How can I demonstrate that a continuous decreasing function in $Bbb R$ has a fixed point?












1












$begingroup$


I am trying to demonstrate that a continuous decreasing function $f: Bbb R → Bbb R$ has a fixed point. I tried to use a function $g(x) = f(x) - x$, which should be a decreasing one, but I don't know how to obtain a point of $g$ that is $0$.






analysis functions continuity







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$endgroup$








  • 1




    $begingroup$
    Intermediate value theorem.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 23:29










  • $begingroup$
    Welcome to Maths SX! No other hypothesis on $f$?
    $endgroup$
    – Bernard
    Dec 2 '18 at 23:32










  • $begingroup$
    @Bernard What for?
    $endgroup$
    – Did
    Dec 2 '18 at 23:53
















1












$begingroup$


I am trying to demonstrate that a continuous decreasing function $f: Bbb R → Bbb R$ has a fixed point. I tried to use a function $g(x) = f(x) - x$, which should be a decreasing one, but I don't know how to obtain a point of $g$ that is $0$.






analysis functions continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Intermediate value theorem.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 23:29










  • $begingroup$
    Welcome to Maths SX! No other hypothesis on $f$?
    $endgroup$
    – Bernard
    Dec 2 '18 at 23:32










  • $begingroup$
    @Bernard What for?
    $endgroup$
    – Did
    Dec 2 '18 at 23:53














1












1








1





$begingroup$


I am trying to demonstrate that a continuous decreasing function $f: Bbb R → Bbb R$ has a fixed point. I tried to use a function $g(x) = f(x) - x$, which should be a decreasing one, but I don't know how to obtain a point of $g$ that is $0$.






analysis functions continuity







share|cite|improve this question











$endgroup$




I am trying to demonstrate that a continuous decreasing function $f: Bbb R → Bbb R$ has a fixed point. I tried to use a function $g(x) = f(x) - x$, which should be a decreasing one, but I don't know how to obtain a point of $g$ that is $0$.






analysis functions continuity




analysis functions continuity






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 3 '18 at 0:41









Jean-Claude Arbaut

14.7k63464




14.7k63464










asked Dec 2 '18 at 23:28









Marco GutiérrezMarco Gutiérrez

153




153








  • 1




    $begingroup$
    Intermediate value theorem.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 23:29










  • $begingroup$
    Welcome to Maths SX! No other hypothesis on $f$?
    $endgroup$
    – Bernard
    Dec 2 '18 at 23:32










  • $begingroup$
    @Bernard What for?
    $endgroup$
    – Did
    Dec 2 '18 at 23:53














  • 1




    $begingroup$
    Intermediate value theorem.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 23:29










  • $begingroup$
    Welcome to Maths SX! No other hypothesis on $f$?
    $endgroup$
    – Bernard
    Dec 2 '18 at 23:32










  • $begingroup$
    @Bernard What for?
    $endgroup$
    – Did
    Dec 2 '18 at 23:53








1




1




$begingroup$
Intermediate value theorem.
$endgroup$
– T. Bongers
Dec 2 '18 at 23:29




$begingroup$
Intermediate value theorem.
$endgroup$
– T. Bongers
Dec 2 '18 at 23:29












$begingroup$
Welcome to Maths SX! No other hypothesis on $f$?
$endgroup$
– Bernard
Dec 2 '18 at 23:32




$begingroup$
Welcome to Maths SX! No other hypothesis on $f$?
$endgroup$
– Bernard
Dec 2 '18 at 23:32












$begingroup$
@Bernard What for?
$endgroup$
– Did
Dec 2 '18 at 23:53




$begingroup$
@Bernard What for?
$endgroup$
– Did
Dec 2 '18 at 23:53










2 Answers
2






active

oldest

votes


















2












$begingroup$

$g(x) geq f(0)-x$ for $x<0$ so $g(x) to infty$ as $x to -infty$. Similarly, $g(x) leq f(0)-x to -infty$ as $x to infty$. By IVP we get $g(x)=x$ for some $x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your example is not sound. For $g(x)=-frac 12 x$ we get $g(x)ge g(0)-ximplies -frac 12 x ge -x implies xge 0$, which is a contradiction with $x<0$.
    $endgroup$
    – I like Serena
    Dec 2 '18 at 23:38










  • $begingroup$
    @IlikeSerena Did you read the question properly? If $g(x)=-frac 1 2 x$ then $f(x)=frac 1 2 x$ but it is given that $f$ is decreasing.
    $endgroup$
    – Kavi Rama Murthy
    Dec 2 '18 at 23:40












  • $begingroup$
    Ah okay. Still, with $g(x)=x$, we don't have $f(x)=x$. Did you intend $g(x)=0$?
    $endgroup$
    – I like Serena
    Dec 2 '18 at 23:44












  • $begingroup$
    @IlikeSerena I really don't understand what you are saying. I am not defining any function. $f$ and $g$ are both given, We cannot have $g(x)=x$ because of the same reason I gave earlier.
    $endgroup$
    – Kavi Rama Murthy
    Dec 2 '18 at 23:52










  • $begingroup$
    I'm saying that you have a typo in your answer, which makes it confusing.
    $endgroup$
    – I like Serena
    Dec 3 '18 at 0:28



















1












$begingroup$

Introducing the fhe function $g:xmapsto f(x)-x$ is not necessary. Instead, consider $a=f(0)$.




