Mixing
All the derivatives of distributions are also distributions, but what about the converse?
$begingroup$
Say you have some linear functional $f$ well defined on $mathscr{D}(mathbb{R})$: then what if for some test function
$phi$ you have
$$
-f(phi') = g(phi)?
$$
If that $g$ defines a distribution, can you say that $f$ is also a distribution?
It would help me solve so many problems much faster.
Intuitvely I'd say yes, because if $g$ is a distribution then we have that $f(phi'')$ is also a 'distribution' (meaning if every test function could be written as the second derivative of a test function then $f$ would be a distribution)
which I'm not very sure of, main reason I came here to post this question.
Comment, clarification and counter examples will all be greatly appreciated. thanks!
functional-analysis soft-question distribution-theory
edited Jan 29 at 16:15
Daniele Tampieri
2,58721022
asked Jan 28 at 18:30
rapidracimrapidracim
1,7291419
$endgroup$
add a comment |
$begingroup$
Say you have some linear functional $f$ well defined on $mathscr{D}(mathbb{R})$: then what if for some test function
$phi$ you have
$$
-f(phi') = g(phi)?
$$
If that $g$ defines a distribution, can you say that $f$ is also a distribution?
It would help me solve so many problems much faster.
Intuitvely I'd say yes, because if $g$ is a distribution then we have that $f(phi'')$ is also a 'distribution' (meaning if every test function could be written as the second derivative of a test function then $f$ would be a distribution)
which I'm not very sure of, main reason I came here to post this question.
Comment, clarification and counter examples will all be greatly appreciated. thanks!
functional-analysis soft-question distribution-theory
edited Jan 29 at 16:15
Daniele Tampieri
2,58721022
asked Jan 28 at 18:30
rapidracimrapidracim
1,7291419
$endgroup$
1
$begingroup$
Let $I(phi)(x) = int_{-infty}^x phi(y)dy$ and $langle G, phi rangle = -langle g, I(phi) - I(phi)(infty) psi rangle $ where $psi in C^infty, psi(x) =0 (x < -1), psi(x) = 1 (x > 1)$. Is $G$ a distribution, what is its derivative ?
$endgroup$
– reuns
Jan 28 at 20:27
add a comment |
$begingroup$
Say you have some linear functional $f$ well defined on $mathscr{D}(mathbb{R})$: then what if for some test function
$phi$ you have
$$
-f(phi') = g(phi)?
$$
If that $g$ defines a distribution, can you say that $f$ is also a distribution?
It would help me solve so many problems much faster.
Intuitvely I'd say yes, because if $g$ is a distribution then we have that $f(phi'')$ is also a 'distribution' (meaning if every test function could be written as the second derivative of a test function then $f$ would be a distribution)
which I'm not very sure of, main reason I came here to post this question.
Comment, clarification and counter examples will all be greatly appreciated. thanks!
functional-analysis soft-question distribution-theory
edited Jan 29 at 16:15
Daniele Tampieri
2,58721022
asked Jan 28 at 18:30
rapidracimrapidracim
1,7291419
$endgroup$
Say you have some linear functional $f$ well defined on $mathscr{D}(mathbb{R})$: then what if for some test function
$phi$ you have
$$
-f(phi') = g(phi)?
$$
If that $g$ defines a distribution, can you say that $f$ is also a distribution?
It would help me solve so many problems much faster.
Intuitvely I'd say yes, because if $g$ is a distribution then we have that $f(phi'')$ is also a 'distribution' (meaning if every test function could be written as the second derivative of a test function then $f$ would be a distribution)
which I'm not very sure of, main reason I came here to post this question.
Comment, clarification and counter examples will all be greatly appreciated. thanks!
functional-analysis soft-question distribution-theory
functional-analysis soft-question distribution-theory
edited Jan 29 at 16:15
Daniele Tampieri
2,58721022
asked Jan 28 at 18:30
rapidracimrapidracim
1,7291419
edited Jan 29 at 16:15
Daniele Tampieri
2,58721022
asked Jan 28 at 18:30
rapidracimrapidracim
1,7291419
edited Jan 29 at 16:15
Daniele Tampieri
2,58721022
edited Jan 29 at 16:15
Daniele Tampieri
2,58721022
edited Jan 29 at 16:15
Daniele Tampieri
2,58721022
2,58721022
asked Jan 28 at 18:30
rapidracimrapidracim
1,7291419
asked Jan 28 at 18:30
rapidracimrapidracim
1,7291419
asked Jan 28 at 18:30
rapidracimrapidracim
1,7291419
1,7291419
1
$begingroup$
Let $I(phi)(x) = int_{-infty}^x phi(y)dy$ and $langle G, phi rangle = -langle g, I(phi) - I(phi)(infty) psi rangle $ where $psi in C^infty, psi(x) =0 (x < -1), psi(x) = 1 (x > 1)$. Is $G$ a distribution, what is its derivative ?
