Are smooth bounded-variation functions in the Sobolev Space $W_1,1$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is this equivalent to bounded variation?Two definitions of “Bounded Variation Function”The convergence in Bounded Variation functionsBounded variation problemThe best constant in Poincare-liked inequality in BV spaceExploring the total variation of a $C^1$ functionDecomposition of function of bounded variationUniformly Lipschitz and bounded variationThe limit of total variation minimizer goes to the averageConvergence of Bounded Variation on Open Subsets
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Are smooth bounded-variation functions in the Sobolev Space $W_1,1$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is this equivalent to bounded variation?Two definitions of “Bounded Variation Function”The convergence in Bounded Variation functionsBounded variation problemThe best constant in Poincare-liked inequality in BV spaceExploring the total variation of a $C^1$ functionDecomposition of function of bounded variationUniformly Lipschitz and bounded variationThe limit of total variation minimizer goes to the averageConvergence of Bounded Variation on Open Subsets
$begingroup$
Let $Omega subseteq mathbbR^n$ open and $f in BV(Omega) cap C^infty(Omega)$. Now I would like to prove that $f in W_1^1(Omega)$. I know that the distributional derivates of $f$ are the classical derivates, but I don't find a way to show that $partial_if in L_1(Omega)$ for $i=1,...,n$. Can maybe someone help me? :)
bounded-variation
edited Mar 24 at 0:39
David G. Stork
12.2k41836
asked Mar 23 at 20:23
Matthis StresemannMatthis Stresemann
61
$endgroup$
add a comment |
$begingroup$
Let $Omega subseteq mathbbR^n$ open and $f in BV(Omega) cap C^infty(Omega)$. Now I would like to prove that $f in W_1^1(Omega)$. I know that the distributional derivates of $f$ are the classical derivates, but I don't find a way to show that $partial_if in L_1(Omega)$ for $i=1,...,n$. Can maybe someone help me? :)
bounded-variation
edited Mar 24 at 0:39
David G. Stork
12.2k41836
asked Mar 23 at 20:23
Matthis StresemannMatthis Stresemann
61
$endgroup$
$begingroup$
What is BV? And why wouldn't you write it out?
$endgroup$
– David G. Stork
Mar 23 at 22:13
1
$begingroup$
Oh sorry, I thought this space would be well-known. BV is the space of functions of bounded variation. For a definiton look here en.wikipedia.org/wiki/Bounded_variation.
$endgroup$
– Matthis Stresemann
Mar 23 at 22:32
add a comment |
$begingroup$
Let $Omega subseteq mathbbR^n$ open and $f in BV(Omega) cap C^infty(Omega)$. Now I would like to prove that $f in W_1^1(Omega)$. I know that the distributional derivates of $f$ are the classical derivates, but I don't find a way to show that $partial_if in L_1(Omega)$ for $i=1,...,n$. Can maybe someone help me? :)
bounded-variation
edited Mar 24 at 0:39
David G. Stork
12.2k41836
asked Mar 23 at 20:23
Matthis StresemannMatthis Stresemann
61
$endgroup$
Let $Omega subseteq mathbbR^n$ open and $f in BV(Omega) cap C^infty(Omega)$. Now I would like to prove that $f in W_1^1(Omega)$. I know that the distributional derivates of $f$ are the classical derivates, but I don't find a way to show that $partial_if in L_1(Omega)$ for $i=1,...,n$. Can maybe someone help me? :)
bounded-variation
bounded-variation
edited Mar 24 at 0:39
David G. Stork
12.2k41836
asked Mar 23 at 20:23
Matthis StresemannMatthis Stresemann
61
edited Mar 24 at 0:39
David G. Stork
12.2k41836
asked Mar 23 at 20:23
Matthis StresemannMatthis Stresemann
61
edited Mar 24 at 0:39
David G. Stork
12.2k41836
edited Mar 24 at 0:39
David G. Stork
12.2k41836
edited Mar 24 at 0:39
David G. Stork
12.2k41836
12.2k41836
asked Mar 23 at 20:23
Matthis StresemannMatthis Stresemann
61
asked Mar 23 at 20:23
Matthis StresemannMatthis Stresemann
61
asked Mar 23 at 20:23
Matthis StresemannMatthis Stresemann
61
61
$begingroup$
What is BV? And why wouldn't you write it out?
