Mixing
Help understanding a little Jacobians lemma
$begingroup$
This was used as lemma in a bigger proof.
Let $A ⊆ ℝ^m$ be an open set, $f = (f_1,…,f_m) : A → ℝ^m$ a $C^1$ function, and $p ∈ A$. If $det[Df(p)] ≠ 0$, then there is an open ball $B ⊆ A$ of center $p$ such that $f|_B$ is injective.
proof:
The function $x ↦ det[Df(x)]$ is continuous, so there is an open ball $B ⊆ A$, of center $p$, such that $x ∈ B ⇒ det[Df(x)] ≠ 0$. Let $a$ and $b$ be two points of $B$. By the mean value theorem we have $f_i(b) - f_i(a) = Df_i(c_i)⋅(b-a)$ for some $c_i ∈ {ta + (1-t)b : t in [0,1]}$, this means that $$f(b) - f(a) = begin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix}⋅(b-a).$$But $det[Df(c_i)] ≠ 0$ so $f(b) = f(a) ⇒ b = a$.
The last step is where I have troubles, it seems to me that we need $$detbegin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix} ≠ 0$$
and it's not clear how this follows from $det[Df(c_i)] ≠ 0$.
calculus linear-algebra multivariable-calculus
asked Feb 2 at 18:22
sawesawe
365
$endgroup$
add a comment |
$begingroup$
This was used as lemma in a bigger proof.
Let $A ⊆ ℝ^m$ be an open set, $f = (f_1,…,f_m) : A → ℝ^m$ a $C^1$ function, and $p ∈ A$. If $det[Df(p)] ≠ 0$, then there is an open ball $B ⊆ A$ of center $p$ such that $f|_B$ is injective.
proof:
The function $x ↦ det[Df(x)]$ is continuous, so there is an open ball $B ⊆ A$, of center $p$, such that $x ∈ B ⇒ det[Df(x)] ≠ 0$. Let $a$ and $b$ be two points of $B$. By the mean value theorem we have $f_i(b) - f_i(a) = Df_i(c_i)⋅(b-a)$ for some $c_i ∈ {ta + (1-t)b : t in [0,1]}$, this means that $$f(b) - f(a) = begin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix}⋅(b-a).$$But $det[Df(c_i)] ≠ 0$ so $f(b) = f(a) ⇒ b = a$.
The last step is where I have troubles, it seems to me that we need $$detbegin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix} ≠ 0$$
and it's not clear how this follows from $det[Df(c_i)] ≠ 0$.
calculus linear-algebra multivariable-calculus
asked Feb 2 at 18:22
sawesawe
365
$endgroup$
$begingroup$
Small correction in the proof: the last line should be $f(a) = f(b)$ implies $a = b$, not the other way around. Also, shouldn't the beginning of the proof read that $x mapsto det(Df(x))$ is continuous (not $x mapsto det(Df(p))$)?
$endgroup$
– Jacob Maibach
Feb 2 at 18:31
$begingroup$
Sorry, another small correction: shouldn't $Df_{1}(c_{i})$ read $Df_{i}(c_{i})$ in the equation?
$endgroup$
– Jacob Maibach
Feb 2 at 18:40
$begingroup$
@Jacob Maibach you are absolutely right, thanks for the corrections.
$endgroup$
– sawe
Feb 2 at 18:48
add a comment |
$begingroup$
This was used as lemma in a bigger proof.
Let $A ⊆ ℝ^m$ be an open set, $f = (f_1,…,f_m) : A → ℝ^m$ a $C^1$ function, and $p ∈ A$. If $det[Df(p)] ≠ 0$, then there is an open ball $B ⊆ A$ of center $p$ such that $f|_B$ is injective.
proof:
The function $x ↦ det[Df(x)]$ is continuous, so there is an open ball $B ⊆ A$, of center $p$, such that $x ∈ B ⇒ det[Df(x)] ≠ 0$. Let $a$ and $b$ be two points of $B$. By the mean value theorem we have $f_i(b) - f_i(a) = Df_i(c_i)⋅(b-a)$ for some $c_i ∈ {ta + (1-t)b : t in [0,1]}$, this means that $$f(b) - f(a) = begin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix}⋅(b-a).$$But $det[Df(c_i)] ≠ 0$ so $f(b) = f(a) ⇒ b = a$.