$endgroup$
– reuns
Jan 28 at 20:27
add a comment |
1
$begingroup$
Let $I(phi)(x) = int_{-infty}^x phi(y)dy$ and $langle G, phi rangle = -langle g, I(phi) - I(phi)(infty) psi rangle $ where $psi in C^infty, psi(x) =0 (x < -1), psi(x) = 1 (x > 1)$. Is $G$ a distribution, what is its derivative ?
$endgroup$
– reuns
Jan 28 at 20:27
1
1
$begingroup$
Let $I(phi)(x) = int_{-infty}^x phi(y)dy$ and $langle G, phi rangle = -langle g, I(phi) - I(phi)(infty) psi rangle $ where $psi in C^infty, psi(x) =0 (x < -1), psi(x) = 1 (x > 1)$. Is $G$ a distribution, what is its derivative ?
$endgroup$
– reuns
Jan 28 at 20:27
$begingroup$
Let $I(phi)(x) = int_{-infty}^x phi(y)dy$ and $langle G, phi rangle = -langle g, I(phi) - I(phi)(infty) psi rangle $ where $psi in C^infty, psi(x) =0 (x < -1), psi(x) = 1 (x > 1)$. Is $G$ a distribution, what is its derivative ?
$endgroup$
– reuns
Jan 28 at 20:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer to the question you are asking is yes: it is a standard property of distributions of one variable.
Precisely, the following theorem is a standard fact of the general theory of distributions:
Theorem ([1], §1.5.3. pp. 26-27) Equation eqref{1}
$$
f^prime=glabel{1}tag{1}
$$
has a solution $finmathscr{D}^prime$ for all distributions $ginmathscr{D}^prime$.
Sketch of proof. We have that eqref{1} is equivalent to the following equation
$$
(f,-varphi^prime)=(f^prime,varphi)=(g,varphi)=Bigg( g,intlimits_{-infty}^x varphi^prime(y),mathrm{d}y Bigg)quad forallvarphiinmathscr{D}(Bbb R).
$$
Thus the functional $f$ is already defined on any test function $varphi$ which is the derivative of some other test functions: now define
$$
varphi_o(x)=varphi(x)-omega(x)intlimits_{-infty}^{+infty}varphi(x),mathrm{d}x quad forall varphiinmathscr{D}(Bbb R),label{p}tag{P}
$$
where $omega(x)$ is an arbitrary test function such that
$$
intlimits_{-infty}^{+infty}omega(x),mathrm{d}x=1.
$$
It can be proved that $varphi_o$ is a test function which is the derivative of another test function, and that if we choose a sequence ${varphi_nu}_{nuinBbb N}$ covering to $0$ in $mathscr{D}(Bbb R)$, then the sequence ${varphi_{onu}}_{nuinBbb N}$ associated to it by the projection eqref{p} converges to to $0$ in $mathscr{D}(Bbb R)$.
Now, defining the antiderivative functional $g^{(-1)}$ as
$$
big(,g^{(-1)},varphibig)=-Bigg( g,intlimits_{-infty}^x varphi_o(y),mathrm{d}y Bigg)quad forallvarphiinmathscr{D}(Bbb R),label{a}tag{A}
$$
the linearity of such functional is an immediate consequence of the linearity of the integral and of the distribution $g$. The continuity follows from the continuity of the map
$$
varphi(x)mapstointlimits_{-infty}^x varphi_o(y),mathrm{d}y.
$$
Precisely, as said above, for any sequence of test functions ${varphi_nu}_{nuinBbb N}$ converging to $0$ in $mathscr{D}(Bbb R)$ the sequence ${varphi_{onu}}_{nuinBbb N}$ converges to $0$ in $mathscr{D}(Bbb R)$ therefore
$$
begin{split}
varphi_{onu}underset{nutoinfty}{longrightarrow} 0 ; &implies intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}yunderset{nutoinfty}{longrightarrow} 0 \
frac{mathrm{d}^kvarphi_{onu}}{mathrm{d}x^k}underset{nutoinfty}{longrightarrow} 0 ; &implies frac{mathrm{d}^k}{mathrm{d}x^k}intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}y=frac{mathrm{d}^{k-1}varphi_{onu}}{mathrm{d}x^{k-1}}underset{nutoinfty}{longrightarrow} 0quad kinBbb N
end{split}text{in }mathscr{D}(Bbb R)
$$
thus $Big{intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}yBig}_{nuinBbb N}$ is a sequence of functions in $mathscr{D}(Bbb R)$ coverging to $0$ in of $mathscr{D}(Bbb R)$.
