Convolution of an unintegrable function and the convolution theorem
1
$begingroup$
1) Does a convolution of an integrable function and an unintegrable function exist?
2) If the answer for the first question is possible, then:
suppose a function $f(x)$ is integrable, and denote by $g(x)=sum_{|n|geq N} e^{inx}$ for a given $N$. Can we use the "convolution theorem" in this case?
(this is a sort of "high pass filter")
It seems to me that in order to do this, we need to change the order of summation and integration, but $g$ diverges.
calculus convolution harmonic-analysis
asked Nov 30 '18 at 22:37
BorisBoris
192
$endgroup$
add a comment |
1
$begingroup$
1) Does a convolution of an integrable function and an unintegrable function exist?
2) If the answer for the first question is possible, then:
suppose a function $f(x)$ is integrable, and denote by $g(x)=sum_{|n|geq N} e^{inx}$ for a given $N$. Can we use the "convolution theorem" in this case?
(this is a sort of "high pass filter")
It seems to me that in order to do this, we need to change the order of summation and integration, but $g$ diverges.
calculus convolution harmonic-analysis
asked Nov 30 '18 at 22:37
BorisBoris
192
$endgroup$
$begingroup$
by unintegrable, what do you exactly mean? the integral is not defined or it is unbounded above? or absolute integrability ? any other notions? what do you mean exactly? based on that the answer is different. for example consider convolution of an absolute integrable function with impulse train which exist, but impulse train is not integrable (the integral is unbounded above).
$endgroup$
– K.K.McDonald
Nov 30 '18 at 22:52
$begingroup$
Can you clarify (2)? Since the sum for $g(x)$ does not converge, that expression is meaningless. So what do you actually want to ask? For part (1), the answer is certainly yes.... for instance, let $f(x)=1$ (not integrable) and $g(x)=e^{-x^2/2}$; the convolution exists. More generally let $f(x)$ be any locally integrable but not integrable function, and let $g(x)$ be the indicator function on $[0,1]$; the convolution again exists.
$endgroup$
– mjqxxxx
Nov 30 '18 at 22:52
$begingroup$
Thank you for your answers and sorry for being unclear. By unintegrable I mean that on a finite interval the integral of the function diverges. To clarify 2: Suppose I define the series I have denoted by $g$ and perform a convolution with a function $f$. This should somehow work like a high pass filter for the frequencies of $f$. My question is whether the convolution is possible?
$endgroup$
– Boris
Nov 30 '18 at 23:22
add a comment |
1
1
1
$begingroup$
1) Does a convolution of an integrable function and an unintegrable function exist?
2) If the answer for the first question is possible, then:
suppose a function $f(x)$ is integrable, and denote by $g(x)=sum_{|n|geq N} e^{inx}$ for a given $N$. Can we use the "convolution theorem" in this case?
(this is a sort of "high pass filter")
It seems to me that in order to do this, we need to change the order of summation and integration, but $g$ diverges.
calculus convolution harmonic-analysis
asked Nov 30 '18 at 22:37
BorisBoris
192
$endgroup$
1) Does a convolution of an integrable function and an unintegrable function exist?
2) If the answer for the first question is possible, then:
suppose a function $f(x)$ is integrable, and denote by $g(x)=sum_{|n|geq N} e^{inx}$ for a given $N$. Can we use the "convolution theorem" in this case?
(this is a sort of "high pass filter")
It seems to me that in order to do this, we need to change the order of summation and integration, but $g$ diverges.
calculus convolution harmonic-analysis
calculus convolution harmonic-analysis
asked Nov 30 '18 at 22:37
BorisBoris
192
asked Nov 30 '18 at 22:37
BorisBoris
192
asked Nov 30 '18 at 22:37
BorisBoris
192
asked Nov 30 '18 at 22:37
BorisBoris
192
asked Nov 30 '18 at 22:37
BorisBoris
192
192
$begingroup$
by unintegrable, what do you exactly mean? the integral is not defined or it is unbounded above? or absolute integrability ? any other notions? what do you mean exactly? based on that the answer is different. for example consider convolution of an absolute integrable function with impulse train which exist, but impulse train is not integrable (the integral is unbounded above).
$endgroup$
– K.K.McDonald
Nov 30 '18 at 22:52
$begingroup$
Can you clarify (2)? Since the sum for $g(x)$ does not converge, that expression is meaningless. So what do you actually want to ask? For part (1), the answer is certainly yes.... for instance, let $f(x)=1$ (not integrable) and $g(x)=e^{-x^2/2}$; the convolution exists. More generally let $f(x)$ be any locally integrable but not integrable function, and let $g(x)$ be the indicator function on $[0,1]$; the convolution again exists.
$endgroup$
– mjqxxxx
Nov 30 '18 at 22:52
$begingroup$
Thank you for your answers and sorry for being unclear. By unintegrable I mean that on a finite interval the integral of the function diverges. To clarify 2: Suppose I define the series I have denoted by $g$ and perform a convolution with a function $f$. This should somehow work like a high pass filter for the frequencies of $f$. My question is whether the convolution is possible?
