Trying to prove Sard's lemma
$begingroup$
This is an exercise of Analysis III of Amann and Escher:
Note: here $XsubsetBbb R^n$ is open.
I was trying to follow the hint, however I cant finish a proof. My work so far:
Let $x_0in C$. Now let $J_0$ the cube centered at $x_0$ with side length $r>0$ such that $J_0subset X$. Then $Phi(J_0)$ is compact and convex, and from the mean value theorem we find that
$$
Phi(x)-Phi(x_0)=int_0^1partialPhi(x_0+t(x-x_0))(x-x_0), dttag1
$$
Hence
$$
|Phi(x)-Phi(x_0)|le |x-x_0|int_0^1|partialPhi(x_0+t(x-x_0))|, dttag2
$$
Then setting $rho(r):=max_{xin J_0}int_0^1|partialPhi(x_0+t(x-x_0))|, dt$ we find that
$$
lambda_n(Phi(J_0))le rho(r)^nlambda_n(J_0)= (rrho(r))^ntag3
$$
Now note that $|partial Phi(x_0)|lemax_{1le kle n}|lambda_k|$, where $lambda_1,ldots,lambda_n$ are the eigenvalues of $partialPhi(x)_{Bbb C}$ (where $partialPhi(x)_{Bbb C}$ is the complexification of $partialPhi(x)$).
However Im stuck here. I cant see that $lim_{rto 0^+}rho(r)=0$, what I see is that $lim_{rto 0^+}rho(r)=|partialPhi(x_0)|$, what is not necessarily zero.
Some help will be appreciated, thank you.
real-analysis measure-theory lebesgue-measure
asked Dec 27 '18 at 3:32
MasacrosoMasacroso
13.1k41747
$endgroup$
add a comment |
$begingroup$
This is an exercise of Analysis III of Amann and Escher:
Note: here $XsubsetBbb R^n$ is open.
I was trying to follow the hint, however I cant finish a proof. My work so far:
Let $x_0in C$. Now let $J_0$ the cube centered at $x_0$ with side length $r>0$ such that $J_0subset X$. Then $Phi(J_0)$ is compact and convex, and from the mean value theorem we find that
$$
Phi(x)-Phi(x_0)=int_0^1partialPhi(x_0+t(x-x_0))(x-x_0), dttag1
$$
Hence
$$
|Phi(x)-Phi(x_0)|le |x-x_0|int_0^1|partialPhi(x_0+t(x-x_0))|, dttag2
$$
Then setting $rho(r):=max_{xin J_0}int_0^1|partialPhi(x_0+t(x-x_0))|, dt$ we find that
$$
lambda_n(Phi(J_0))le rho(r)^nlambda_n(J_0)= (rrho(r))^ntag3
$$
Now note that $|partial Phi(x_0)|lemax_{1le kle n}|lambda_k|$, where $lambda_1,ldots,lambda_n$ are the eigenvalues of $partialPhi(x)_{Bbb C}$ (where $partialPhi(x)_{Bbb C}$ is the complexification of $partialPhi(x)$).
However Im stuck here. I cant see that $lim_{rto 0^+}rho(r)=0$, what I see is that $lim_{rto 0^+}rho(r)=|partialPhi(x_0)|$, what is not necessarily zero.
Some help will be appreciated, thank you.
real-analysis measure-theory lebesgue-measure
asked Dec 27 '18 at 3:32
MasacrosoMasacroso
13.1k41747
$endgroup$
$begingroup$
You need to use that $|partial Phi(x_0) - partial Phi (x_0 + t (x-x_0))|$ gets small as $tto 0$, and then you need to estimate the measure of $partial Phi (x_0) (J_0)$.
$endgroup$
– PhoemueX
Dec 27 '18 at 16:48
$begingroup$
@PhoemueX I dont follow what you says... Of course $|partialPhi(x_0)-partialPhi(x_0+t(x-x_0)|to 0$ as $tto 0$ because $partialPhi$ is continuous. But this can be said of any continuous derivative. By the other hand is clear that $lambda_n(partialPhi(x_0)Bbb R^n)=0$ because $partialPhi(x_0)$ is not injective. However I dont see the relation of this things with my question
$endgroup$
– Masacroso
Dec 28 '18 at 6:41
$begingroup$
Sorry if this is a silly question, but why do you say "$partialPhi(x_0)$ is not injective"? Is not that denoting a matrix with values, i.e. the Jacobian matrix evaluated at $x = x_0$ and hence it is non-sense to talk about injective, surjective, etc?