The last step is where I have troubles, it seems to me that we need $$detbegin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix} ≠ 0$$
and it's not clear how this follows from $det[Df(c_i)] ≠ 0$.
calculus linear-algebra multivariable-calculus
asked Feb 2 at 18:22
sawesawe
365
$endgroup$
This was used as lemma in a bigger proof.
Let $A ⊆ ℝ^m$ be an open set, $f = (f_1,…,f_m) : A → ℝ^m$ a $C^1$ function, and $p ∈ A$. If $det[Df(p)] ≠ 0$, then there is an open ball $B ⊆ A$ of center $p$ such that $f|_B$ is injective.
proof:
The function $x ↦ det[Df(x)]$ is continuous, so there is an open ball $B ⊆ A$, of center $p$, such that $x ∈ B ⇒ det[Df(x)] ≠ 0$. Let $a$ and $b$ be two points of $B$. By the mean value theorem we have $f_i(b) - f_i(a) = Df_i(c_i)⋅(b-a)$ for some $c_i ∈ {ta + (1-t)b : t in [0,1]}$, this means that $$f(b) - f(a) = begin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix}⋅(b-a).$$But $det[Df(c_i)] ≠ 0$ so $f(b) = f(a) ⇒ b = a$.
The last step is where I have troubles, it seems to me that we need $$detbegin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix} ≠ 0$$
and it's not clear how this follows from $det[Df(c_i)] ≠ 0$.
calculus linear-algebra multivariable-calculus
calculus linear-algebra multivariable-calculus
asked Feb 2 at 18:22
sawesawe
365
asked Feb 2 at 18:22
sawesawe
365
edited Feb 2 at 19:39
sawe
asked Feb 2 at 18:22
sawesawe
365
asked Feb 2 at 18:22
sawesawe
365
asked Feb 2 at 18:22
sawesawe
365
365
$begingroup$
Small correction in the proof: the last line should be $f(a) = f(b)$ implies $a = b$, not the other way around. Also, shouldn't the beginning of the proof read that $x mapsto det(Df(x))$ is continuous (not $x mapsto det(Df(p))$)?
$endgroup$
– Jacob Maibach
Feb 2 at 18:31
$begingroup$
Sorry, another small correction: shouldn't $Df_{1}(c_{i})$ read $Df_{i}(c_{i})$ in the equation?
$endgroup$
– Jacob Maibach
Feb 2 at 18:40
$begingroup$
@Jacob Maibach you are absolutely right, thanks for the corrections.
$endgroup$
– sawe
Feb 2 at 18:48
add a comment |
$begingroup$
Small correction in the proof: the last line should be $f(a) = f(b)$ implies $a = b$, not the other way around. Also, shouldn't the beginning of the proof read that $x mapsto det(Df(x))$ is continuous (not $x mapsto det(Df(p))$)?
$endgroup$
– Jacob Maibach
Feb 2 at 18:31
$begingroup$
Sorry, another small correction: shouldn't $Df_{1}(c_{i})$ read $Df_{i}(c_{i})$ in the equation?
$endgroup$
– Jacob Maibach
Feb 2 at 18:40
$begingroup$
@Jacob Maibach you are absolutely right, thanks for the corrections.
$endgroup$
– sawe
Feb 2 at 18:48
$begingroup$
Small correction in the proof: the last line should be $f(a) = f(b)$ implies $a = b$, not the other way around. Also, shouldn't the beginning of the proof read that $x mapsto det(Df(x))$ is continuous (not $x mapsto det(Df(p))$)?
