity
– Nuntractatuses Amável
Nov 27 at 7:17
No, because there is no bound near 0.
– T. Bongers
Nov 27 at 6:59
No, because there is no bound near 0.
– T. Bongers
Nov 27 at 6:59
0 is not included, because it is when $|x| to + infty$. So there is a $x_0 > 0 $ such that if $|x| geq x_0$, $f(x) = O(frac{1}{x^alpha})$. So, inside $[-x_0, x_0]$, $f$ is limited, and decreases slower than $frac{1}{x^alpha}$ outside the interval.
– Nuntractatuses Amável
Nov 27 at 7:14
0 is not included, because it is when $|x| to + infty$. So there is a $x_0 > 0 $ such that if $|x| geq x_0$, $f(x) = O(frac{1}{x^alpha})$. So, inside $[-x_0, x_0]$, $f$ is limited, and decreases slower than $frac{1}{x^alpha}$ outside the interval.
– Nuntractatuses Amável
Nov 27 at 7:14
1
1
If you assume continuity or bounds on f, sure. But that's not a consequence of the big O estimate you wrote down.
– T. Bongers
Nov 27 at 7:16
If you assume continuity or bounds on f, sure. But that's not a consequence of the big O estimate you wrote down.
– T. Bongers
Nov 27 at 7:16
Yes, forgot to add continuity
– Nuntractatuses Amável
Nov 27 at 7:17
Yes, forgot to add continuity
– Nuntractatuses Amável
Nov 27 at 7:17
add a comment |
1 Answer
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$(1+|x|^{alpha} )|f(x)| leq 2|x|^{alpha} |f(x)|$ so $|f(x)| leq frac {2|x|^{alpha} |f(x)|} {(1+|x|^{alpha} )}$ for $|x| >1$. I hope this answers your question. (There are several errors in the statement; for example you have written 'for all $x in mathbb R$' in the definition which is not what you meant).
answered Nov 27 at 7:32
Kavi Rama Murthy
49.8k31854
If $f$ is of moderate decrease, doesn't that mean that, for all $x in mathbb{R}$, $|f(x)| leq frac{M}{1 + x^alpha}$? Please, point out the errors for me. Maybe I should not have used $f$ for the definition and for the function of the problem. But I really don't see the errors.
– Nuntractatuses Amável
Nov 27 at 7:37
1
Is $x^{alpha}$ defined for $x<0$?. A natural definition of moderate decrease should say $|f(x)| leq frac M {1+|x|^{alpha}}$ for $|x|$ sufficiently large, for some $M in (0,infty)$ and some $alpha >0$.
– Kavi Rama Murthy
Nov 27 at 7:40
I see your point. The definition in my textbook says for all $x$ and and $alpha = 2$. Also, we should have $alpha > 1$ so you can integrate over $mathbb{R}$. I edited the question according to your corrections. Thank you.
– Nuntractatuses Amável
Nov 27 at 7:48
$alpha >0$ was a typo in my comment. Including all $x$ means that the author wants some control for bounded $x$ also but it is not concerned with how the function decreases as $|x| to infty$.
– Kavi Rama Murthy
Nov 27 at 7:52
add a comment |
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$(1+|x|^{alpha} )|f(x)| leq 2|x|^{alpha} |f(x)|$ so $|f(x)| leq frac {2|x|^{alpha} |f(x)|} {(1+|x|^{alpha} )}$ for $|x| >1$. I hope this answers your question. (There are several errors in the statement; for example you have written 'for all $x in mathbb R$' in the definition which is not what you meant).
answered Nov 27 at 7:32
Kavi Rama Murthy
49.8k31854
If $f$ is of moderate decrease, doesn't that mean that, for all $x in mathbb{R}$, $|f(x)| leq frac{M}{1 + x^alpha}$? Please, point out the errors for me. Maybe I should not have used $f$ for the definition and for the function of the problem. But I really don't see the errors.
– Nuntractatuses Amável
Nov 27 at 7:37
1
Is $x^{alpha}$ defined for $x<0$?. A natural definition of moderate decrease should say $|f(x)| leq frac M {1+|x|^{alpha}}$ for $|x|$ sufficiently large, for some $M in (0,infty)$ and some $alpha >0$.
