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
edited Dec 3 '18 at 0:41
Jean-Claude Arbaut
14.7k63464
asked Dec 2 '18 at 23:28
Marco GutiérrezMarco Gutiérrez
153
$endgroup$
add a comment |
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
edited Dec 3 '18 at 0:41
Jean-Claude Arbaut
14.7k63464
asked Dec 2 '18 at 23:28
Marco GutiérrezMarco Gutiérrez
153
$endgroup$
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
add a comment |
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
edited Dec 3 '18 at 0:41
Jean-Claude Arbaut
14.7k63464
asked Dec 2 '18 at 23:28
Marco GutiérrezMarco Gutiérrez
153
$endgroup$
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
edited Dec 3 '18 at 0:41
Jean-Claude Arbaut
14.7k63464
asked Dec 2 '18 at 23:28
Marco GutiérrezMarco Gutiérrez
153
edited Dec 3 '18 at 0:41
Jean-Claude Arbaut
14.7k63464
asked Dec 2 '18 at 23:28
Marco GutiérrezMarco Gutiérrez
153
edited Dec 3 '18 at 0:41
Jean-Claude Arbaut
14.7k63464
edited Dec 3 '18 at 0:41
Jean-Claude Arbaut
14.7k63464
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
asked Dec 2 '18 at 23:28
Marco GutiérrezMarco Gutiérrez
153
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
add a comment |
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
add a comment |
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 $x$.
answered Dec 2 '18 at 23:33
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
$begingroup$
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
|
show 1 more comment
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 $f(x)=x$.
answered Dec 2 '18 at 23:53
DidDid
247k23222457
$endgroup$
$begingroup$
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
add a comment |
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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 $x$.
answered Dec 2 '18 at 23:33
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
$begingroup$
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
|
show 1 more comment
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 $x$.
answered Dec 2 '18 at 23:33
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
$begingroup$
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
|
show 1 more comment
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 $x$.
answered Dec 2 '18 at 23:33
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
$g(x) geq f(0)-x$ 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 $x$.
answered Dec 2 '18 at 23:33
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
edited Dec 3 '18 at 0:30
answered Dec 2 '18 at 23:33
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
answered Dec 2 '18 at 23:33
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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 $f(x)=x$.
answered Dec 2 '18 at 23:53
DidDid
247k23222457
$endgroup$
$begingroup$
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
add a comment |
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 $f(x)=x$.
answered Dec 2 '18 at 23:53
DidDid
247k23222457
$endgroup$
$begingroup$
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
add a comment |
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 $f(x)=x$.
answered Dec 2 '18 at 23:53
DidDid
247k23222457
$endgroup$
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 $f(x)=x$.
answered Dec 2 '18 at 23:53
DidDid
247k23222457
answered Dec 2 '18 at 23:53
DidDid
247k23222457
answered Dec 2 '18 at 23:53
DidDid
247k23222457
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
add a comment |
$begingroup$
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
add a comment |
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1
$begingroup$
Intermediate value theorem.
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– 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