What does the integral of a delta distribution even mean?
$begingroup$
Formally, we define $delta(phi)=phi(0)$ where $phi$ comes from a suitable class of test function. Based on this, the expression $int_{-infty}^{infty} delta(x) dx$ seems completely meaningless and I'm unsure how to attribute meaning to integrals involving $delta$. I feel like I'm missing something important here - help!!
functional-analysis
asked Jan 14 at 16:39
DrefainDrefain
736
$endgroup$
add a comment |
$begingroup$
Formally, we define $delta(phi)=phi(0)$ where $phi$ comes from a suitable class of test function. Based on this, the expression $int_{-infty}^{infty} delta(x) dx$ seems completely meaningless and I'm unsure how to attribute meaning to integrals involving $delta$. I feel like I'm missing something important here - help!!
functional-analysis
asked Jan 14 at 16:39
DrefainDrefain
736
$endgroup$
$begingroup$
Well, a delta-distribution ist just defined to be integrable. So the value of the integral $int_{-infty}^{infty} delta(x) dx = 1$ is taken by definition. The wider purpose of delta-distribution is rather "picking values" as in $int_{-infty}^{infty} f(x) delta(x-x_0) dx = f(x_0)$. Also, it is well suited for analysis in transformations, as it has nice properties under transformations: e.g. the spectral integral under Fourier transformations. I'm not sure where your question futher directs to, i.e. which properties of the delta-distribution you are interested in.
$endgroup$
– Andreas
Jan 14 at 17:01
add a comment |
$begingroup$
Formally, we define $delta(phi)=phi(0)$ where $phi$ comes from a suitable class of test function. Based on this, the expression $int_{-infty}^{infty} delta(x) dx$ seems completely meaningless and I'm unsure how to attribute meaning to integrals involving $delta$. I feel like I'm missing something important here - help!!
functional-analysis
asked Jan 14 at 16:39
DrefainDrefain
736
$endgroup$
Formally, we define $delta(phi)=phi(0)$ where $phi$ comes from a suitable class of test function. Based on this, the expression $int_{-infty}^{infty} delta(x) dx$ seems completely meaningless and I'm unsure how to attribute meaning to integrals involving $delta$. I feel like I'm missing something important here - help!!
functional-analysis
functional-analysis
asked Jan 14 at 16:39
DrefainDrefain
736
asked Jan 14 at 16:39
DrefainDrefain
736
asked Jan 14 at 16:39
DrefainDrefain
736
asked Jan 14 at 16:39
DrefainDrefain
736
asked Jan 14 at 16:39
DrefainDrefain
736
736
$begingroup$
Well, a delta-distribution ist just defined to be integrable. So the value of the integral $int_{-infty}^{infty} delta(x) dx = 1$ is taken by definition. The wider purpose of delta-distribution is rather "picking values" as in $int_{-infty}^{infty} f(x) delta(x-x_0) dx = f(x_0)$. Also, it is well suited for analysis in transformations, as it has nice properties under transformations: e.g. the spectral integral under Fourier transformations. I'm not sure where your question futher directs to, i.e. which properties of the delta-distribution you are interested in.
$endgroup$
– Andreas
Jan 14 at 17:01
add a comment |
$begingroup$
Well, a delta-distribution ist just defined to be integrable. So the value of the integral $int_{-infty}^{infty} delta(x) dx = 1$ is taken by definition. The wider purpose of delta-distribution is rather "picking values" as in $int_{-infty}^{infty} f(x) delta(x-x_0) dx = f(x_0)$. Also, it is well suited for analysis in transformations, as it has nice properties under transformations: e.g. the spectral integral under Fourier transformations. I'm not sure where your question futher directs to, i.e. which properties of the delta-distribution you are interested in.
$endgroup$
– Andreas
Jan 14 at 17:01
$begingroup$
Well, a delta-distribution ist just defined to be integrable. So the value of the integral $int_{-infty}^{infty} delta(x) dx = 1$ is taken by definition. The wider purpose of delta-distribution is rather "picking values" as in $int_{-infty}^{infty} f(x) delta(x-x_0) dx = f(x_0)$. Also, it is well suited for analysis in transformations, as it has nice properties under transformations: e.g. the spectral integral under Fourier transformations. I'm not sure where your question futher directs to, i.e. which properties of the delta-distribution you are interested in.
$endgroup$
– Andreas
Jan 14 at 17:01
$begingroup$
Well, a delta-distribution ist just defined to be integrable. So the value of the integral $int_{-infty}^{infty} delta(x) dx = 1$ is taken by definition. The wider purpose of delta-distribution is rather "picking values" as in $int_{-infty}^{infty} f(x) delta(x-x_0) dx = f(x_0)$. Also, it is well suited for analysis in transformations, as it has nice properties under transformations: e.g. the spectral integral under Fourier transformations. I'm not sure where your question futher directs to, i.e. which properties of the delta-distribution you are interested in.