$endgroup$
– David G. Stork
Mar 23 at 22:13
1
$begingroup$
Oh sorry, I thought this space would be well-known. BV is the space of functions of bounded variation. For a definiton look here en.wikipedia.org/wiki/Bounded_variation.
$endgroup$
– Matthis Stresemann
Mar 23 at 22:32
add a comment |
$begingroup$
What is BV? And why wouldn't you write it out?
$endgroup$
– David G. Stork
Mar 23 at 22:13
1
$begingroup$
Oh sorry, I thought this space would be well-known. BV is the space of functions of bounded variation. For a definiton look here en.wikipedia.org/wiki/Bounded_variation.
$endgroup$
– Matthis Stresemann
Mar 23 at 22:32
$begingroup$
What is BV? And why wouldn't you write it out?
$endgroup$
– David G. Stork
Mar 23 at 22:13
$begingroup$
What is BV? And why wouldn't you write it out?
$endgroup$
– David G. Stork
Mar 23 at 22:13
1
1
$begingroup$
Oh sorry, I thought this space would be well-known. BV is the space of functions of bounded variation. For a definiton look here en.wikipedia.org/wiki/Bounded_variation.
$endgroup$
– Matthis Stresemann
Mar 23 at 22:32
$begingroup$
Oh sorry, I thought this space would be well-known. BV is the space of functions of bounded variation. For a definiton look here en.wikipedia.org/wiki/Bounded_variation.
$endgroup$
– Matthis Stresemann
Mar 23 at 22:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In order to correctly answer the question, let's state explicitly the two definitions lying at the roots of the theory of BV functions (see for example [1], chapter 1, pp. 3-4):
Definition 1. Let $Omegasubseteq Bbb R^n$ be an open set and let $fin L^1(Omega)$. The total variation of f (in $Omega$) is the following non negative quantity
$$
V(f,Omega)=supleftintlimits_Omega! f(x)operatornamedivmathbfg(x),mathrmdx,Big
$$
Definition 2. A function $fin L^1(Omega)$ is said to have bounded variation if $V(f,Omega)<infty$. $BV(Omega)$ is defined as the space of all functions in $L^1(Omega)$ of bounded variation.
Now consider $fin BV(Omega)cap C^1(Omega)$: then, for every $mathbfg(x)=in C_0^1(Omega,Bbb R^n)$ such that $;|mathbfg(x)|le1$ on $Omega$, we have that
$$
beginsplit
intlimits_Omega! f(x)operatornamedivmathbfg(x),mathrmdx
&= - intlimits_Omega sum_i=1^n fracpartial f(x)partial x_icdot g_i(x),mathrmdx\
&= -intlimits_Omegalangleoperatornamegrad!f(x),mathbfg(x)rangle,mathrmdx.\
endsplit
$$
But this implies also that
$$
V(f,Omega)=intlimits_Omegavertoperatornamegrad!f(x)vert,mathrmdx.
$$
To see this, it is sufficient to consider a sequence of functions $mathbfg(x)$ whose compact supports $operatornamesuppmathbfg$ increasingly cover the set $Omega$ and note that, by the Cauchy-Schwartz inequality
$$
beginsplit
-langleoperatornamegrad!f(x) ,mathbfg(x)rangle
&le |langleoperatornamegrad!f(x) ,mathbfg(x)rangle|= left| sum_i=1^n fracpartial f(x)partial x_icdot g_i(x)right|\
&le left( sum_i=1^n left|fracpartial f(x)partial x_iright|^2right)^frac12left( sum_i=1^n g_i(x)^2right)^frac12\
\
&= vertoperatornamegrad!f(x)vert|mathbfg(x)|levertoperatornamegrad!f(x)vert
endsplit
$$
and thus
$$
-intlimits_operatornamesuppmathbfglangleoperatornamegrad!f(x),mathbfg(x)rangle,mathrmdxle intlimits_operatornamesuppmathbfgvertoperatornamegrad!f(x)vert,mathrmdx
$$
This, since $fin BV(Omega)$, implies
$$
intlimits_Omegavertoperatornamegrad!f(x)vert,mathrmdx<inftyiff operatornamegrad!f(x)in L^1(Omega,Bbb R^n)
$$
The reasoning is exactly the same if $fin BV(Omega)cap C^infty(Omega)$.
