If $f'(x)not = 1$ for all real numbers $x$, then $f$ has at most one fixed point [duplicate]
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This question already has an answer here:
Show that $f$ has at most one fixed point
3 answers
A number $a$ is called a fixed point of a function $f$ if $f(a)=a$.Prove that if $f'(x)not = 1$ for all real numbers $x$, then $f$ has at most one fixed point.
This is an exercise in Stewart's calculus textbook.
Do we need to suppose $f$ is continuous? Or the assumption, $f'(x) not = 1$ for all real numbers $x$, guarantees that $f$ is continuous? What if $f'$ is not defined at some points? For example, when $f$ is an increasing piecewise function which has infinitely many intersections with the line $y=x$, $f$ has infinitely many fixed points and $f'(x) not =1$ for all real $x$.
real-analysis calculus analysis proof-writing
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marked as duplicate by Martin R, Leucippus, Lord Shark the Unknown, Chinnapparaj R, KReiser Dec 25 '18 at 7:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Show that $f$ has at most one fixed point
3 answers
A number $a$ is called a fixed point of a function $f$ if $f(a)=a$.Prove that if $f'(x)not = 1$ for all real numbers $x$, then $f$ has at most one fixed point.
This is an exercise in Stewart's calculus textbook.
Do we need to suppose $f$ is continuous? Or the assumption, $f'(x) not = 1$ for all real numbers $x$, guarantees that $f$ is continuous? What if $f'$ is not defined at some points? For example, when $f$ is an increasing piecewise function which has infinitely many intersections with the line $y=x$, $f$ has infinitely many fixed points and $f'(x) not =1$ for all real $x$.
real-analysis calculus analysis proof-writing
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marked as duplicate by Martin R, Leucippus, Lord Shark the Unknown, Chinnapparaj R, KReiser Dec 25 '18 at 7:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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You're right, this isn't true if $f$ isn't differentiable. Continuity isn't even enough - a piecewise function that alternates between slopes $1/2$ and $2$ can intersect the diagonal infinitely often.
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– dbx
Dec 24 '18 at 21:10
2
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Mean Value Theorem
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– Will Jagy
Dec 24 '18 at 21:17
add a comment |
$begingroup$
This question already has an answer here:
Show that $f$ has at most one fixed point
3 answers
A number $a$ is called a fixed point of a function $f$ if $f(a)=a$.Prove that if $f'(x)not = 1$ for all real numbers $x$, then $f$ has at most one fixed point.
This is an exercise in Stewart's calculus textbook.
Do we need to suppose $f$ is continuous? Or the assumption, $f'(x) not = 1$ for all real numbers $x$, guarantees that $f$ is continuous? What if $f'$ is not defined at some points? For example, when $f$ is an increasing piecewise function which has infinitely many intersections with the line $y=x$, $f$ has infinitely many fixed points and $f'(x) not =1$ for all real $x$.
real-analysis calculus analysis proof-writing
$endgroup$
This question already has an answer here:
Show that $f$ has at most one fixed point
3 answers
A number $a$ is called a fixed point of a function $f$ if $f(a)=a$.Prove that if $f'(x)not = 1$ for all real numbers $x$, then $f$ has at most one fixed point.
This is an exercise in Stewart's calculus textbook.
Do we need to suppose $f$ is continuous? Or the assumption, $f'(x) not = 1$ for all real numbers $x$, guarantees that $f$ is continuous? What if $f'$ is not defined at some points? For example, when $f$ is an increasing piecewise function which has infinitely many intersections with the line $y=x$, $f$ has infinitely many fixed points and $f'(x) not =1$ for all real $x$.
This question already has an answer here:
Show that $f$ has at most one fixed point
3 answers
real-analysis calculus analysis proof-writing
real-analysis calculus analysis proof-writing
edited Dec 24 '18 at 21:14
user398843
asked Dec 24 '18 at 21:07
user398843user398843
650216
650216
marked as duplicate by Martin R, Leucippus, Lord Shark the Unknown, Chinnapparaj R, KReiser Dec 25 '18 at 7:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Leucippus, Lord Shark the Unknown, Chinnapparaj R, KReiser Dec 25 '18 at 7:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
You're right, this isn't true if $f$ isn't differentiable. Continuity isn't even enough - a piecewise function that alternates between slopes $1/2$ and $2$ can intersect the diagonal infinitely often.
