distribution associated with a discontinuous function
Let $fcolonmathbb{R}tomathbb{R}$ be such that, for every $ninmathbb{Z}$, $f$ is differentiable on $left(n,n+1right)$ and $n$ is a discontinuity of first kind of $f$. We define
$$T_f(phi)=int_{mathbb{R}}f(x)phi(x)dx,quadtext{where }phitext{ is a test function}.$$
I understand that, when $f$ is continuously differentiable, $T_f$ is well defined and the derivative of $T_f$ is just $T_{f'}$. However, these two do not seem clear in this case.
My attempt: Since $n$ and $n+1$ are discontinuities of first kind of $f$, we can define a new function, say $g_{n}$ such that $g_{n}$ is continuous on $[n,n+1]$ and equal to $f$ a.e. Thus $f$ is (Lebesgue) integrable on $[n,n+1]$. We then conclude that $f$ is locally integrable on $mathbb{R}$ and therefore $T_{f}$ is well defined.
My question: 1) Is the above reasoning (to show that $T_f$ is well defined) correct? And:
2) What would the derivative of $T_f$ look like in this case? I think it is not $T_{f'}$ as $f'$ is not defined at $n$.
Any help/hint is highly appreciated.
functional-analysis derivatives distribution-theory topological-vector-spaces discontinuous-functions
asked Nov 27 at 3:59
weirdo
420210
add a comment |
Let $fcolonmathbb{R}tomathbb{R}$ be such that, for every $ninmathbb{Z}$, $f$ is differentiable on $left(n,n+1right)$ and $n$ is a discontinuity of first kind of $f$. We define
$$T_f(phi)=int_{mathbb{R}}f(x)phi(x)dx,quadtext{where }phitext{ is a test function}.$$
I understand that, when $f$ is continuously differentiable, $T_f$ is well defined and the derivative of $T_f$ is just $T_{f'}$. However, these two do not seem clear in this case.
My attempt: Since $n$ and $n+1$ are discontinuities of first kind of $f$, we can define a new function, say $g_{n}$ such that $g_{n}$ is continuous on $[n,n+1]$ and equal to $f$ a.e. Thus $f$ is (Lebesgue) integrable on $[n,n+1]$. We then conclude that $f$ is locally integrable on $mathbb{R}$ and therefore $T_{f}$ is well defined.
My question: 1) Is the above reasoning (to show that $T_f$ is well defined) correct? And:
2) What would the derivative of $T_f$ look like in this case? I think it is not $T_{f'}$ as $f'$ is not defined at $n$.
Any help/hint is highly appreciated.
functional-analysis derivatives distribution-theory topological-vector-spaces discontinuous-functions
asked Nov 27 at 3:59
weirdo
420210
add a comment |
Let $fcolonmathbb{R}tomathbb{R}$ be such that, for every $ninmathbb{Z}$, $f$ is differentiable on $left(n,n+1right)$ and $n$ is a discontinuity of first kind of $f$. We define
$$T_f(phi)=int_{mathbb{R}}f(x)phi(x)dx,quadtext{where }phitext{ is a test function}.$$
I understand that, when $f$ is continuously differentiable, $T_f$ is well defined and the derivative of $T_f$ is just $T_{f'}$. However, these two do not seem clear in this case.
My attempt: Since $n$ and $n+1$ are discontinuities of first kind of $f$, we can define a new function, say $g_{n}$ such that $g_{n}$ is continuous on $[n,n+1]$ and equal to $f$ a.e. Thus $f$ is (Lebesgue) integrable on $[n,n+1]$. We then conclude that $f$ is locally integrable on $mathbb{R}$ and therefore $T_{f}$ is well defined.
My question: 1) Is the above reasoning (to show that $T_f$ is well defined) correct? And:
2) What would the derivative of $T_f$ look like in this case? I think it is not $T_{f'}$ as $f'$ is not defined at $n$.
Any help/hint is highly appreciated.
functional-analysis derivatives distribution-theory topological-vector-spaces discontinuous-functions
asked Nov 27 at 3:59
weirdo
420210
Let $fcolonmathbb{R}tomathbb{R}$ be such that, for every $ninmathbb{Z}$, $f$ is differentiable on $left(n,n+1right)$ and $n$ is a discontinuity of first kind of $f$. We define
$$T_f(phi)=int_{mathbb{R}}f(x)phi(x)dx,quadtext{where }phitext{ is a test function}.$$
I understand that, when $f$ is continuously differentiable, $T_f$ is well defined and the derivative of $T_f$ is just $T_{f'}$. However, these two do not seem clear in this case.