[1] Enrico Giusti (1984), Minimal surfaces and functions of bounded variations, Monographs in Mathematics, 80, Basel–Boston–Stuttgart: Birkhäuser Verlag, pp. XII+240, ISBN 978-0-8176-3153-6, MR 0775682, Zbl 0545.49018
answered Mar 24 at 22:24
Daniele TampieriDaniele Tampieri
2,75721022
$endgroup$
1
$begingroup$
Thank you for this long answer. I guess you wanna use the monotone convergence Thm. to prove the last equality. Therefore, you need a sequence of non-negative measurable functions. But $f$ and $partial_i f$ don't have to be non-negative, so I wonder how you exactly construct this approximating sequence. Do you split $f$ in its positive and negative part or what do you do?
$endgroup$
– Matthis Stresemann
Mar 25 at 14:42
$begingroup$
@MatthisStresemann the matter is simpler. I’ll post you some further notes later, since I am currently involved in a Work problem.
$endgroup$
– Daniele Tampieri
Mar 25 at 14:55
$begingroup$
@MatthisStresemann. I apologize for not being able to refine my answer last evening: I will try to do it this one. However, just a few remarks: the trick which proves the last equality is simply to consider classes of test functions $mathbf g$ whose $i$-th components are positive when $partial_i f>0$ and negative when $partial_i f<0$. Also remember that we are simply searching for the $sup$, not for the equality $limsup=liminf=lim$ i.e. we are not calculating a limit.
$endgroup$
– Daniele Tampieri
Mar 26 at 6:48
$begingroup$
Sorry, but I still don't see it. Could you eventually give me the exact definition of the sequence $g$?
$endgroup$
– Matthis Stresemann
Mar 26 at 12:39
1
$begingroup$
I think I have found another way to show that $f in W_1^1$.Therefore, you don't need to construct a sequence like $g$. But you can still present your solution if you want.
$endgroup$
– Matthis Stresemann
Mar 27 at 12:47
|
show 3 more comments
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1 Answer
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1 Answer
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$begingroup$
In order to correctly answer the question, let's state explicitly the two definitions lying at the roots of the theory of BV functions (see for example [1], chapter 1, pp. 3-4):
Definition 1. Let $Omegasubseteq Bbb R^n$ be an open set and let $fin L^1(Omega)$. The total variation of f (in $Omega$) is the following non negative quantity
$$
V(f,Omega)=supleftintlimits_Omega! f(x)operatornamedivmathbfg(x),mathrmdx,Big
$$
Definition 2. A function $fin L^1(Omega)$ is said to have bounded variation if $V(f,Omega)<infty$. $BV(Omega)$ is defined as the space of all functions in $L^1(Omega)$ of bounded variation.
Now consider $fin BV(Omega)cap C^1(Omega)$: then, for every $mathbfg(x)=in C_0^1(Omega,Bbb R^n)$ such that $;|mathbfg(x)|le1$ on $Omega$, we have that
$$
beginsplit
intlimits_Omega! f(x)operatornamedivmathbfg(x),mathrmdx
&= - intlimits_Omega sum_i=1^n fracpartial f(x)partial x_icdot g_i(x),mathrmdx\
&= -intlimits_Omegalangleoperatornamegrad!f(x),mathbfg(x)rangle,mathrmdx.\
endsplit
$$
But this implies also that
$$
V(f,Omega)=intlimits_Omegavertoperatornamegrad!f(x)vert,mathrmdx.
