Map $f: mathbbR to mathbbR$ with certain contraction property has fixed point
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Suppose $f: mathbbR to mathbbR$ is continuous such that for any
real $x$,
$|f(x) - f(f(x))| leq frac12 |f(x) -x|$.
Must $f$ have a fixed point?
The question seems to invite an eventual application of the standard contraction mapping theorem. But this approach has not led me to showing there is a fixed point.
Is it true that there is a fixed point, and if so why?
real-analysis
asked Aug 26 at 23:42
CuriousKid7
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up vote
4
down vote
favorite
Suppose $f: mathbbR to mathbbR$ is continuous such that for any
real $x$,
$|f(x) - f(f(x))| leq frac12 |f(x) -x|$.
Must $f$ have a fixed point?
The question seems to invite an eventual application of the standard contraction mapping theorem. But this approach has not led me to showing there is a fixed point.
Is it true that there is a fixed point, and if so why?
real-analysis
asked Aug 26 at 23:42
CuriousKid7
1,573617
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Suppose $f: mathbbR to mathbbR$ is continuous such that for any
real $x$,
$|f(x) - f(f(x))| leq frac12 |f(x) -x|$.
Must $f$ have a fixed point?
The question seems to invite an eventual application of the standard contraction mapping theorem. But this approach has not led me to showing there is a fixed point.
Is it true that there is a fixed point, and if so why?
real-analysis
asked Aug 26 at 23:42
CuriousKid7
1,573617
Suppose $f: mathbbR to mathbbR$ is continuous such that for any
real $x$,
$|f(x) - f(f(x))| leq frac12 |f(x) -x|$.
Must $f$ have a fixed point?
The question seems to invite an eventual application of the standard contraction mapping theorem. But this approach has not led me to showing there is a fixed point.
Is it true that there is a fixed point, and if so why?
real-analysis
asked Aug 26 at 23:42
CuriousKid7
1,573617
asked Aug 26 at 23:42
CuriousKid7
1,573617
asked Aug 26 at 23:42
CuriousKid7
1,573617
asked Aug 26 at 23:42
CuriousKid7
1,573617
1,573617
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Let $x_i$ be the sequence defined by $x_0=0$ and $x_i+1=f(x_i)$. Then the inequality implies that
$$|x_i+1-x_i+2|<frac12|x_i+1-x_i|.$$
Therefore $|x_i+1-x_i|<2^-i|x_1-x_0| rightarrow 0$. For all $m>n>N$, one has
$$|x_m-x_n|leq |x_m-x_m-1|+ldots +|x_n+1-x_n|<(2^-m+1+ldots+2^-n)|x_1-x_0|<2^-N+1|x_1-x_0|$$
It follows that $x_i_i=1^infty$ is a Cauchy sequence, hence convergent to some $xin mathbbR$. Since $x_i+1=f(x_i)$, by taking the limit, we get $x=f(x)$.
answered Aug 26 at 23:49
Marco
1,55917
Why is it Cauchy?
– CuriousKid7
Aug 27 at 0:03
A Cauchy sequence $x_n$ has the property that for every $epsilon>0$ there exists $N$ such that $m,n>N Rightarrow |x_m-x_n|<epsilon$. In $mathbbR$, every Cauchy sequence is convergent.
– Marco
Aug 27 at 0:07
I understand, but I only see that $d(x_i, x_i+1) to 0$, which is not enough to show it's Cauchy.
– CuriousKid7
Aug 27 at 0:30
@CuriousKid7 a sequence whose subsequent terms go to $0$ exponentially fast is Cauchy.
– Robert Wolfe
Aug 27 at 3:28
add a comment |Â
up vote
2
down vote
To further improve on Marcos answer:
the sequence is Cauchy, since for all $n<m$, we have
$$
|x_m - x_n| leq Sigma_i=n^m-1|x_i+1-x_n| leq 2^-n+1 |x_1-x_0|
$$
answered Aug 27 at 1:43
enedil
520313
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $x_i$ be the sequence defined by $x_0=0$ and $x_i+1=f(x_i)$. Then the inequality implies that
$$|x_i+1-x_i+2|<frac12|x_i+1-x_i|.$$
Therefore $|x_i+1-x_i|<2^-i|x_1-x_0| rightarrow 0$. For all $m>n>N$, one has
$$|x_m-x_n|leq |x_m-x_m-1|+ldots +|x_n+1-x_n|<(2^-m+1+ldots+2^-n)|x_1-x_0|<2^-N+1|x_1-x_0|$$
It follows that $x_i_i=1^infty$ is a Cauchy sequence, hence convergent to some $xin mathbbR$. Since $x_i+1=f(x_i)$, by taking the limit, we get $x=f(x)$.
answered Aug 26 at 23:49
Marco
1,55917
Why is it Cauchy?