$endgroup$
– Boris
Nov 30 '18 at 23:22
add a comment |
$begingroup$
by unintegrable, what do you exactly mean? the integral is not defined or it is unbounded above? or absolute integrability ? any other notions? what do you mean exactly? based on that the answer is different. for example consider convolution of an absolute integrable function with impulse train which exist, but impulse train is not integrable (the integral is unbounded above).
$endgroup$
– K.K.McDonald
Nov 30 '18 at 22:52
$begingroup$
Can you clarify (2)? Since the sum for $g(x)$ does not converge, that expression is meaningless. So what do you actually want to ask? For part (1), the answer is certainly yes.... for instance, let $f(x)=1$ (not integrable) and $g(x)=e^{-x^2/2}$; the convolution exists. More generally let $f(x)$ be any locally integrable but not integrable function, and let $g(x)$ be the indicator function on $[0,1]$; the convolution again exists.
$endgroup$
– mjqxxxx
Nov 30 '18 at 22:52
$begingroup$
Thank you for your answers and sorry for being unclear. By unintegrable I mean that on a finite interval the integral of the function diverges. To clarify 2: Suppose I define the series I have denoted by $g$ and perform a convolution with a function $f$. This should somehow work like a high pass filter for the frequencies of $f$. My question is whether the convolution is possible?
$endgroup$
– Boris
Nov 30 '18 at 23:22
$begingroup$
by unintegrable, what do you exactly mean? the integral is not defined or it is unbounded above? or absolute integrability ? any other notions? what do you mean exactly? based on that the answer is different. for example consider convolution of an absolute integrable function with impulse train which exist, but impulse train is not integrable (the integral is unbounded above).
$endgroup$
– K.K.McDonald
Nov 30 '18 at 22:52
$begingroup$
by unintegrable, what do you exactly mean? the integral is not defined or it is unbounded above? or absolute integrability ? any other notions? what do you mean exactly? based on that the answer is different. for example consider convolution of an absolute integrable function with impulse train which exist, but impulse train is not integrable (the integral is unbounded above).
$endgroup$
– K.K.McDonald
Nov 30 '18 at 22:52
$begingroup$
Can you clarify (2)? Since the sum for $g(x)$ does not converge, that expression is meaningless. So what do you actually want to ask? For part (1), the answer is certainly yes.... for instance, let $f(x)=1$ (not integrable) and $g(x)=e^{-x^2/2}$; the convolution exists. More generally let $f(x)$ be any locally integrable but not integrable function, and let $g(x)$ be the indicator function on $[0,1]$; the convolution again exists.
$endgroup$
– mjqxxxx
Nov 30 '18 at 22:52
$begingroup$
Can you clarify (2)? Since the sum for $g(x)$ does not converge, that expression is meaningless. So what do you actually want to ask? For part (1), the answer is certainly yes.... for instance, let $f(x)=1$ (not integrable) and $g(x)=e^{-x^2/2}$; the convolution exists. More generally let $f(x)$ be any locally integrable but not integrable function, and let $g(x)$ be the indicator function on $[0,1]$; the convolution again exists.
$endgroup$
– mjqxxxx
Nov 30 '18 at 22:52
$begingroup$
Thank you for your answers and sorry for being unclear. By unintegrable I mean that on a finite interval the integral of the function diverges. To clarify 2: Suppose I define the series I have denoted by $g$ and perform a convolution with a function $f$. This should somehow work like a high pass filter for the frequencies of $f$. My question is whether the convolution is possible?
$endgroup$
– Boris
Nov 30 '18 at 23:22
$begingroup$
Thank you for your answers and sorry for being unclear. By unintegrable I mean that on a finite interval the integral of the function diverges. To clarify 2: Suppose I define the series I have denoted by $g$ and perform a convolution with a function $f$. This should somehow work like a high pass filter for the frequencies of $f$. My question is whether the convolution is possible?
$endgroup$
– Boris
Nov 30 '18 at 23:22
add a comment |
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$begingroup$
by unintegrable, what do you exactly mean? the integral is not defined or it is unbounded above? or absolute integrability ? any other notions? what do you mean exactly? based on that the answer is different. for example consider convolution of an absolute integrable function with impulse train which exist, but impulse train is not integrable (the integral is unbounded above).
$endgroup$
– K.K.McDonald
Nov 30 '18 at 22:52
$begingroup$
Can you clarify (2)? Since the sum for $g(x)$ does not converge, that expression is meaningless. So what do you actually want to ask? For part (1), the answer is certainly yes.... for instance, let $f(x)=1$ (not integrable) and $g(x)=e^{-x^2/2}$; the convolution exists. More generally let $f(x)$ be any locally integrable but not integrable function, and let $g(x)$ be the indicator function on $[0,1]$; the convolution again exists.
$endgroup$
– mjqxxxx
Nov 30 '18 at 22:52
$begingroup$
Thank you for your answers and sorry for being unclear. By unintegrable I mean that on a finite interval the integral of the function diverges. To clarify 2: Suppose I define the series I have denoted by $g$ and perform a convolution with a function $f$. This should somehow work like a high pass filter for the frequencies of $f$. My question is whether the convolution is possible?
$endgroup$
– Boris
Nov 30 '18 at 23:22