Fixed point function
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Given a function $g(x)$ defined over $Omega = [a,b]$ with the following properties:
$Omega$ stable by $g$ : $g(Omega) subset Omega iff forall x in Omega, g(Omega) in Omega$
$exists K,K<1, forall x forall y, |g(x)-g(y)|<K|x-y|$
How do I show that the function $g(x)-x$ changes sign over $Omega$ and deduct that exists a fixed point over $Omega$? I am inclined to use the fact that $(g(a)-a).(g(b)-b)<0$, but I am not quite sure how to get there.
functions fixedpoints
asked Nov 14 at 11:12
Vinícius Lopes Simões
428211
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up vote
1
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Given a function $g(x)$ defined over $Omega = [a,b]$ with the following properties:
$Omega$ stable by $g$ : $g(Omega) subset Omega iff forall x in Omega, g(Omega) in Omega$
$exists K,K<1, forall x forall y, |g(x)-g(y)|<K|x-y|$
How do I show that the function $g(x)-x$ changes sign over $Omega$ and deduct that exists a fixed point over $Omega$? I am inclined to use the fact that $(g(a)-a).(g(b)-b)<0$, but I am not quite sure how to get there.
functions fixedpoints
asked Nov 14 at 11:12
Vinícius Lopes Simões
428211
1
Hint: use the Banach fixed point theorem. mathonline.wikidot.com/banach-s-fixed-point-theorem
– hamza boulahia
Nov 14 at 11:18
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a function $g(x)$ defined over $Omega = [a,b]$ with the following properties:
$Omega$ stable by $g$ : $g(Omega) subset Omega iff forall x in Omega, g(Omega) in Omega$
$exists K,K<1, forall x forall y, |g(x)-g(y)|<K|x-y|$
How do I show that the function $g(x)-x$ changes sign over $Omega$ and deduct that exists a fixed point over $Omega$? I am inclined to use the fact that $(g(a)-a).(g(b)-b)<0$, but I am not quite sure how to get there.
functions fixedpoints
asked Nov 14 at 11:12
Vinícius Lopes Simões
428211
Given a function $g(x)$ defined over $Omega = [a,b]$ with the following properties:
$Omega$ stable by $g$ : $g(Omega) subset Omega iff forall x in Omega, g(Omega) in Omega$
$exists K,K<1, forall x forall y, |g(x)-g(y)|<K|x-y|$
How do I show that the function $g(x)-x$ changes sign over $Omega$ and deduct that exists a fixed point over $Omega$? I am inclined to use the fact that $(g(a)-a).(g(b)-b)<0$, but I am not quite sure how to get there.
functions fixedpoints
functions fixedpoints
asked Nov 14 at 11:12
Vinícius Lopes Simões
428211
asked Nov 14 at 11:12
Vinícius Lopes Simões
428211
asked Nov 14 at 11:12
Vinícius Lopes Simões
428211
asked Nov 14 at 11:12
Vinícius Lopes Simões
428211
asked Nov 14 at 11:12
Vinícius Lopes Simões
428211
428211
1
Hint: use the Banach fixed point theorem. mathonline.wikidot.com/banach-s-fixed-point-theorem
– hamza boulahia
Nov 14 at 11:18
add a comment |
1
Hint: use the Banach fixed point theorem. mathonline.wikidot.com/banach-s-fixed-point-theorem
– hamza boulahia
Nov 14 at 11:18
1
1
Hint: use the Banach fixed point theorem. mathonline.wikidot.com/banach-s-fixed-point-theorem
– hamza boulahia
Nov 14 at 11:18
Hint: use the Banach fixed point theorem. mathonline.wikidot.com/banach-s-fixed-point-theorem
– hamza boulahia
Nov 14 at 11:18
add a comment |
1 Answer
1
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2
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$g(a) in [a,b] implies g(a) ge a$. If $g(a)=a$, then $a$ is a fixed point.
$g(b) in [a,b] implies g(b) le b$. If $g(b)=b$, then $b$ is a fixed point.
Otherwise, $g(a)-a>0$ and $g(b)-b<0$.
Finally, $|g(x)-g(y)|<K|x-y|$ implies that $g$ is continuous and you can apply the intermediate value theorem to $f(x)=g(x)-x$, as planned.
answered Nov 14 at 11:23
lhf
161k9164383
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
$g(a) in [a,b] implies g(a) ge a$. If $g(a)=a$, then $a$ is a fixed point.
$g(b) in [a,b] implies g(b) le b$. If $g(b)=b$, then $b$ is a fixed point.
Otherwise, $g(a)-a>0$ and $g(b)-b<0$.
Finally, $|g(x)-g(y)|<K|x-y|$ implies that $g$ is continuous and you can apply the intermediate value theorem to $f(x)=g(x)-x$, as planned.
answered Nov 14 at 11:23
lhf
161k9164383
add a comment |
up vote
2
down vote
accepted
$g(a) in [a,b] implies g(a) ge a$. If $g(a)=a$, then $a$ is a fixed point.
$g(b) in [a,b] implies g(b) le b$. If $g(b)=b$, then $b$ is a fixed point.
Otherwise, $g(a)-a>0$ and $g(b)-b<0$.
Finally, $|g(x)-g(y)|<K|x-y|$ implies that $g$ is continuous and you can apply the intermediate value theorem to $f(x)=g(x)-x$, as planned.
answered Nov 14 at 11:23
lhf
161k9164383
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$g(a) in [a,b] implies g(a) ge a$. If $g(a)=a$, then $a$ is a fixed point.
$g(b) in [a,b] implies g(b) le b$. If $g(b)=b$, then $b$ is a fixed point.
Otherwise, $g(a)-a>0$ and $g(b)-b<0$.
Finally, $|g(x)-g(y)|<K|x-y|$ implies that $g$ is continuous and you can apply the intermediate value theorem to $f(x)=g(x)-x$, as planned.
answered Nov 14 at 11:23
lhf
161k9164383
$g(a) in [a,b] implies g(a) ge a$. If $g(a)=a$, then $a$ is a fixed point.
$g(b) in [a,b] implies g(b) le b$. If $g(b)=b$, then $b$ is a fixed point.
Otherwise, $g(a)-a>0$ and $g(b)-b<0$.
Finally, $|g(x)-g(y)|<K|x-y|$ implies that $g$ is continuous and you can apply the intermediate value theorem to $f(x)=g(x)-x$, as planned.
answered Nov 14 at 11:23
lhf
161k9164383
answered Nov 14 at 11:23
lhf
161k9164383
answered Nov 14 at 11:23
lhf
161k9164383
answered Nov 14 at 11:23
lhf
161k9164383
161k9164383
add a comment |
add a comment |
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Hint: use the Banach fixed point theorem. mathonline.wikidot.com/banach-s-fixed-point-theorem
– hamza boulahia
Nov 14 at 11:18