Bogachev change of variables theorem.
I am reading the Bogachev's change of variable theorem proof. And I am stuck in an argument contained in it:
Let $F:Usubset mathbb{R}^nto mathbb{R}^n$ an injective $C¹$ map. Given $epsilon>$ there exist $delta>0$ s.t $forall x,y in U,~|x-y|<delta$ implies
$$
F(x)-F(y)=F^{'}(x)(y-x)+r(x,y)
$$
where $|r(x,y)|leq epsilon |x-y|$.
Here is where I am stucked:
Claim: Let $Q$ be a cube with center $x_0$ and diameter less than $delta.$ If $|det F'(x_0)|leq sqrt{epsilon}$ then we have that
$$
lambda (L(Q))-lambda(F(Q))leq sqrt{epsilon}lambda (Q)
$$
where $L(x)=F^{'}(x_0)(x-x_0)+F(x_0)$ and $lambda$ is the $n$-dimensional lebesgue measure.
Can someone give me some help?
Comment: I think the argument pass over to give some inferior bound for
$text{diam}( F(U))$ by using in some way the inequality
$|F(x)-F(y)|geq |F^{'}(x_0)(x-y)|-|r(x,y|$.
measure-theory lebesgue-integral lebesgue-measure
asked Nov 18 at 15:35
Eduardo
551112
add a comment |
I am reading the Bogachev's change of variable theorem proof. And I am stuck in an argument contained in it:
Let $F:Usubset mathbb{R}^nto mathbb{R}^n$ an injective $C¹$ map. Given $epsilon>$ there exist $delta>0$ s.t $forall x,y in U,~|x-y|<delta$ implies
$$
F(x)-F(y)=F^{'}(x)(y-x)+r(x,y)
$$
where $|r(x,y)|leq epsilon |x-y|$.
Here is where I am stucked:
Claim: Let $Q$ be a cube with center $x_0$ and diameter less than $delta.$ If $|det F'(x_0)|leq sqrt{epsilon}$ then we have that
$$
lambda (L(Q))-lambda(F(Q))leq sqrt{epsilon}lambda (Q)
$$
where $L(x)=F^{'}(x_0)(x-x_0)+F(x_0)$ and $lambda$ is the $n$-dimensional lebesgue measure.
Can someone give me some help?
Comment: I think the argument pass over to give some inferior bound for
$text{diam}( F(U))$ by using in some way the inequality
$|F(x)-F(y)|geq |F^{'}(x_0)(x-y)|-|r(x,y|$.
measure-theory lebesgue-integral lebesgue-measure
asked Nov 18 at 15:35
Eduardo
551112
add a comment |
I am reading the Bogachev's change of variable theorem proof. And I am stuck in an argument contained in it:
Let $F:Usubset mathbb{R}^nto mathbb{R}^n$ an injective $C¹$ map. Given $epsilon>$ there exist $delta>0$ s.t $forall x,y in U,~|x-y|<delta$ implies
$$
F(x)-F(y)=F^{'}(x)(y-x)+r(x,y)
$$
where $|r(x,y)|leq epsilon |x-y|$.
Here is where I am stucked:
Claim: Let $Q$ be a cube with center $x_0$ and diameter less than $delta.$ If $|det F'(x_0)|leq sqrt{epsilon}$ then we have that
$$
lambda (L(Q))-lambda(F(Q))leq sqrt{epsilon}lambda (Q)
$$
where $L(x)=F^{'}(x_0)(x-x_0)+F(x_0)$ and $lambda$ is the $n$-dimensional lebesgue measure.
Can someone give me some help?
Comment: I think the argument pass over to give some inferior bound for
$text{diam}( F(U))$ by using in some way the inequality
$|F(x)-F(y)|geq |F^{'}(x_0)(x-y)|-|r(x,y|$.
measure-theory lebesgue-integral lebesgue-measure
asked Nov 18 at 15:35
Eduardo
551112
I am reading the Bogachev's change of variable theorem proof. And I am stuck in an argument contained in it:
Let $F:Usubset mathbb{R}^nto mathbb{R}^n$ an injective $C¹$ map. Given $epsilon>$ there exist $delta>0$ s.t $forall x,y in U,~|x-y|<delta$ implies
$$
F(x)-F(y)=F^{'}(x)(y-x)+r(x,y)
$$
where $|r(x,y)|leq epsilon |x-y|$.
Here is where I am stucked:
Claim: Let $Q$ be a cube with center $x_0$ and diameter less than $delta.$ If $|det F'(x_0)|leq sqrt{epsilon}$ then we have that
$$
lambda (L(Q))-lambda(F(Q))leq sqrt{epsilon}lambda (Q)
$$
where $L(x)=F^{'}(x_0)(x-x_0)+F(x_0)$ and $lambda$ is the $n$-dimensional lebesgue measure.
Can someone give me some help?
Comment: I think the argument pass over to give some inferior bound for
$text{diam}( F(U))$ by using in some way the inequality
$|F(x)-F(y)|geq |F^{'}(x_0)(x-y)|-|r(x,y|$.
measure-theory lebesgue-integral lebesgue-measure
measure-theory lebesgue-integral lebesgue-measure
asked Nov 18 at 15:35
Eduardo
551112
asked Nov 18 at 15:35
Eduardo
551112
edited Nov 18 at 16:22
asked Nov 18 at 15:35
Eduardo
551112
asked Nov 18 at 15:35
Eduardo
551112
asked Nov 18 at 15:35
Eduardo
551112
551112
add a comment |
add a comment |
1 Answer
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I don't know enough measure theory but since $lambda(L(Q))=det L cdot lambda(Q)$, you could try to prove that $det L le frac{lambda(F(Q))}{lambda(Q)} + |det F'(x_0)|$.
answered Nov 18 at 17:06
lzralbu
560412
add a comment |
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1 Answer
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1 Answer
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I don't know enough measure theory but since $lambda(L(Q))=det L cdot lambda(Q)$, you could try to prove that $det L le frac{lambda(F(Q))}{lambda(Q)} + |det F'(x_0)|$.
answered Nov 18 at 17:06
lzralbu
560412
add a comment |
I don't know enough measure theory but since $lambda(L(Q))=det L cdot lambda(Q)$, you could try to prove that $det L le frac{lambda(F(Q))}{lambda(Q)} + |det F'(x_0)|$.
answered Nov 18 at 17:06
lzralbu
560412
add a comment |
I don't know enough measure theory but since $lambda(L(Q))=det L cdot lambda(Q)$, you could try to prove that $det L le frac{lambda(F(Q))}{lambda(Q)} + |det F'(x_0)|$.
answered Nov 18 at 17:06
lzralbu
560412
I don't know enough measure theory but since $lambda(L(Q))=det L cdot lambda(Q)$, you could try to prove that $det L le frac{lambda(F(Q))}{lambda(Q)} + |det F'(x_0)|$.
answered Nov 18 at 17:06
lzralbu
560412
answered Nov 18 at 17:06
lzralbu
560412
answered Nov 18 at 17:06
lzralbu
560412
answered Nov 18 at 17:06
lzralbu
560412
560412
add a comment |
add a comment |
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