$endgroup$
– dbx
Dec 24 '18 at 21:10
2
$begingroup$
Mean Value Theorem
$endgroup$
– Will Jagy
Dec 24 '18 at 21:17
add a comment |
1
$begingroup$
You're right, this isn't true if $f$ isn't differentiable. Continuity isn't even enough - a piecewise function that alternates between slopes $1/2$ and $2$ can intersect the diagonal infinitely often.
$endgroup$
– dbx
Dec 24 '18 at 21:10
2
$begingroup$
Mean Value Theorem
$endgroup$
– Will Jagy
Dec 24 '18 at 21:17
1
1
$begingroup$
You're right, this isn't true if $f$ isn't differentiable. Continuity isn't even enough - a piecewise function that alternates between slopes $1/2$ and $2$ can intersect the diagonal infinitely often.
$endgroup$
– dbx
Dec 24 '18 at 21:10
$begingroup$
You're right, this isn't true if $f$ isn't differentiable. Continuity isn't even enough - a piecewise function that alternates between slopes $1/2$ and $2$ can intersect the diagonal infinitely often.
$endgroup$
– dbx
Dec 24 '18 at 21:10
2
2
$begingroup$
Mean Value Theorem
$endgroup$
– Will Jagy
Dec 24 '18 at 21:17
$begingroup$
Mean Value Theorem
$endgroup$
– Will Jagy
Dec 24 '18 at 21:17
add a comment |
5 Answers
5
active
oldest
votes
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Suppose $f(x)$ has two fixed points $a$ and $b$:
$f(a) = a, ; f(b) = b; tag 1$
we may without loss of generality assume that
$a < b; tag 2$
then by the mean value theorem there exists $c in (a, b)$ with
$b - a = f(b) - f(a) = f'(c)(b - a), tag 3$
and we thus find
$f'(c) = dfrac{b - a}{b - a} = 1, tag 4$
which contradicts
$f'(x) ne 1, forall x in Bbb R; tag 5$
therefore, $f(x)$ has at most one fixed point.
A little commentary on our OP user398843's closing questions: the assertion (5) precludes the possibility that $f'(x)$ is not defined for some $x in Bbb R$; saying
$f'(x) ne 1$ means that $f'(x)$ exists, that is, is defined, and that it's value is not equal to one; the statement is meaningless if $f'(x)$ is undefined. Since differentiability at a point implies continuity at that point, $f(x)$ is continuous. The key lies in the title, "$f'(x) ne 1$ for all real numbers $x$ . . ."
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1
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You last paragraph is exactly what I wanted to know.
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– user398843
Dec 24 '18 at 21:26
1
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@user398843: glad to help out, and thanks for the "acceptance". Cheers!
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– Robert Lewis
Dec 24 '18 at 21:27
add a comment |
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Let $x,y$ be two distinct fixed points of a differentiable function $f:(a,b)toBbb R$ with $x,yin(a,b)$. Then by a Lagrange MVT we get $$1=frac{x-y}{x-y}=frac{f(x)-f(y)}{x-y}=f'(c)$$ for some $c$ between $x$ and $y$. A contradiction.
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add a comment |
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If $f$ has two fixed points, $a$ and $b$, then $frac{f(b)-f(a)}{b-a}=1$. Now apply the mean value theorem.
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add a comment |
$begingroup$
Let $$g(x)=f(x)-x$$
assume there are two fixed points $a$ and $b$ for $f$.
$g$ is differentiable at $[a,b]$ and $g(a)=g(b)=0$, thus by Rolle's theorem
there exists $c$ such that $$g'(c)=0=f'(c)-1$$
this contradicts the hypothese $f'(c)ne 1$.
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add a comment |
$begingroup$
$f'(x)not =1$ , $x in mathbb{R}.$
1) $f'(x)>1$, $xin mathbb{R}$, or
2)$f'(x)<1$, $x in mathbb{R}.$
$F(x):=f(x)-x;$
1)Let $f'(x)>1$.
$F'(x)=f'(x)-1>0.$
$F$ is strictly increasing, i.e.
$F$ has at most one zero.
2) Let $f'(x) <1$.
$F'(x)=f'(x)-1 <0$, $F$ is strictly decreasing, has at most one zero.
3) See comment by Jens Schwaiger.Thanks .
