Is this function involving indicator function Lipschitz?
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Is this function $$x1_{{x>0}}(x)$$ Lipschitz? It's not differential so mean value cant be used here.
lipschitz-functions
asked Nov 22 at 14:30
Vaolter
6851827
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up vote
0
down vote
favorite
Is this function $$x1_{{x>0}}(x)$$ Lipschitz? It's not differential so mean value cant be used here.
lipschitz-functions
asked Nov 22 at 14:30
Vaolter
6851827
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is this function $$x1_{{x>0}}(x)$$ Lipschitz? It's not differential so mean value cant be used here.
lipschitz-functions
asked Nov 22 at 14:30
Vaolter
6851827
Is this function $$x1_{{x>0}}(x)$$ Lipschitz? It's not differential so mean value cant be used here.
lipschitz-functions
lipschitz-functions
asked Nov 22 at 14:30
Vaolter
6851827
asked Nov 22 at 14:30
Vaolter
6851827
asked Nov 22 at 14:30
Vaolter
6851827
asked Nov 22 at 14:30
Vaolter
6851827
asked Nov 22 at 14:30
Vaolter
6851827
6851827
add a comment |
add a comment |
1 Answer
1
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2
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Yes. You only need $|f(x)-f(y)| le K |x-y|$, and $K=1$ will do.
answered Nov 22 at 14:41
Richard Martin
1,6328
please explain your reasoning, I cant see why.
– Vaolter
Nov 22 at 14:45
1
@Vaolter: Have you sketched a graph of your function?
– Henning Makholm
Nov 22 at 14:47
Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
– Richard Martin
Nov 22 at 14:48
Yes. Or draw it.
– Richard Martin
Nov 22 at 14:48
@Richard Martin, Ok, thank you, I have checked it.
– Vaolter
Nov 22 at 15:13
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes. You only need $|f(x)-f(y)| le K |x-y|$, and $K=1$ will do.
answered Nov 22 at 14:41
Richard Martin
1,6328
please explain your reasoning, I cant see why.
– Vaolter
Nov 22 at 14:45
1
@Vaolter: Have you sketched a graph of your function?
– Henning Makholm
Nov 22 at 14:47
Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
– Richard Martin
Nov 22 at 14:48
Yes. Or draw it.
– Richard Martin
Nov 22 at 14:48
@Richard Martin, Ok, thank you, I have checked it.
– Vaolter
Nov 22 at 15:13
add a comment |
up vote
2
down vote
accepted
Yes. You only need $|f(x)-f(y)| le K |x-y|$, and $K=1$ will do.
answered Nov 22 at 14:41
Richard Martin
1,6328
please explain your reasoning, I cant see why.
– Vaolter
Nov 22 at 14:45
1
@Vaolter: Have you sketched a graph of your function?
– Henning Makholm
Nov 22 at 14:47
Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
– Richard Martin
Nov 22 at 14:48
Yes. Or draw it.
– Richard Martin
Nov 22 at 14:48
@Richard Martin, Ok, thank you, I have checked it.
– Vaolter
Nov 22 at 15:13
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes. You only need $|f(x)-f(y)| le K |x-y|$, and $K=1$ will do.
answered Nov 22 at 14:41
Richard Martin
1,6328
Yes. You only need $|f(x)-f(y)| le K |x-y|$, and $K=1$ will do.
answered Nov 22 at 14:41
Richard Martin
1,6328
answered Nov 22 at 14:41
Richard Martin
1,6328
answered Nov 22 at 14:41
Richard Martin
1,6328
answered Nov 22 at 14:41
Richard Martin
1,6328
1,6328
please explain your reasoning, I cant see why.
– Vaolter
Nov 22 at 14:45
1
@Vaolter: Have you sketched a graph of your function?
– Henning Makholm
Nov 22 at 14:47
Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
– Richard Martin
Nov 22 at 14:48
Yes. Or draw it.
– Richard Martin
Nov 22 at 14:48
@Richard Martin, Ok, thank you, I have checked it.
– Vaolter
Nov 22 at 15:13
add a comment |
please explain your reasoning, I cant see why.
– Vaolter
Nov 22 at 14:45
1
@Vaolter: Have you sketched a graph of your function?
– Henning Makholm
Nov 22 at 14:47
Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
– Richard Martin
Nov 22 at 14:48
Yes. Or draw it.
– Richard Martin
Nov 22 at 14:48
@Richard Martin, Ok, thank you, I have checked it.
– Vaolter
Nov 22 at 15:13
please explain your reasoning, I cant see why.
– Vaolter
Nov 22 at 14:45
please explain your reasoning, I cant see why.
– Vaolter
Nov 22 at 14:45
1
1
@Vaolter: Have you sketched a graph of your function?
– Henning Makholm
Nov 22 at 14:47
@Vaolter: Have you sketched a graph of your function?
– Henning Makholm
Nov 22 at 14:47
Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
– Richard Martin
Nov 22 at 14:48
Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
– Richard Martin
Nov 22 at 14:48
Yes. Or draw it.
– Richard Martin
Nov 22 at 14:48
Yes. Or draw it.
– Richard Martin
Nov 22 at 14:48
@Richard Martin, Ok, thank you, I have checked it.
– Vaolter
Nov 22 at 15:13
@Richard Martin, Ok, thank you, I have checked it.
– Vaolter
Nov 22 at 15:13
add a comment |
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