This implies that the antiderivative eqref{a} of a a distribution is a distribution: then
$$
f=g^{(-1)} +C label{2}tag{2}
$$
where $C$ is an arbitrary constant is the sought for primitive of $f$. $blacksquare$
Notes
- The development offered above is only a sketch of the standard proof offered by Shilov, since the rigorous proof justifying all steps, particularly the properties of $varphi_o(x)$, while entirely elementary, is nevertheless not too short.
- Vladimirov ([1], §2.2 pp. 27-29) offers a similar proof of the solvability of eqref{1} in $mathscr{D}^prime$: he works in $mathscr{D}^primebig(]a,b[big)$ for arbitrary $a,binoverline{Bbb R}$ with $a<b$ (i.e. including $a=-infty$ and $b=+infty$).
- Erik Talvila used the solution of eqref{1} to define a generalized integral which is more general of the ones o Henstock and Kurzweil. Recently also Ricardo Estrada and Jasson Vindas proposed a definition of distributional integral based on a similar concept.
[1] G. E. Shilov (1968), Generalized functions and partial differential equations, Mathematics and Its Applications, Vol. 7, (English)
New York-London-Paris: Gordon and Breach Science Publishers, XII+345, MR0230129, Zbl 0177.36302.
[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
answered Jan 28 at 22:16
Daniele TampieriDaniele Tampieri
2,58721022
$endgroup$
$begingroup$
You meant $varphi mapsto int_{-infty}^. varphi_o(y)dy$ is continuous in the $C^infty_c(mathbb{R})$ topology, not only $varphi mapsto varphi_o$, this is what makes $varphi mapsto langle g,int_{-infty}^. varphi_o(y)dy rangle$ a distribution
$endgroup$
– reuns
Jan 28 at 23:38
$begingroup$
@reuns. Exactly: I added a section on this to my answer.
$endgroup$
– Daniele Tampieri
Jan 29 at 9:53
$begingroup$
@reuns. I also apologize for not noting your comment to the question. Probably I started editing my answer, then had lunch and finished answering later, without refreshing the post.
$endgroup$
– Daniele Tampieri
Jan 29 at 10:22
add a comment |
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1 Answer
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$begingroup$
The answer to the question you are asking is yes: it is a standard property of distributions of one variable.
Precisely, the following theorem is a standard fact of the general theory of distributions:
Theorem ([1], §1.5.3. pp. 26-27) Equation eqref{1}
$$
f^prime=glabel{1}tag{1}
$$
has a solution $finmathscr{D}^prime$ for all distributions $ginmathscr{D}^prime$.
Sketch of proof. We have that eqref{1} is equivalent to the following equation
$$
(f,-varphi^prime)=(f^prime,varphi)=(g,varphi)=Bigg( g,intlimits_{-infty}^x varphi^prime(y),mathrm{d}y Bigg)quad forallvarphiinmathscr{D}(Bbb R).
$$
Thus the functional $f$ is already defined on any test function $varphi$ which is the derivative of some other test functions: now define
$$
varphi_o(x)=varphi(x)-omega(x)intlimits_{-infty}^{+infty}varphi(x),mathrm{d}x quad forall varphiinmathscr{D}(Bbb R),label{p}tag{P}
$$
where $omega(x)$ is an arbitrary test function such that
$$
intlimits_{-infty}^{+infty}omega(x),mathrm{d}x=1.
$$
It can be proved that $varphi_o$ is a test function which is the derivative of another test function, and that if we choose a sequence ${varphi_nu}_{nuinBbb N}$ covering to $0$ in $mathscr{D}(Bbb R)$, then the sequence ${varphi_{onu}}_{nuinBbb N}$ associated to it by the projection eqref{p} converges to to $0$ in $mathscr{D}(Bbb R)$.
