Contradiction with Banach Fixed Point Theorem
$begingroup$
I am trying to find the fixed point of the function $g(x) = e^{-x}$. Wolfram Alpha tells me that this fixed point is approximately $x approx 0,567$. However, if I apply the Banach fixed point theorem, I can prove that $g(x)$ has a fixed point in the interval $[2,infty)$. I reasoned as follows:
Banach fixed point theorem:
Let $X$ be a Banach space, $D subseteq X$ a closed interval and $T:D rightarrow D$ a contraction, which means that $T$ is Lipschitz continuous with Lipschitz constant $L < 1$:
begin{equation}
Vert T(u) - T(v) Vert_X leq L Vert u-v Vert_X text{ } forall u,v in D.
end{equation}
Now $X = (mathbb{R}, Vert cdot Vert_1)$ is a Banach space and $D = [2,infty)$ is a closed interval in $X$. Now, I show that $g(x)$ is a contraction: without loss of generality, assume $x < y$. Then
begin{equation}
Vert g_1(x) - g_1(y) Vert_1 = |e^{-x} - e^{-y}| = e^{-x} - e^{-y} = e^{-x}(1-e^{x-y}).
end{equation}
Since $e^a geq 1+a text{ } forall a in mathbb{R}$, I obtain
begin{equation}
begin{split}
Vert g_1(x) - g_1(y) Vert leq e^{-x}(1-(1+x-y)) = e^{-x}(y-x) \
leq e^{-2}(y-x) = e^{-2}|x-y| = e^{-2} Vert x-y Vert_1 = L Vert x-y Vert_1,
end{split}
end{equation}
where $L = e^{-2}$ < 1.
So the conditions of the Banach fixed point theorem are satisfied and $g(x)$ has a fixed point in the interval $[2, infty)$. However... This is not true!
Can anyone tell me what is going wrong here?
Thank you very much in advance!
fixed-point-theorems
asked Mar 4 at 8:06
LunaLuna
536
$endgroup$
add a comment |
$begingroup$
I am trying to find the fixed point of the function $g(x) = e^{-x}$. Wolfram Alpha tells me that this fixed point is approximately $x approx 0,567$. However, if I apply the Banach fixed point theorem, I can prove that $g(x)$ has a fixed point in the interval $[2,infty)$. I reasoned as follows:
Banach fixed point theorem:
Let $X$ be a Banach space, $D subseteq X$ a closed interval and $T:D rightarrow D$ a contraction, which means that $T$ is Lipschitz continuous with Lipschitz constant $L < 1$:
begin{equation}
Vert T(u) - T(v) Vert_X leq L Vert u-v Vert_X text{ } forall u,v in D.
end{equation}
Now $X = (mathbb{R}, Vert cdot Vert_1)$ is a Banach space and $D = [2,infty)$ is a closed interval in $X$. Now, I show that $g(x)$ is a contraction: without loss of generality, assume $x < y$. Then
begin{equation}
Vert g_1(x) - g_1(y) Vert_1 = |e^{-x} - e^{-y}| = e^{-x} - e^{-y} = e^{-x}(1-e^{x-y}).
end{equation}
Since $e^a geq 1+a text{ } forall a in mathbb{R}$, I obtain
begin{equation}
begin{split}
Vert g_1(x) - g_1(y) Vert leq e^{-x}(1-(1+x-y)) = e^{-x}(y-x) \
leq e^{-2}(y-x) = e^{-2}|x-y| = e^{-2} Vert x-y Vert_1 = L Vert x-y Vert_1,
end{split}
end{equation}
where $L = e^{-2}$ < 1.
So the conditions of the Banach fixed point theorem are satisfied and $g(x)$ has a fixed point in the interval $[2, infty)$. However... This is not true!
Can anyone tell me what is going wrong here?
Thank you very much in advance!
fixed-point-theorems
asked Mar 4 at 8:06
LunaLuna
536
$endgroup$
$begingroup$
The important line in the theorem is $T : D to D$. This is the hypothesis you are missing.
