Mixing
What is the total variation of a dirac delta function $delta(x)$?
$begingroup$
What is the total variation of a dirac delta function $delta(x)$? My guess is that it is something like $infty$. If not defined, what would be the best way to define?
real-analysis definition dirac-delta bounded-variation
asked Aug 16 '16 at 13:11
Rajesh DachirajuRajesh Dachiraju
1,08942870
$endgroup$
add a comment |
$begingroup$
What is the total variation of a dirac delta function $delta(x)$? My guess is that it is something like $infty$. If not defined, what would be the best way to define?
real-analysis definition dirac-delta bounded-variation
asked Aug 16 '16 at 13:11
Rajesh DachirajuRajesh Dachiraju
1,08942870
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Pedro Tamaroff♦
Sep 1 '16 at 13:30
$begingroup$
@PedroTamaroff : When you move things to "chat", the MathJax within the comments doesn't get rendered.
$endgroup$
– Michael Hardy
Sep 4 '16 at 1:09
$begingroup$
@MichaelHardy You have to use ChatJax or its variants.
$endgroup$
– Pedro Tamaroff♦
Sep 4 '16 at 1:14
$begingroup$
Here, when you say dirac do you mean that it takes values $1$ and $0$ or values $0$ and $infty$? I'm guessing the later...?
$endgroup$
– AIM_BLB
Jan 31 at 23:51
add a comment |
$begingroup$
What is the total variation of a dirac delta function $delta(x)$? My guess is that it is something like $infty$. If not defined, what would be the best way to define?
real-analysis definition dirac-delta bounded-variation
asked Aug 16 '16 at 13:11
Rajesh DachirajuRajesh Dachiraju
1,08942870
$endgroup$
What is the total variation of a dirac delta function $delta(x)$? My guess is that it is something like $infty$. If not defined, what would be the best way to define?
real-analysis definition dirac-delta bounded-variation
real-analysis definition dirac-delta bounded-variation
asked Aug 16 '16 at 13:11
Rajesh DachirajuRajesh Dachiraju
1,08942870
asked Aug 16 '16 at 13:11
Rajesh DachirajuRajesh Dachiraju
1,08942870
asked Aug 16 '16 at 13:11
Rajesh DachirajuRajesh Dachiraju
1,08942870
asked Aug 16 '16 at 13:11
Rajesh DachirajuRajesh Dachiraju
1,08942870
asked Aug 16 '16 at 13:11
Rajesh DachirajuRajesh Dachiraju
1,08942870
1,08942870
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Pedro Tamaroff♦
Sep 1 '16 at 13:30
$begingroup$
@PedroTamaroff : When you move things to "chat", the MathJax within the comments doesn't get rendered.
$endgroup$
– Michael Hardy
Sep 4 '16 at 1:09
$begingroup$
@MichaelHardy You have to use ChatJax or its variants.
$endgroup$
– Pedro Tamaroff♦
Sep 4 '16 at 1:14
$begingroup$
Here, when you say dirac do you mean that it takes values $1$ and $0$ or values $0$ and $infty$? I'm guessing the later...?
$endgroup$
– AIM_BLB
Jan 31 at 23:51
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Pedro Tamaroff♦
Sep 1 '16 at 13:30
$begingroup$
@PedroTamaroff : When you move things to "chat", the MathJax within the comments doesn't get rendered.
$endgroup$
– Michael Hardy
Sep 4 '16 at 1:09
$begingroup$
@MichaelHardy You have to use ChatJax or its variants.
$endgroup$
– Pedro Tamaroff♦
Sep 4 '16 at 1:14
$begingroup$
Here, when you say dirac do you mean that it takes values $1$ and $0$ or values $0$ and $infty$? I'm guessing the later...?
$endgroup$
– AIM_BLB
Jan 31 at 23:51
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Pedro Tamaroff♦
Sep 1 '16 at 13:30
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Pedro Tamaroff♦
Sep 1 '16 at 13:30
$begingroup$
@PedroTamaroff : When you move things to "chat", the MathJax within the comments doesn't get rendered.
