Metric on real line
$begingroup$
Let $(X,d)$ be a metric space and fix some $x_{0}in X$. Show that the function $f(x) = d(x_{0},x)$ is Lipschitz continuous?
My problem wasn't with solving the question originally I did that using the triangle inequality and showing $lvert f(x) - f(y)rvertle d(x,y)$, but the definition for Lipschitz continuity is $d_Y(f(x),f(y))le K d_X(x,y)$ where $K > 0$, why can we assume that $d_Y$ is the the euclidean metric?
metric-spaces lipschitz-functions
edited Dec 17 '18 at 20:00
user10354138
7,3772925
asked Dec 17 '18 at 19:51
Scosh_lrScosh_lr
516
$endgroup$
add a comment |
$begingroup$
Let $(X,d)$ be a metric space and fix some $x_{0}in X$. Show that the function $f(x) = d(x_{0},x)$ is Lipschitz continuous?
My problem wasn't with solving the question originally I did that using the triangle inequality and showing $lvert f(x) - f(y)rvertle d(x,y)$, but the definition for Lipschitz continuity is $d_Y(f(x),f(y))le K d_X(x,y)$ where $K > 0$, why can we assume that $d_Y$ is the the euclidean metric?
metric-spaces lipschitz-functions
edited Dec 17 '18 at 20:00
user10354138
7,3772925
asked Dec 17 '18 at 19:51
Scosh_lrScosh_lr
516
$endgroup$
1
$begingroup$
It's implicit in the problem: $f:Xtomathbb R$, is a real function, so it is natural to use the standard metric on $mathbb R$, which is $d_{text{Eucl}}(a,b)=|a-b|$.
$endgroup$
– Federico
Dec 17 '18 at 19:54
add a comment |
$begingroup$
Let $(X,d)$ be a metric space and fix some $x_{0}in X$. Show that the function $f(x) = d(x_{0},x)$ is Lipschitz continuous?
My problem wasn't with solving the question originally I did that using the triangle inequality and showing $lvert f(x) - f(y)rvertle d(x,y)$, but the definition for Lipschitz continuity is $d_Y(f(x),f(y))le K d_X(x,y)$ where $K > 0$, why can we assume that $d_Y$ is the the euclidean metric?
metric-spaces lipschitz-functions
edited Dec 17 '18 at 20:00
user10354138
7,3772925
asked Dec 17 '18 at 19:51
Scosh_lrScosh_lr
516
$endgroup$
Let $(X,d)$ be a metric space and fix some $x_{0}in X$. Show that the function $f(x) = d(x_{0},x)$ is Lipschitz continuous?
My problem wasn't with solving the question originally I did that using the triangle inequality and showing $lvert f(x) - f(y)rvertle d(x,y)$, but the definition for Lipschitz continuity is $d_Y(f(x),f(y))le K d_X(x,y)$ where $K > 0$, why can we assume that $d_Y$ is the the euclidean metric?
metric-spaces lipschitz-functions
metric-spaces lipschitz-functions
edited Dec 17 '18 at 20:00
user10354138
7,3772925
asked Dec 17 '18 at 19:51
Scosh_lrScosh_lr
516
edited Dec 17 '18 at 20:00
user10354138
7,3772925
asked Dec 17 '18 at 19:51
Scosh_lrScosh_lr
516
edited Dec 17 '18 at 20:00
user10354138
7,3772925
edited Dec 17 '18 at 20:00
user10354138
7,3772925
edited Dec 17 '18 at 20:00
user10354138
7,3772925
7,3772925
asked Dec 17 '18 at 19:51
Scosh_lrScosh_lr
516
asked Dec 17 '18 at 19:51
Scosh_lrScosh_lr
516
asked Dec 17 '18 at 19:51
Scosh_lrScosh_lr
516
516
1
$begingroup$
It's implicit in the problem: $f:Xtomathbb R$, is a real function, so it is natural to use the standard metric on $mathbb R$, which is $d_{text{Eucl}}(a,b)=|a-b|$.
$endgroup$
– Federico
Dec 17 '18 at 19:54
add a comment |
1
$begingroup$
It's implicit in the problem: $f:Xtomathbb R$, is a real function, so it is natural to use the standard metric on $mathbb R$, which is $d_{text{Eucl}}(a,b)=|a-b|$.