My attempt: Since $n$ and $n+1$ are discontinuities of first kind of $f$, we can define a new function, say $g_{n}$ such that $g_{n}$ is continuous on $[n,n+1]$ and equal to $f$ a.e. Thus $f$ is (Lebesgue) integrable on $[n,n+1]$. We then conclude that $f$ is locally integrable on $mathbb{R}$ and therefore $T_{f}$ is well defined.
My question: 1) Is the above reasoning (to show that $T_f$ is well defined) correct? And:
2) What would the derivative of $T_f$ look like in this case? I think it is not $T_{f'}$ as $f'$ is not defined at $n$.
Any help/hint is highly appreciated.
functional-analysis derivatives distribution-theory topological-vector-spaces discontinuous-functions
functional-analysis derivatives distribution-theory topological-vector-spaces discontinuous-functions
asked Nov 27 at 3:59
weirdo
420210
asked Nov 27 at 3:59
weirdo
420210
asked Nov 27 at 3:59
weirdo
420210
asked Nov 27 at 3:59
weirdo
420210
asked Nov 27 at 3:59
weirdo
420210
420210
add a comment |
add a comment |
2 Answers
2
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oldest
votes
If $f$ is absolutely continuous on each segment $[n, n + 1]$ when defined to take the values $f(n + 0)$ and $f((n + 1) - 0)$ at the endpoints, integration by parts shows that the distributional derivative is
$$f'(x) + sum_n (f(n + 0) - f(n - 0)) delta(x - n),$$
where $f'(x)$ is the ordinary derivative. If the conditions for integration by parts do not hold, the distributional derivative may not be representable in this form. Also, if the test functions do not have finite support, it is necessary to add a condition on how fast $f$ can grow.
answered Dec 1 at 6:42
Maxim
4,5031219
add a comment |
It is important to use the fact that $f$ has left and right limits at integer points. Otherwise $f$ need not be locally integrable and $T_f$ is not a well defined distribution. The fact that these limits exist implies that it is bounded in $(n,n+1)$ and hence it is locally integrable. Any locally integrable function defines a distribution. In 2) there is no guarantee that $f'$ is even locally integrable so $T_f'$ cannot be written in terms of $f'$ without extra assumptions.
answered Nov 27 at 6:18
Kavi Rama Murthy
49.8k31854
For the Heaviside function $f(x)=1$ for $xge 0$ and $=0$ else, the derivative is $0$ a.e. but $T_f'=delta_0$ and hence different from $T_{f'}$.
– Jochen
Nov 27 at 7:44
@Jochen Thank you. I have omitted the last part of my answer.
– Kavi Rama Murthy
Nov 27 at 7:46
@KaviRamaMurthy So my attempt to show that $f$ is locally integrable was correct? I was not sure which result we can use to deduce that, if $f$ is integrable on every interval $[n,n+1]$, then $f$ is integrable on each of the finite union of these sets...Perhaps I forgot something in measure theory
– weirdo
Nov 27 at 16:41
1
@weirdo $fI_{cup E_i}=sum fI_{E_i}$ if $E_i$'s are disjoint. Finite sum of integrable functions is integrable.
– Kavi Rama Murthy
Nov 27 at 23:08
add a comment |
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2 Answers
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2 Answers
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If $f$ is absolutely continuous on each segment $[n, n + 1]$ when defined to take the values $f(n + 0)$ and $f((n + 1) - 0)$ at the endpoints, integration by parts shows that the distributional derivative is
$$f'(x) + sum_n (f(n + 0) - f(n - 0)) delta(x - n),$$
where $f'(x)$ is the ordinary derivative. If the conditions for integration by parts do not hold, the distributional derivative may not be representable in this form. Also, if the test functions do not have finite support, it is necessary to add a condition on how fast $f$ can grow.
answered Dec 1 at 6:42
Maxim
4,5031219
add a comment |
If $f$ is absolutely continuous on each segment $[n, n + 1]$ when defined to take the values $f(n + 0)$ and $f((n + 1) - 0)$ at the endpoints, integration by parts shows that the distributional derivative is
$$f'(x) + sum_n (f(n + 0) - f(n - 0)) delta(x - n),$$
where $f'(x)$ is the ordinary derivative. If the conditions for integration by parts do not hold, the distributional derivative may not be representable in this form. Also, if the test functions do not have finite support, it is necessary to add a condition on how fast $f$ can grow.