$endgroup$
– Andreas
Jan 14 at 17:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For a function $fin L^1_{text{loc}}(Bbb R)$, we can associate with it the distribution $T_fin mathcal D'(Bbb R)$ given by
$$
langle T_f, varphirangle := int_{-infty}^infty f(x)varphi(x) ,dx.
$$
The Dirac distribution is defined as
$$
langle delta, varphirangle := varphi(0)
$$
and is not associated with any $fin L^1_{text{loc}}(Bbb R)$ (but can be associated with an atomic measure). However, some people (the physicists) prefer to abuse the notation and write
$$
langle delta, varphirangle =: int_{-infty}^infty delta(x)varphi(x) ,dx
$$
anyway, as if $delta$ is a function. Hence we have
$$
int_{-infty}^infty delta(x),dx = langle delta, 1rangle = 1.
$$
Note that we can use the constant function $1$ as an input because $delta$ is not only a distribution, but a distribution with compact support, hence it acts on any $psiin C^infty(Bbb R)$.
To add more details about my final remark, we have the following inclusion
$$
mathcal D subset mathcal S subset mathcal E
$$
where $mathcal D$ is the space of smooth and compactly supported test functions, $mathcal S$ the space of Schwartz functions and $mathcal E$ the space of smooth functions (each with its appropriate topology). Their dual spaces can be identified with subspaces of $mathcal D'$, i.e.
$$
mathcal E' subset mathcal S' subset mathcal D'
$$
where $mathcal E'$ is the space of distributions with compact support and $mathcal S'$ the space of tempered distributions.
The Dirac distribution $delta$ belongs to $mathcal E'$ and hence $mathcal S'$, which is the "correct" space to do Fourier transform. This is what @Andreas meant in his comment about $delta$ interacting well with Fourier transform.
answered Jan 14 at 16:59
BigbearZzzBigbearZzz
9,08421753
$endgroup$
$begingroup$
This is quite clear now; I liked the part where you showed that you can extend the definition of $delta$ to make sense even for functions in $C^{infty}$; this is not something I had considered but since $delta$ does indeed have compact support this makes sense. Thanks!
$endgroup$
– Drefain
Jan 14 at 17:12
1
$begingroup$
@Drefain Seeing that you're interested, I edited the post to add some further details regarding distribution with compact support. Hope it helps.
$endgroup$
– BigbearZzz
Jan 15 at 22:14
add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
For a function $fin L^1_{text{loc}}(Bbb R)$, we can associate with it the distribution $T_fin mathcal D'(Bbb R)$ given by
$$
langle T_f, varphirangle := int_{-infty}^infty f(x)varphi(x) ,dx.
$$
The Dirac distribution is defined as
$$
langle delta, varphirangle := varphi(0)
$$
and is not associated with any $fin L^1_{text{loc}}(Bbb R)$ (but can be associated with an atomic measure). However, some people (the physicists) prefer to abuse the notation and write
$$
langle delta, varphirangle =: int_{-infty}^infty delta(x)varphi(x) ,dx
$$
anyway, as if $delta$ is a function. Hence we have
$$
int_{-infty}^infty delta(x),dx = langle delta, 1rangle = 1.
$$
Note that we can use the constant function $1$ as an input because $delta$ is not only a distribution, but a distribution with compact support, hence it acts on any $psiin C^infty(Bbb R)$.
To add more details about my final remark, we have the following inclusion
$$
mathcal D subset mathcal S subset mathcal E
$$
where $mathcal D$ is the space of smooth and compactly supported test functions, $mathcal S$ the space of Schwartz functions and $mathcal E$ the space of smooth functions (each with its appropriate topology). Their dual spaces can be identified with subspaces of $mathcal D'$, i.e.
$$
mathcal E' subset mathcal S' subset mathcal D'
$$
where $mathcal E'$ is the space of distributions with compact support and $mathcal S'$ the space of tempered distributions.
The Dirac distribution $delta$ belongs to $mathcal E'$ and hence $mathcal S'$, which is the "correct" space to do Fourier transform. This is what @Andreas meant in his comment about $delta$ interacting well with Fourier transform.
answered Jan 14 at 16:59
BigbearZzzBigbearZzz
9,08421753
$endgroup$
$begingroup$
This is quite clear now; I liked the part where you showed that you can extend the definition of $delta$ to make sense even for functions in $C^{infty}$; this is not something I had considered but since $delta$ does indeed have compact support this makes sense. Thanks!