– Kavi Rama Murthy
Nov 27 at 7:40
I see your point. The definition in my textbook says for all $x$ and and $alpha = 2$. Also, we should have $alpha > 1$ so you can integrate over $mathbb{R}$. I edited the question according to your corrections. Thank you.
– Nuntractatuses Amável
Nov 27 at 7:48
$alpha >0$ was a typo in my comment. Including all $x$ means that the author wants some control for bounded $x$ also but it is not concerned with how the function decreases as $|x| to infty$.
– Kavi Rama Murthy
Nov 27 at 7:52
add a comment |
1
$(1+|x|^{alpha} )|f(x)| leq 2|x|^{alpha} |f(x)|$ so $|f(x)| leq frac {2|x|^{alpha} |f(x)|} {(1+|x|^{alpha} )}$ for $|x| >1$. I hope this answers your question. (There are several errors in the statement; for example you have written 'for all $x in mathbb R$' in the definition which is not what you meant).
answered Nov 27 at 7:32
Kavi Rama Murthy
49.8k31854
If $f$ is of moderate decrease, doesn't that mean that, for all $x in mathbb{R}$, $|f(x)| leq frac{M}{1 + x^alpha}$? Please, point out the errors for me. Maybe I should not have used $f$ for the definition and for the function of the problem. But I really don't see the errors.
– Nuntractatuses Amável
Nov 27 at 7:37
1
Is $x^{alpha}$ defined for $x<0$?. A natural definition of moderate decrease should say $|f(x)| leq frac M {1+|x|^{alpha}}$ for $|x|$ sufficiently large, for some $M in (0,infty)$ and some $alpha >0$.
– Kavi Rama Murthy
Nov 27 at 7:40
I see your point. The definition in my textbook says for all $x$ and and $alpha = 2$. Also, we should have $alpha > 1$ so you can integrate over $mathbb{R}$. I edited the question according to your corrections. Thank you.
– Nuntractatuses Amável
Nov 27 at 7:48
$alpha >0$ was a typo in my comment. Including all $x$ means that the author wants some control for bounded $x$ also but it is not concerned with how the function decreases as $|x| to infty$.
– Kavi Rama Murthy
Nov 27 at 7:52
add a comment |
1
1
1
$(1+|x|^{alpha} )|f(x)| leq 2|x|^{alpha} |f(x)|$ so $|f(x)| leq frac {2|x|^{alpha} |f(x)|} {(1+|x|^{alpha} )}$ for $|x| >1$. I hope this answers your question. (There are several errors in the statement; for example you have written 'for all $x in mathbb R$' in the definition which is not what you meant).
answered Nov 27 at 7:32
Kavi Rama Murthy
49.8k31854
$(1+|x|^{alpha} )|f(x)| leq 2|x|^{alpha} |f(x)|$ so $|f(x)| leq frac {2|x|^{alpha} |f(x)|} {(1+|x|^{alpha} )}$ for $|x| >1$. I hope this answers your question. (There are several errors in the statement; for example you have written 'for all $x in mathbb R$' in the definition which is not what you meant).
answered Nov 27 at 7:32
Kavi Rama Murthy
49.8k31854
answered Nov 27 at 7:32
Kavi Rama Murthy
49.8k31854
answered Nov 27 at 7:32
Kavi Rama Murthy
49.8k31854
answered Nov 27 at 7:32
Kavi Rama Murthy
49.8k31854
49.8k31854
If $f$ is of moderate decrease, doesn't that mean that, for all $x in mathbb{R}$, $|f(x)| leq frac{M}{1 + x^alpha}$? Please, point out the errors for me. Maybe I should not have used $f$ for the definition and for the function of the problem. But I really don't see the errors.
– Nuntractatuses Amável
Nov 27 at 7:37
1
Is $x^{alpha}$ defined for $x<0$?. A natural definition of moderate decrease should say $|f(x)| leq frac M {1+|x|^{alpha}}$ for $|x|$ sufficiently large, for some $M in (0,infty)$ and some $alpha >0$.
– Kavi Rama Murthy
Nov 27 at 7:40
I see your point. The definition in my textbook says for all $x$ and and $alpha = 2$. Also, we should have $alpha > 1$ so you can integrate over $mathbb{R}$. I edited the question according to your corrections. Thank you.
– Nuntractatuses Amável
Nov 27 at 7:48
$alpha >0$ was a typo in my comment. Including all $x$ means that the author wants some control for bounded $x$ also but it is not concerned with how the function decreases as $|x| to infty$.