$endgroup$
– Jacob Maibach
Feb 2 at 18:31
$begingroup$
Small correction in the proof: the last line should be $f(a) = f(b)$ implies $a = b$, not the other way around. Also, shouldn't the beginning of the proof read that $x mapsto det(Df(x))$ is continuous (not $x mapsto det(Df(p))$)?
$endgroup$
– Jacob Maibach
Feb 2 at 18:31
$begingroup$
Sorry, another small correction: shouldn't $Df_{1}(c_{i})$ read $Df_{i}(c_{i})$ in the equation?
$endgroup$
– Jacob Maibach
Feb 2 at 18:40
$begingroup$
Sorry, another small correction: shouldn't $Df_{1}(c_{i})$ read $Df_{i}(c_{i})$ in the equation?
$endgroup$
– Jacob Maibach
Feb 2 at 18:40
$begingroup$
@Jacob Maibach you are absolutely right, thanks for the corrections.
$endgroup$
– sawe
Feb 2 at 18:48
$begingroup$
@Jacob Maibach you are absolutely right, thanks for the corrections.
$endgroup$
– sawe
Feb 2 at 18:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are right, the last step needs to be justified.
I'm denoting with $|⋅|$ the max norm, so $|[a_{ij}]| = max_{i,j}|a_{ij}|$, and I'll call $T$ the matrix $begin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix}$.
Being $det[Df(p)] ≠ 0$, we can write:
$$|b-a| = |Df(p)^{-1}Df(p)(b-a)| ≤ m|Df(p)^{-1}||Df(p)(b-a)| = σ^{-1}|Df(p)(b-a)| ⇒ \|Df(p)(b-a)| ≥ σ|b-a|.$$
The function $x ↦ Df(x)$ is continuous, so we can choose the radius of $U$ is small enough that: $$x ∈ U ⇒ |Df(p)-Df(x)| < σ/m ⇒ |Df_i(p)-Df_i(c_i)| < σ/m ⇒ |Df(p)-T| < σ/m.$$
Now, if $b ≠ a$, we have:
$$σ|b-a| - |T(b-a)| ≤ |Df(p)(b-a)| - |T(b-a)| ≤ |[Df(p) - T](b-a)| ≤ \m|Df(p) - T||b-a| < σ|b-a| ⇒ |f(b) - f(a)| = |T(b-a)| > 0.$$
answered Feb 3 at 9:17
Markus SteinerMarkus Steiner
1036
$endgroup$
add a comment |
$begingroup$
The proof does not seem to make sense because of an issue it has. First, note
that for the last step, the equation should read
$$
f(b) - f(a) =
begin{bmatrix}
Df_{1}(c) \
Df_{2}(c) \
dots \
Df_{m}(c)
end{bmatrix}
(b - a)
$$
where $c$ is the vector with components $c_{i}$. That is, each $f_{i}$ is a function from $R^{m}$ to $R^{1}$, so its linearization should have $R^{m}$ as the domain. Therefore, the previous claim
$$ f_{i}(b) - f_{i}(a) = Df_{i}(c_{i})(b - a) $$
needs to be corrected somehow. That should be sufficient to fix the proof.
answered Feb 2 at 18:53
Jacob MaibachJacob Maibach
1,4902917
$endgroup$
$begingroup$
Sorry, but $c_i$ is meant to be a vector on the segment $[a,b]$, not the component of a vector. The function $f$ is multivariate, when we apply the mean value theorem for each component, we get different $c_i$.
$endgroup$
– sawe
Feb 2 at 19:00
$begingroup$
@sawe but why does the proof state that $c_{i} in [a, b]$?
$endgroup$
– Jacob Maibach
Feb 2 at 19:29
$begingroup$
@sawe it seems that line needs to be clarified then, since it does not make sense as written. There is no general multivariate mean value theorem, so it must be a more complicated application of the single-variable MVT.