Assume $f'(x) >1$ for some $x$, and $f'(y)<1$ for some $y$, say, $x <y$, then by Darboux' Intermediate Value Property, there is a $z in (x,y)$ s.t. $f'(z)=1$. Contradiction.
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It seems that you have still to exclude the case that $f'(x)<1$ and $f'(y)>1$ for some $x,y$. This can be done by using the fact that the derivative of a function has the Darboux property.
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– Jens Schwaiger
Dec 24 '18 at 23:07
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Jens.Yes, thanks.Assumed f' is continuos(tacitly).
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– Peter Szilas
Dec 25 '18 at 7:21
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Jens.Hopefully fixed.Greetings.
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– Peter Szilas
Dec 25 '18 at 7:55
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose $f(x)$ has two fixed points $a$ and $b$:
$f(a) = a, ; f(b) = b; tag 1$
we may without loss of generality assume that
$a < b; tag 2$
then by the mean value theorem there exists $c in (a, b)$ with
$b - a = f(b) - f(a) = f'(c)(b - a), tag 3$
and we thus find
$f'(c) = dfrac{b - a}{b - a} = 1, tag 4$
which contradicts
$f'(x) ne 1, forall x in Bbb R; tag 5$
therefore, $f(x)$ has at most one fixed point.
A little commentary on our OP user398843's closing questions: the assertion (5) precludes the possibility that $f'(x)$ is not defined for some $x in Bbb R$; saying
$f'(x) ne 1$ means that $f'(x)$ exists, that is, is defined, and that it's value is not equal to one; the statement is meaningless if $f'(x)$ is undefined. Since differentiability at a point implies continuity at that point, $f(x)$ is continuous. The key lies in the title, "$f'(x) ne 1$ for all real numbers $x$ . . ."
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1
$begingroup$
You last paragraph is exactly what I wanted to know.
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– user398843
Dec 24 '18 at 21:26
1
$begingroup$
@user398843: glad to help out, and thanks for the "acceptance". Cheers!
$endgroup$
– Robert Lewis
Dec 24 '18 at 21:27
add a comment |
$begingroup$
Suppose $f(x)$ has two fixed points $a$ and $b$:
$f(a) = a, ; f(b) = b; tag 1$
we may without loss of generality assume that
$a < b; tag 2$
then by the mean value theorem there exists $c in (a, b)$ with
$b - a = f(b) - f(a) = f'(c)(b - a), tag 3$
and we thus find
$f'(c) = dfrac{b - a}{b - a} = 1, tag 4$
which contradicts
$f'(x) ne 1, forall x in Bbb R; tag 5$
therefore, $f(x)$ has at most one fixed point.
A little commentary on our OP user398843's closing questions: the assertion (5) precludes the possibility that $f'(x)$ is not defined for some $x in Bbb R$; saying
$f'(x) ne 1$ means that $f'(x)$ exists, that is, is defined, and that it's value is not equal to one; the statement is meaningless if $f'(x)$ is undefined. Since differentiability at a point implies continuity at that point, $f(x)$ is continuous. The key lies in the title, "$f'(x) ne 1$ for all real numbers $x$ . . ."
$endgroup$
1
$begingroup$
You last paragraph is exactly what I wanted to know.
$endgroup$
– user398843
Dec 24 '18 at 21:26
1
$begingroup$
@user398843: glad to help out, and thanks for the "acceptance". Cheers!
$endgroup$
– Robert Lewis
Dec 24 '18 at 21:27
add a comment |
$begingroup$
Suppose $f(x)$ has two fixed points $a$ and $b$:
$f(a) = a, ; f(b) = b; tag 1$
we may without loss of generality assume that
$a < b; tag 2$
then by the mean value theorem there exists $c in (a, b)$ with
$b - a = f(b) - f(a) = f'(c)(b - a), tag 3$
and we thus find
$f'(c) = dfrac{b - a}{b - a} = 1, tag 4$
which contradicts
$f'(x) ne 1, forall x in Bbb R; tag 5$
therefore, $f(x)$ has at most one fixed point.
A little commentary on our OP user398843's closing questions: the assertion (5) precludes the possibility that $f'(x)$ is not defined for some $x in Bbb R$; saying
$f'(x) ne 1$ means that $f'(x)$ exists, that is, is defined, and that it's value is not equal to one; the statement is meaningless if $f'(x)$ is undefined. Since differentiability at a point implies continuity at that point, $f(x)$ is continuous. The key lies in the title, "$f'(x) ne 1$ for all real numbers $x$ . . ."