$endgroup$
– Drefain
Jan 14 at 17:12
1
$begingroup$
@Drefain Seeing that you're interested, I edited the post to add some further details regarding distribution with compact support. Hope it helps.
$endgroup$
– BigbearZzz
Jan 15 at 22:14
add a comment |
$begingroup$
For a function $fin L^1_{text{loc}}(Bbb R)$, we can associate with it the distribution $T_fin mathcal D'(Bbb R)$ given by
$$
langle T_f, varphirangle := int_{-infty}^infty f(x)varphi(x) ,dx.
$$
The Dirac distribution is defined as
$$
langle delta, varphirangle := varphi(0)
$$
and is not associated with any $fin L^1_{text{loc}}(Bbb R)$ (but can be associated with an atomic measure). However, some people (the physicists) prefer to abuse the notation and write
$$
langle delta, varphirangle =: int_{-infty}^infty delta(x)varphi(x) ,dx
$$
anyway, as if $delta$ is a function. Hence we have
$$
int_{-infty}^infty delta(x),dx = langle delta, 1rangle = 1.
$$
Note that we can use the constant function $1$ as an input because $delta$ is not only a distribution, but a distribution with compact support, hence it acts on any $psiin C^infty(Bbb R)$.
To add more details about my final remark, we have the following inclusion
$$
mathcal D subset mathcal S subset mathcal E
$$
where $mathcal D$ is the space of smooth and compactly supported test functions, $mathcal S$ the space of Schwartz functions and $mathcal E$ the space of smooth functions (each with its appropriate topology). Their dual spaces can be identified with subspaces of $mathcal D'$, i.e.
$$
mathcal E' subset mathcal S' subset mathcal D'
$$
where $mathcal E'$ is the space of distributions with compact support and $mathcal S'$ the space of tempered distributions.
The Dirac distribution $delta$ belongs to $mathcal E'$ and hence $mathcal S'$, which is the "correct" space to do Fourier transform. This is what @Andreas meant in his comment about $delta$ interacting well with Fourier transform.
answered Jan 14 at 16:59
BigbearZzzBigbearZzz
9,08421753
$endgroup$
$begingroup$
This is quite clear now; I liked the part where you showed that you can extend the definition of $delta$ to make sense even for functions in $C^{infty}$; this is not something I had considered but since $delta$ does indeed have compact support this makes sense. Thanks!
$endgroup$
– Drefain
Jan 14 at 17:12
1
$begingroup$
@Drefain Seeing that you're interested, I edited the post to add some further details regarding distribution with compact support. Hope it helps.
$endgroup$
– BigbearZzz
Jan 15 at 22:14
add a comment |
$begingroup$
For a function $fin L^1_{text{loc}}(Bbb R)$, we can associate with it the distribution $T_fin mathcal D'(Bbb R)$ given by
$$
langle T_f, varphirangle := int_{-infty}^infty f(x)varphi(x) ,dx.
$$
The Dirac distribution is defined as
$$
langle delta, varphirangle := varphi(0)
$$
and is not associated with any $fin L^1_{text{loc}}(Bbb R)$ (but can be associated with an atomic measure). However, some people (the physicists) prefer to abuse the notation and write
$$
langle delta, varphirangle =: int_{-infty}^infty delta(x)varphi(x) ,dx
$$
anyway, as if $delta$ is a function. Hence we have
$$
int_{-infty}^infty delta(x),dx = langle delta, 1rangle = 1.
$$
Note that we can use the constant function $1$ as an input because $delta$ is not only a distribution, but a distribution with compact support, hence it acts on any $psiin C^infty(Bbb R)$.
To add more details about my final remark, we have the following inclusion
$$
mathcal D subset mathcal S subset mathcal E
$$
where $mathcal D$ is the space of smooth and compactly supported test functions, $mathcal S$ the space of Schwartz functions and $mathcal E$ the space of smooth functions (each with its appropriate topology). Their dual spaces can be identified with subspaces of $mathcal D'$, i.e.
$$
mathcal E' subset mathcal S' subset mathcal D'
$$
where $mathcal E'$ is the space of distributions with compact support and $mathcal S'$ the space of tempered distributions.
