Mixing
Inuitive Meaning of Lipchitz Continuity
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Intuitively, a function $f$ from an interval to the real numbers being Lipchitz continuous means that its derivative is bounded.
How does this intuitive meaning correspond for Lipchitz continuous functions whose domains and ranges might be of dimension greater than $1$?
real-analysis analysis
asked Jan 8 at 18:36
LinearGuyLinearGuy
13711
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add a comment |
$begingroup$
Intuitively, a function $f$ from an interval to the real numbers being Lipchitz continuous means that its derivative is bounded.
How does this intuitive meaning correspond for Lipchitz continuous functions whose domains and ranges might be of dimension greater than $1$?
real-analysis analysis
asked Jan 8 at 18:36
LinearGuyLinearGuy
13711
$endgroup$
$begingroup$
Does this help? math.stackexchange.com/q/2374289/525520
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– T. Fo
Jan 8 at 18:41
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@T.Ford Not really
$endgroup$
– LinearGuy
Jan 8 at 18:43
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Ya sry, I guess not, everything in there is dim 1
$endgroup$
– T. Fo
Jan 8 at 18:44
1
$begingroup$
If you use the induced norm, then the same interpretation can be used for higher dimensional domains.
$endgroup$
– copper.hat
Jan 8 at 18:54
add a comment |
$begingroup$
Intuitively, a function $f$ from an interval to the real numbers being Lipchitz continuous means that its derivative is bounded.
How does this intuitive meaning correspond for Lipchitz continuous functions whose domains and ranges might be of dimension greater than $1$?
real-analysis analysis
asked Jan 8 at 18:36
LinearGuyLinearGuy
13711
$endgroup$
Intuitively, a function $f$ from an interval to the real numbers being Lipchitz continuous means that its derivative is bounded.
How does this intuitive meaning correspond for Lipchitz continuous functions whose domains and ranges might be of dimension greater than $1$?
real-analysis analysis
real-analysis analysis
asked Jan 8 at 18:36
LinearGuyLinearGuy
13711
asked Jan 8 at 18:36
LinearGuyLinearGuy
13711
asked Jan 8 at 18:36
LinearGuyLinearGuy
13711
asked Jan 8 at 18:36
LinearGuyLinearGuy
13711
asked Jan 8 at 18:36
LinearGuyLinearGuy
13711
13711
$begingroup$
Does this help? math.stackexchange.com/q/2374289/525520
$endgroup$
– T. Fo
Jan 8 at 18:41
$begingroup$
@T.Ford Not really
$endgroup$
– LinearGuy
Jan 8 at 18:43
$begingroup$
Ya sry, I guess not, everything in there is dim 1
$endgroup$
– T. Fo
Jan 8 at 18:44
1
$begingroup$
If you use the induced norm, then the same interpretation can be used for higher dimensional domains.
$endgroup$
– copper.hat
Jan 8 at 18:54
add a comment |
$begingroup$
Does this help? math.stackexchange.com/q/2374289/525520
$endgroup$
– T. Fo
Jan 8 at 18:41
$begingroup$
@T.Ford Not really
$endgroup$
– LinearGuy
Jan 8 at 18:43
$begingroup$
Ya sry, I guess not, everything in there is dim 1
$endgroup$
– T. Fo
Jan 8 at 18:44
1
$begingroup$
If you use the induced norm, then the same interpretation can be used for higher dimensional domains.
$endgroup$
– copper.hat
Jan 8 at 18:54
$begingroup$
Does this help? math.stackexchange.com/q/2374289/525520
$endgroup$
– T. Fo
Jan 8 at 18:41
$begingroup$
Does this help? math.stackexchange.com/q/2374289/525520
$endgroup$
– T. Fo
Jan 8 at 18:41
$begingroup$
@T.Ford Not really
$endgroup$
– LinearGuy
Jan 8 at 18:43
$begingroup$
@T.Ford Not really
$endgroup$
– LinearGuy
Jan 8 at 18:43
$begingroup$
Ya sry, I guess not, everything in there is dim 1
$endgroup$
– T. Fo
Jan 8 at 18:44
$begingroup$
Ya sry, I guess not, everything in there is dim 1
$endgroup$
– T. Fo
Jan 8 at 18:44
1
1
$begingroup$
If you use the induced norm, then the same interpretation can be used for higher dimensional domains.
$endgroup$
– copper.hat
Jan 8 at 18:54
$begingroup$
If you use the induced norm, then the same interpretation can be used for higher dimensional domains.
$endgroup$
– copper.hat
Jan 8 at 18:54
add a comment |
1 Answer
1
active
oldest
votes
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The meaning of "Lipschitz continuity" is much simpler than the meaning of "continuity". "Lipschitz continuity" of a function $f$ says that there is a $C>0$ with $$|f(x)-f(y)|leq C|x-y|qquadbigl(=C,d(x,y)bigr)$$
for all $x$, $y$ in the domain of $f$. No dangling with particular points, $epsilon$s and $delta$s.