$$
To see this, it is sufficient to consider a sequence of functions $mathbfg(x)$ whose compact supports $operatornamesuppmathbfg$ increasingly cover the set $Omega$ and note that, by the Cauchy-Schwartz inequality
$$
beginsplit
-langleoperatornamegrad!f(x) ,mathbfg(x)rangle
&le |langleoperatornamegrad!f(x) ,mathbfg(x)rangle|= left| sum_i=1^n fracpartial f(x)partial x_icdot g_i(x)right|\
&le left( sum_i=1^n left|fracpartial f(x)partial x_iright|^2right)^frac12left( sum_i=1^n g_i(x)^2right)^frac12\
\
&= vertoperatornamegrad!f(x)vert|mathbfg(x)|levertoperatornamegrad!f(x)vert
endsplit
$$
and thus
$$
-intlimits_operatornamesuppmathbfglangleoperatornamegrad!f(x),mathbfg(x)rangle,mathrmdxle intlimits_operatornamesuppmathbfgvertoperatornamegrad!f(x)vert,mathrmdx
$$
This, since $fin BV(Omega)$, implies
$$
intlimits_Omegavertoperatornamegrad!f(x)vert,mathrmdx<inftyiff operatornamegrad!f(x)in L^1(Omega,Bbb R^n)
$$
The reasoning is exactly the same if $fin BV(Omega)cap C^infty(Omega)$.
[1] Enrico Giusti (1984), Minimal surfaces and functions of bounded variations, Monographs in Mathematics, 80, Basel–Boston–Stuttgart: Birkhäuser Verlag, pp. XII+240, ISBN 978-0-8176-3153-6, MR 0775682, Zbl 0545.49018
answered Mar 24 at 22:24
Daniele TampieriDaniele Tampieri
2,75721022
$endgroup$
1
$begingroup$
Thank you for this long answer. I guess you wanna use the monotone convergence Thm. to prove the last equality. Therefore, you need a sequence of non-negative measurable functions. But $f$ and $partial_i f$ don't have to be non-negative, so I wonder how you exactly construct this approximating sequence. Do you split $f$ in its positive and negative part or what do you do?
$endgroup$
– Matthis Stresemann
Mar 25 at 14:42
$begingroup$
@MatthisStresemann the matter is simpler. I’ll post you some further notes later, since I am currently involved in a Work problem.
$endgroup$
– Daniele Tampieri
Mar 25 at 14:55
$begingroup$
@MatthisStresemann. I apologize for not being able to refine my answer last evening: I will try to do it this one. However, just a few remarks: the trick which proves the last equality is simply to consider classes of test functions $mathbf g$ whose $i$-th components are positive when $partial_i f>0$ and negative when $partial_i f<0$. Also remember that we are simply searching for the $sup$, not for the equality $limsup=liminf=lim$ i.e. we are not calculating a limit.
$endgroup$
– Daniele Tampieri
Mar 26 at 6:48
$begingroup$
Sorry, but I still don't see it. Could you eventually give me the exact definition of the sequence $g$?
$endgroup$
– Matthis Stresemann
Mar 26 at 12:39
1
$begingroup$
I think I have found another way to show that $f in W_1^1$.Therefore, you don't need to construct a sequence like $g$. But you can still present your solution if you want.
$endgroup$
– Matthis Stresemann
Mar 27 at 12:47
|
show 3 more comments
$begingroup$
In order to correctly answer the question, let's state explicitly the two definitions lying at the roots of the theory of BV functions (see for example [1], chapter 1, pp. 3-4):
Definition 1. Let $Omegasubseteq Bbb R^n$ be an open set and let $fin L^1(Omega)$. The total variation of f (in $Omega$) is the following non negative quantity
$$
V(f,Omega)=supleftintlimits_Omega! f(x)operatornamedivmathbfg(x),mathrmdx,Big
$$
Definition 2. A function $fin L^1(Omega)$ is said to have bounded variation if $V(f,Omega)<infty$. $BV(Omega)$ is defined as the space of all functions in $L^1(Omega)$ of bounded variation.
Now consider $fin BV(Omega)cap C^1(Omega)$: then, for every $mathbfg(x)=in C_0^1(Omega,Bbb R^n)$ such that $;|mathbfg(x)|le1$ on $Omega$, we have that
$$
beginsplit
intlimits_Omega! f(x)operatornamedivmathbfg(x),mathrmdx
&= - intlimits_Omega sum_i=1^n fracpartial f(x)partial x_icdot g_i(x),mathrmdx\
&= -intlimits_Omegalangleoperatornamegrad!f(x),mathbfg(x)rangle,mathrmdx.\
endsplit
$$
But this implies also that
$$
V(f,Omega)=intlimits_Omegavertoperatornamegrad!f(x)vert,mathrmdx.