$endgroup$
– Jacob Maibach
Feb 2 at 19:33
1
$begingroup$
It is the notation that was used for segments, [a,b] is not an interval. I'll clarify in the op, because this can lead to misunderstandings.
$endgroup$
– sawe
Feb 2 at 19:37
add a comment |
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2 Answers
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$begingroup$
You are right, the last step needs to be justified.
I'm denoting with $|⋅|$ the max norm, so $|[a_{ij}]| = max_{i,j}|a_{ij}|$, and I'll call $T$ the matrix $begin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix}$.
Being $det[Df(p)] ≠ 0$, we can write:
$$|b-a| = |Df(p)^{-1}Df(p)(b-a)| ≤ m|Df(p)^{-1}||Df(p)(b-a)| = σ^{-1}|Df(p)(b-a)| ⇒ \|Df(p)(b-a)| ≥ σ|b-a|.$$
The function $x ↦ Df(x)$ is continuous, so we can choose the radius of $U$ is small enough that: $$x ∈ U ⇒ |Df(p)-Df(x)| < σ/m ⇒ |Df_i(p)-Df_i(c_i)| < σ/m ⇒ |Df(p)-T| < σ/m.$$
Now, if $b ≠ a$, we have:
$$σ|b-a| - |T(b-a)| ≤ |Df(p)(b-a)| - |T(b-a)| ≤ |[Df(p) - T](b-a)| ≤ \m|Df(p) - T||b-a| < σ|b-a| ⇒ |f(b) - f(a)| = |T(b-a)| > 0.$$
answered Feb 3 at 9:17
Markus SteinerMarkus Steiner
1036
$endgroup$
add a comment |
$begingroup$
You are right, the last step needs to be justified.
I'm denoting with $|⋅|$ the max norm, so $|[a_{ij}]| = max_{i,j}|a_{ij}|$, and I'll call $T$ the matrix $begin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix}$.
Being $det[Df(p)] ≠ 0$, we can write:
$$|b-a| = |Df(p)^{-1}Df(p)(b-a)| ≤ m|Df(p)^{-1}||Df(p)(b-a)| = σ^{-1}|Df(p)(b-a)| ⇒ \|Df(p)(b-a)| ≥ σ|b-a|.$$
The function $x ↦ Df(x)$ is continuous, so we can choose the radius of $U$ is small enough that: $$x ∈ U ⇒ |Df(p)-Df(x)| < σ/m ⇒ |Df_i(p)-Df_i(c_i)| < σ/m ⇒ |Df(p)-T| < σ/m.$$
Now, if $b ≠ a$, we have:
$$σ|b-a| - |T(b-a)| ≤ |Df(p)(b-a)| - |T(b-a)| ≤ |[Df(p) - T](b-a)| ≤ \m|Df(p) - T||b-a| < σ|b-a| ⇒ |f(b) - f(a)| = |T(b-a)| > 0.$$
answered Feb 3 at 9:17
Markus SteinerMarkus Steiner
1036
$endgroup$
add a comment |
$begingroup$
You are right, the last step needs to be justified.
I'm denoting with $|⋅|$ the max norm, so $|[a_{ij}]| = max_{i,j}|a_{ij}|$, and I'll call $T$ the matrix $begin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix}$.
Being $det[Df(p)] ≠ 0$, we can write:
$$|b-a| = |Df(p)^{-1}Df(p)(b-a)| ≤ m|Df(p)^{-1}||Df(p)(b-a)| = σ^{-1}|Df(p)(b-a)| ⇒ \|Df(p)(b-a)| ≥ σ|b-a|.$$
The function $x ↦ Df(x)$ is continuous, so we can choose the radius of $U$ is small enough that: $$x ∈ U ⇒ |Df(p)-Df(x)| < σ/m ⇒ |Df_i(p)-Df_i(c_i)| < σ/m ⇒ |Df(p)-T| < σ/m.$$
Now, if $b ≠ a$, we have:
$$σ|b-a| - |T(b-a)| ≤ |Df(p)(b-a)| - |T(b-a)| ≤ |[Df(p) - T](b-a)| ≤ \m|Df(p) - T||b-a| < σ|b-a| ⇒ |f(b) - f(a)| = |T(b-a)| > 0.$$
answered Feb 3 at 9:17
Markus SteinerMarkus Steiner
1036
$endgroup$
You are right, the last step needs to be justified.