Now, defining the antiderivative functional $g^{(-1)}$ as
$$
big(,g^{(-1)},varphibig)=-Bigg( g,intlimits_{-infty}^x varphi_o(y),mathrm{d}y Bigg)quad forallvarphiinmathscr{D}(Bbb R),label{a}tag{A}
$$
the linearity of such functional is an immediate consequence of the linearity of the integral and of the distribution $g$. The continuity follows from the continuity of the map
$$
varphi(x)mapstointlimits_{-infty}^x varphi_o(y),mathrm{d}y.
$$
Precisely, as said above, for any sequence of test functions ${varphi_nu}_{nuinBbb N}$ converging to $0$ in $mathscr{D}(Bbb R)$ the sequence ${varphi_{onu}}_{nuinBbb N}$ converges to $0$ in $mathscr{D}(Bbb R)$ therefore
$$
begin{split}
varphi_{onu}underset{nutoinfty}{longrightarrow} 0 ; &implies intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}yunderset{nutoinfty}{longrightarrow} 0 \
frac{mathrm{d}^kvarphi_{onu}}{mathrm{d}x^k}underset{nutoinfty}{longrightarrow} 0 ; &implies frac{mathrm{d}^k}{mathrm{d}x^k}intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}y=frac{mathrm{d}^{k-1}varphi_{onu}}{mathrm{d}x^{k-1}}underset{nutoinfty}{longrightarrow} 0quad kinBbb N
end{split}text{in }mathscr{D}(Bbb R)
$$
thus $Big{intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}yBig}_{nuinBbb N}$ is a sequence of functions in $mathscr{D}(Bbb R)$ coverging to $0$ in of $mathscr{D}(Bbb R)$.
This implies that the antiderivative eqref{a} of a a distribution is a distribution: then
$$
f=g^{(-1)} +C label{2}tag{2}
$$
where $C$ is an arbitrary constant is the sought for primitive of $f$. $blacksquare$
Notes
- The development offered above is only a sketch of the standard proof offered by Shilov, since the rigorous proof justifying all steps, particularly the properties of $varphi_o(x)$, while entirely elementary, is nevertheless not too short.
- Vladimirov ([1], §2.2 pp. 27-29) offers a similar proof of the solvability of eqref{1} in $mathscr{D}^prime$: he works in $mathscr{D}^primebig(]a,b[big)$ for arbitrary $a,binoverline{Bbb R}$ with $a<b$ (i.e. including $a=-infty$ and $b=+infty$).
- Erik Talvila used the solution of eqref{1} to define a generalized integral which is more general of the ones o Henstock and Kurzweil. Recently also Ricardo Estrada and Jasson Vindas proposed a definition of distributional integral based on a similar concept.
[1] G. E. Shilov (1968), Generalized functions and partial differential equations, Mathematics and Its Applications, Vol. 7, (English)
New York-London-Paris: Gordon and Breach Science Publishers, XII+345, MR0230129, Zbl 0177.36302.
[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
answered Jan 28 at 22:16
Daniele TampieriDaniele Tampieri
2,58721022
$endgroup$
$begingroup$
You meant $varphi mapsto int_{-infty}^. varphi_o(y)dy$ is continuous in the $C^infty_c(mathbb{R})$ topology, not only $varphi mapsto varphi_o$, this is what makes $varphi mapsto langle g,int_{-infty}^. varphi_o(y)dy rangle$ a distribution
$endgroup$
– reuns
Jan 28 at 23:38
$begingroup$
@reuns. Exactly: I added a section on this to my answer.
$endgroup$
– Daniele Tampieri
Jan 29 at 9:53
$begingroup$
@reuns. I also apologize for not noting your comment to the question. Probably I started editing my answer, then had lunch and finished answering later, without refreshing the post.
$endgroup$
– Daniele Tampieri
Jan 29 at 10:22
add a comment |
$begingroup$
The answer to the question you are asking is yes: it is a standard property of distributions of one variable.
Precisely, the following theorem is a standard fact of the general theory of distributions:
Theorem ([1], §1.5.3. pp. 26-27) Equation eqref{1}
$$
f^prime=glabel{1}tag{1}
$$
has a solution $finmathscr{D}^prime$ for all distributions $ginmathscr{D}^prime$.