$$
To see this, it is sufficient to consider a sequence of functions $mathbfg(x)$ whose compact supports $operatornamesuppmathbfg$ increasingly cover the set $Omega$ and note that, by the Cauchy-Schwartz inequality
$$
beginsplit
-langleoperatornamegrad!f(x) ,mathbfg(x)rangle
&le |langleoperatornamegrad!f(x) ,mathbfg(x)rangle|= left| sum_i=1^n fracpartial f(x)partial x_icdot g_i(x)right|\
&le left( sum_i=1^n left|fracpartial f(x)partial x_iright|^2right)^frac12left( sum_i=1^n g_i(x)^2right)^frac12\
\
&= vertoperatornamegrad!f(x)vert|mathbfg(x)|levertoperatornamegrad!f(x)vert
endsplit
$$
and thus
$$
-intlimits_operatornamesuppmathbfglangleoperatornamegrad!f(x),mathbfg(x)rangle,mathrmdxle intlimits_operatornamesuppmathbfgvertoperatornamegrad!f(x)vert,mathrmdx
$$
This, since $fin BV(Omega)$, implies
$$
intlimits_Omegavertoperatornamegrad!f(x)vert,mathrmdx<inftyiff operatornamegrad!f(x)in L^1(Omega,Bbb R^n)
$$
The reasoning is exactly the same if $fin BV(Omega)cap C^infty(Omega)$.
[1] Enrico Giusti (1984), Minimal surfaces and functions of bounded variations, Monographs in Mathematics, 80, Basel–Boston–Stuttgart: Birkhäuser Verlag, pp. XII+240, ISBN 978-0-8176-3153-6, MR 0775682, Zbl 0545.49018
answered Mar 24 at 22:24
Daniele TampieriDaniele Tampieri
2,75721022
$endgroup$
1
$begingroup$
Thank you for this long answer. I guess you wanna use the monotone convergence Thm. to prove the last equality. Therefore, you need a sequence of non-negative measurable functions. But $f$ and $partial_i f$ don't have to be non-negative, so I wonder how you exactly construct this approximating sequence. Do you split $f$ in its positive and negative part or what do you do?
$endgroup$
– Matthis Stresemann
Mar 25 at 14:42
$begingroup$
@MatthisStresemann the matter is simpler. I’ll post you some further notes later, since I am currently involved in a Work problem.
$endgroup$
– Daniele Tampieri
Mar 25 at 14:55
$begingroup$
@MatthisStresemann. I apologize for not being able to refine my answer last evening: I will try to do it this one. However, just a few remarks: the trick which proves the last equality is simply to consider classes of test functions $mathbf g$ whose $i$-th components are positive when $partial_i f>0$ and negative when $partial_i f<0$. Also remember that we are simply searching for the $sup$, not for the equality $limsup=liminf=lim$ i.e. we are not calculating a limit.
$endgroup$
– Daniele Tampieri
Mar 26 at 6:48
$begingroup$
Sorry, but I still don't see it. Could you eventually give me the exact definition of the sequence $g$?
$endgroup$
– Matthis Stresemann
Mar 26 at 12:39
1
$begingroup$
I think I have found another way to show that $f in W_1^1$.Therefore, you don't need to construct a sequence like $g$. But you can still present your solution if you want.
$endgroup$
– Matthis Stresemann
Mar 27 at 12:47
|
show 3 more comments
$begingroup$
In order to correctly answer the question, let's state explicitly the two definitions lying at the roots of the theory of BV functions (see for example [1], chapter 1, pp. 3-4):
Definition 1. Let $Omegasubseteq Bbb R^n$ be an open set and let $fin L^1(Omega)$. The total variation of f (in $Omega$) is the following non negative quantity
$$
V(f,Omega)=supleftintlimits_Omega! f(x)operatornamedivmathbfg(x),mathrmdx,Big
$$
Definition 2. A function $fin L^1(Omega)$ is said to have bounded variation if $V(f,Omega)<infty$. $BV(Omega)$ is defined as the space of all functions in $L^1(Omega)$ of bounded variation.