$endgroup$
– Nate Eldredge
Mar 5 at 3:21
add a comment |
$begingroup$
I am trying to find the fixed point of the function $g(x) = e^{-x}$. Wolfram Alpha tells me that this fixed point is approximately $x approx 0,567$. However, if I apply the Banach fixed point theorem, I can prove that $g(x)$ has a fixed point in the interval $[2,infty)$. I reasoned as follows:
Banach fixed point theorem:
Let $X$ be a Banach space, $D subseteq X$ a closed interval and $T:D rightarrow D$ a contraction, which means that $T$ is Lipschitz continuous with Lipschitz constant $L < 1$:
begin{equation}
Vert T(u) - T(v) Vert_X leq L Vert u-v Vert_X text{ } forall u,v in D.
end{equation}
Now $X = (mathbb{R}, Vert cdot Vert_1)$ is a Banach space and $D = [2,infty)$ is a closed interval in $X$. Now, I show that $g(x)$ is a contraction: without loss of generality, assume $x < y$. Then
begin{equation}
Vert g_1(x) - g_1(y) Vert_1 = |e^{-x} - e^{-y}| = e^{-x} - e^{-y} = e^{-x}(1-e^{x-y}).
end{equation}
Since $e^a geq 1+a text{ } forall a in mathbb{R}$, I obtain
begin{equation}
begin{split}
Vert g_1(x) - g_1(y) Vert leq e^{-x}(1-(1+x-y)) = e^{-x}(y-x) \
leq e^{-2}(y-x) = e^{-2}|x-y| = e^{-2} Vert x-y Vert_1 = L Vert x-y Vert_1,
end{split}
end{equation}
where $L = e^{-2}$ < 1.
So the conditions of the Banach fixed point theorem are satisfied and $g(x)$ has a fixed point in the interval $[2, infty)$. However... This is not true!
Can anyone tell me what is going wrong here?
Thank you very much in advance!
fixed-point-theorems
asked Mar 4 at 8:06
LunaLuna
536
$endgroup$
I am trying to find the fixed point of the function $g(x) = e^{-x}$. Wolfram Alpha tells me that this fixed point is approximately $x approx 0,567$. However, if I apply the Banach fixed point theorem, I can prove that $g(x)$ has a fixed point in the interval $[2,infty)$. I reasoned as follows:
Banach fixed point theorem:
Let $X$ be a Banach space, $D subseteq X$ a closed interval and $T:D rightarrow D$ a contraction, which means that $T$ is Lipschitz continuous with Lipschitz constant $L < 1$:
begin{equation}
Vert T(u) - T(v) Vert_X leq L Vert u-v Vert_X text{ } forall u,v in D.
end{equation}
Now $X = (mathbb{R}, Vert cdot Vert_1)$ is a Banach space and $D = [2,infty)$ is a closed interval in $X$. Now, I show that $g(x)$ is a contraction: without loss of generality, assume $x < y$. Then
begin{equation}
Vert g_1(x) - g_1(y) Vert_1 = |e^{-x} - e^{-y}| = e^{-x} - e^{-y} = e^{-x}(1-e^{x-y}).
end{equation}
Since $e^a geq 1+a text{ } forall a in mathbb{R}$, I obtain
begin{equation}
begin{split}
Vert g_1(x) - g_1(y) Vert leq e^{-x}(1-(1+x-y)) = e^{-x}(y-x) \
leq e^{-2}(y-x) = e^{-2}|x-y| = e^{-2} Vert x-y Vert_1 = L Vert x-y Vert_1,
end{split}
end{equation}
where $L = e^{-2}$ < 1.
So the conditions of the Banach fixed point theorem are satisfied and $g(x)$ has a fixed point in the interval $[2, infty)$. However... This is not true!
Can anyone tell me what is going wrong here?
Thank you very much in advance!
fixed-point-theorems
fixed-point-theorems
asked Mar 4 at 8:06
LunaLuna
536
asked Mar 4 at 8:06
LunaLuna
536
asked Mar 4 at 8:06
LunaLuna
536
asked Mar 4 at 8:06
LunaLuna
536
asked Mar 4 at 8:06
LunaLuna
536
536
$begingroup$
The important line in the theorem is $T : D to D$. This is the hypothesis you are missing.
$endgroup$
– Nate Eldredge
Mar 5 at 3:21
add a comment |
$begingroup$
The important line in the theorem is $T : D to D$. This is the hypothesis you are missing.
$endgroup$
– Nate Eldredge
Mar 5 at 3:21
$begingroup$
The important line in the theorem is $T : D to D$. This is the hypothesis you are missing.
$endgroup$
– Nate Eldredge
Mar 5 at 3:21
$begingroup$
The important line in the theorem is $T : D to D$. This is the hypothesis you are missing.
