Kdtyjb

Reason Why Dimensional Travelling Would be Restricted

Most significant research articles for practical investors with research perspectives

What's the purpose of these copper coils with resistors inside them in A Yamaha RX-V396RDS amplifier?

How can atoms be electrically neutral when there is a difference in the positions of the charges?

Has the Isbell–Freyd criterion ever been used to check that a category is concretisable?

Find the next monthly expiration date

How to mitigate "bandwagon attacking" from players?

What can I substitute for soda pop in a sweet pork recipe?

Is divide-by-zero a security vulnerability?

Is there a German word for “analytics”?

What is this waxed root vegetable?

"Murder!" The knight said

Understanding Kramnik's play in game 1 of Candidates 2018

Exponential growth/decay formula: what happened to the other constant of integration?

Why do members of Congress in committee hearings ask witnesses the same question multiple times?

If nine coins are tossed, what is the probability that the number of heads is even?

What is the difference between throw e and throw new Exception(e)?

What's the difference between a cart and a wagon?

Skis versus snow shoes - when to choose which for travelling the backcountry?

Logistics of a hovering watercraft in a fantasy setting

Linear regression when Y is bounded and discrete

Replacement ford fiesta radiator has extra hose

I am on the US no-fly list. What can I do in order to be allowed on flights which go through US airspace?

Can you 'upgrade' leather armor to studded leather armor without purchasing the new armor directly?

Contradiction with Banach Fixed Point Theorem

Understanding the Banach fixed point theoremgeneralization of Banach fixed-point theorem on short maps?Banach Fixed Point Theorem. Measurable version.Leray-Schauder fixed point theoremNewton's Method and Banach Fixed Point TheoremSolution of an equation involving Banach fixed point.Banach fixed point theoremBanach Fixed Pointfixed point of a contraction on a closed, convex spaceBanach fixed-point theorem

6

$begingroup$

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

share|cite|improve this question

asked 17 hours ago

LunaLuna

486

$endgroup$

add a comment | 

6

$begingroup$

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

share|cite|improve this question

asked 17 hours ago

LunaLuna

486

$endgroup$

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

share|cite|improve this question

asked 17 hours ago

LunaLuna

486

$endgroup$

share|cite|improve this question

asked 17 hours ago

LunaLuna

486

share|cite|improve this question

asked 17 hours ago

LunaLuna

486

asked 17 hours ago

LunaLuna

486

asked 17 hours ago

LunaLuna

486

2 Answers
2

active

oldest

votes

37

$begingroup$

$e^{-x}$ does not map $[2,infty)$ into itself.

share|cite|improve this answer

answered 17 hours ago

Kavi Rama MurthyKavi Rama Murthy

64.5k42665

$endgroup$

• 
  
  
  

  
  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
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

$e^{-2}$ 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.

share|cite|improve this answer

edited 4 hours ago

answered 9 hours ago

AcccumulationAcccumulation

7,0472619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  4 hours ago
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3134607%2fcontradiction-with-banach-fixed-point-theorem%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

37

$begingroup$

$e^{-x}$ does not map $[2,infty)$ into itself.

share|cite|improve this answer

answered 17 hours ago

Kavi Rama MurthyKavi Rama Murthy

64.5k42665

$endgroup$

• 
  
  
  

  
  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
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

 | 
show 2 more comments

37

$begingroup$

$e^{-x}$ does not map $[2,infty)$ into itself.

share|cite|improve this answer

answered 17 hours ago

Kavi Rama MurthyKavi Rama Murthy

64.5k42665

$endgroup$

• 
  
  
  

  
  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
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

$e^{-x}$ does not map $[2,infty)$ into itself.

share|cite|improve this answer

answered 17 hours ago

Kavi Rama MurthyKavi Rama Murthy

64.5k42665

$endgroup$

share|cite|improve this answer

answered 17 hours ago

Kavi Rama MurthyKavi Rama Murthy

64.5k42665

answered 17 hours ago

Kavi Rama MurthyKavi Rama Murthy

64.5k42665

answered 17 hours ago

Kavi Rama MurthyKavi Rama Murthy

64.5k42665

1

$begingroup$

$e^{-2}$ 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.

share|cite|improve this answer

edited 4 hours ago

answered 9 hours ago

AcccumulationAcccumulation

7,0472619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  4 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

$e^{-2}$ 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.

share|cite|improve this answer

edited 4 hours ago

answered 9 hours ago

AcccumulationAcccumulation

7,0472619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  4 hours ago
  

  
  
  
  

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.

share|cite|improve this answer

edited 4 hours ago

answered 9 hours ago

AcccumulationAcccumulation

7,0472619

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 9 hours ago

AcccumulationAcccumulation

7,0472619

answered 9 hours ago

AcccumulationAcccumulation

7,0472619

answered 9 hours ago

AcccumulationAcccumulation

7,0472619