$endgroup$
– Michael Hardy
Sep 4 '16 at 1:09
$begingroup$
@PedroTamaroff : When you move things to "chat", the MathJax within the comments doesn't get rendered.
$endgroup$
– Michael Hardy
Sep 4 '16 at 1:09
$begingroup$
@MichaelHardy You have to use ChatJax or its variants.
$endgroup$
– Pedro Tamaroff♦
Sep 4 '16 at 1:14
$begingroup$
@MichaelHardy You have to use ChatJax or its variants.
$endgroup$
– Pedro Tamaroff♦
Sep 4 '16 at 1:14
$begingroup$
Here, when you say dirac do you mean that it takes values $1$ and $0$ or values $0$ and $infty$? I'm guessing the later...?
$endgroup$
– AIM_BLB
Jan 31 at 23:51
$begingroup$
Here, when you say dirac do you mean that it takes values $1$ and $0$ or values $0$ and $infty$? I'm guessing the later...?
$endgroup$
– AIM_BLB
Jan 31 at 23:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $(f_1,f_2,f_3,ldots)$ be any sequence of functions of bounded variation such that, given any continuous function $g$,$$lim_{ntoinfty}int f_ng=g(0).$$(Of course, there are many such sequences, but the argument is independent of that.) Then$$lim_{ntoinfty}mathrm{TV}(f_n)=infty.$$This is one way to interpret your question, and then the answer is indeed $infty$.
answered Mar 16 '18 at 21:08
Toby BartelsToby Bartels
679516
$endgroup$
$begingroup$
I think pretty much the only other interpretation that makes any sense is to think of $delta$ as a measure and then consider its total variation. (This interpretation gives a different result of course.)
$endgroup$
– Ian
Mar 16 '18 at 21:12
1
$begingroup$
Yes, but that seems to me to be misinterpreting the original question, which seems to want $delta$ to be a generalized function. You can formalize some generalized functions as measures, but in this formalism, an ordinary function $f$ (if integrable) is represented by the signed measure $A mapsto int_A f$, and the total variation of this measure is not the total variation of $f$. So this is not giving us a notion of the total variation of a generalized function, but rather a notion of the total variation of an indefinite integral of a generalized function.
$endgroup$
– Toby Bartels
Mar 18 '18 at 20:58
add a comment |
$begingroup$
To clarify this, lets agree on the definitions of which we're using. Let $(X,Sigma)$ be a measurable space.
For any measure $mu$ on $(X,Sigma)$ define, the upper and lower variation (respectively) as follows
$$
overline{mathrm{W}}(mu,E)=supleft{mu(A)mid AinSigmatext{ and }Asubset E right}qquadforall EinSigma\
underline{mathrm{W}}(mu,E)=infleft{mu(A)mid AinSigmatext{ and }Asubset E right}qquadforall EinSigma.
$$
From this, we may define the total variation (extended-value) "norm" of a measure as follows
$$
|mu|_{TV}triangleq sup_{E in Sigma} left(
overline{mathrm{W}}(mu,E) + left| underline{mathrm{W}}(mu,E)right|
right)
= overline{mathrm{W}}(mu,X) + left| underline{mathrm{W}}(mu,X)right|
.
$$
Let's make some computations:
Definition 1: Probability
If the Dirac delta refers to the degenerate probability measure, defined by:
$$delta_B(A)triangleq begin{cases}
1 : & A cap Bneqemptyset\
0 : & mbox{else}\
end{cases},
$$
then
$$
overline{mathrm{W}}(mu,X)=supleft{delta(A)mid AinSigmatext{ and }Asubset E right}=
delta(B)=1.
$$
Similarly the lower variation is minimized by $B^c$ (possibly empty) and takes value $0$. Therefore,
$$
|delta_B|_{TV}=1<infty.
$$
Definition 2: Physics
Fix some $x in X$ and define the Dirac delta (generalized) function as
$$
delta_x(A)triangleq begin{cases}
infty : & x in A\
0 : & x notin A
end{cases}
,$$
The analogous argument shows that the upper variation on $X$ is $infty$, and is maximized by ${x}$. Likewise, the lower variation is minimized to value $0$ (if $# X>1$). Hence
$$
|delta|_{TV}=infty
.
