Multidimensionally Lipschitz continuous iff Lipschitz continuous in every coordinate
Definition. A function $f$ defined on a set $S subseteq mathbb R$ is said to be Lipschitz continuous on $S$ if there exists an $L geq 0$ such that $$|f(x_1) - f(x_2)| le L|x_1 - x_2|$$ for all $x_1$ and $x_2$ in $S$ such that $x_1 ne x_2$.
Consider some function $f: mathbb R^n to mathbb R^n$ taking several variables $x^{(k)}$ with $1 < k leq n$ as argument. Let $f_i: mathbb R^n to mathbb R$ be the function of $i$-th coordinate of the function value of $f$. Are the following statements true?
- If $f$ is Lipschitz continuous, then $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$.
- If $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$, then $f$ is Lipschitz continuous.
multivariable-calculus continuity lipschitz-functions
asked Nov 25 '18 at 10:03
meisterluk
14610
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Definition. A function $f$ defined on a set $S subseteq mathbb R$ is said to be Lipschitz continuous on $S$ if there exists an $L geq 0$ such that $$|f(x_1) - f(x_2)| le L|x_1 - x_2|$$ for all $x_1$ and $x_2$ in $S$ such that $x_1 ne x_2$.
Consider some function $f: mathbb R^n to mathbb R^n$ taking several variables $x^{(k)}$ with $1 < k leq n$ as argument. Let $f_i: mathbb R^n to mathbb R$ be the function of $i$-th coordinate of the function value of $f$. Are the following statements true?
- If $f$ is Lipschitz continuous, then $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$.
- If $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$, then $f$ is Lipschitz continuous.
multivariable-calculus continuity lipschitz-functions
asked Nov 25 '18 at 10:03
meisterluk
14610
add a comment |
Definition. A function $f$ defined on a set $S subseteq mathbb R$ is said to be Lipschitz continuous on $S$ if there exists an $L geq 0$ such that $$|f(x_1) - f(x_2)| le L|x_1 - x_2|$$ for all $x_1$ and $x_2$ in $S$ such that $x_1 ne x_2$.
Consider some function $f: mathbb R^n to mathbb R^n$ taking several variables $x^{(k)}$ with $1 < k leq n$ as argument. Let $f_i: mathbb R^n to mathbb R$ be the function of $i$-th coordinate of the function value of $f$. Are the following statements true?
- If $f$ is Lipschitz continuous, then $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$.
- If $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$, then $f$ is Lipschitz continuous.
multivariable-calculus continuity lipschitz-functions
asked Nov 25 '18 at 10:03
meisterluk
14610
Definition. A function $f$ defined on a set $S subseteq mathbb R$ is said to be Lipschitz continuous on $S$ if there exists an $L geq 0$ such that $$|f(x_1) - f(x_2)| le L|x_1 - x_2|$$ for all $x_1$ and $x_2$ in $S$ such that $x_1 ne x_2$.
Consider some function $f: mathbb R^n to mathbb R^n$ taking several variables $x^{(k)}$ with $1 < k leq n$ as argument. Let $f_i: mathbb R^n to mathbb R$ be the function of $i$-th coordinate of the function value of $f$. Are the following statements true?
- If $f$ is Lipschitz continuous, then $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$.
- If $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$, then $f$ is Lipschitz continuous.
multivariable-calculus continuity lipschitz-functions
multivariable-calculus continuity lipschitz-functions
asked Nov 25 '18 at 10:03
meisterluk
14610
asked Nov 25 '18 at 10:03
meisterluk
14610
asked Nov 25 '18 at 10:03
meisterluk
14610
asked Nov 25 '18 at 10:03
meisterluk
14610
asked Nov 25 '18 at 10:03
meisterluk
14610
14610
add a comment |
add a comment |
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$f$ is K-Lipschitz continuous implies $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$:
$left | f_i(x)-f_i(y) right |leq left | f(x)-f(y) right |leq Kleft | x-y right |$
Each coordinate $f_i$ is $K_i$-Lipschitz continuous implies $f$ is Lipschitz-continuous:
$f=sum_{i=1}^{n}f_imathbf{e_i}$ where $mathbf{e_i}=left (0,...,0,1,0,...,0 right )$ where the $1$ is at the $i$-th position.