I'm denoting with $|⋅|$ the max norm, so $|[a_{ij}]| = max_{i,j}|a_{ij}|$, and I'll call $T$ the matrix $begin{bmatrix}Df_1(c_1)\vdots\Df_m(c_m)end{bmatrix}$.
Being $det[Df(p)] ≠ 0$, we can write:
$$|b-a| = |Df(p)^{-1}Df(p)(b-a)| ≤ m|Df(p)^{-1}||Df(p)(b-a)| = σ^{-1}|Df(p)(b-a)| ⇒ \|Df(p)(b-a)| ≥ σ|b-a|.$$
The function $x ↦ Df(x)$ is continuous, so we can choose the radius of $U$ is small enough that: $$x ∈ U ⇒ |Df(p)-Df(x)| < σ/m ⇒ |Df_i(p)-Df_i(c_i)| < σ/m ⇒ |Df(p)-T| < σ/m.$$
Now, if $b ≠ a$, we have:
$$σ|b-a| - |T(b-a)| ≤ |Df(p)(b-a)| - |T(b-a)| ≤ |[Df(p) - T](b-a)| ≤ \m|Df(p) - T||b-a| < σ|b-a| ⇒ |f(b) - f(a)| = |T(b-a)| > 0.$$
answered Feb 3 at 9:17
Markus SteinerMarkus Steiner
1036
answered Feb 3 at 9:17
Markus SteinerMarkus Steiner
1036
answered Feb 3 at 9:17
Markus SteinerMarkus Steiner
1036
answered Feb 3 at 9:17
Markus SteinerMarkus Steiner
1036
1036
add a comment |
add a comment |
$begingroup$
The proof does not seem to make sense because of an issue it has. First, note
that for the last step, the equation should read
$$
f(b) - f(a) =
begin{bmatrix}
Df_{1}(c) \
Df_{2}(c) \
dots \
Df_{m}(c)
end{bmatrix}
(b - a)
$$
where $c$ is the vector with components $c_{i}$. That is, each $f_{i}$ is a function from $R^{m}$ to $R^{1}$, so its linearization should have $R^{m}$ as the domain. Therefore, the previous claim
$$ f_{i}(b) - f_{i}(a) = Df_{i}(c_{i})(b - a) $$
needs to be corrected somehow. That should be sufficient to fix the proof.
answered Feb 2 at 18:53
Jacob MaibachJacob Maibach
1,4902917
$endgroup$
$begingroup$
Sorry, but $c_i$ is meant to be a vector on the segment $[a,b]$, not the component of a vector. The function $f$ is multivariate, when we apply the mean value theorem for each component, we get different $c_i$.
$endgroup$
– sawe
Feb 2 at 19:00
$begingroup$
@sawe but why does the proof state that $c_{i} in [a, b]$?
$endgroup$
– Jacob Maibach
Feb 2 at 19:29
$begingroup$
@sawe it seems that line needs to be clarified then, since it does not make sense as written. There is no general multivariate mean value theorem, so it must be a more complicated application of the single-variable MVT.
$endgroup$
– Jacob Maibach
Feb 2 at 19:33
1
$begingroup$
It is the notation that was used for segments, [a,b] is not an interval. I'll clarify in the op, because this can lead to misunderstandings.