Sketch of proof. We have that eqref{1} is equivalent to the following equation
$$
(f,-varphi^prime)=(f^prime,varphi)=(g,varphi)=Bigg( g,intlimits_{-infty}^x varphi^prime(y),mathrm{d}y Bigg)quad forallvarphiinmathscr{D}(Bbb R).
$$
Thus the functional $f$ is already defined on any test function $varphi$ which is the derivative of some other test functions: now define
$$
varphi_o(x)=varphi(x)-omega(x)intlimits_{-infty}^{+infty}varphi(x),mathrm{d}x quad forall varphiinmathscr{D}(Bbb R),label{p}tag{P}
$$
where $omega(x)$ is an arbitrary test function such that
$$
intlimits_{-infty}^{+infty}omega(x),mathrm{d}x=1.
$$
It can be proved that $varphi_o$ is a test function which is the derivative of another test function, and that if we choose a sequence ${varphi_nu}_{nuinBbb N}$ covering to $0$ in $mathscr{D}(Bbb R)$, then the sequence ${varphi_{onu}}_{nuinBbb N}$ associated to it by the projection eqref{p} converges to to $0$ in $mathscr{D}(Bbb R)$.
Now, defining the antiderivative functional $g^{(-1)}$ as
$$
big(,g^{(-1)},varphibig)=-Bigg( g,intlimits_{-infty}^x varphi_o(y),mathrm{d}y Bigg)quad forallvarphiinmathscr{D}(Bbb R),label{a}tag{A}
$$
the linearity of such functional is an immediate consequence of the linearity of the integral and of the distribution $g$. The continuity follows from the continuity of the map
$$
varphi(x)mapstointlimits_{-infty}^x varphi_o(y),mathrm{d}y.
$$
Precisely, as said above, for any sequence of test functions ${varphi_nu}_{nuinBbb N}$ converging to $0$ in $mathscr{D}(Bbb R)$ the sequence ${varphi_{onu}}_{nuinBbb N}$ converges to $0$ in $mathscr{D}(Bbb R)$ therefore
$$
begin{split}
varphi_{onu}underset{nutoinfty}{longrightarrow} 0 ; &implies intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}yunderset{nutoinfty}{longrightarrow} 0 \
frac{mathrm{d}^kvarphi_{onu}}{mathrm{d}x^k}underset{nutoinfty}{longrightarrow} 0 ; &implies frac{mathrm{d}^k}{mathrm{d}x^k}intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}y=frac{mathrm{d}^{k-1}varphi_{onu}}{mathrm{d}x^{k-1}}underset{nutoinfty}{longrightarrow} 0quad kinBbb N
end{split}text{in }mathscr{D}(Bbb R)
$$
thus $Big{intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}yBig}_{nuinBbb N}$ is a sequence of functions in $mathscr{D}(Bbb R)$ coverging to $0$ in of $mathscr{D}(Bbb R)$.
This implies that the antiderivative eqref{a} of a a distribution is a distribution: then
$$
f=g^{(-1)} +C label{2}tag{2}
$$
where $C$ is an arbitrary constant is the sought for primitive of $f$. $blacksquare$
Notes
- The development offered above is only a sketch of the standard proof offered by Shilov, since the rigorous proof justifying all steps, particularly the properties of $varphi_o(x)$, while entirely elementary, is nevertheless not too short.
- Vladimirov ([1], §2.2 pp. 27-29) offers a similar proof of the solvability of eqref{1} in $mathscr{D}^prime$: he works in $mathscr{D}^primebig(]a,b[big)$ for arbitrary $a,binoverline{Bbb R}$ with $a<b$ (i.e. including $a=-infty$ and $b=+infty$).
- Erik Talvila used the solution of eqref{1} to define a generalized integral which is more general of the ones o Henstock and Kurzweil. Recently also Ricardo Estrada and Jasson Vindas proposed a definition of distributional integral based on a similar concept.
[1] G. E. Shilov (1968), Generalized functions and partial differential equations, Mathematics and Its Applications, Vol. 7, (English)
New York-London-Paris: Gordon and Breach Science Publishers, XII+345, MR0230129, Zbl 0177.36302.
[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
answered Jan 28 at 22:16
Daniele TampieriDaniele Tampieri
2,58721022
$endgroup$
$begingroup$
You meant $varphi mapsto int_{-infty}^. varphi_o(y)dy$ is continuous in the $C^infty_c(mathbb{R})$ topology, not only $varphi mapsto varphi_o$, this is what makes $varphi mapsto langle g,int_{-infty}^. varphi_o(y)dy rangle$ a distribution
$endgroup$
– reuns
Jan 28 at 23:38
$begingroup$
@reuns. Exactly: I added a section on this to my answer.