$$
So it really depends on what you mean by "Dirac Delta" function.
answered Feb 1 at 0:08
AIM_BLBAIM_BLB
2,5392820
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $(f_1,f_2,f_3,ldots)$ be any sequence of functions of bounded variation such that, given any continuous function $g$,$$lim_{ntoinfty}int f_ng=g(0).$$(Of course, there are many such sequences, but the argument is independent of that.) Then$$lim_{ntoinfty}mathrm{TV}(f_n)=infty.$$This is one way to interpret your question, and then the answer is indeed $infty$.
answered Mar 16 '18 at 21:08
Toby BartelsToby Bartels
679516
$endgroup$
$begingroup$
I think pretty much the only other interpretation that makes any sense is to think of $delta$ as a measure and then consider its total variation. (This interpretation gives a different result of course.)
$endgroup$
– Ian
Mar 16 '18 at 21:12
1
$begingroup$
Yes, but that seems to me to be misinterpreting the original question, which seems to want $delta$ to be a generalized function. You can formalize some generalized functions as measures, but in this formalism, an ordinary function $f$ (if integrable) is represented by the signed measure $A mapsto int_A f$, and the total variation of this measure is not the total variation of $f$. So this is not giving us a notion of the total variation of a generalized function, but rather a notion of the total variation of an indefinite integral of a generalized function.
$endgroup$
– Toby Bartels
Mar 18 '18 at 20:58
add a comment |
$begingroup$
Let $(f_1,f_2,f_3,ldots)$ be any sequence of functions of bounded variation such that, given any continuous function $g$,$$lim_{ntoinfty}int f_ng=g(0).$$(Of course, there are many such sequences, but the argument is independent of that.) Then$$lim_{ntoinfty}mathrm{TV}(f_n)=infty.$$This is one way to interpret your question, and then the answer is indeed $infty$.
answered Mar 16 '18 at 21:08
Toby BartelsToby Bartels
679516
$endgroup$
$begingroup$
I think pretty much the only other interpretation that makes any sense is to think of $delta$ as a measure and then consider its total variation. (This interpretation gives a different result of course.)
$endgroup$
– Ian
Mar 16 '18 at 21:12
1
$begingroup$
Yes, but that seems to me to be misinterpreting the original question, which seems to want $delta$ to be a generalized function. You can formalize some generalized functions as measures, but in this formalism, an ordinary function $f$ (if integrable) is represented by the signed measure $A mapsto int_A f$, and the total variation of this measure is not the total variation of $f$. So this is not giving us a notion of the total variation of a generalized function, but rather a notion of the total variation of an indefinite integral of a generalized function.
$endgroup$
– Toby Bartels
Mar 18 '18 at 20:58
add a comment |
$begingroup$
Let $(f_1,f_2,f_3,ldots)$ be any sequence of functions of bounded variation such that, given any continuous function $g$,$$lim_{ntoinfty}int f_ng=g(0).$$(Of course, there are many such sequences, but the argument is independent of that.) Then$$lim_{ntoinfty}mathrm{TV}(f_n)=infty.$$This is one way to interpret your question, and then the answer is indeed $infty$.
answered Mar 16 '18 at 21:08
Toby BartelsToby Bartels
679516
$endgroup$
Let $(f_1,f_2,f_3,ldots)$ be any sequence of functions of bounded variation such that, given any continuous function $g$,$$lim_{ntoinfty}int f_ng=g(0).$$(Of course, there are many such sequences, but the argument is independent of that.) Then$$lim_{ntoinfty}mathrm{TV}(f_n)=infty.$$This is one way to interpret your question, and then the answer is indeed $infty$.
answered Mar 16 '18 at 21:08
Toby BartelsToby Bartels
679516
answered Mar 16 '18 at 21:08
Toby BartelsToby Bartels
679516
answered Mar 16 '18 at 21:08
Toby BartelsToby Bartels
679516
answered Mar 16 '18 at 21:08
Toby BartelsToby Bartels
679516
679516
$begingroup$
I think pretty much the only other interpretation that makes any sense is to think of $delta$ as a measure and then consider its total variation. (This interpretation gives a different result of course.)