answered Dec 1 at 6:42
Maxim
4,5031219
add a comment |
If $f$ is absolutely continuous on each segment $[n, n + 1]$ when defined to take the values $f(n + 0)$ and $f((n + 1) - 0)$ at the endpoints, integration by parts shows that the distributional derivative is
$$f'(x) + sum_n (f(n + 0) - f(n - 0)) delta(x - n),$$
where $f'(x)$ is the ordinary derivative. If the conditions for integration by parts do not hold, the distributional derivative may not be representable in this form. Also, if the test functions do not have finite support, it is necessary to add a condition on how fast $f$ can grow.
answered Dec 1 at 6:42
Maxim
4,5031219
If $f$ is absolutely continuous on each segment $[n, n + 1]$ when defined to take the values $f(n + 0)$ and $f((n + 1) - 0)$ at the endpoints, integration by parts shows that the distributional derivative is
$$f'(x) + sum_n (f(n + 0) - f(n - 0)) delta(x - n),$$
where $f'(x)$ is the ordinary derivative. If the conditions for integration by parts do not hold, the distributional derivative may not be representable in this form. Also, if the test functions do not have finite support, it is necessary to add a condition on how fast $f$ can grow.
answered Dec 1 at 6:42
Maxim
4,5031219
edited Dec 7 at 15:39
answered Dec 1 at 6:42
Maxim
4,5031219
answered Dec 1 at 6:42
Maxim
4,5031219
answered Dec 1 at 6:42
Maxim
4,5031219
4,5031219
add a comment |
add a comment |
It is important to use the fact that $f$ has left and right limits at integer points. Otherwise $f$ need not be locally integrable and $T_f$ is not a well defined distribution. The fact that these limits exist implies that it is bounded in $(n,n+1)$ and hence it is locally integrable. Any locally integrable function defines a distribution. In 2) there is no guarantee that $f'$ is even locally integrable so $T_f'$ cannot be written in terms of $f'$ without extra assumptions.
answered Nov 27 at 6:18
Kavi Rama Murthy
49.8k31854
For the Heaviside function $f(x)=1$ for $xge 0$ and $=0$ else, the derivative is $0$ a.e. but $T_f'=delta_0$ and hence different from $T_{f'}$.
– Jochen
Nov 27 at 7:44
@Jochen Thank you. I have omitted the last part of my answer.
– Kavi Rama Murthy
Nov 27 at 7:46
@KaviRamaMurthy So my attempt to show that $f$ is locally integrable was correct? I was not sure which result we can use to deduce that, if $f$ is integrable on every interval $[n,n+1]$, then $f$ is integrable on each of the finite union of these sets...Perhaps I forgot something in measure theory
– weirdo
Nov 27 at 16:41
1
@weirdo $fI_{cup E_i}=sum fI_{E_i}$ if $E_i$'s are disjoint. Finite sum of integrable functions is integrable.
– Kavi Rama Murthy
Nov 27 at 23:08
add a comment |
It is important to use the fact that $f$ has left and right limits at integer points. Otherwise $f$ need not be locally integrable and $T_f$ is not a well defined distribution. The fact that these limits exist implies that it is bounded in $(n,n+1)$ and hence it is locally integrable. Any locally integrable function defines a distribution. In 2) there is no guarantee that $f'$ is even locally integrable so $T_f'$ cannot be written in terms of $f'$ without extra assumptions.
answered Nov 27 at 6:18
Kavi Rama Murthy
49.8k31854
For the Heaviside function $f(x)=1$ for $xge 0$ and $=0$ else, the derivative is $0$ a.e. but $T_f'=delta_0$ and hence different from $T_{f'}$.
– Jochen
Nov 27 at 7:44
@Jochen Thank you. I have omitted the last part of my answer.
– Kavi Rama Murthy
Nov 27 at 7:46
@KaviRamaMurthy So my attempt to show that $f$ is locally integrable was correct? I was not sure which result we can use to deduce that, if $f$ is integrable on every interval $[n,n+1]$, then $f$ is integrable on each of the finite union of these sets...Perhaps I forgot something in measure theory
– weirdo
Nov 27 at 16:41
1
@weirdo $fI_{cup E_i}=sum fI_{E_i}$ if $E_i$'s are disjoint. Finite sum of integrable functions is integrable.