$endgroup$
– Nate Eldredge
Mar 5 at 3:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$e^{-x}$ does not map $[2,infty)$ into itself.
answered Mar 4 at 8:08
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
1
$begingroup$
The phrase "map into" is often used to denote injections (and "map onto" used to denote surjections), and this is injective.
$endgroup$
– Acccumulation
Mar 4 at 16:02
6
$begingroup$
@Acccumulation - the phrase "map into" does not, and should not, mean "injection". "into" in standard English usage indicates inclusion, not uniqueness. Attaching such a specialized meaning to it in Mathematics simply invites confusion to readers. And while I obviously cannot say that no author has used such a thing, I personally am not familiar with any text where "map into" is given such a meaning.
$endgroup$
– Paul Sinclair
Mar 4 at 17:46
$begingroup$
@PaulSinclair "An "into" or "one-to-one" function, also called an "injection" (which, as you can guess, is French for "throwing into") moves the domain A INTO B" mathforum.org/library/drmath/view/52454.html
$endgroup$
– Acccumulation
Mar 4 at 17:53
4
$begingroup$
@Accumulation - So there is one person who is willing to make such an association in a forum post. However, I still say this is far from conventional. The texts I am familiar with do not use "into" to mean "injection", but simply when the function is not known to be surjective.
$endgroup$
– Paul Sinclair
Mar 4 at 18:03
4
$begingroup$
Regardless, the main point of this answer is that the range of $e^{-x}$ is not in $[2,infty)$. Whether "into" means "one-to-one" or not, $e^{-x}$ does not map $[2,infty)$ into itself because the range is disjoint from the domain.
$endgroup$
– Teepeemm
Mar 4 at 18:40
|
show 3 more comments
$begingroup$
$e^{-2}$ is not in the interval $[2,infty)$, so the image of the interval is not contained within the interval; i.e. $g$ does not map the interval to itself. In fact, the interval and its image are disjoint, so there can't possibly be a fixed point. One way of thinking of the Banach fixed-point theorem is that if you have an interval that is mapped to itself, then you can find a a sub-interval that is mapped to that sub-interval, and a sub-sub-interval of that sub-interval that is mapped to that sub-sub-interval, and so on, and the limit of that process is a single point that's mapped to itself. Once you get an interval that isn't mapped to itself, though, that interval doesn't necessarily have a fixed point, even though the space as a whole does. If you don't check that an interval is mapped to itself, you would find that the BFPT requires every interval to contain a fixed point, which is clearly absurd.
answered Mar 4 at 16:00
AcccumulationAcccumulation
7,3052619
$endgroup$
$begingroup$
@FedericoPoloni You're right; the article on the BFPT says that L has to be in the interval [0,1), which I missed.
$endgroup$
– Acccumulation
Mar 4 at 21:23
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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votes
active
oldest
votes
$begingroup$
$e^{-x}$ does not map $[2,infty)$ into itself.
answered Mar 4 at 8:08
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
1
$begingroup$
The phrase "map into" is often used to denote injections (and "map onto" used to denote surjections), and this is injective.
$endgroup$
– Acccumulation
Mar 4 at 16:02
6
$begingroup$
@Acccumulation - the phrase "map into" does not, and should not, mean "injection". "into" in standard English usage indicates inclusion, not uniqueness. Attaching such a specialized meaning to it in Mathematics simply invites confusion to readers. And while I obviously cannot say that no author has used such a thing, I personally am not familiar with any text where "map into" is given such a meaning.
$endgroup$
– Paul Sinclair
Mar 4 at 17:46
$begingroup$
@PaulSinclair "An "into" or "one-to-one" function, also called an "injection" (which, as you can guess, is French for "throwing into") moves the domain A INTO B" mathforum.org/library/drmath/view/52454.html
$endgroup$
– Acccumulation
Mar 4 at 17:53
4
$begingroup$
@Accumulation - So there is one person who is willing to make such an association in a forum post. However, I still say this is far from conventional. The texts I am familiar with do not use "into" to mean "injection", but simply when the function is not known to be surjective.
$endgroup$
– Paul Sinclair
Mar 4 at 18:03
4
$begingroup$
Regardless, the main point of this answer is that the range of $e^{-x}$ is not in $[2,infty)$. Whether "into" means "one-to-one" or not, $e^{-x}$ does not map $[2,infty)$ into itself because the range is disjoint from the domain.