$endgroup$
– Ian
Mar 16 '18 at 21:12
1
$begingroup$
Yes, but that seems to me to be misinterpreting the original question, which seems to want $delta$ to be a generalized function. You can formalize some generalized functions as measures, but in this formalism, an ordinary function $f$ (if integrable) is represented by the signed measure $A mapsto int_A f$, and the total variation of this measure is not the total variation of $f$. So this is not giving us a notion of the total variation of a generalized function, but rather a notion of the total variation of an indefinite integral of a generalized function.
$endgroup$
– Toby Bartels
Mar 18 '18 at 20:58
add a comment |
$begingroup$
I think pretty much the only other interpretation that makes any sense is to think of $delta$ as a measure and then consider its total variation. (This interpretation gives a different result of course.)
$endgroup$
– Ian
Mar 16 '18 at 21:12
1
$begingroup$
Yes, but that seems to me to be misinterpreting the original question, which seems to want $delta$ to be a generalized function. You can formalize some generalized functions as measures, but in this formalism, an ordinary function $f$ (if integrable) is represented by the signed measure $A mapsto int_A f$, and the total variation of this measure is not the total variation of $f$. So this is not giving us a notion of the total variation of a generalized function, but rather a notion of the total variation of an indefinite integral of a generalized function.
$endgroup$
– Toby Bartels
Mar 18 '18 at 20:58
$begingroup$
I think pretty much the only other interpretation that makes any sense is to think of $delta$ as a measure and then consider its total variation. (This interpretation gives a different result of course.)
$endgroup$
– Ian
Mar 16 '18 at 21:12
$begingroup$
I think pretty much the only other interpretation that makes any sense is to think of $delta$ as a measure and then consider its total variation. (This interpretation gives a different result of course.)
$endgroup$
– Ian
Mar 16 '18 at 21:12
1
1
$begingroup$
Yes, but that seems to me to be misinterpreting the original question, which seems to want $delta$ to be a generalized function. You can formalize some generalized functions as measures, but in this formalism, an ordinary function $f$ (if integrable) is represented by the signed measure $A mapsto int_A f$, and the total variation of this measure is not the total variation of $f$. So this is not giving us a notion of the total variation of a generalized function, but rather a notion of the total variation of an indefinite integral of a generalized function.
$endgroup$
– Toby Bartels
Mar 18 '18 at 20:58
$begingroup$
Yes, but that seems to me to be misinterpreting the original question, which seems to want $delta$ to be a generalized function. You can formalize some generalized functions as measures, but in this formalism, an ordinary function $f$ (if integrable) is represented by the signed measure $A mapsto int_A f$, and the total variation of this measure is not the total variation of $f$. So this is not giving us a notion of the total variation of a generalized function, but rather a notion of the total variation of an indefinite integral of a generalized function.
$endgroup$
– Toby Bartels
Mar 18 '18 at 20:58
add a comment |
$begingroup$
To clarify this, lets agree on the definitions of which we're using. Let $(X,Sigma)$ be a measurable space.
For any measure $mu$ on $(X,Sigma)$ define, the upper and lower variation (respectively) as follows
$$
overline{mathrm{W}}(mu,E)=supleft{mu(A)mid AinSigmatext{ and }Asubset E right}qquadforall EinSigma\
underline{mathrm{W}}(mu,E)=infleft{mu(A)mid AinSigmatext{ and }Asubset E right}qquadforall EinSigma.
$$
From this, we may define the total variation (extended-value) "norm" of a measure as follows
$$
|mu|_{TV}triangleq sup_{E in Sigma} left(
overline{mathrm{W}}(mu,E) + left| underline{mathrm{W}}(mu,E)right|
right)
= overline{mathrm{W}}(mu,X) + left| underline{mathrm{W}}(mu,X)right|
.
$$
Let's make some computations:
Definition 1: Probability
If the Dirac delta refers to the degenerate probability measure, defined by:
$$delta_B(A)triangleq begin{cases}
1 : & A cap Bneqemptyset\
0 : & mbox{else}\
end{cases},
$$
then
$$
overline{mathrm{W}}(mu,X)=supleft{delta(A)mid AinSigmatext{ and }Asubset E right}=
delta(B)=1.
