Reverse fixed point conclusion
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0
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If I have a function such as:
$$f:M rightarrow mathbb{R} $$
where $M$ is any metric space denoted by :
$$(M,d)$$
$$f(x) =d(x,y) $$ where $y in M$ is a fixed point.
I am trying to show that this function satisfies the Lipschitz condition.
https://en.wikipedia.org/wiki/Lipschitz_continuity
$$frac{d(f(x),f(z))}{d(x,z)}leq K$$ for $K geq 0$
Currently I am stuck at two things.
First, the distance representation is not quite clear for me , I am not sure if I can use here $d(x,y) = |x-y|$?
I only learnt two fixed points theorems, the Banach fixed point and the Brouwer fixed point.
Secondly, If I have a fixed point $y$ :
it could be a Banach case(unique fixed point), if $M$ is a complete metric space and If $f$ is a contraction i.e. $0 leq K <1$
Otherwise, it is not unique fixed point.
I am not sure what approach should I use here? contradiction or attempt some sort of reverse fixed point iteration direct proof?
real-analysis metric-spaces fixed-point-theorems
asked Nov 19 at 18:45
JKM
6415
add a comment |
up vote
0
down vote
favorite
If I have a function such as:
$$f:M rightarrow mathbb{R} $$
where $M$ is any metric space denoted by :
$$(M,d)$$
$$f(x) =d(x,y) $$ where $y in M$ is a fixed point.
I am trying to show that this function satisfies the Lipschitz condition.
https://en.wikipedia.org/wiki/Lipschitz_continuity
$$frac{d(f(x),f(z))}{d(x,z)}leq K$$ for $K geq 0$
Currently I am stuck at two things.
First, the distance representation is not quite clear for me , I am not sure if I can use here $d(x,y) = |x-y|$?
I only learnt two fixed points theorems, the Banach fixed point and the Brouwer fixed point.
Secondly, If I have a fixed point $y$ :
it could be a Banach case(unique fixed point), if $M$ is a complete metric space and If $f$ is a contraction i.e. $0 leq K <1$
Otherwise, it is not unique fixed point.
I am not sure what approach should I use here? contradiction or attempt some sort of reverse fixed point iteration direct proof?
real-analysis metric-spaces fixed-point-theorems
asked Nov 19 at 18:45
JKM
6415
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If I have a function such as:
$$f:M rightarrow mathbb{R} $$
where $M$ is any metric space denoted by :
$$(M,d)$$
$$f(x) =d(x,y) $$ where $y in M$ is a fixed point.
I am trying to show that this function satisfies the Lipschitz condition.
https://en.wikipedia.org/wiki/Lipschitz_continuity
$$frac{d(f(x),f(z))}{d(x,z)}leq K$$ for $K geq 0$
Currently I am stuck at two things.
First, the distance representation is not quite clear for me , I am not sure if I can use here $d(x,y) = |x-y|$?
I only learnt two fixed points theorems, the Banach fixed point and the Brouwer fixed point.
Secondly, If I have a fixed point $y$ :
it could be a Banach case(unique fixed point), if $M$ is a complete metric space and If $f$ is a contraction i.e. $0 leq K <1$
Otherwise, it is not unique fixed point.
I am not sure what approach should I use here? contradiction or attempt some sort of reverse fixed point iteration direct proof?
real-analysis metric-spaces fixed-point-theorems
asked Nov 19 at 18:45
JKM
6415
If I have a function such as:
$$f:M rightarrow mathbb{R} $$
where $M$ is any metric space denoted by :
$$(M,d)$$
$$f(x) =d(x,y) $$ where $y in M$ is a fixed point.
I am trying to show that this function satisfies the Lipschitz condition.
https://en.wikipedia.org/wiki/Lipschitz_continuity
$$frac{d(f(x),f(z))}{d(x,z)}leq K$$ for $K geq 0$
Currently I am stuck at two things.
First, the distance representation is not quite clear for me , I am not sure if I can use here $d(x,y) = |x-y|$?
I only learnt two fixed points theorems, the Banach fixed point and the Brouwer fixed point.
Secondly, If I have a fixed point $y$ :
it could be a Banach case(unique fixed point), if $M$ is a complete metric space and If $f$ is a contraction i.e. $0 leq K <1$
Otherwise, it is not unique fixed point.