– CuriousKid7
Aug 27 at 0:03
A Cauchy sequence $x_n$ has the property that for every $epsilon>0$ there exists $N$ such that $m,n>N Rightarrow |x_m-x_n|<epsilon$. In $mathbbR$, every Cauchy sequence is convergent.
– Marco
Aug 27 at 0:07
I understand, but I only see that $d(x_i, x_i+1) to 0$, which is not enough to show it's Cauchy.
– CuriousKid7
Aug 27 at 0:30
@CuriousKid7 a sequence whose subsequent terms go to $0$ exponentially fast is Cauchy.
– Robert Wolfe
Aug 27 at 3:28
add a comment |Â
up vote
3
down vote
accepted
Let $x_i$ be the sequence defined by $x_0=0$ and $x_i+1=f(x_i)$. Then the inequality implies that
$$|x_i+1-x_i+2|<frac12|x_i+1-x_i|.$$
Therefore $|x_i+1-x_i|<2^-i|x_1-x_0| rightarrow 0$. For all $m>n>N$, one has
$$|x_m-x_n|leq |x_m-x_m-1|+ldots +|x_n+1-x_n|<(2^-m+1+ldots+2^-n)|x_1-x_0|<2^-N+1|x_1-x_0|$$
It follows that $x_i_i=1^infty$ is a Cauchy sequence, hence convergent to some $xin mathbbR$. Since $x_i+1=f(x_i)$, by taking the limit, we get $x=f(x)$.
answered Aug 26 at 23:49
Marco
1,55917
Why is it Cauchy?
– CuriousKid7
Aug 27 at 0:03
A Cauchy sequence $x_n$ has the property that for every $epsilon>0$ there exists $N$ such that $m,n>N Rightarrow |x_m-x_n|<epsilon$. In $mathbbR$, every Cauchy sequence is convergent.
– Marco
Aug 27 at 0:07
I understand, but I only see that $d(x_i, x_i+1) to 0$, which is not enough to show it's Cauchy.
– CuriousKid7
Aug 27 at 0:30
@CuriousKid7 a sequence whose subsequent terms go to $0$ exponentially fast is Cauchy.
– Robert Wolfe
Aug 27 at 3:28
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $x_i$ be the sequence defined by $x_0=0$ and $x_i+1=f(x_i)$. Then the inequality implies that
$$|x_i+1-x_i+2|<frac12|x_i+1-x_i|.$$
Therefore $|x_i+1-x_i|<2^-i|x_1-x_0| rightarrow 0$. For all $m>n>N$, one has
$$|x_m-x_n|leq |x_m-x_m-1|+ldots +|x_n+1-x_n|<(2^-m+1+ldots+2^-n)|x_1-x_0|<2^-N+1|x_1-x_0|$$
It follows that $x_i_i=1^infty$ is a Cauchy sequence, hence convergent to some $xin mathbbR$. Since $x_i+1=f(x_i)$, by taking the limit, we get $x=f(x)$.
answered Aug 26 at 23:49
Marco
1,55917
Let $x_i$ be the sequence defined by $x_0=0$ and $x_i+1=f(x_i)$. Then the inequality implies that
$$|x_i+1-x_i+2|<frac12|x_i+1-x_i|.$$
Therefore $|x_i+1-x_i|<2^-i|x_1-x_0| rightarrow 0$. For all $m>n>N$, one has
$$|x_m-x_n|leq |x_m-x_m-1|+ldots +|x_n+1-x_n|<(2^-m+1+ldots+2^-n)|x_1-x_0|<2^-N+1|x_1-x_0|$$
It follows that $x_i_i=1^infty$ is a Cauchy sequence, hence convergent to some $xin mathbbR$. Since $x_i+1=f(x_i)$, by taking the limit, we get $x=f(x)$.
answered Aug 26 at 23:49
Marco
1,55917
edited Aug 27 at 1:04
answered Aug 26 at 23:49
Marco
1,55917
answered Aug 26 at 23:49
Marco
1,55917
answered Aug 26 at 23:49
Marco
1,55917
1,55917
Why is it Cauchy?