$endgroup$
– sawe
Feb 2 at 19:37
add a comment |
$begingroup$
The proof does not seem to make sense because of an issue it has. First, note
that for the last step, the equation should read
$$
f(b) - f(a) =
begin{bmatrix}
Df_{1}(c) \
Df_{2}(c) \
dots \
Df_{m}(c)
end{bmatrix}
(b - a)
$$
where $c$ is the vector with components $c_{i}$. That is, each $f_{i}$ is a function from $R^{m}$ to $R^{1}$, so its linearization should have $R^{m}$ as the domain. Therefore, the previous claim
$$ f_{i}(b) - f_{i}(a) = Df_{i}(c_{i})(b - a) $$
needs to be corrected somehow. That should be sufficient to fix the proof.
answered Feb 2 at 18:53
Jacob MaibachJacob Maibach
1,4902917
$endgroup$
$begingroup$
Sorry, but $c_i$ is meant to be a vector on the segment $[a,b]$, not the component of a vector. The function $f$ is multivariate, when we apply the mean value theorem for each component, we get different $c_i$.
$endgroup$
– sawe
Feb 2 at 19:00
$begingroup$
@sawe but why does the proof state that $c_{i} in [a, b]$?
$endgroup$
– Jacob Maibach
Feb 2 at 19:29
$begingroup$
@sawe it seems that line needs to be clarified then, since it does not make sense as written. There is no general multivariate mean value theorem, so it must be a more complicated application of the single-variable MVT.
$endgroup$
– Jacob Maibach
Feb 2 at 19:33
1
$begingroup$
It is the notation that was used for segments, [a,b] is not an interval. I'll clarify in the op, because this can lead to misunderstandings.
$endgroup$
– sawe
Feb 2 at 19:37
add a comment |
$begingroup$
The proof does not seem to make sense because of an issue it has. First, note
that for the last step, the equation should read
$$
f(b) - f(a) =
begin{bmatrix}
Df_{1}(c) \
Df_{2}(c) \
dots \
Df_{m}(c)
end{bmatrix}
(b - a)
$$
where $c$ is the vector with components $c_{i}$. That is, each $f_{i}$ is a function from $R^{m}$ to $R^{1}$, so its linearization should have $R^{m}$ as the domain. Therefore, the previous claim
$$ f_{i}(b) - f_{i}(a) = Df_{i}(c_{i})(b - a) $$
needs to be corrected somehow. That should be sufficient to fix the proof.
answered Feb 2 at 18:53
Jacob MaibachJacob Maibach
1,4902917
$endgroup$
The proof does not seem to make sense because of an issue it has. First, note
that for the last step, the equation should read
$$
f(b) - f(a) =
begin{bmatrix}
Df_{1}(c) \
Df_{2}(c) \
dots \
Df_{m}(c)
end{bmatrix}
(b - a)
$$
where $c$ is the vector with components $c_{i}$. That is, each $f_{i}$ is a function from $R^{m}$ to $R^{1}$, so its linearization should have $R^{m}$ as the domain. Therefore, the previous claim
$$ f_{i}(b) - f_{i}(a) = Df_{i}(c_{i})(b - a) $$
needs to be corrected somehow. That should be sufficient to fix the proof.
answered Feb 2 at 18:53
Jacob MaibachJacob Maibach
1,4902917
answered Feb 2 at 18:53
Jacob MaibachJacob Maibach
1,4902917
answered Feb 2 at 18:53
Jacob MaibachJacob Maibach
1,4902917
answered Feb 2 at 18:53
Jacob MaibachJacob Maibach
1,4902917
1,4902917
$begingroup$
Sorry, but $c_i$ is meant to be a vector on the segment $[a,b]$, not the component of a vector. The function $f$ is multivariate, when we apply the mean value theorem for each component, we get different $c_i$.
$endgroup$
– sawe
Feb 2 at 19:00
$begingroup$
@sawe but why does the proof state that $c_{i} in [a, b]$?