– Kavi Rama Murthy
Nov 27 at 7:52
add a comment |
If $f$ is of moderate decrease, doesn't that mean that, for all $x in mathbb{R}$, $|f(x)| leq frac{M}{1 + x^alpha}$? Please, point out the errors for me. Maybe I should not have used $f$ for the definition and for the function of the problem. But I really don't see the errors.
– Nuntractatuses Amável
Nov 27 at 7:37
1
Is $x^{alpha}$ defined for $x<0$?. A natural definition of moderate decrease should say $|f(x)| leq frac M {1+|x|^{alpha}}$ for $|x|$ sufficiently large, for some $M in (0,infty)$ and some $alpha >0$.
– Kavi Rama Murthy
Nov 27 at 7:40
I see your point. The definition in my textbook says for all $x$ and and $alpha = 2$. Also, we should have $alpha > 1$ so you can integrate over $mathbb{R}$. I edited the question according to your corrections. Thank you.
– Nuntractatuses Amável
Nov 27 at 7:48
$alpha >0$ was a typo in my comment. Including all $x$ means that the author wants some control for bounded $x$ also but it is not concerned with how the function decreases as $|x| to infty$.
– Kavi Rama Murthy
Nov 27 at 7:52
If $f$ is of moderate decrease, doesn't that mean that, for all $x in mathbb{R}$, $|f(x)| leq frac{M}{1 + x^alpha}$? Please, point out the errors for me. Maybe I should not have used $f$ for the definition and for the function of the problem. But I really don't see the errors.
– Nuntractatuses Amável
Nov 27 at 7:37
If $f$ is of moderate decrease, doesn't that mean that, for all $x in mathbb{R}$, $|f(x)| leq frac{M}{1 + x^alpha}$? Please, point out the errors for me. Maybe I should not have used $f$ for the definition and for the function of the problem. But I really don't see the errors.
– Nuntractatuses Amável
Nov 27 at 7:37
1
1
Is $x^{alpha}$ defined for $x<0$?. A natural definition of moderate decrease should say $|f(x)| leq frac M {1+|x|^{alpha}}$ for $|x|$ sufficiently large, for some $M in (0,infty)$ and some $alpha >0$.
– Kavi Rama Murthy
Nov 27 at 7:40
Is $x^{alpha}$ defined for $x<0$?. A natural definition of moderate decrease should say $|f(x)| leq frac M {1+|x|^{alpha}}$ for $|x|$ sufficiently large, for some $M in (0,infty)$ and some $alpha >0$.
– Kavi Rama Murthy
Nov 27 at 7:40
I see your point. The definition in my textbook says for all $x$ and and $alpha = 2$. Also, we should have $alpha > 1$ so you can integrate over $mathbb{R}$. I edited the question according to your corrections. Thank you.
– Nuntractatuses Amável
Nov 27 at 7:48
I see your point. The definition in my textbook says for all $x$ and and $alpha = 2$. Also, we should have $alpha > 1$ so you can integrate over $mathbb{R}$. I edited the question according to your corrections. Thank you.
– Nuntractatuses Amável
Nov 27 at 7:48
$alpha >0$ was a typo in my comment. Including all $x$ means that the author wants some control for bounded $x$ also but it is not concerned with how the function decreases as $|x| to infty$.
– Kavi Rama Murthy
Nov 27 at 7:52
$alpha >0$ was a typo in my comment. Including all $x$ means that the author wants some control for bounded $x$ also but it is not concerned with how the function decreases as $|x| to infty$.
– Kavi Rama Murthy
Nov 27 at 7:52
add a comment |
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No, because there is no bound near 0.
– T. Bongers
Nov 27 at 6:59
0 is not included, because it is when $|x| to + infty$. So there is a $x_0 > 0 $ such that if $|x| geq x_0$, $f(x) = O(frac{1}{x^alpha})$. So, inside $[-x_0, x_0]$, $f$ is limited, and decreases slower than $frac{1}{x^alpha}$ outside the interval.
– Nuntractatuses Amável
Nov 27 at 7:14
1
If you assume continuity or bounds on f, sure. But that's not a consequence of the big O estimate you wrote down.
– T. Bongers
Nov 27 at 7:16
Yes, forgot to add continuity
– Nuntractatuses Amável
Nov 27 at 7:17