$endgroup$
– Teepeemm
Mar 4 at 18:40
|
show 3 more comments
$begingroup$
$e^{-x}$ does not map $[2,infty)$ into itself.
answered Mar 4 at 8:08
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
1
$begingroup$
The phrase "map into" is often used to denote injections (and "map onto" used to denote surjections), and this is injective.
$endgroup$
– Acccumulation
Mar 4 at 16:02
6
$begingroup$
@Acccumulation - the phrase "map into" does not, and should not, mean "injection". "into" in standard English usage indicates inclusion, not uniqueness. Attaching such a specialized meaning to it in Mathematics simply invites confusion to readers. And while I obviously cannot say that no author has used such a thing, I personally am not familiar with any text where "map into" is given such a meaning.
$endgroup$
– Paul Sinclair
Mar 4 at 17:46
$begingroup$
@PaulSinclair "An "into" or "one-to-one" function, also called an "injection" (which, as you can guess, is French for "throwing into") moves the domain A INTO B" mathforum.org/library/drmath/view/52454.html
$endgroup$
– Acccumulation
Mar 4 at 17:53
4
$begingroup$
@Accumulation - So there is one person who is willing to make such an association in a forum post. However, I still say this is far from conventional. The texts I am familiar with do not use "into" to mean "injection", but simply when the function is not known to be surjective.
$endgroup$
– Paul Sinclair
Mar 4 at 18:03
4
$begingroup$
Regardless, the main point of this answer is that the range of $e^{-x}$ is not in $[2,infty)$. Whether "into" means "one-to-one" or not, $e^{-x}$ does not map $[2,infty)$ into itself because the range is disjoint from the domain.
$endgroup$
– Teepeemm
Mar 4 at 18:40
|
show 3 more comments
$begingroup$
$e^{-x}$ does not map $[2,infty)$ into itself.
answered Mar 4 at 8:08
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
$e^{-x}$ does not map $[2,infty)$ into itself.
answered Mar 4 at 8:08
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Mar 4 at 8:08
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Mar 4 at 8:08
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Mar 4 at 8:08
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
72.6k53170
1
$begingroup$
The phrase "map into" is often used to denote injections (and "map onto" used to denote surjections), and this is injective.
$endgroup$
– Acccumulation
Mar 4 at 16:02
6
$begingroup$
@Acccumulation - the phrase "map into" does not, and should not, mean "injection". "into" in standard English usage indicates inclusion, not uniqueness. Attaching such a specialized meaning to it in Mathematics simply invites confusion to readers. And while I obviously cannot say that no author has used such a thing, I personally am not familiar with any text where "map into" is given such a meaning.
$endgroup$
– Paul Sinclair
Mar 4 at 17:46
$begingroup$
@PaulSinclair "An "into" or "one-to-one" function, also called an "injection" (which, as you can guess, is French for "throwing into") moves the domain A INTO B" mathforum.org/library/drmath/view/52454.html
$endgroup$
– Acccumulation
Mar 4 at 17:53
4
$begingroup$
@Accumulation - So there is one person who is willing to make such an association in a forum post. However, I still say this is far from conventional. The texts I am familiar with do not use "into" to mean "injection", but simply when the function is not known to be surjective.
$endgroup$
– Paul Sinclair
Mar 4 at 18:03
4
$begingroup$
Regardless, the main point of this answer is that the range of $e^{-x}$ is not in $[2,infty)$. Whether "into" means "one-to-one" or not, $e^{-x}$ does not map $[2,infty)$ into itself because the range is disjoint from the domain.
$endgroup$
– Teepeemm
Mar 4 at 18:40
|
show 3 more comments
1
$begingroup$
The phrase "map into" is often used to denote injections (and "map onto" used to denote surjections), and this is injective.
$endgroup$
– Acccumulation
Mar 4 at 16:02
6
$begingroup$
@Acccumulation - the phrase "map into" does not, and should not, mean "injection". "into" in standard English usage indicates inclusion, not uniqueness. Attaching such a specialized meaning to it in Mathematics simply invites confusion to readers. And while I obviously cannot say that no author has used such a thing, I personally am not familiar with any text where "map into" is given such a meaning.
$endgroup$
– Paul Sinclair
Mar 4 at 17:46
$begingroup$
@PaulSinclair "An "into" or "one-to-one" function, also called an "injection" (which, as you can guess, is French for "throwing into") moves the domain A INTO B" mathforum.org/library/drmath/view/52454.html
$endgroup$
– Acccumulation
Mar 4 at 17:53
4
$begingroup$
@Accumulation - So there is one person who is willing to make such an association in a forum post. However, I still say this is far from conventional. The texts I am familiar with do not use "into" to mean "injection", but simply when the function is not known to be surjective.