$endgroup$
– Jacob Maibach
Feb 2 at 19:29
$begingroup$
@sawe it seems that line needs to be clarified then, since it does not make sense as written. There is no general multivariate mean value theorem, so it must be a more complicated application of the single-variable MVT.
$endgroup$
– Jacob Maibach
Feb 2 at 19:33
1
$begingroup$
It is the notation that was used for segments, [a,b] is not an interval. I'll clarify in the op, because this can lead to misunderstandings.
$endgroup$
– sawe
Feb 2 at 19:37
add a comment |
$begingroup$
Sorry, but $c_i$ is meant to be a vector on the segment $[a,b]$, not the component of a vector. The function $f$ is multivariate, when we apply the mean value theorem for each component, we get different $c_i$.
$endgroup$
– sawe
Feb 2 at 19:00
$begingroup$
@sawe but why does the proof state that $c_{i} in [a, b]$?
$endgroup$
– Jacob Maibach
Feb 2 at 19:29
$begingroup$
@sawe it seems that line needs to be clarified then, since it does not make sense as written. There is no general multivariate mean value theorem, so it must be a more complicated application of the single-variable MVT.
$endgroup$
– Jacob Maibach
Feb 2 at 19:33
1
$begingroup$
It is the notation that was used for segments, [a,b] is not an interval. I'll clarify in the op, because this can lead to misunderstandings.
$endgroup$
– sawe
Feb 2 at 19:37
$begingroup$
Sorry, but $c_i$ is meant to be a vector on the segment $[a,b]$, not the component of a vector. The function $f$ is multivariate, when we apply the mean value theorem for each component, we get different $c_i$.
$endgroup$
– sawe
Feb 2 at 19:00
$begingroup$
Sorry, but $c_i$ is meant to be a vector on the segment $[a,b]$, not the component of a vector. The function $f$ is multivariate, when we apply the mean value theorem for each component, we get different $c_i$.
$endgroup$
– sawe
Feb 2 at 19:00
$begingroup$
@sawe but why does the proof state that $c_{i} in [a, b]$?
$endgroup$
– Jacob Maibach
Feb 2 at 19:29
$begingroup$
@sawe but why does the proof state that $c_{i} in [a, b]$?
$endgroup$
– Jacob Maibach
Feb 2 at 19:29
$begingroup$
@sawe it seems that line needs to be clarified then, since it does not make sense as written. There is no general multivariate mean value theorem, so it must be a more complicated application of the single-variable MVT.
$endgroup$
– Jacob Maibach
Feb 2 at 19:33
$begingroup$
@sawe it seems that line needs to be clarified then, since it does not make sense as written. There is no general multivariate mean value theorem, so it must be a more complicated application of the single-variable MVT.
$endgroup$
– Jacob Maibach
Feb 2 at 19:33
1
1
$begingroup$
It is the notation that was used for segments, [a,b] is not an interval. I'll clarify in the op, because this can lead to misunderstandings.
$endgroup$
– sawe
Feb 2 at 19:37
$begingroup$
It is the notation that was used for segments, [a,b] is not an interval. I'll clarify in the op, because this can lead to misunderstandings.
$endgroup$
– sawe
Feb 2 at 19:37
add a comment |
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$begingroup$
Small correction in the proof: the last line should be $f(a) = f(b)$ implies $a = b$, not the other way around. Also, shouldn't the beginning of the proof read that $x mapsto det(Df(x))$ is continuous (not $x mapsto det(Df(p))$)?
$endgroup$
– Jacob Maibach
Feb 2 at 18:31
$begingroup$
Sorry, another small correction: shouldn't $Df_{1}(c_{i})$ read $Df_{i}(c_{i})$ in the equation?
$endgroup$
– Jacob Maibach
Feb 2 at 18:40
$begingroup$
@Jacob Maibach you are absolutely right, thanks for the corrections.
$endgroup$
– sawe
Feb 2 at 18:48