Mixing
Multivariable function equivalent definition
Say a function $f: B(x;r) rightarrow mathbb{R}^q$ is a continuously differentiable function with $|Jf(x)| leq c$ for all $x in B(x;r)$. I want to show that $|f(x_1) - f(x_2)| leq c|x_1-x_2|$ whenever $|x_1-x_2| < r$.
I have difficulty using the integration of the jacobian $Jf(x)$ to prove it, since it is not one-dimensional.
multivariable-calculus vector-analysis jacobian
asked Nov 20 '18 at 17:10
Neyo Yang
614
add a comment |
Say a function $f: B(x;r) rightarrow mathbb{R}^q$ is a continuously differentiable function with $|Jf(x)| leq c$ for all $x in B(x;r)$. I want to show that $|f(x_1) - f(x_2)| leq c|x_1-x_2|$ whenever $|x_1-x_2| < r$.
I have difficulty using the integration of the jacobian $Jf(x)$ to prove it, since it is not one-dimensional.
multivariable-calculus vector-analysis jacobian
asked Nov 20 '18 at 17:10
Neyo Yang
614
add a comment |
Say a function $f: B(x;r) rightarrow mathbb{R}^q$ is a continuously differentiable function with $|Jf(x)| leq c$ for all $x in B(x;r)$. I want to show that $|f(x_1) - f(x_2)| leq c|x_1-x_2|$ whenever $|x_1-x_2| < r$.
I have difficulty using the integration of the jacobian $Jf(x)$ to prove it, since it is not one-dimensional.
multivariable-calculus vector-analysis jacobian
asked Nov 20 '18 at 17:10
Neyo Yang
614
Say a function $f: B(x;r) rightarrow mathbb{R}^q$ is a continuously differentiable function with $|Jf(x)| leq c$ for all $x in B(x;r)$. I want to show that $|f(x_1) - f(x_2)| leq c|x_1-x_2|$ whenever $|x_1-x_2| < r$.
I have difficulty using the integration of the jacobian $Jf(x)$ to prove it, since it is not one-dimensional.
multivariable-calculus vector-analysis jacobian
multivariable-calculus vector-analysis jacobian
asked Nov 20 '18 at 17:10
Neyo Yang
614
asked Nov 20 '18 at 17:10
Neyo Yang
614
asked Nov 20 '18 at 17:10
Neyo Yang
614
asked Nov 20 '18 at 17:10
Neyo Yang
614
asked Nov 20 '18 at 17:10
Neyo Yang
614
614
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Since $f$ is $C^1$ you have
$f(x_1)-f(x_2) = int_0^1 Jf(x_2+t(x_1-x_2)) (x_1-x_2) dt$ and so
$|f(x_1)-f(x_2)| le (int_0^1 c dt) |x_1-x_2|$.
The result is true if $f$ is just differentiable, but the proof needs a little more finesse.
answered Nov 20 '18 at 17:28
copper.hat
126k559159
add a comment |
This is wrong. Let $a>1$, and define $f:>{mathbb R}^2to{mathbb R}^2$ by
$$f(x,y):=left(ax,{yover a}right) .$$
Then $J_fequiv1$, but $$bigl|f(x,0)-f(x',0)bigr|=a|x-x'|>1cdotbigl|(x,0)-x',0)bigr| .$$
answered Nov 20 '18 at 18:57
Christian Blatter
172k7112326
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
Since $f$ is $C^1$ you have
$f(x_1)-f(x_2) = int_0^1 Jf(x_2+t(x_1-x_2)) (x_1-x_2) dt$ and so
$|f(x_1)-f(x_2)| le (int_0^1 c dt) |x_1-x_2|$.
The result is true if $f$ is just differentiable, but the proof needs a little more finesse.
answered Nov 20 '18 at 17:28
copper.hat
126k559159
add a comment |
Since $f$ is $C^1$ you have
$f(x_1)-f(x_2) = int_0^1 Jf(x_2+t(x_1-x_2)) (x_1-x_2) dt$ and so
$|f(x_1)-f(x_2)| le (int_0^1 c dt) |x_1-x_2|$.
The result is true if $f$ is just differentiable, but the proof needs a little more finesse.
answered Nov 20 '18 at 17:28
copper.hat
126k559159
add a comment |
Since $f$ is $C^1$ you have
$f(x_1)-f(x_2) = int_0^1 Jf(x_2+t(x_1-x_2)) (x_1-x_2) dt$ and so
$|f(x_1)-f(x_2)| le (int_0^1 c dt) |x_1-x_2|$.
The result is true if $f$ is just differentiable, but the proof needs a little more finesse.
answered Nov 20 '18 at 17:28
copper.hat
126k559159
Since $f$ is $C^1$ you have
$f(x_1)-f(x_2) = int_0^1 Jf(x_2+t(x_1-x_2)) (x_1-x_2) dt$ and so
$|f(x_1)-f(x_2)| le (int_0^1 c dt) |x_1-x_2|$.
The result is true if $f$ is just differentiable, but the proof needs a little more finesse.
answered Nov 20 '18 at 17:28
copper.hat
126k559159
answered Nov 20 '18 at 17:28
copper.hat
126k559159
answered Nov 20 '18 at 17:28
copper.hat
126k559159
answered Nov 20 '18 at 17:28
copper.hat
126k559159
126k559159
add a comment |
add a comment |
This is wrong. Let $a>1$, and define $f:>{mathbb R}^2to{mathbb R}^2$ by
$$f(x,y):=left(ax,{yover a}right) .$$
Then $J_fequiv1$, but $$bigl|f(x,0)-f(x',0)bigr|=a|x-x'|>1cdotbigl|(x,0)-x',0)bigr| .$$
answered Nov 20 '18 at 18:57
Christian Blatter
172k7112326
add a comment |
This is wrong. Let $a>1$, and define $f:>{mathbb R}^2to{mathbb R}^2$ by
$$f(x,y):=left(ax,{yover a}right) .$$
Then $J_fequiv1$, but $$bigl|f(x,0)-f(x',0)bigr|=a|x-x'|>1cdotbigl|(x,0)-x',0)bigr| .$$
answered Nov 20 '18 at 18:57
Christian Blatter
172k7112326
add a comment |
This is wrong. Let $a>1$, and define $f:>{mathbb R}^2to{mathbb R}^2$ by
$$f(x,y):=left(ax,{yover a}right) .$$
Then $J_fequiv1$, but $$bigl|f(x,0)-f(x',0)bigr|=a|x-x'|>1cdotbigl|(x,0)-x',0)bigr| .$$
answered Nov 20 '18 at 18:57
Christian Blatter
172k7112326
This is wrong. Let $a>1$, and define $f:>{mathbb R}^2to{mathbb R}^2$ by
$$f(x,y):=left(ax,{yover a}right) .$$
Then $J_fequiv1$, but $$bigl|f(x,0)-f(x',0)bigr|=a|x-x'|>1cdotbigl|(x,0)-x',0)bigr| .$$
answered Nov 20 '18 at 18:57
Christian Blatter
172k7112326
answered Nov 20 '18 at 18:57
Christian Blatter
172k7112326
answered Nov 20 '18 at 18:57
Christian Blatter
172k7112326
answered Nov 20 '18 at 18:57
Christian Blatter
172k7112326
172k7112326
add a comment |
add a comment |
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