$endgroup$
– manooooh
Dec 29 '18 at 9:06
1
$begingroup$
@manooooh ok, I see what you are asking. The Fréchet derivative at a point of some differentiable function is a linear function, that can be represented by a matrix acting in $Bbb R^n$ (the Jacobian matrix). In this case we knows that $partialPhi(x_0)notinmathcal L{rm aut}(Bbb R^n)$, that is, it is a linear function in $Bbb R^n$ that is not an automorphism. This imply that $partialPhi(x_0)$ is not surjective, and because it is linear, then it is neither injective. This mean that the Jacobian matrix (that is a square matrix $ntimes n$) that represent this function is not invertible.
$endgroup$
– Masacroso
Dec 29 '18 at 10:56
add a comment |
$begingroup$
This is an exercise of Analysis III of Amann and Escher:
Note: here $XsubsetBbb R^n$ is open.
I was trying to follow the hint, however I cant finish a proof. My work so far:
Let $x_0in C$. Now let $J_0$ the cube centered at $x_0$ with side length $r>0$ such that $J_0subset X$. Then $Phi(J_0)$ is compact and convex, and from the mean value theorem we find that
$$
Phi(x)-Phi(x_0)=int_0^1partialPhi(x_0+t(x-x_0))(x-x_0), dttag1
$$
Hence
$$
|Phi(x)-Phi(x_0)|le |x-x_0|int_0^1|partialPhi(x_0+t(x-x_0))|, dttag2
$$
Then setting $rho(r):=max_{xin J_0}int_0^1|partialPhi(x_0+t(x-x_0))|, dt$ we find that
$$
lambda_n(Phi(J_0))le rho(r)^nlambda_n(J_0)= (rrho(r))^ntag3
$$
Now note that $|partial Phi(x_0)|lemax_{1le kle n}|lambda_k|$, where $lambda_1,ldots,lambda_n$ are the eigenvalues of $partialPhi(x)_{Bbb C}$ (where $partialPhi(x)_{Bbb C}$ is the complexification of $partialPhi(x)$).
However Im stuck here. I cant see that $lim_{rto 0^+}rho(r)=0$, what I see is that $lim_{rto 0^+}rho(r)=|partialPhi(x_0)|$, what is not necessarily zero.
Some help will be appreciated, thank you.
real-analysis measure-theory lebesgue-measure
asked Dec 27 '18 at 3:32
MasacrosoMasacroso
13.1k41747
$endgroup$
This is an exercise of Analysis III of Amann and Escher:
Note: here $XsubsetBbb R^n$ is open.
I was trying to follow the hint, however I cant finish a proof. My work so far:
Let $x_0in C$. Now let $J_0$ the cube centered at $x_0$ with side length $r>0$ such that $J_0subset X$. Then $Phi(J_0)$ is compact and convex, and from the mean value theorem we find that
$$
Phi(x)-Phi(x_0)=int_0^1partialPhi(x_0+t(x-x_0))(x-x_0), dttag1
$$
Hence
$$
|Phi(x)-Phi(x_0)|le |x-x_0|int_0^1|partialPhi(x_0+t(x-x_0))|, dttag2
$$
Then setting $rho(r):=max_{xin J_0}int_0^1|partialPhi(x_0+t(x-x_0))|, dt$ we find that
$$
lambda_n(Phi(J_0))le rho(r)^nlambda_n(J_0)= (rrho(r))^ntag3
$$
Now note that $|partial Phi(x_0)|lemax_{1le kle n}|lambda_k|$, where $lambda_1,ldots,lambda_n$ are the eigenvalues of $partialPhi(x)_{Bbb C}$ (where $partialPhi(x)_{Bbb C}$ is the complexification of $partialPhi(x)$).
However Im stuck here. I cant see that $lim_{rto 0^+}rho(r)=0$, what I see is that $lim_{rto 0^+}rho(r)=|partialPhi(x_0)|$, what is not necessarily zero.
Some help will be appreciated, thank you.
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Dec 27 '18 at 3:32
MasacrosoMasacroso
13.1k41747
asked Dec 27 '18 at 3:32
MasacrosoMasacroso
13.1k41747
edited Dec 27 '18 at 3:45
Masacroso
asked Dec 27 '18 at 3:32
MasacrosoMasacroso
13.1k41747
asked Dec 27 '18 at 3:32
MasacrosoMasacroso
13.1k41747
asked Dec 27 '18 at 3:32
MasacrosoMasacroso
13.1k41747
13.1k41747
$begingroup$
You need to use that $|partial Phi(x_0) - partial Phi (x_0 + t (x-x_0))|$ gets small as $tto 0$, and then you need to estimate the measure of $partial Phi (x_0) (J_0)$.