$endgroup$
– Paul Sinclair
Mar 4 at 18:03
4
$begingroup$
Regardless, the main point of this answer is that the range of $e^{-x}$ is not in $[2,infty)$. Whether "into" means "one-to-one" or not, $e^{-x}$ does not map $[2,infty)$ into itself because the range is disjoint from the domain.
$endgroup$
– Teepeemm
Mar 4 at 18:40
1
1
$begingroup$
The phrase "map into" is often used to denote injections (and "map onto" used to denote surjections), and this is injective.
$endgroup$
– Acccumulation
Mar 4 at 16:02
$begingroup$
The phrase "map into" is often used to denote injections (and "map onto" used to denote surjections), and this is injective.
$endgroup$
– Acccumulation
Mar 4 at 16:02
6
6
$begingroup$
@Acccumulation - the phrase "map into" does not, and should not, mean "injection". "into" in standard English usage indicates inclusion, not uniqueness. Attaching such a specialized meaning to it in Mathematics simply invites confusion to readers. And while I obviously cannot say that no author has used such a thing, I personally am not familiar with any text where "map into" is given such a meaning.
$endgroup$
– Paul Sinclair
Mar 4 at 17:46
$begingroup$
@Acccumulation - the phrase "map into" does not, and should not, mean "injection". "into" in standard English usage indicates inclusion, not uniqueness. Attaching such a specialized meaning to it in Mathematics simply invites confusion to readers. And while I obviously cannot say that no author has used such a thing, I personally am not familiar with any text where "map into" is given such a meaning.
$endgroup$
– Paul Sinclair
Mar 4 at 17:46
$begingroup$
@PaulSinclair "An "into" or "one-to-one" function, also called an "injection" (which, as you can guess, is French for "throwing into") moves the domain A INTO B" mathforum.org/library/drmath/view/52454.html
$endgroup$
– Acccumulation
Mar 4 at 17:53
$begingroup$
@PaulSinclair "An "into" or "one-to-one" function, also called an "injection" (which, as you can guess, is French for "throwing into") moves the domain A INTO B" mathforum.org/library/drmath/view/52454.html
$endgroup$
– Acccumulation
Mar 4 at 17:53
4
4
$begingroup$
@Accumulation - So there is one person who is willing to make such an association in a forum post. However, I still say this is far from conventional. The texts I am familiar with do not use "into" to mean "injection", but simply when the function is not known to be surjective.
$endgroup$
– Paul Sinclair
Mar 4 at 18:03
$begingroup$
@Accumulation - So there is one person who is willing to make such an association in a forum post. However, I still say this is far from conventional. The texts I am familiar with do not use "into" to mean "injection", but simply when the function is not known to be surjective.
$endgroup$
– Paul Sinclair
Mar 4 at 18:03
4
4
$begingroup$
Regardless, the main point of this answer is that the range of $e^{-x}$ is not in $[2,infty)$. Whether "into" means "one-to-one" or not, $e^{-x}$ does not map $[2,infty)$ into itself because the range is disjoint from the domain.
$endgroup$
– Teepeemm
Mar 4 at 18:40
$begingroup$
Regardless, the main point of this answer is that the range of $e^{-x}$ is not in $[2,infty)$. Whether "into" means "one-to-one" or not, $e^{-x}$ does not map $[2,infty)$ into itself because the range is disjoint from the domain.
$endgroup$
– Teepeemm
Mar 4 at 18:40
|
show 3 more comments
$begingroup$
$e^{-2}$ is not in the interval $[2,infty)$, so the image of the interval is not contained within the interval; i.e. $g$ does not map the interval to itself. In fact, the interval and its image are disjoint, so there can't possibly be a fixed point. One way of thinking of the Banach fixed-point theorem is that if you have an interval that is mapped to itself, then you can find a a sub-interval that is mapped to that sub-interval, and a sub-sub-interval of that sub-interval that is mapped to that sub-sub-interval, and so on, and the limit of that process is a single point that's mapped to itself. Once you get an interval that isn't mapped to itself, though, that interval doesn't necessarily have a fixed point, even though the space as a whole does. If you don't check that an interval is mapped to itself, you would find that the BFPT requires every interval to contain a fixed point, which is clearly absurd.
answered Mar 4 at 16:00
AcccumulationAcccumulation
7,3052619
$endgroup$
$begingroup$
@FedericoPoloni You're right; the article on the BFPT says that L has to be in the interval [0,1), which I missed.