– CuriousKid7
Aug 27 at 0:03
A Cauchy sequence $x_n$ has the property that for every $epsilon>0$ there exists $N$ such that $m,n>N Rightarrow |x_m-x_n|<epsilon$. In $mathbbR$, every Cauchy sequence is convergent.
– Marco
Aug 27 at 0:07
I understand, but I only see that $d(x_i, x_i+1) to 0$, which is not enough to show it's Cauchy.
– CuriousKid7
Aug 27 at 0:30
@CuriousKid7 a sequence whose subsequent terms go to $0$ exponentially fast is Cauchy.
– Robert Wolfe
Aug 27 at 3:28
add a comment |Â
Why is it Cauchy?
– CuriousKid7
Aug 27 at 0:03
A Cauchy sequence $x_n$ has the property that for every $epsilon>0$ there exists $N$ such that $m,n>N Rightarrow |x_m-x_n|<epsilon$. In $mathbbR$, every Cauchy sequence is convergent.
– Marco
Aug 27 at 0:07
I understand, but I only see that $d(x_i, x_i+1) to 0$, which is not enough to show it's Cauchy.
– CuriousKid7
Aug 27 at 0:30
@CuriousKid7 a sequence whose subsequent terms go to $0$ exponentially fast is Cauchy.
– Robert Wolfe
Aug 27 at 3:28
Why is it Cauchy?
– CuriousKid7
Aug 27 at 0:03
Why is it Cauchy?
– CuriousKid7
Aug 27 at 0:03
A Cauchy sequence $x_n$ has the property that for every $epsilon>0$ there exists $N$ such that $m,n>N Rightarrow |x_m-x_n|<epsilon$. In $mathbbR$, every Cauchy sequence is convergent.
– Marco
Aug 27 at 0:07
A Cauchy sequence $x_n$ has the property that for every $epsilon>0$ there exists $N$ such that $m,n>N Rightarrow |x_m-x_n|<epsilon$. In $mathbbR$, every Cauchy sequence is convergent.
– Marco
Aug 27 at 0:07
I understand, but I only see that $d(x_i, x_i+1) to 0$, which is not enough to show it's Cauchy.
– CuriousKid7
Aug 27 at 0:30
I understand, but I only see that $d(x_i, x_i+1) to 0$, which is not enough to show it's Cauchy.
– CuriousKid7
Aug 27 at 0:30
@CuriousKid7 a sequence whose subsequent terms go to $0$ exponentially fast is Cauchy.
– Robert Wolfe
Aug 27 at 3:28
@CuriousKid7 a sequence whose subsequent terms go to $0$ exponentially fast is Cauchy.
– Robert Wolfe
Aug 27 at 3:28
add a comment |Â
up vote
2
down vote
To further improve on Marcos answer:
the sequence is Cauchy, since for all $n<m$, we have
$$
|x_m - x_n| leq Sigma_i=n^m-1|x_i+1-x_n| leq 2^-n+1 |x_1-x_0|
$$
answered Aug 27 at 1:43
enedil
520313
add a comment |Â
up vote
2
down vote
To further improve on Marcos answer:
the sequence is Cauchy, since for all $n<m$, we have
$$
|x_m - x_n| leq Sigma_i=n^m-1|x_i+1-x_n| leq 2^-n+1 |x_1-x_0|
$$
answered Aug 27 at 1:43
enedil
520313
add a comment |Â
up vote
2
down vote
up vote
2
down vote
To further improve on Marcos answer:
the sequence is Cauchy, since for all $n<m$, we have
$$
|x_m - x_n| leq Sigma_i=n^m-1|x_i+1-x_n| leq 2^-n+1 |x_1-x_0|
$$
answered Aug 27 at 1:43
enedil
520313
To further improve on Marcos answer:
the sequence is Cauchy, since for all $n<m$, we have
$$
|x_m - x_n| leq Sigma_i=n^m-1|x_i+1-x_n| leq 2^-n+1 |x_1-x_0|
$$
answered Aug 27 at 1:43
enedil
520313
edited Aug 27 at 2:09
answered Aug 27 at 1:43
enedil
520313
answered Aug 27 at 1:43
enedil
520313
answered Aug 27 at 1:43
enedil
520313
520313
add a comment |Â
add a comment |Â
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