$endgroup$
– PhoemueX
Dec 27 '18 at 16:48
$begingroup$
@PhoemueX I dont follow what you says... Of course $|partialPhi(x_0)-partialPhi(x_0+t(x-x_0)|to 0$ as $tto 0$ because $partialPhi$ is continuous. But this can be said of any continuous derivative. By the other hand is clear that $lambda_n(partialPhi(x_0)Bbb R^n)=0$ because $partialPhi(x_0)$ is not injective. However I dont see the relation of this things with my question
$endgroup$
– Masacroso
Dec 28 '18 at 6:41
$begingroup$
Sorry if this is a silly question, but why do you say "$partialPhi(x_0)$ is not injective"? Is not that denoting a matrix with values, i.e. the Jacobian matrix evaluated at $x = x_0$ and hence it is non-sense to talk about injective, surjective, etc?
$endgroup$
– manooooh
Dec 29 '18 at 9:06
1
$begingroup$
@manooooh ok, I see what you are asking. The Fréchet derivative at a point of some differentiable function is a linear function, that can be represented by a matrix acting in $Bbb R^n$ (the Jacobian matrix). In this case we knows that $partialPhi(x_0)notinmathcal L{rm aut}(Bbb R^n)$, that is, it is a linear function in $Bbb R^n$ that is not an automorphism. This imply that $partialPhi(x_0)$ is not surjective, and because it is linear, then it is neither injective. This mean that the Jacobian matrix (that is a square matrix $ntimes n$) that represent this function is not invertible.
$endgroup$
– Masacroso
Dec 29 '18 at 10:56
add a comment |
$begingroup$
You need to use that $|partial Phi(x_0) - partial Phi (x_0 + t (x-x_0))|$ gets small as $tto 0$, and then you need to estimate the measure of $partial Phi (x_0) (J_0)$.
$endgroup$
– PhoemueX
Dec 27 '18 at 16:48
$begingroup$
@PhoemueX I dont follow what you says... Of course $|partialPhi(x_0)-partialPhi(x_0+t(x-x_0)|to 0$ as $tto 0$ because $partialPhi$ is continuous. But this can be said of any continuous derivative. By the other hand is clear that $lambda_n(partialPhi(x_0)Bbb R^n)=0$ because $partialPhi(x_0)$ is not injective. However I dont see the relation of this things with my question
$endgroup$
– Masacroso
Dec 28 '18 at 6:41
$begingroup$
Sorry if this is a silly question, but why do you say "$partialPhi(x_0)$ is not injective"? Is not that denoting a matrix with values, i.e. the Jacobian matrix evaluated at $x = x_0$ and hence it is non-sense to talk about injective, surjective, etc?
$endgroup$
– manooooh
Dec 29 '18 at 9:06
1
$begingroup$
@manooooh ok, I see what you are asking. The Fréchet derivative at a point of some differentiable function is a linear function, that can be represented by a matrix acting in $Bbb R^n$ (the Jacobian matrix). In this case we knows that $partialPhi(x_0)notinmathcal L{rm aut}(Bbb R^n)$, that is, it is a linear function in $Bbb R^n$ that is not an automorphism. This imply that $partialPhi(x_0)$ is not surjective, and because it is linear, then it is neither injective. This mean that the Jacobian matrix (that is a square matrix $ntimes n$) that represent this function is not invertible.
$endgroup$
– Masacroso
Dec 29 '18 at 10:56
$begingroup$
You need to use that $|partial Phi(x_0) - partial Phi (x_0 + t (x-x_0))|$ gets small as $tto 0$, and then you need to estimate the measure of $partial Phi (x_0) (J_0)$.
$endgroup$
– PhoemueX
Dec 27 '18 at 16:48
$begingroup$
You need to use that $|partial Phi(x_0) - partial Phi (x_0 + t (x-x_0))|$ gets small as $tto 0$, and then you need to estimate the measure of $partial Phi (x_0) (J_0)$.