$endgroup$
Suppose $f(x)$ has two fixed points $a$ and $b$:
$f(a) = a, ; f(b) = b; tag 1$
we may without loss of generality assume that
$a < b; tag 2$
then by the mean value theorem there exists $c in (a, b)$ with
$b - a = f(b) - f(a) = f'(c)(b - a), tag 3$
and we thus find
$f'(c) = dfrac{b - a}{b - a} = 1, tag 4$
which contradicts
$f'(x) ne 1, forall x in Bbb R; tag 5$
therefore, $f(x)$ has at most one fixed point.
A little commentary on our OP user398843's closing questions: the assertion (5) precludes the possibility that $f'(x)$ is not defined for some $x in Bbb R$; saying
$f'(x) ne 1$ means that $f'(x)$ exists, that is, is defined, and that it's value is not equal to one; the statement is meaningless if $f'(x)$ is undefined. Since differentiability at a point implies continuity at that point, $f(x)$ is continuous. The key lies in the title, "$f'(x) ne 1$ for all real numbers $x$ . . ."
answered Dec 24 '18 at 21:23
Robert LewisRobert Lewis
46.9k23067
46.9k23067
1
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You last paragraph is exactly what I wanted to know.
$endgroup$
– user398843
Dec 24 '18 at 21:26
1
$begingroup$
@user398843: glad to help out, and thanks for the "acceptance". Cheers!
$endgroup$
– Robert Lewis
Dec 24 '18 at 21:27
add a comment |
1
$begingroup$
You last paragraph is exactly what I wanted to know.
$endgroup$
– user398843
Dec 24 '18 at 21:26
1
$begingroup$
@user398843: glad to help out, and thanks for the "acceptance". Cheers!
$endgroup$
– Robert Lewis
Dec 24 '18 at 21:27
1
1
$begingroup$
You last paragraph is exactly what I wanted to know.
$endgroup$
– user398843
Dec 24 '18 at 21:26
$begingroup$
You last paragraph is exactly what I wanted to know.
$endgroup$
– user398843
Dec 24 '18 at 21:26
1
1
$begingroup$
@user398843: glad to help out, and thanks for the "acceptance". Cheers!
$endgroup$
– Robert Lewis
Dec 24 '18 at 21:27
$begingroup$
@user398843: glad to help out, and thanks for the "acceptance". Cheers!
$endgroup$
– Robert Lewis
Dec 24 '18 at 21:27
add a comment |
$begingroup$
Let $x,y$ be two distinct fixed points of a differentiable function $f:(a,b)toBbb R$ with $x,yin(a,b)$. Then by a Lagrange MVT we get $$1=frac{x-y}{x-y}=frac{f(x)-f(y)}{x-y}=f'(c)$$ for some $c$ between $x$ and $y$. A contradiction.
$endgroup$
add a comment |
$begingroup$
Let $x,y$ be two distinct fixed points of a differentiable function $f:(a,b)toBbb R$ with $x,yin(a,b)$. Then by a Lagrange MVT we get $$1=frac{x-y}{x-y}=frac{f(x)-f(y)}{x-y}=f'(c)$$ for some $c$ between $x$ and $y$. A contradiction.
$endgroup$
add a comment |
$begingroup$
Let $x,y$ be two distinct fixed points of a differentiable function $f:(a,b)toBbb R$ with $x,yin(a,b)$. Then by a Lagrange MVT we get $$1=frac{x-y}{x-y}=frac{f(x)-f(y)}{x-y}=f'(c)$$ for some $c$ between $x$ and $y$. A contradiction.
$endgroup$
Let $x,y$ be two distinct fixed points of a differentiable function $f:(a,b)toBbb R$ with $x,yin(a,b)$. Then by a Lagrange MVT we get $$1=frac{x-y}{x-y}=frac{f(x)-f(y)}{x-y}=f'(c)$$ for some $c$ between $x$ and $y$. A contradiction.
answered Dec 24 '18 at 21:19
szw1710szw1710
6,4551223
6,4551223
add a comment |
add a comment |
$begingroup$
If $f$ has two fixed points, $a$ and $b$, then $frac{f(b)-f(a)}{b-a}=1$. Now apply the mean value theorem.