Now consider $fin BV(Omega)cap C^1(Omega)$: then, for every $mathbfg(x)=in C_0^1(Omega,Bbb R^n)$ such that $;|mathbfg(x)|le1$ on $Omega$, we have that
$$
beginsplit
intlimits_Omega! f(x)operatornamedivmathbfg(x),mathrmdx
&= - intlimits_Omega sum_i=1^n fracpartial f(x)partial x_icdot g_i(x),mathrmdx\
&= -intlimits_Omegalangleoperatornamegrad!f(x),mathbfg(x)rangle,mathrmdx.\
endsplit
$$
But this implies also that
$$
V(f,Omega)=intlimits_Omegavertoperatornamegrad!f(x)vert,mathrmdx.
$$
To see this, it is sufficient to consider a sequence of functions $mathbfg(x)$ whose compact supports $operatornamesuppmathbfg$ increasingly cover the set $Omega$ and note that, by the Cauchy-Schwartz inequality
$$
beginsplit
-langleoperatornamegrad!f(x) ,mathbfg(x)rangle
&le |langleoperatornamegrad!f(x) ,mathbfg(x)rangle|= left| sum_i=1^n fracpartial f(x)partial x_icdot g_i(x)right|\
&le left( sum_i=1^n left|fracpartial f(x)partial x_iright|^2right)^frac12left( sum_i=1^n g_i(x)^2right)^frac12\
\
&= vertoperatornamegrad!f(x)vert|mathbfg(x)|levertoperatornamegrad!f(x)vert
endsplit
$$
and thus
$$
-intlimits_operatornamesuppmathbfglangleoperatornamegrad!f(x),mathbfg(x)rangle,mathrmdxle intlimits_operatornamesuppmathbfgvertoperatornamegrad!f(x)vert,mathrmdx
$$
This, since $fin BV(Omega)$, implies
$$
intlimits_Omegavertoperatornamegrad!f(x)vert,mathrmdx<inftyiff operatornamegrad!f(x)in L^1(Omega,Bbb R^n)
$$
The reasoning is exactly the same if $fin BV(Omega)cap C^infty(Omega)$.
[1] Enrico Giusti (1984), Minimal surfaces and functions of bounded variations, Monographs in Mathematics, 80, Basel–Boston–Stuttgart: Birkhäuser Verlag, pp. XII+240, ISBN 978-0-8176-3153-6, MR 0775682, Zbl 0545.49018
answered Mar 24 at 22:24
Daniele TampieriDaniele Tampieri
2,75721022
$endgroup$
In order to correctly answer the question, let's state explicitly the two definitions lying at the roots of the theory of BV functions (see for example [1], chapter 1, pp. 3-4):
Definition 1. Let $Omegasubseteq Bbb R^n$ be an open set and let $fin L^1(Omega)$. The total variation of f (in $Omega$) is the following non negative quantity
$$
V(f,Omega)=supleftintlimits_Omega! f(x)operatornamedivmathbfg(x),mathrmdx,Big
$$
Definition 2. A function $fin L^1(Omega)$ is said to have bounded variation if $V(f,Omega)<infty$. $BV(Omega)$ is defined as the space of all functions in $L^1(Omega)$ of bounded variation.
Now consider $fin BV(Omega)cap C^1(Omega)$: then, for every $mathbfg(x)=in C_0^1(Omega,Bbb R^n)$ such that $;|mathbfg(x)|le1$ on $Omega$, we have that
$$
beginsplit
intlimits_Omega! f(x)operatornamedivmathbfg(x),mathrmdx
&= - intlimits_Omega sum_i=1^n fracpartial f(x)partial x_icdot g_i(x),mathrmdx\
&= -intlimits_Omegalangleoperatornamegrad!f(x),mathbfg(x)rangle,mathrmdx.\
endsplit
$$
But this implies also that
$$
V(f,Omega)=intlimits_Omegavertoperatornamegrad!f(x)vert,mathrmdx.