$endgroup$
– PhoemueX
Dec 27 '18 at 16:48
$begingroup$
@PhoemueX I dont follow what you says... Of course $|partialPhi(x_0)-partialPhi(x_0+t(x-x_0)|to 0$ as $tto 0$ because $partialPhi$ is continuous. But this can be said of any continuous derivative. By the other hand is clear that $lambda_n(partialPhi(x_0)Bbb R^n)=0$ because $partialPhi(x_0)$ is not injective. However I dont see the relation of this things with my question
$endgroup$
– Masacroso
Dec 28 '18 at 6:41
$begingroup$
@PhoemueX I dont follow what you says... Of course $|partialPhi(x_0)-partialPhi(x_0+t(x-x_0)|to 0$ as $tto 0$ because $partialPhi$ is continuous. But this can be said of any continuous derivative. By the other hand is clear that $lambda_n(partialPhi(x_0)Bbb R^n)=0$ because $partialPhi(x_0)$ is not injective. However I dont see the relation of this things with my question
$endgroup$
– Masacroso
Dec 28 '18 at 6:41
$begingroup$
Sorry if this is a silly question, but why do you say "$partialPhi(x_0)$ is not injective"? Is not that denoting a matrix with values, i.e. the Jacobian matrix evaluated at $x = x_0$ and hence it is non-sense to talk about injective, surjective, etc?
$endgroup$
– manooooh
Dec 29 '18 at 9:06
$begingroup$
Sorry if this is a silly question, but why do you say "$partialPhi(x_0)$ is not injective"? Is not that denoting a matrix with values, i.e. the Jacobian matrix evaluated at $x = x_0$ and hence it is non-sense to talk about injective, surjective, etc?
$endgroup$
– manooooh
Dec 29 '18 at 9:06
1
1
$begingroup$
@manooooh ok, I see what you are asking. The Fréchet derivative at a point of some differentiable function is a linear function, that can be represented by a matrix acting in $Bbb R^n$ (the Jacobian matrix). In this case we knows that $partialPhi(x_0)notinmathcal L{rm aut}(Bbb R^n)$, that is, it is a linear function in $Bbb R^n$ that is not an automorphism. This imply that $partialPhi(x_0)$ is not surjective, and because it is linear, then it is neither injective. This mean that the Jacobian matrix (that is a square matrix $ntimes n$) that represent this function is not invertible.
$endgroup$
– Masacroso
Dec 29 '18 at 10:56
$begingroup$
@manooooh ok, I see what you are asking. The Fréchet derivative at a point of some differentiable function is a linear function, that can be represented by a matrix acting in $Bbb R^n$ (the Jacobian matrix). In this case we knows that $partialPhi(x_0)notinmathcal L{rm aut}(Bbb R^n)$, that is, it is a linear function in $Bbb R^n$ that is not an automorphism. This imply that $partialPhi(x_0)$ is not surjective, and because it is linear, then it is neither injective. This mean that the Jacobian matrix (that is a square matrix $ntimes n$) that represent this function is not invertible.
$endgroup$
– Masacroso
Dec 29 '18 at 10:56
add a comment |
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$begingroup$
You need to use that $|partial Phi(x_0) - partial Phi (x_0 + t (x-x_0))|$ gets small as $tto 0$, and then you need to estimate the measure of $partial Phi (x_0) (J_0)$.
$endgroup$
– PhoemueX
Dec 27 '18 at 16:48
$begingroup$
@PhoemueX I dont follow what you says... Of course $|partialPhi(x_0)-partialPhi(x_0+t(x-x_0)|to 0$ as $tto 0$ because $partialPhi$ is continuous. But this can be said of any continuous derivative. By the other hand is clear that $lambda_n(partialPhi(x_0)Bbb R^n)=0$ because $partialPhi(x_0)$ is not injective. However I dont see the relation of this things with my question
$endgroup$
– Masacroso
Dec 28 '18 at 6:41
$begingroup$
Sorry if this is a silly question, but why do you say "$partialPhi(x_0)$ is not injective"? Is not that denoting a matrix with values, i.e. the Jacobian matrix evaluated at $x = x_0$ and hence it is non-sense to talk about injective, surjective, etc?
$endgroup$
– manooooh
Dec 29 '18 at 9:06
1
$begingroup$
@manooooh ok, I see what you are asking. The Fréchet derivative at a point of some differentiable function is a linear function, that can be represented by a matrix acting in $Bbb R^n$ (the Jacobian matrix). In this case we knows that $partialPhi(x_0)notinmathcal L{rm aut}(Bbb R^n)$, that is, it is a linear function in $Bbb R^n$ that is not an automorphism. This imply that $partialPhi(x_0)$ is not surjective, and because it is linear, then it is neither injective. This mean that the Jacobian matrix (that is a square matrix $ntimes n$) that represent this function is not invertible.
$endgroup$
– Masacroso
Dec 29 '18 at 10:56