I am not sure what approach should I use here? contradiction or attempt some sort of reverse fixed point iteration direct proof?
real-analysis metric-spaces fixed-point-theorems
real-analysis metric-spaces fixed-point-theorems
asked Nov 19 at 18:45
JKM
6415
asked Nov 19 at 18:45
JKM
6415
edited Nov 19 at 18:51
asked Nov 19 at 18:45
JKM
6415
asked Nov 19 at 18:45
JKM
6415
asked Nov 19 at 18:45
JKM
6415
6415
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
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accepted
The triangle inequality implies that, for all $x_1,x_2in M$,
$$
d(x_1,x_2)+d(x_2,y)ge d(x_1,y) quadtext{and}quad
d(x_2,x_1)+d(x_1,y)ge d(x_2,y)
$$
hence
$$
d(x_1,x_2)ge d(x_1,y)-d(x_2,y) quadtext{and}quad
d(x_1,x_2)ge d(x_2,y)-d(x_1,y)
$$
and therefore
$$
|,f(x_1)-f(x_2)|=big|,d(x_1,y)-d(x_2,y)big|le d(x_1,x_2).
$$
So, $f$ is Lipschitz, with $K=1$.
answered Nov 19 at 18:59
Yiorgos S. Smyrlis
62.2k1383162
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The triangle inequality implies that, for all $x_1,x_2in M$,
$$
d(x_1,x_2)+d(x_2,y)ge d(x_1,y) quadtext{and}quad
d(x_2,x_1)+d(x_1,y)ge d(x_2,y)
$$
hence
$$
d(x_1,x_2)ge d(x_1,y)-d(x_2,y) quadtext{and}quad
d(x_1,x_2)ge d(x_2,y)-d(x_1,y)
$$
and therefore
$$
|,f(x_1)-f(x_2)|=big|,d(x_1,y)-d(x_2,y)big|le d(x_1,x_2).
$$
So, $f$ is Lipschitz, with $K=1$.
answered Nov 19 at 18:59
Yiorgos S. Smyrlis
62.2k1383162
add a comment |
up vote
1
down vote
accepted
The triangle inequality implies that, for all $x_1,x_2in M$,
$$
d(x_1,x_2)+d(x_2,y)ge d(x_1,y) quadtext{and}quad
d(x_2,x_1)+d(x_1,y)ge d(x_2,y)
$$
hence
$$
d(x_1,x_2)ge d(x_1,y)-d(x_2,y) quadtext{and}quad
d(x_1,x_2)ge d(x_2,y)-d(x_1,y)
$$
and therefore
$$
|,f(x_1)-f(x_2)|=big|,d(x_1,y)-d(x_2,y)big|le d(x_1,x_2).
$$
So, $f$ is Lipschitz, with $K=1$.
answered Nov 19 at 18:59
Yiorgos S. Smyrlis
62.2k1383162
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The triangle inequality implies that, for all $x_1,x_2in M$,
$$
d(x_1,x_2)+d(x_2,y)ge d(x_1,y) quadtext{and}quad
d(x_2,x_1)+d(x_1,y)ge d(x_2,y)
$$
hence
$$
d(x_1,x_2)ge d(x_1,y)-d(x_2,y) quadtext{and}quad
d(x_1,x_2)ge d(x_2,y)-d(x_1,y)
$$
and therefore
$$
|,f(x_1)-f(x_2)|=big|,d(x_1,y)-d(x_2,y)big|le d(x_1,x_2).
$$
So, $f$ is Lipschitz, with $K=1$.
answered Nov 19 at 18:59
Yiorgos S. Smyrlis
62.2k1383162
The triangle inequality implies that, for all $x_1,x_2in M$,
$$
d(x_1,x_2)+d(x_2,y)ge d(x_1,y) quadtext{and}quad
d(x_2,x_1)+d(x_1,y)ge d(x_2,y)
$$
hence
$$
d(x_1,x_2)ge d(x_1,y)-d(x_2,y) quadtext{and}quad
d(x_1,x_2)ge d(x_2,y)-d(x_1,y)
$$
and therefore
$$
|,f(x_1)-f(x_2)|=big|,d(x_1,y)-d(x_2,y)big|le d(x_1,x_2).
$$
So, $f$ is Lipschitz, with $K=1$.
answered Nov 19 at 18:59
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 19 at 18:59
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 19 at 18:59
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 19 at 18:59
Yiorgos S. Smyrlis
62.2k1383162
62.2k1383162
add a comment |
add a comment |
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