$$
To see this, it is sufficient to consider a sequence of functions $mathbfg(x)$ whose compact supports $operatornamesuppmathbfg$ increasingly cover the set $Omega$ and note that, by the Cauchy-Schwartz inequality
$$
beginsplit
-langleoperatornamegrad!f(x) ,mathbfg(x)rangle
&le |langleoperatornamegrad!f(x) ,mathbfg(x)rangle|= left| sum_i=1^n fracpartial f(x)partial x_icdot g_i(x)right|\
&le left( sum_i=1^n left|fracpartial f(x)partial x_iright|^2right)^frac12left( sum_i=1^n g_i(x)^2right)^frac12\
\
&= vertoperatornamegrad!f(x)vert|mathbfg(x)|levertoperatornamegrad!f(x)vert
endsplit
$$
and thus
$$
-intlimits_operatornamesuppmathbfglangleoperatornamegrad!f(x),mathbfg(x)rangle,mathrmdxle intlimits_operatornamesuppmathbfgvertoperatornamegrad!f(x)vert,mathrmdx
$$
This, since $fin BV(Omega)$, implies
$$
intlimits_Omegavertoperatornamegrad!f(x)vert,mathrmdx<inftyiff operatornamegrad!f(x)in L^1(Omega,Bbb R^n)
$$
The reasoning is exactly the same if $fin BV(Omega)cap C^infty(Omega)$.
[1] Enrico Giusti (1984), Minimal surfaces and functions of bounded variations, Monographs in Mathematics, 80, Basel–Boston–Stuttgart: Birkhäuser Verlag, pp. XII+240, ISBN 978-0-8176-3153-6, MR 0775682, Zbl 0545.49018
answered Mar 24 at 22:24
Daniele TampieriDaniele Tampieri
2,75721022
edited Apr 1 at 10:44
answered Mar 24 at 22:24
Daniele TampieriDaniele Tampieri
2,75721022
answered Mar 24 at 22:24
Daniele TampieriDaniele Tampieri
2,75721022
answered Mar 24 at 22:24
Daniele TampieriDaniele Tampieri
2,75721022
2,75721022
1
$begingroup$
Thank you for this long answer. I guess you wanna use the monotone convergence Thm. to prove the last equality. Therefore, you need a sequence of non-negative measurable functions. But $f$ and $partial_i f$ don't have to be non-negative, so I wonder how you exactly construct this approximating sequence. Do you split $f$ in its positive and negative part or what do you do?
$endgroup$
– Matthis Stresemann
Mar 25 at 14:42
$begingroup$
@MatthisStresemann the matter is simpler. I’ll post you some further notes later, since I am currently involved in a Work problem.
$endgroup$
– Daniele Tampieri
Mar 25 at 14:55
$begingroup$
@MatthisStresemann. I apologize for not being able to refine my answer last evening: I will try to do it this one. However, just a few remarks: the trick which proves the last equality is simply to consider classes of test functions $mathbf g$ whose $i$-th components are positive when $partial_i f>0$ and negative when $partial_i f<0$. Also remember that we are simply searching for the $sup$, not for the equality $limsup=liminf=lim$ i.e. we are not calculating a limit.
$endgroup$
– Daniele Tampieri
Mar 26 at 6:48
$begingroup$
Sorry, but I still don't see it. Could you eventually give me the exact definition of the sequence $g$?
$endgroup$
– Matthis Stresemann
Mar 26 at 12:39
1
$begingroup$
I think I have found another way to show that $f in W_1^1$.Therefore, you don't need to construct a sequence like $g$. But you can still present your solution if you want.
$endgroup$
– Matthis Stresemann
Mar 27 at 12:47
|
show 3 more comments
1
$begingroup$
Thank you for this long answer. I guess you wanna use the monotone convergence Thm. to prove the last equality. Therefore, you need a sequence of non-negative measurable functions. But $f$ and $partial_i f$ don't have to be non-negative, so I wonder how you exactly construct this approximating sequence. Do you split $f$ in its positive and negative part or what do you do?
$endgroup$
– Matthis Stresemann
Mar 25 at 14:42
$begingroup$
@MatthisStresemann the matter is simpler. I’ll post you some further notes later, since I am currently involved in a Work problem.
$endgroup$
– Daniele Tampieri
Mar 25 at 14:55
$begingroup$
@MatthisStresemann. I apologize for not being able to refine my answer last evening: I will try to do it this one. However, just a few remarks: the trick which proves the last equality is simply to consider classes of test functions $mathbf g$ whose $i$-th components are positive when $partial_i f>0$ and negative when $partial_i f<0$. Also remember that we are simply searching for the $sup$, not for the equality $limsup=liminf=lim$ i.e. we are not calculating a limit.