  • If $a=0$, we are done.

  • If $a>0$, note that $f(a)leqslant f(0)=a$ since $a>0$ and $f$ is nonincreasing on $[0,a]$. Thus, $f(0)>0$ and $f(a)leqslant a$. By the intermediate value theorem, there exists some $x$ in $(0,a]$ such that $f(x)=x$.

  • If $a<0$, note that $f(a)geqslant f(0)=a$ since $a<0$ and $f$ is nonincreasing on $[a,0]$. Thus, $f(0)<0$ and $f(a)geqslant a$. By the intermediate value theorem, there exists some $x$ in $[a,0)$ such that $f(x)=x$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are implicitly using $g(x)=f(x)-x$. Otherwise, from $a=f(0)>0$ and $f(a)le a$, the IVT would only tell you $f$ takes all values in $[f(a),f(0)]$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:38













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

$g(x) geq f(0)-x$ for $x<0$ so $g(x) to infty$ as $x to -infty$. Similarly, $g(x) leq f(0)-x to -infty$ as $x to infty$. By IVP we get $g(x)=x$ for some $x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your example is not sound. For $g(x)=-frac 12 x$ we get $g(x)ge g(0)-ximplies -frac 12 x ge -x implies xge 0$, which is a contradiction with $x<0$.
    $endgroup$
    – I like Serena
    Dec 2 '18 at 23:38










  • $begingroup$
    @IlikeSerena Did you read the question properly? If $g(x)=-frac 1 2 x$ then $f(x)=frac 1 2 x$ but it is given that $f$ is decreasing.
    $endgroup$
    – Kavi Rama Murthy
    Dec 2 '18 at 23:40












  • $begingroup$
    Ah okay. Still, with $g(x)=x$, we don't have $f(x)=x$. Did you intend $g(x)=0$?
    $endgroup$
    – I like Serena
    Dec 2 '18 at 23:44












  • $begingroup$
    @IlikeSerena I really don't understand what you are saying. I am not defining any function. $f$ and $g$ are both given, We cannot have $g(x)=x$ because of the same reason I gave earlier.
    $endgroup$
    – Kavi Rama Murthy
    Dec 2 '18 at 23:52










  • $begingroup$
    I'm saying that you have a typo in your answer, which makes it confusing.
    $endgroup$
    – I like Serena
    Dec 3 '18 at 0:28
















2












$begingroup$

$g(x) geq f(0)-x$ for $x<0$ so $g(x) to infty$ as $x to -infty$. Similarly, $g(x) leq f(0)-x to -infty$ as $x to infty$. By IVP we get $g(x)=x$ for some $x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your example is not sound. For $g(x)=-frac 12 x$ we get $g(x)ge g(0)-ximplies -frac 12 x ge -x implies xge 0$, which is a contradiction with $x<0$.
    $endgroup$
    – I like Serena
    Dec 2 '18 at 23:38










  • $begingroup$
    @IlikeSerena Did you read the question properly? If $g(x)=-frac 1 2 x$ then $f(x)=frac 1 2 x$ but it is given that $f$ is decreasing.
    $endgroup$
    – Kavi Rama Murthy
    Dec 2 '18 at 23:40












  • $begingroup$
    Ah okay. Still, with $g(x)=x$, we don't have $f(x)=x$. Did you intend $g(x)=0$?
    $endgroup$
    – I like Serena
    Dec 2 '18 at 23:44












  • $begingroup$
    @IlikeSerena I really don't understand what you are saying. I am not defining any function. $f$ and $g$ are both given, We cannot have $g(x)=x$ because of the same reason I gave earlier.
    $endgroup$
    – Kavi Rama Murthy
    Dec 2 '18 at 23:52










  • $begingroup$
    I'm saying that you have a typo in your answer, which makes it confusing.
    $endgroup$
    – I like Serena
    Dec 3 '18 at 0:28














2












2








2





$begingroup$

$g(x) geq f(0)-x$ for $x<0$ so $g(x) to infty$ as $x to -infty$. Similarly, $g(x) leq f(0)-x to -infty$ as $x to infty$. By IVP we get $g(x)=x$ for some $x$.