Note that $left | f_1(x)mathbf{e_1}+f_2(x)mathbf{e_2}-f_1(y)mathbf{e_1}-f_2(y)mathbf{e_2} right |leq left | f_1(x)-f_1(y) right |+left |f_2(x)-f_2(y) right |leq L_1left | x-y right |+L_2left | x-y right |leq (L_1+L_2)left | x-y right |$
The result follows by induction.
answered Nov 25 '18 at 10:27
John11
1,0061821
So it is. Thank you very much for showing it formally :)
– meisterluk
Nov 26 '18 at 10:14
add a comment |
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1 Answer
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$f$ is K-Lipschitz continuous implies $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$:
$left | f_i(x)-f_i(y) right |leq left | f(x)-f(y) right |leq Kleft | x-y right |$
Each coordinate $f_i$ is $K_i$-Lipschitz continuous implies $f$ is Lipschitz-continuous:
$f=sum_{i=1}^{n}f_imathbf{e_i}$ where $mathbf{e_i}=left (0,...,0,1,0,...,0 right )$ where the $1$ is at the $i$-th position.
Note that $left | f_1(x)mathbf{e_1}+f_2(x)mathbf{e_2}-f_1(y)mathbf{e_1}-f_2(y)mathbf{e_2} right |leq left | f_1(x)-f_1(y) right |+left |f_2(x)-f_2(y) right |leq L_1left | x-y right |+L_2left | x-y right |leq (L_1+L_2)left | x-y right |$
The result follows by induction.
answered Nov 25 '18 at 10:27
John11
1,0061821
So it is. Thank you very much for showing it formally :)
– meisterluk
Nov 26 '18 at 10:14
add a comment |
$f$ is K-Lipschitz continuous implies $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$:
$left | f_i(x)-f_i(y) right |leq left | f(x)-f(y) right |leq Kleft | x-y right |$
Each coordinate $f_i$ is $K_i$-Lipschitz continuous implies $f$ is Lipschitz-continuous:
$f=sum_{i=1}^{n}f_imathbf{e_i}$ where $mathbf{e_i}=left (0,...,0,1,0,...,0 right )$ where the $1$ is at the $i$-th position.
Note that $left | f_1(x)mathbf{e_1}+f_2(x)mathbf{e_2}-f_1(y)mathbf{e_1}-f_2(y)mathbf{e_2} right |leq left | f_1(x)-f_1(y) right |+left |f_2(x)-f_2(y) right |leq L_1left | x-y right |+L_2left | x-y right |leq (L_1+L_2)left | x-y right |$
The result follows by induction.
answered Nov 25 '18 at 10:27
John11
1,0061821
So it is. Thank you very much for showing it formally :)
– meisterluk
Nov 26 '18 at 10:14
add a comment |
$f$ is K-Lipschitz continuous implies $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$:
$left | f_i(x)-f_i(y) right |leq left | f(x)-f(y) right |leq Kleft | x-y right |$
Each coordinate $f_i$ is $K_i$-Lipschitz continuous implies $f$ is Lipschitz-continuous:
$f=sum_{i=1}^{n}f_imathbf{e_i}$ where $mathbf{e_i}=left (0,...,0,1,0,...,0 right )$ where the $1$ is at the $i$-th position.
Note that $left | f_1(x)mathbf{e_1}+f_2(x)mathbf{e_2}-f_1(y)mathbf{e_1}-f_2(y)mathbf{e_2} right |leq left | f_1(x)-f_1(y) right |+left |f_2(x)-f_2(y) right |leq L_1left | x-y right |+L_2left | x-y right |leq (L_1+L_2)left | x-y right |$
The result follows by induction.
answered Nov 25 '18 at 10:27
John11
1,0061821
$f$ is K-Lipschitz continuous implies $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$:
$left | f_i(x)-f_i(y) right |leq left | f(x)-f(y) right |leq Kleft | x-y right |$
Each coordinate $f_i$ is $K_i$-Lipschitz continuous implies $f$ is Lipschitz-continuous:
$f=sum_{i=1}^{n}f_imathbf{e_i}$ where $mathbf{e_i}=left (0,...,0,1,0,...,0 right )$ where the $1$ is at the $i$-th position.
Note that $left | f_1(x)mathbf{e_1}+f_2(x)mathbf{e_2}-f_1(y)mathbf{e_1}-f_2(y)mathbf{e_2} right |leq left | f_1(x)-f_1(y) right |+left |f_2(x)-f_2(y) right |leq L_1left | x-y right |+L_2left | x-y right |leq (L_1+L_2)left | x-y right |$
The result follows by induction.
answered Nov 25 '18 at 10:27
John11
1,0061821
answered Nov 25 '18 at 10:27
John11
1,0061821
answered Nov 25 '18 at 10:27
John11
1,0061821
answered Nov 25 '18 at 10:27
John11
1,0061821
1,0061821
So it is. Thank you very much for showing it formally :)
– meisterluk
Nov 26 '18 at 10:14
add a comment |
So it is. Thank you very much for showing it formally :)
– meisterluk
Nov 26 '18 at 10:14
So it is. Thank you very much for showing it formally :)
– meisterluk
Nov 26 '18 at 10:14
So it is. Thank you very much for showing it formally :)
– meisterluk
Nov 26 '18 at 10:14
add a comment |
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