In particular, if $f$ is Lipschitz continuous with some $C$, then for any $x_0$ in the domain of $f$, and any given $epsilon>0$, we can guarantee that
$$|f(x)-f(x_0)|leqepsilonqquadleft(|x-x_0|leq{epsilon over C}right) ;$$
hence $f$ is automatically uniformly continuous on its domain.
answered Jan 8 at 19:33
Christian BlatterChristian Blatter
173k7113326
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The meaning of "Lipschitz continuity" is much simpler than the meaning of "continuity". "Lipschitz continuity" of a function $f$ says that there is a $C>0$ with $$|f(x)-f(y)|leq C|x-y|qquadbigl(=C,d(x,y)bigr)$$
for all $x$, $y$ in the domain of $f$. No dangling with particular points, $epsilon$s and $delta$s.
In particular, if $f$ is Lipschitz continuous with some $C$, then for any $x_0$ in the domain of $f$, and any given $epsilon>0$, we can guarantee that
$$|f(x)-f(x_0)|leqepsilonqquadleft(|x-x_0|leq{epsilon over C}right) ;$$
hence $f$ is automatically uniformly continuous on its domain.
answered Jan 8 at 19:33
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
$begingroup$
The meaning of "Lipschitz continuity" is much simpler than the meaning of "continuity". "Lipschitz continuity" of a function $f$ says that there is a $C>0$ with $$|f(x)-f(y)|leq C|x-y|qquadbigl(=C,d(x,y)bigr)$$
for all $x$, $y$ in the domain of $f$. No dangling with particular points, $epsilon$s and $delta$s.
In particular, if $f$ is Lipschitz continuous with some $C$, then for any $x_0$ in the domain of $f$, and any given $epsilon>0$, we can guarantee that
$$|f(x)-f(x_0)|leqepsilonqquadleft(|x-x_0|leq{epsilon over C}right) ;$$
hence $f$ is automatically uniformly continuous on its domain.
answered Jan 8 at 19:33
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
$begingroup$
The meaning of "Lipschitz continuity" is much simpler than the meaning of "continuity". "Lipschitz continuity" of a function $f$ says that there is a $C>0$ with $$|f(x)-f(y)|leq C|x-y|qquadbigl(=C,d(x,y)bigr)$$
for all $x$, $y$ in the domain of $f$. No dangling with particular points, $epsilon$s and $delta$s.
In particular, if $f$ is Lipschitz continuous with some $C$, then for any $x_0$ in the domain of $f$, and any given $epsilon>0$, we can guarantee that
$$|f(x)-f(x_0)|leqepsilonqquadleft(|x-x_0|leq{epsilon over C}right) ;$$
hence $f$ is automatically uniformly continuous on its domain.
answered Jan 8 at 19:33
Christian BlatterChristian Blatter
173k7113326
$endgroup$
The meaning of "Lipschitz continuity" is much simpler than the meaning of "continuity". "Lipschitz continuity" of a function $f$ says that there is a $C>0$ with $$|f(x)-f(y)|leq C|x-y|qquadbigl(=C,d(x,y)bigr)$$
for all $x$, $y$ in the domain of $f$. No dangling with particular points, $epsilon$s and $delta$s.
In particular, if $f$ is Lipschitz continuous with some $C$, then for any $x_0$ in the domain of $f$, and any given $epsilon>0$, we can guarantee that
$$|f(x)-f(x_0)|leqepsilonqquadleft(|x-x_0|leq{epsilon over C}right) ;$$
hence $f$ is automatically uniformly continuous on its domain.
answered Jan 8 at 19:33
Christian BlatterChristian Blatter
173k7113326
answered Jan 8 at 19:33
Christian BlatterChristian Blatter
173k7113326
answered Jan 8 at 19:33
Christian BlatterChristian Blatter
173k7113326
answered Jan 8 at 19:33
Christian BlatterChristian Blatter
173k7113326
173k7113326
add a comment |
add a comment |
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$begingroup$
Does this help? math.stackexchange.com/q/2374289/525520
$endgroup$
– T. Fo
Jan 8 at 18:41
$begingroup$
@T.Ford Not really
$endgroup$
– LinearGuy
Jan 8 at 18:43
$begingroup$
Ya sry, I guess not, everything in there is dim 1
$endgroup$
– T. Fo
Jan 8 at 18:44
1
$begingroup$
If you use the induced norm, then the same interpretation can be used for higher dimensional domains.
$endgroup$
– copper.hat
Jan 8 at 18:54