The Dirac distribution $delta$ belongs to $mathcal E'$ and hence $mathcal S'$, which is the "correct" space to do Fourier transform. This is what @Andreas meant in his comment about $delta$ interacting well with Fourier transform.
answered Jan 14 at 16:59
BigbearZzzBigbearZzz
9,08421753
$endgroup$
For a function $fin L^1_{text{loc}}(Bbb R)$, we can associate with it the distribution $T_fin mathcal D'(Bbb R)$ given by
$$
langle T_f, varphirangle := int_{-infty}^infty f(x)varphi(x) ,dx.
$$
The Dirac distribution is defined as
$$
langle delta, varphirangle := varphi(0)
$$
and is not associated with any $fin L^1_{text{loc}}(Bbb R)$ (but can be associated with an atomic measure). However, some people (the physicists) prefer to abuse the notation and write
$$
langle delta, varphirangle =: int_{-infty}^infty delta(x)varphi(x) ,dx
$$
anyway, as if $delta$ is a function. Hence we have
$$
int_{-infty}^infty delta(x),dx = langle delta, 1rangle = 1.
$$
Note that we can use the constant function $1$ as an input because $delta$ is not only a distribution, but a distribution with compact support, hence it acts on any $psiin C^infty(Bbb R)$.
To add more details about my final remark, we have the following inclusion
$$
mathcal D subset mathcal S subset mathcal E
$$
where $mathcal D$ is the space of smooth and compactly supported test functions, $mathcal S$ the space of Schwartz functions and $mathcal E$ the space of smooth functions (each with its appropriate topology). Their dual spaces can be identified with subspaces of $mathcal D'$, i.e.
$$
mathcal E' subset mathcal S' subset mathcal D'
$$
where $mathcal E'$ is the space of distributions with compact support and $mathcal S'$ the space of tempered distributions.
The Dirac distribution $delta$ belongs to $mathcal E'$ and hence $mathcal S'$, which is the "correct" space to do Fourier transform. This is what @Andreas meant in his comment about $delta$ interacting well with Fourier transform.
answered Jan 14 at 16:59
BigbearZzzBigbearZzz
9,08421753
edited Jan 15 at 23:22
answered Jan 14 at 16:59
BigbearZzzBigbearZzz
9,08421753
answered Jan 14 at 16:59
BigbearZzzBigbearZzz
9,08421753
answered Jan 14 at 16:59
BigbearZzzBigbearZzz
9,08421753
9,08421753
$begingroup$
This is quite clear now; I liked the part where you showed that you can extend the definition of $delta$ to make sense even for functions in $C^{infty}$; this is not something I had considered but since $delta$ does indeed have compact support this makes sense. Thanks!
$endgroup$
– Drefain
Jan 14 at 17:12
1
$begingroup$
@Drefain Seeing that you're interested, I edited the post to add some further details regarding distribution with compact support. Hope it helps.
$endgroup$
– BigbearZzz
Jan 15 at 22:14
add a comment |
$begingroup$
This is quite clear now; I liked the part where you showed that you can extend the definition of $delta$ to make sense even for functions in $C^{infty}$; this is not something I had considered but since $delta$ does indeed have compact support this makes sense. Thanks!
$endgroup$
– Drefain
Jan 14 at 17:12
1
$begingroup$
@Drefain Seeing that you're interested, I edited the post to add some further details regarding distribution with compact support. Hope it helps.
$endgroup$
– BigbearZzz
Jan 15 at 22:14
$begingroup$
This is quite clear now; I liked the part where you showed that you can extend the definition of $delta$ to make sense even for functions in $C^{infty}$; this is not something I had considered but since $delta$ does indeed have compact support this makes sense. Thanks!
$endgroup$
– Drefain
Jan 14 at 17:12
$begingroup$
This is quite clear now; I liked the part where you showed that you can extend the definition of $delta$ to make sense even for functions in $C^{infty}$; this is not something I had considered but since $delta$ does indeed have compact support this makes sense. Thanks!
$endgroup$
– Drefain
Jan 14 at 17:12
1
1
$begingroup$
@Drefain Seeing that you're interested, I edited the post to add some further details regarding distribution with compact support. Hope it helps.
$endgroup$
– BigbearZzz
Jan 15 at 22:14
$begingroup$
@Drefain Seeing that you're interested, I edited the post to add some further details regarding distribution with compact support. Hope it helps.
$endgroup$
– BigbearZzz
Jan 15 at 22:14
add a comment |
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$begingroup$
Well, a delta-distribution ist just defined to be integrable. So the value of the integral $int_{-infty}^{infty} delta(x) dx = 1$ is taken by definition. The wider purpose of delta-distribution is rather "picking values" as in $int_{-infty}^{infty} f(x) delta(x-x_0) dx = f(x_0)$. Also, it is well suited for analysis in transformations, as it has nice properties under transformations: e.g. the spectral integral under Fourier transformations. I'm not sure where your question futher directs to, i.e. which properties of the delta-distribution you are interested in.
$endgroup$
– Andreas
Jan 14 at 17:01