$endgroup$
– Acccumulation
Mar 4 at 21:23
add a comment |
$begingroup$
$e^{-2}$ is not in the interval $[2,infty)$, so the image of the interval is not contained within the interval; i.e. $g$ does not map the interval to itself. In fact, the interval and its image are disjoint, so there can't possibly be a fixed point. One way of thinking of the Banach fixed-point theorem is that if you have an interval that is mapped to itself, then you can find a a sub-interval that is mapped to that sub-interval, and a sub-sub-interval of that sub-interval that is mapped to that sub-sub-interval, and so on, and the limit of that process is a single point that's mapped to itself. Once you get an interval that isn't mapped to itself, though, that interval doesn't necessarily have a fixed point, even though the space as a whole does. If you don't check that an interval is mapped to itself, you would find that the BFPT requires every interval to contain a fixed point, which is clearly absurd.
answered Mar 4 at 16:00
AcccumulationAcccumulation
7,3052619
$endgroup$
$begingroup$
@FedericoPoloni You're right; the article on the BFPT says that L has to be in the interval [0,1), which I missed.
$endgroup$
– Acccumulation
Mar 4 at 21:23
add a comment |
$begingroup$
$e^{-2}$ is not in the interval $[2,infty)$, so the image of the interval is not contained within the interval; i.e. $g$ does not map the interval to itself. In fact, the interval and its image are disjoint, so there can't possibly be a fixed point. One way of thinking of the Banach fixed-point theorem is that if you have an interval that is mapped to itself, then you can find a a sub-interval that is mapped to that sub-interval, and a sub-sub-interval of that sub-interval that is mapped to that sub-sub-interval, and so on, and the limit of that process is a single point that's mapped to itself. Once you get an interval that isn't mapped to itself, though, that interval doesn't necessarily have a fixed point, even though the space as a whole does. If you don't check that an interval is mapped to itself, you would find that the BFPT requires every interval to contain a fixed point, which is clearly absurd.
answered Mar 4 at 16:00
AcccumulationAcccumulation
7,3052619
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$e^{-2}$ is not in the interval $[2,infty)$, so the image of the interval is not contained within the interval; i.e. $g$ does not map the interval to itself. In fact, the interval and its image are disjoint, so there can't possibly be a fixed point. One way of thinking of the Banach fixed-point theorem is that if you have an interval that is mapped to itself, then you can find a a sub-interval that is mapped to that sub-interval, and a sub-sub-interval of that sub-interval that is mapped to that sub-sub-interval, and so on, and the limit of that process is a single point that's mapped to itself. Once you get an interval that isn't mapped to itself, though, that interval doesn't necessarily have a fixed point, even though the space as a whole does. If you don't check that an interval is mapped to itself, you would find that the BFPT requires every interval to contain a fixed point, which is clearly absurd.
answered Mar 4 at 16:00
AcccumulationAcccumulation
7,3052619
edited Mar 4 at 21:23
answered Mar 4 at 16:00
AcccumulationAcccumulation
7,3052619
answered Mar 4 at 16:00
AcccumulationAcccumulation
7,3052619
answered Mar 4 at 16:00
AcccumulationAcccumulation
7,3052619
7,3052619
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@FedericoPoloni You're right; the article on the BFPT says that L has to be in the interval [0,1), which I missed.
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– Acccumulation
Mar 4 at 21:23
add a comment |
$begingroup$
@FedericoPoloni You're right; the article on the BFPT says that L has to be in the interval [0,1), which I missed.
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– Acccumulation
Mar 4 at 21:23
$begingroup$
@FedericoPoloni You're right; the article on the BFPT says that L has to be in the interval [0,1), which I missed.
$endgroup$
– Acccumulation
Mar 4 at 21:23
$begingroup$
@FedericoPoloni You're right; the article on the BFPT says that L has to be in the interval [0,1), which I missed.
$endgroup$
– Acccumulation
Mar 4 at 21:23
add a comment |
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$begingroup$
The important line in the theorem is $T : D to D$. This is the hypothesis you are missing.
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– Nate Eldredge
Mar 5 at 3:21