$$
Similarly the lower variation is minimized by $B^c$ (possibly empty) and takes value $0$. Therefore,
$$
|delta_B|_{TV}=1<infty.
$$
Definition 2: Physics
Fix some $x in X$ and define the Dirac delta (generalized) function as
$$
delta_x(A)triangleq begin{cases}
infty : & x in A\
0 : & x notin A
end{cases}
,$$
The analogous argument shows that the upper variation on $X$ is $infty$, and is maximized by ${x}$. Likewise, the lower variation is minimized to value $0$ (if $# X>1$). Hence
$$
|delta|_{TV}=infty
.
$$
So it really depends on what you mean by "Dirac Delta" function.
answered Feb 1 at 0:08
AIM_BLBAIM_BLB
2,5392820
$endgroup$
add a comment |
$begingroup$
To clarify this, lets agree on the definitions of which we're using. Let $(X,Sigma)$ be a measurable space.
For any measure $mu$ on $(X,Sigma)$ define, the upper and lower variation (respectively) as follows
$$
overline{mathrm{W}}(mu,E)=supleft{mu(A)mid AinSigmatext{ and }Asubset E right}qquadforall EinSigma\
underline{mathrm{W}}(mu,E)=infleft{mu(A)mid AinSigmatext{ and }Asubset E right}qquadforall EinSigma.
$$
From this, we may define the total variation (extended-value) "norm" of a measure as follows
$$
|mu|_{TV}triangleq sup_{E in Sigma} left(
overline{mathrm{W}}(mu,E) + left| underline{mathrm{W}}(mu,E)right|
right)
= overline{mathrm{W}}(mu,X) + left| underline{mathrm{W}}(mu,X)right|
.
$$
Let's make some computations:
Definition 1: Probability
If the Dirac delta refers to the degenerate probability measure, defined by:
$$delta_B(A)triangleq begin{cases}
1 : & A cap Bneqemptyset\
0 : & mbox{else}\
end{cases},
$$
then
$$
overline{mathrm{W}}(mu,X)=supleft{delta(A)mid AinSigmatext{ and }Asubset E right}=
delta(B)=1.
$$
Similarly the lower variation is minimized by $B^c$ (possibly empty) and takes value $0$. Therefore,
$$
|delta_B|_{TV}=1<infty.
$$
Definition 2: Physics
Fix some $x in X$ and define the Dirac delta (generalized) function as
$$
delta_x(A)triangleq begin{cases}
infty : & x in A\
0 : & x notin A
end{cases}
,$$
The analogous argument shows that the upper variation on $X$ is $infty$, and is maximized by ${x}$. Likewise, the lower variation is minimized to value $0$ (if $# X>1$). Hence
$$
|delta|_{TV}=infty
.
$$
So it really depends on what you mean by "Dirac Delta" function.
answered Feb 1 at 0:08
AIM_BLBAIM_BLB
2,5392820
$endgroup$
add a comment |
$begingroup$
To clarify this, lets agree on the definitions of which we're using. Let $(X,Sigma)$ be a measurable space.
For any measure $mu$ on $(X,Sigma)$ define, the upper and lower variation (respectively) as follows
$$
overline{mathrm{W}}(mu,E)=supleft{mu(A)mid AinSigmatext{ and }Asubset E right}qquadforall EinSigma\
underline{mathrm{W}}(mu,E)=infleft{mu(A)mid AinSigmatext{ and }Asubset E right}qquadforall EinSigma.
$$
From this, we may define the total variation (extended-value) "norm" of a measure as follows
$$
|mu|_{TV}triangleq sup_{E in Sigma} left(
overline{mathrm{W}}(mu,E) + left| underline{mathrm{W}}(mu,E)right|
right)
= overline{mathrm{W}}(mu,X) + left| underline{mathrm{W}}(mu,X)right|
.