share|cite|improve this answer











$endgroup$



$g(x) geq f(0)-x$ for $x<0$ so $g(x) to infty$ as $x to -infty$. Similarly, $g(x) leq f(0)-x to -infty$ as $x to infty$. By IVP we get $g(x)=x$ for some $x$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 0:30

























answered Dec 2 '18 at 23:33









Kavi Rama MurthyKavi Rama Murthy

53.9k32055




53.9k32055












  • $begingroup$
    Your example is not sound. For $g(x)=-frac 12 x$ we get $g(x)ge g(0)-ximplies -frac 12 x ge -x implies xge 0$, which is a contradiction with $x<0$.
    $endgroup$
    – I like Serena
    Dec 2 '18 at 23:38










  • $begingroup$
    @IlikeSerena Did you read the question properly? If $g(x)=-frac 1 2 x$ then $f(x)=frac 1 2 x$ but it is given that $f$ is decreasing.
    $endgroup$
    – Kavi Rama Murthy
    Dec 2 '18 at 23:40












  • $begingroup$
    Ah okay. Still, with $g(x)=x$, we don't have $f(x)=x$. Did you intend $g(x)=0$?
    $endgroup$
    – I like Serena
    Dec 2 '18 at 23:44












  • $begingroup$
    @IlikeSerena I really don't understand what you are saying. I am not defining any function. $f$ and $g$ are both given, We cannot have $g(x)=x$ because of the same reason I gave earlier.
    $endgroup$
    – Kavi Rama Murthy
    Dec 2 '18 at 23:52










  • $begingroup$
    I'm saying that you have a typo in your answer, which makes it confusing.
    $endgroup$
    – I like Serena
    Dec 3 '18 at 0:28


















  • $begingroup$
    Your example is not sound. For $g(x)=-frac 12 x$ we get $g(x)ge g(0)-ximplies -frac 12 x ge -x implies xge 0$, which is a contradiction with $x<0$.
    $endgroup$
    – I like Serena
    Dec 2 '18 at 23:38










  • $begingroup$
    @IlikeSerena Did you read the question properly? If $g(x)=-frac 1 2 x$ then $f(x)=frac 1 2 x$ but it is given that $f$ is decreasing.
    $endgroup$
    – Kavi Rama Murthy
    Dec 2 '18 at 23:40












  • $begingroup$
    Ah okay. Still, with $g(x)=x$, we don't have $f(x)=x$. Did you intend $g(x)=0$?
    $endgroup$
    – I like Serena
    Dec 2 '18 at 23:44












  • $begingroup$
    @IlikeSerena I really don't understand what you are saying. I am not defining any function. $f$ and $g$ are both given, We cannot have $g(x)=x$ because of the same reason I gave earlier.
    $endgroup$
    – Kavi Rama Murthy
    Dec 2 '18 at 23:52










  • $begingroup$
    I'm saying that you have a typo in your answer, which makes it confusing.
    $endgroup$
    – I like Serena
    Dec 3 '18 at 0:28
















$begingroup$
Your example is not sound. For $g(x)=-frac 12 x$ we get $g(x)ge g(0)-ximplies -frac 12 x ge -x implies xge 0$, which is a contradiction with $x<0$.
$endgroup$
– I like Serena
Dec 2 '18 at 23:38




$begingroup$
Your example is not sound. For $g(x)=-frac 12 x$ we get $g(x)ge g(0)-ximplies -frac 12 x ge -x implies xge 0$, which is a contradiction with $x<0$.
$endgroup$
– I like Serena
Dec 2 '18 at 23:38












$begingroup$
@IlikeSerena Did you read the question properly? If $g(x)=-frac 1 2 x$ then $f(x)=frac 1 2 x$ but it is given that $f$ is decreasing.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:40






$begingroup$
@IlikeSerena Did you read the question properly? If $g(x)=-frac 1 2 x$ then $f(x)=frac 1 2 x$ but it is given that $f$ is decreasing.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:40














$begingroup$
Ah okay. Still, with $g(x)=x$, we don't have $f(x)=x$. Did you intend $g(x)=0$?
$endgroup$
– I like Serena
Dec 2 '18 at 23:44






$begingroup$
Ah okay. Still, with $g(x)=x$, we don't have $f(x)=x$. Did you intend $g(x)=0$?
$endgroup$
– I like Serena
Dec 2 '18 at 23:44














$begingroup$
@IlikeSerena I really don't understand what you are saying. I am not defining any function. $f$ and $g$ are both given, We cannot have $g(x)=x$ because of the same reason I gave earlier.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:52




$begingroup$
@IlikeSerena I really don't understand what you are saying. I am not defining any function. $f$ and $g$ are both given, We cannot have $g(x)=x$ because of the same reason I gave earlier.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:52












$begingroup$
I'm saying that you have a typo in your answer, which makes it confusing.
$endgroup$
– I like Serena
Dec 3 '18 at 0:28




$begingroup$
I'm saying that you have a typo in your answer, which makes it confusing.
$endgroup$
– I like Serena
Dec 3 '18 at 0:28











1












$begingroup$

Introducing the fhe function $g:xmapsto f(x)-x$ is not necessary. Instead, consider $a=f(0)$.