$endgroup$
– Federico
Dec 17 '18 at 19:54
1
1
$begingroup$
It's implicit in the problem: $f:Xtomathbb R$, is a real function, so it is natural to use the standard metric on $mathbb R$, which is $d_{text{Eucl}}(a,b)=|a-b|$.
$endgroup$
– Federico
Dec 17 '18 at 19:54
$begingroup$
It's implicit in the problem: $f:Xtomathbb R$, is a real function, so it is natural to use the standard metric on $mathbb R$, which is $d_{text{Eucl}}(a,b)=|a-b|$.
$endgroup$
– Federico
Dec 17 '18 at 19:54
add a comment |
1 Answer
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$begingroup$
It is always the metric of your image space you need to look at. And since there was no other metric on $mathbb{R}$ given, which is apparently the image space of $f$, one implicitly means the standard metric, i.e. the Euclidean one.
answered Dec 17 '18 at 20:05
YoungMathYoungMath
190110
$endgroup$
1
$begingroup$
And then we just take $K=1$ as well.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:44
1
$begingroup$
Yeah, but I thought he/she already got that.
$endgroup$
– YoungMath
Dec 18 '18 at 11:51
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
It is always the metric of your image space you need to look at. And since there was no other metric on $mathbb{R}$ given, which is apparently the image space of $f$, one implicitly means the standard metric, i.e. the Euclidean one.
answered Dec 17 '18 at 20:05
YoungMathYoungMath
190110
$endgroup$
1
$begingroup$
And then we just take $K=1$ as well.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:44
1
$begingroup$
Yeah, but I thought he/she already got that.
$endgroup$
– YoungMath
Dec 18 '18 at 11:51
add a comment |
$begingroup$
It is always the metric of your image space you need to look at. And since there was no other metric on $mathbb{R}$ given, which is apparently the image space of $f$, one implicitly means the standard metric, i.e. the Euclidean one.
answered Dec 17 '18 at 20:05
YoungMathYoungMath
190110
$endgroup$
1
$begingroup$
And then we just take $K=1$ as well.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:44
1
$begingroup$
Yeah, but I thought he/she already got that.
$endgroup$
– YoungMath
Dec 18 '18 at 11:51
add a comment |
$begingroup$
It is always the metric of your image space you need to look at. And since there was no other metric on $mathbb{R}$ given, which is apparently the image space of $f$, one implicitly means the standard metric, i.e. the Euclidean one.
answered Dec 17 '18 at 20:05
YoungMathYoungMath
190110
$endgroup$
It is always the metric of your image space you need to look at. And since there was no other metric on $mathbb{R}$ given, which is apparently the image space of $f$, one implicitly means the standard metric, i.e. the Euclidean one.
answered Dec 17 '18 at 20:05
YoungMathYoungMath
190110
answered Dec 17 '18 at 20:05
YoungMathYoungMath
190110
answered Dec 17 '18 at 20:05
YoungMathYoungMath
190110
answered Dec 17 '18 at 20:05
YoungMathYoungMath
190110
190110
1
$begingroup$
And then we just take $K=1$ as well.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:44
1
$begingroup$
Yeah, but I thought he/she already got that.
$endgroup$
– YoungMath
Dec 18 '18 at 11:51
add a comment |
1
$begingroup$
And then we just take $K=1$ as well.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:44
1
$begingroup$
Yeah, but I thought he/she already got that.
$endgroup$
– YoungMath
Dec 18 '18 at 11:51
1
1
$begingroup$
And then we just take $K=1$ as well.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:44
$begingroup$
And then we just take $K=1$ as well.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:44
1
1
$begingroup$
Yeah, but I thought he/she already got that.
$endgroup$
– YoungMath
Dec 18 '18 at 11:51
$begingroup$
Yeah, but I thought he/she already got that.
$endgroup$
– YoungMath
Dec 18 '18 at 11:51
add a comment |
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$begingroup$
It's implicit in the problem: $f:Xtomathbb R$, is a real function, so it is natural to use the standard metric on $mathbb R$, which is $d_{text{Eucl}}(a,b)=|a-b|$.
$endgroup$
– Federico
Dec 17 '18 at 19:54