$endgroup$
– Daniele Tampieri
Jan 29 at 9:53
$begingroup$
@reuns. I also apologize for not noting your comment to the question. Probably I started editing my answer, then had lunch and finished answering later, without refreshing the post.
$endgroup$
– Daniele Tampieri
Jan 29 at 10:22
add a comment |
$begingroup$
The answer to the question you are asking is yes: it is a standard property of distributions of one variable.
Precisely, the following theorem is a standard fact of the general theory of distributions:
Theorem ([1], §1.5.3. pp. 26-27) Equation eqref{1}
$$
f^prime=glabel{1}tag{1}
$$
has a solution $finmathscr{D}^prime$ for all distributions $ginmathscr{D}^prime$.
Sketch of proof. We have that eqref{1} is equivalent to the following equation
$$
(f,-varphi^prime)=(f^prime,varphi)=(g,varphi)=Bigg( g,intlimits_{-infty}^x varphi^prime(y),mathrm{d}y Bigg)quad forallvarphiinmathscr{D}(Bbb R).
$$
Thus the functional $f$ is already defined on any test function $varphi$ which is the derivative of some other test functions: now define
$$
varphi_o(x)=varphi(x)-omega(x)intlimits_{-infty}^{+infty}varphi(x),mathrm{d}x quad forall varphiinmathscr{D}(Bbb R),label{p}tag{P}
$$
where $omega(x)$ is an arbitrary test function such that
$$
intlimits_{-infty}^{+infty}omega(x),mathrm{d}x=1.
$$
It can be proved that $varphi_o$ is a test function which is the derivative of another test function, and that if we choose a sequence ${varphi_nu}_{nuinBbb N}$ covering to $0$ in $mathscr{D}(Bbb R)$, then the sequence ${varphi_{onu}}_{nuinBbb N}$ associated to it by the projection eqref{p} converges to to $0$ in $mathscr{D}(Bbb R)$.
Now, defining the antiderivative functional $g^{(-1)}$ as
$$
big(,g^{(-1)},varphibig)=-Bigg( g,intlimits_{-infty}^x varphi_o(y),mathrm{d}y Bigg)quad forallvarphiinmathscr{D}(Bbb R),label{a}tag{A}
$$
the linearity of such functional is an immediate consequence of the linearity of the integral and of the distribution $g$. The continuity follows from the continuity of the map
$$
varphi(x)mapstointlimits_{-infty}^x varphi_o(y),mathrm{d}y.
$$
Precisely, as said above, for any sequence of test functions ${varphi_nu}_{nuinBbb N}$ converging to $0$ in $mathscr{D}(Bbb R)$ the sequence ${varphi_{onu}}_{nuinBbb N}$ converges to $0$ in $mathscr{D}(Bbb R)$ therefore
$$
begin{split}
varphi_{onu}underset{nutoinfty}{longrightarrow} 0 ; &implies intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}yunderset{nutoinfty}{longrightarrow} 0 \
frac{mathrm{d}^kvarphi_{onu}}{mathrm{d}x^k}underset{nutoinfty}{longrightarrow} 0 ; &implies frac{mathrm{d}^k}{mathrm{d}x^k}intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}y=frac{mathrm{d}^{k-1}varphi_{onu}}{mathrm{d}x^{k-1}}underset{nutoinfty}{longrightarrow} 0quad kinBbb N
end{split}text{in }mathscr{D}(Bbb R)
$$
thus $Big{intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}yBig}_{nuinBbb N}$ is a sequence of functions in $mathscr{D}(Bbb R)$ coverging to $0$ in of $mathscr{D}(Bbb R)$.
This implies that the antiderivative eqref{a} of a a distribution is a distribution: then
$$
f=g^{(-1)} +C label{2}tag{2}
$$
where $C$ is an arbitrary constant is the sought for primitive of $f$. $blacksquare$
Notes
- The development offered above is only a sketch of the standard proof offered by Shilov, since the rigorous proof justifying all steps, particularly the properties of $varphi_o(x)$, while entirely elementary, is nevertheless not too short.
- Vladimirov ([1], §2.2 pp. 27-29) offers a similar proof of the solvability of eqref{1} in $mathscr{D}^prime$: he works in $mathscr{D}^primebig(]a,b[big)$ for arbitrary $a,binoverline{Bbb R}$ with $a<b$ (i.e. including $a=-infty$ and $b=+infty$).
- Erik Talvila used the solution of eqref{1} to define a generalized integral which is more general of the ones o Henstock and Kurzweil. Recently also Ricardo Estrada and Jasson Vindas proposed a definition of distributional integral based on a similar concept.