  • If $a=0$, we are done.

  • If $a>0$, note that $f(a)leqslant f(0)=a$ since $a>0$ and $f$ is nonincreasing on $[0,a]$. Thus, $f(0)>0$ and $f(a)leqslant a$. By the intermediate value theorem, there exists some $x$ in $(0,a]$ such that $f(x)=x$.

  • If $a<0$, note that $f(a)geqslant f(0)=a$ since $a<0$ and $f$ is nonincreasing on $[a,0]$. Thus, $f(0)<0$ and $f(a)geqslant a$. By the intermediate value theorem, there exists some $x$ in $[a,0)$ such that $f(x)=x$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are implicitly using $g(x)=f(x)-x$. Otherwise, from $a=f(0)>0$ and $f(a)le a$, the IVT would only tell you $f$ takes all values in $[f(a),f(0)]$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:38


















1












$begingroup$

Introducing the fhe function $g:xmapsto f(x)-x$ is not necessary. Instead, consider $a=f(0)$.




  • If $a=0$, we are done.

  • If $a>0$, note that $f(a)leqslant f(0)=a$ since $a>0$ and $f$ is nonincreasing on $[0,a]$. Thus, $f(0)>0$ and $f(a)leqslant a$. By the intermediate value theorem, there exists some $x$ in $(0,a]$ such that $f(x)=x$.

  • If $a<0$, note that $f(a)geqslant f(0)=a$ since $a<0$ and $f$ is nonincreasing on $[a,0]$. Thus, $f(0)<0$ and $f(a)geqslant a$. By the intermediate value theorem, there exists some $x$ in $[a,0)$ such that $f(x)=x$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are implicitly using $g(x)=f(x)-x$. Otherwise, from $a=f(0)>0$ and $f(a)le a$, the IVT would only tell you $f$ takes all values in $[f(a),f(0)]$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:38
















1












1








1





$begingroup$

Introducing the fhe function $g:xmapsto f(x)-x$ is not necessary. Instead, consider $a=f(0)$.




  • If $a=0$, we are done.

  • If $a>0$, note that $f(a)leqslant f(0)=a$ since $a>0$ and $f$ is nonincreasing on $[0,a]$. Thus, $f(0)>0$ and $f(a)leqslant a$. By the intermediate value theorem, there exists some $x$ in $(0,a]$ such that $f(x)=x$.

  • If $a<0$, note that $f(a)geqslant f(0)=a$ since $a<0$ and $f$ is nonincreasing on $[a,0]$. Thus, $f(0)<0$ and $f(a)geqslant a$. By the intermediate value theorem, there exists some $x$ in $[a,0)$ such that $f(x)=x$.






share|cite|improve this answer









$endgroup$



Introducing the fhe function $g:xmapsto f(x)-x$ is not necessary. Instead, consider $a=f(0)$.




  • If $a=0$, we are done.

  • If $a>0$, note that $f(a)leqslant f(0)=a$ since $a>0$ and $f$ is nonincreasing on $[0,a]$. Thus, $f(0)>0$ and $f(a)leqslant a$. By the intermediate value theorem, there exists some $x$ in $(0,a]$ such that $f(x)=x$.

  • If $a<0$, note that $f(a)geqslant f(0)=a$ since $a<0$ and $f$ is nonincreasing on $[a,0]$. Thus, $f(0)<0$ and $f(a)geqslant a$. By the intermediate value theorem, there exists some $x$ in $[a,0)$ such that $f(x)=x$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 23:53









DidDid

247k23222457




247k23222457












  • $begingroup$
    You are implicitly using $g(x)=f(x)-x$. Otherwise, from $a=f(0)>0$ and $f(a)le a$, the IVT would only tell you $f$ takes all values in $[f(a),f(0)]$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:38




















  • $begingroup$
    You are implicitly using $g(x)=f(x)-x$. Otherwise, from $a=f(0)>0$ and $f(a)le a$, the IVT would only tell you $f$ takes all values in $[f(a),f(0)]$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:38


















$begingroup$
You are implicitly using $g(x)=f(x)-x$. Otherwise, from $a=f(0)>0$ and $f(a)le a$, the IVT would only tell you $f$ takes all values in $[f(a),f(0)]$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:38






$begingroup$
You are implicitly using $g(x)=f(x)-x$. Otherwise, from $a=f(0)>0$ and $f(a)le a$, the IVT would only tell you $f$ takes all values in $[f(a),f(0)]$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:38




















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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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