$endgroup$
– Daniele Tampieri
Mar 26 at 6:48
$begingroup$
Sorry, but I still don't see it. Could you eventually give me the exact definition of the sequence $g$?
$endgroup$
– Matthis Stresemann
Mar 26 at 12:39
1
$begingroup$
I think I have found another way to show that $f in W_1^1$.Therefore, you don't need to construct a sequence like $g$. But you can still present your solution if you want.
$endgroup$
– Matthis Stresemann
Mar 27 at 12:47
1
1
$begingroup$
Thank you for this long answer. I guess you wanna use the monotone convergence Thm. to prove the last equality. Therefore, you need a sequence of non-negative measurable functions. But $f$ and $partial_i f$ don't have to be non-negative, so I wonder how you exactly construct this approximating sequence. Do you split $f$ in its positive and negative part or what do you do?
$endgroup$
– Matthis Stresemann
Mar 25 at 14:42
$begingroup$
Thank you for this long answer. I guess you wanna use the monotone convergence Thm. to prove the last equality. Therefore, you need a sequence of non-negative measurable functions. But $f$ and $partial_i f$ don't have to be non-negative, so I wonder how you exactly construct this approximating sequence. Do you split $f$ in its positive and negative part or what do you do?
$endgroup$
– Matthis Stresemann
Mar 25 at 14:42
$begingroup$
@MatthisStresemann the matter is simpler. I’ll post you some further notes later, since I am currently involved in a Work problem.
$endgroup$
– Daniele Tampieri
Mar 25 at 14:55
$begingroup$
@MatthisStresemann the matter is simpler. I’ll post you some further notes later, since I am currently involved in a Work problem.
$endgroup$
– Daniele Tampieri
Mar 25 at 14:55
$begingroup$
@MatthisStresemann. I apologize for not being able to refine my answer last evening: I will try to do it this one. However, just a few remarks: the trick which proves the last equality is simply to consider classes of test functions $mathbf g$ whose $i$-th components are positive when $partial_i f>0$ and negative when $partial_i f<0$. Also remember that we are simply searching for the $sup$, not for the equality $limsup=liminf=lim$ i.e. we are not calculating a limit.
$endgroup$
– Daniele Tampieri
Mar 26 at 6:48
$begingroup$
@MatthisStresemann. I apologize for not being able to refine my answer last evening: I will try to do it this one. However, just a few remarks: the trick which proves the last equality is simply to consider classes of test functions $mathbf g$ whose $i$-th components are positive when $partial_i f>0$ and negative when $partial_i f<0$. Also remember that we are simply searching for the $sup$, not for the equality $limsup=liminf=lim$ i.e. we are not calculating a limit.
$endgroup$
– Daniele Tampieri
Mar 26 at 6:48
$begingroup$
Sorry, but I still don't see it. Could you eventually give me the exact definition of the sequence $g$?
$endgroup$
– Matthis Stresemann
Mar 26 at 12:39
$begingroup$
Sorry, but I still don't see it. Could you eventually give me the exact definition of the sequence $g$?
$endgroup$
– Matthis Stresemann
Mar 26 at 12:39
1
1
$begingroup$
I think I have found another way to show that $f in W_1^1$.Therefore, you don't need to construct a sequence like $g$. But you can still present your solution if you want.
$endgroup$
– Matthis Stresemann
Mar 27 at 12:47
$begingroup$
I think I have found another way to show that $f in W_1^1$.Therefore, you don't need to construct a sequence like $g$. But you can still present your solution if you want.
$endgroup$
– Matthis Stresemann
Mar 27 at 12:47
|
show 3 more comments
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$begingroup$
What is BV? And why wouldn't you write it out?
$endgroup$
– David G. Stork
Mar 23 at 22:13
1
$begingroup$
Oh sorry, I thought this space would be well-known. BV is the space of functions of bounded variation. For a definiton look here en.wikipedia.org/wiki/Bounded_variation.
$endgroup$
– Matthis Stresemann
Mar 23 at 22:32