[1] G. E. Shilov (1968), Generalized functions and partial differential equations, Mathematics and Its Applications, Vol. 7, (English)
New York-London-Paris: Gordon and Breach Science Publishers, XII+345, MR0230129, Zbl 0177.36302.
[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
answered Jan 28 at 22:16
Daniele TampieriDaniele Tampieri
2,58721022
$endgroup$
The answer to the question you are asking is yes: it is a standard property of distributions of one variable.
Precisely, the following theorem is a standard fact of the general theory of distributions:
Theorem ([1], §1.5.3. pp. 26-27) Equation eqref{1}
$$
f^prime=glabel{1}tag{1}
$$
has a solution $finmathscr{D}^prime$ for all distributions $ginmathscr{D}^prime$.
Sketch of proof. We have that eqref{1} is equivalent to the following equation
$$
(f,-varphi^prime)=(f^prime,varphi)=(g,varphi)=Bigg( g,intlimits_{-infty}^x varphi^prime(y),mathrm{d}y Bigg)quad forallvarphiinmathscr{D}(Bbb R).
$$
Thus the functional $f$ is already defined on any test function $varphi$ which is the derivative of some other test functions: now define
$$
varphi_o(x)=varphi(x)-omega(x)intlimits_{-infty}^{+infty}varphi(x),mathrm{d}x quad forall varphiinmathscr{D}(Bbb R),label{p}tag{P}
$$
where $omega(x)$ is an arbitrary test function such that
$$
intlimits_{-infty}^{+infty}omega(x),mathrm{d}x=1.
$$
It can be proved that $varphi_o$ is a test function which is the derivative of another test function, and that if we choose a sequence ${varphi_nu}_{nuinBbb N}$ covering to $0$ in $mathscr{D}(Bbb R)$, then the sequence ${varphi_{onu}}_{nuinBbb N}$ associated to it by the projection eqref{p} converges to to $0$ in $mathscr{D}(Bbb R)$.
Now, defining the antiderivative functional $g^{(-1)}$ as
$$
big(,g^{(-1)},varphibig)=-Bigg( g,intlimits_{-infty}^x varphi_o(y),mathrm{d}y Bigg)quad forallvarphiinmathscr{D}(Bbb R),label{a}tag{A}
$$
the linearity of such functional is an immediate consequence of the linearity of the integral and of the distribution $g$. The continuity follows from the continuity of the map
$$
varphi(x)mapstointlimits_{-infty}^x varphi_o(y),mathrm{d}y.
$$
Precisely, as said above, for any sequence of test functions ${varphi_nu}_{nuinBbb N}$ converging to $0$ in $mathscr{D}(Bbb R)$ the sequence ${varphi_{onu}}_{nuinBbb N}$ converges to $0$ in $mathscr{D}(Bbb R)$ therefore
$$
begin{split}
varphi_{onu}underset{nutoinfty}{longrightarrow} 0 ; &implies intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}yunderset{nutoinfty}{longrightarrow} 0 \
frac{mathrm{d}^kvarphi_{onu}}{mathrm{d}x^k}underset{nutoinfty}{longrightarrow} 0 ; &implies frac{mathrm{d}^k}{mathrm{d}x^k}intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}y=frac{mathrm{d}^{k-1}varphi_{onu}}{mathrm{d}x^{k-1}}underset{nutoinfty}{longrightarrow} 0quad kinBbb N
end{split}text{in }mathscr{D}(Bbb R)
$$
thus $Big{intlimits_{-infty}^x varphi_{onu}(y),mathrm{d}yBig}_{nuinBbb N}$ is a sequence of functions in $mathscr{D}(Bbb R)$ coverging to $0$ in of $mathscr{D}(Bbb R)$.
This implies that the antiderivative eqref{a} of a a distribution is a distribution: then
$$
f=g^{(-1)} +C label{2}tag{2}
$$
where $C$ is an arbitrary constant is the sought for primitive of $f$. $blacksquare$
Notes
- The development offered above is only a sketch of the standard proof offered by Shilov, since the rigorous proof justifying all steps, particularly the properties of $varphi_o(x)$, while entirely elementary, is nevertheless not too short.
- Vladimirov ([1], §2.2 pp. 27-29) offers a similar proof of the solvability of eqref{1} in $mathscr{D}^prime$: he works in $mathscr{D}^primebig(]a,b[big)$ for arbitrary $a,binoverline{Bbb R}$ with $a<b$ (i.e. including $a=-infty$ and $b=+infty$).