– Kavi Rama Murthy
Nov 27 at 23:08
add a comment |
It is important to use the fact that $f$ has left and right limits at integer points. Otherwise $f$ need not be locally integrable and $T_f$ is not a well defined distribution. The fact that these limits exist implies that it is bounded in $(n,n+1)$ and hence it is locally integrable. Any locally integrable function defines a distribution. In 2) there is no guarantee that $f'$ is even locally integrable so $T_f'$ cannot be written in terms of $f'$ without extra assumptions.
answered Nov 27 at 6:18
Kavi Rama Murthy
49.8k31854
It is important to use the fact that $f$ has left and right limits at integer points. Otherwise $f$ need not be locally integrable and $T_f$ is not a well defined distribution. The fact that these limits exist implies that it is bounded in $(n,n+1)$ and hence it is locally integrable. Any locally integrable function defines a distribution. In 2) there is no guarantee that $f'$ is even locally integrable so $T_f'$ cannot be written in terms of $f'$ without extra assumptions.
answered Nov 27 at 6:18
Kavi Rama Murthy
49.8k31854
edited Nov 27 at 7:46
answered Nov 27 at 6:18
Kavi Rama Murthy
49.8k31854
answered Nov 27 at 6:18
Kavi Rama Murthy
49.8k31854
answered Nov 27 at 6:18
Kavi Rama Murthy
49.8k31854
49.8k31854
For the Heaviside function $f(x)=1$ for $xge 0$ and $=0$ else, the derivative is $0$ a.e. but $T_f'=delta_0$ and hence different from $T_{f'}$.
– Jochen
Nov 27 at 7:44
@Jochen Thank you. I have omitted the last part of my answer.
– Kavi Rama Murthy
Nov 27 at 7:46
@KaviRamaMurthy So my attempt to show that $f$ is locally integrable was correct? I was not sure which result we can use to deduce that, if $f$ is integrable on every interval $[n,n+1]$, then $f$ is integrable on each of the finite union of these sets...Perhaps I forgot something in measure theory
– weirdo
Nov 27 at 16:41
1
@weirdo $fI_{cup E_i}=sum fI_{E_i}$ if $E_i$'s are disjoint. Finite sum of integrable functions is integrable.
– Kavi Rama Murthy
Nov 27 at 23:08
add a comment |
For the Heaviside function $f(x)=1$ for $xge 0$ and $=0$ else, the derivative is $0$ a.e. but $T_f'=delta_0$ and hence different from $T_{f'}$.
– Jochen
Nov 27 at 7:44
@Jochen Thank you. I have omitted the last part of my answer.
– Kavi Rama Murthy
Nov 27 at 7:46
@KaviRamaMurthy So my attempt to show that $f$ is locally integrable was correct? I was not sure which result we can use to deduce that, if $f$ is integrable on every interval $[n,n+1]$, then $f$ is integrable on each of the finite union of these sets...Perhaps I forgot something in measure theory
– weirdo
Nov 27 at 16:41
1
@weirdo $fI_{cup E_i}=sum fI_{E_i}$ if $E_i$'s are disjoint. Finite sum of integrable functions is integrable.
– Kavi Rama Murthy
Nov 27 at 23:08
For the Heaviside function $f(x)=1$ for $xge 0$ and $=0$ else, the derivative is $0$ a.e. but $T_f'=delta_0$ and hence different from $T_{f'}$.
– Jochen
Nov 27 at 7:44
For the Heaviside function $f(x)=1$ for $xge 0$ and $=0$ else, the derivative is $0$ a.e. but $T_f'=delta_0$ and hence different from $T_{f'}$.
– Jochen
Nov 27 at 7:44
@Jochen Thank you. I have omitted the last part of my answer.
– Kavi Rama Murthy
Nov 27 at 7:46
@Jochen Thank you. I have omitted the last part of my answer.
– Kavi Rama Murthy
Nov 27 at 7:46
@KaviRamaMurthy So my attempt to show that $f$ is locally integrable was correct? I was not sure which result we can use to deduce that, if $f$ is integrable on every interval $[n,n+1]$, then $f$ is integrable on each of the finite union of these sets...Perhaps I forgot something in measure theory
– weirdo
Nov 27 at 16:41
@KaviRamaMurthy So my attempt to show that $f$ is locally integrable was correct? I was not sure which result we can use to deduce that, if $f$ is integrable on every interval $[n,n+1]$, then $f$ is integrable on each of the finite union of these sets...Perhaps I forgot something in measure theory
– weirdo
Nov 27 at 16:41
1
1
@weirdo $fI_{cup E_i}=sum fI_{E_i}$ if $E_i$'s are disjoint. Finite sum of integrable functions is integrable.
– Kavi Rama Murthy
Nov 27 at 23:08
@weirdo $fI_{cup E_i}=sum fI_{E_i}$ if $E_i$'s are disjoint. Finite sum of integrable functions is integrable.
– Kavi Rama Murthy
Nov 27 at 23:08
add a comment |
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