$$
Let's make some computations:
Definition 1: Probability
If the Dirac delta refers to the degenerate probability measure, defined by:
$$delta_B(A)triangleq begin{cases}
1 : & A cap Bneqemptyset\
0 : & mbox{else}\
end{cases},
$$
then
$$
overline{mathrm{W}}(mu,X)=supleft{delta(A)mid AinSigmatext{ and }Asubset E right}=
delta(B)=1.
$$
Similarly the lower variation is minimized by $B^c$ (possibly empty) and takes value $0$. Therefore,
$$
|delta_B|_{TV}=1<infty.
$$
Definition 2: Physics
Fix some $x in X$ and define the Dirac delta (generalized) function as
$$
delta_x(A)triangleq begin{cases}
infty : & x in A\
0 : & x notin A
end{cases}
,$$
The analogous argument shows that the upper variation on $X$ is $infty$, and is maximized by ${x}$. Likewise, the lower variation is minimized to value $0$ (if $# X>1$). Hence
$$
|delta|_{TV}=infty
.
$$
So it really depends on what you mean by "Dirac Delta" function.
answered Feb 1 at 0:08
AIM_BLBAIM_BLB
2,5392820
$endgroup$
To clarify this, lets agree on the definitions of which we're using. Let $(X,Sigma)$ be a measurable space.
For any measure $mu$ on $(X,Sigma)$ define, the upper and lower variation (respectively) as follows
$$
overline{mathrm{W}}(mu,E)=supleft{mu(A)mid AinSigmatext{ and }Asubset E right}qquadforall EinSigma\
underline{mathrm{W}}(mu,E)=infleft{mu(A)mid AinSigmatext{ and }Asubset E right}qquadforall EinSigma.
$$
From this, we may define the total variation (extended-value) "norm" of a measure as follows
$$
|mu|_{TV}triangleq sup_{E in Sigma} left(
overline{mathrm{W}}(mu,E) + left| underline{mathrm{W}}(mu,E)right|
right)
= overline{mathrm{W}}(mu,X) + left| underline{mathrm{W}}(mu,X)right|
.
$$
Let's make some computations:
Definition 1: Probability
If the Dirac delta refers to the degenerate probability measure, defined by:
$$delta_B(A)triangleq begin{cases}
1 : & A cap Bneqemptyset\
0 : & mbox{else}\
end{cases},
$$
then
$$
overline{mathrm{W}}(mu,X)=supleft{delta(A)mid AinSigmatext{ and }Asubset E right}=
delta(B)=1.
$$
Similarly the lower variation is minimized by $B^c$ (possibly empty) and takes value $0$. Therefore,
$$
|delta_B|_{TV}=1<infty.
$$
Definition 2: Physics
Fix some $x in X$ and define the Dirac delta (generalized) function as
$$
delta_x(A)triangleq begin{cases}
infty : & x in A\
0 : & x notin A
end{cases}
,$$
The analogous argument shows that the upper variation on $X$ is $infty$, and is maximized by ${x}$. Likewise, the lower variation is minimized to value $0$ (if $# X>1$). Hence
$$
|delta|_{TV}=infty
.
$$
So it really depends on what you mean by "Dirac Delta" function.
answered Feb 1 at 0:08
AIM_BLBAIM_BLB
2,5392820
answered Feb 1 at 0:08
AIM_BLBAIM_BLB
2,5392820
answered Feb 1 at 0:08
AIM_BLBAIM_BLB
2,5392820
answered Feb 1 at 0:08
AIM_BLBAIM_BLB
2,5392820
2,5392820
add a comment |
add a comment |
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Comments are not for extended discussion; this conversation has been moved to chat.
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– Pedro Tamaroff♦
Sep 1 '16 at 13:30
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@PedroTamaroff : When you move things to "chat", the MathJax within the comments doesn't get rendered.
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– Michael Hardy
Sep 4 '16 at 1:09
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@MichaelHardy You have to use ChatJax or its variants.
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– Pedro Tamaroff♦
Sep 4 '16 at 1:14
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Here, when you say dirac do you mean that it takes values $1$ and $0$ or values $0$ and $infty$? I'm guessing the later...?
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– AIM_BLB
Jan 31 at 23:51