- Erik Talvila used the solution of eqref{1} to define a generalized integral which is more general of the ones o Henstock and Kurzweil. Recently also Ricardo Estrada and Jasson Vindas proposed a definition of distributional integral based on a similar concept.
[1] G. E. Shilov (1968), Generalized functions and partial differential equations, Mathematics and Its Applications, Vol. 7, (English)
New York-London-Paris: Gordon and Breach Science Publishers, XII+345, MR0230129, Zbl 0177.36302.
[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.
answered Jan 28 at 22:16
Daniele TampieriDaniele Tampieri
2,58721022
edited Jan 29 at 13:40
answered Jan 28 at 22:16
Daniele TampieriDaniele Tampieri
2,58721022
answered Jan 28 at 22:16
Daniele TampieriDaniele Tampieri
2,58721022
answered Jan 28 at 22:16
Daniele TampieriDaniele Tampieri
2,58721022
2,58721022
$begingroup$
You meant $varphi mapsto int_{-infty}^. varphi_o(y)dy$ is continuous in the $C^infty_c(mathbb{R})$ topology, not only $varphi mapsto varphi_o$, this is what makes $varphi mapsto langle g,int_{-infty}^. varphi_o(y)dy rangle$ a distribution
$endgroup$
– reuns
Jan 28 at 23:38
$begingroup$
@reuns. Exactly: I added a section on this to my answer.
$endgroup$
– Daniele Tampieri
Jan 29 at 9:53
$begingroup$
@reuns. I also apologize for not noting your comment to the question. Probably I started editing my answer, then had lunch and finished answering later, without refreshing the post.
$endgroup$
– Daniele Tampieri
Jan 29 at 10:22
add a comment |
$begingroup$
You meant $varphi mapsto int_{-infty}^. varphi_o(y)dy$ is continuous in the $C^infty_c(mathbb{R})$ topology, not only $varphi mapsto varphi_o$, this is what makes $varphi mapsto langle g,int_{-infty}^. varphi_o(y)dy rangle$ a distribution
$endgroup$
– reuns
Jan 28 at 23:38
$begingroup$
@reuns. Exactly: I added a section on this to my answer.
$endgroup$
– Daniele Tampieri
Jan 29 at 9:53
$begingroup$
@reuns. I also apologize for not noting your comment to the question. Probably I started editing my answer, then had lunch and finished answering later, without refreshing the post.
$endgroup$
– Daniele Tampieri
Jan 29 at 10:22
$begingroup$
You meant $varphi mapsto int_{-infty}^. varphi_o(y)dy$ is continuous in the $C^infty_c(mathbb{R})$ topology, not only $varphi mapsto varphi_o$, this is what makes $varphi mapsto langle g,int_{-infty}^. varphi_o(y)dy rangle$ a distribution
$endgroup$
– reuns
Jan 28 at 23:38
$begingroup$
You meant $varphi mapsto int_{-infty}^. varphi_o(y)dy$ is continuous in the $C^infty_c(mathbb{R})$ topology, not only $varphi mapsto varphi_o$, this is what makes $varphi mapsto langle g,int_{-infty}^. varphi_o(y)dy rangle$ a distribution
$endgroup$
– reuns
Jan 28 at 23:38
$begingroup$
@reuns. Exactly: I added a section on this to my answer.
$endgroup$
– Daniele Tampieri
Jan 29 at 9:53
$begingroup$
@reuns. Exactly: I added a section on this to my answer.
$endgroup$
– Daniele Tampieri
Jan 29 at 9:53
$begingroup$
@reuns. I also apologize for not noting your comment to the question. Probably I started editing my answer, then had lunch and finished answering later, without refreshing the post.
$endgroup$
– Daniele Tampieri
Jan 29 at 10:22
$begingroup$
@reuns. I also apologize for not noting your comment to the question. Probably I started editing my answer, then had lunch and finished answering later, without refreshing the post.
$endgroup$
– Daniele Tampieri
Jan 29 at 10:22
add a comment |
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$begingroup$
Let $I(phi)(x) = int_{-infty}^x phi(y)dy$ and $langle G, phi rangle = -langle g, I(phi) - I(phi)(infty) psi rangle $ where $psi in C^infty, psi(x) =0 (x < -1), psi(x) = 1 (x > 1)$. Is $G$ a distribution, what is its derivative ?
$endgroup$
– reuns
Jan 28 at 20:27