$endgroup$
add a comment |
$begingroup$
If $f$ has two fixed points, $a$ and $b$, then $frac{f(b)-f(a)}{b-a}=1$. Now apply the mean value theorem.
$endgroup$
add a comment |
$begingroup$
If $f$ has two fixed points, $a$ and $b$, then $frac{f(b)-f(a)}{b-a}=1$. Now apply the mean value theorem.
$endgroup$
If $f$ has two fixed points, $a$ and $b$, then $frac{f(b)-f(a)}{b-a}=1$. Now apply the mean value theorem.
answered Dec 24 '18 at 21:20
Chris CusterChris Custer
13.7k3827
13.7k3827
add a comment |
add a comment |
$begingroup$
Let $$g(x)=f(x)-x$$
assume there are two fixed points $a$ and $b$ for $f$.
$g$ is differentiable at $[a,b]$ and $g(a)=g(b)=0$, thus by Rolle's theorem
there exists $c$ such that $$g'(c)=0=f'(c)-1$$
this contradicts the hypothese $f'(c)ne 1$.
$endgroup$
add a comment |
$begingroup$
Let $$g(x)=f(x)-x$$
assume there are two fixed points $a$ and $b$ for $f$.
$g$ is differentiable at $[a,b]$ and $g(a)=g(b)=0$, thus by Rolle's theorem
there exists $c$ such that $$g'(c)=0=f'(c)-1$$
this contradicts the hypothese $f'(c)ne 1$.
$endgroup$
add a comment |
$begingroup$
Let $$g(x)=f(x)-x$$
assume there are two fixed points $a$ and $b$ for $f$.
$g$ is differentiable at $[a,b]$ and $g(a)=g(b)=0$, thus by Rolle's theorem
there exists $c$ such that $$g'(c)=0=f'(c)-1$$
this contradicts the hypothese $f'(c)ne 1$.
$endgroup$
Let $$g(x)=f(x)-x$$
assume there are two fixed points $a$ and $b$ for $f$.
$g$ is differentiable at $[a,b]$ and $g(a)=g(b)=0$, thus by Rolle's theorem
there exists $c$ such that $$g'(c)=0=f'(c)-1$$
this contradicts the hypothese $f'(c)ne 1$.
edited Dec 24 '18 at 21:38
answered Dec 24 '18 at 21:20
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
add a comment |
add a comment |
$begingroup$
$f'(x)not =1$ , $x in mathbb{R}.$
1) $f'(x)>1$, $xin mathbb{R}$, or
2)$f'(x)<1$, $x in mathbb{R}.$
$F(x):=f(x)-x;$
1)Let $f'(x)>1$.
$F'(x)=f'(x)-1>0.$
$F$ is strictly increasing, i.e.
$F$ has at most one zero.
2) Let $f'(x) <1$.
$F'(x)=f'(x)-1 <0$, $F$ is strictly decreasing, has at most one zero.
3) See comment by Jens Schwaiger.Thanks .
Assume $f'(x) >1$ for some $x$, and $f'(y)<1$ for some $y$, say, $x <y$, then by Darboux' Intermediate Value Property, there is a $z in (x,y)$ s.t. $f'(z)=1$. Contradiction.
$endgroup$
$begingroup$
It seems that you have still to exclude the case that $f'(x)<1$ and $f'(y)>1$ for some $x,y$. This can be done by using the fact that the derivative of a function has the Darboux property.
$endgroup$
– Jens Schwaiger
Dec 24 '18 at 23:07
$begingroup$
Jens.Yes, thanks.Assumed f' is continuos(tacitly).
$endgroup$
– Peter Szilas
Dec 25 '18 at 7:21
$begingroup$
Jens.Hopefully fixed.Greetings.
$endgroup$
– Peter Szilas
Dec 25 '18 at 7:55
add a comment |
$begingroup$
$f'(x)not =1$ , $x in mathbb{R}.$
1) $f'(x)>1$, $xin mathbb{R}$, or
2)$f'(x)<1$, $x in mathbb{R}.$
$F(x):=f(x)-x;$
1)Let $f'(x)>1$.
$F'(x)=f'(x)-1>0.$
$F$ is strictly increasing, i.e.
$F$ has at most one zero.
2) Let $f'(x) <1$.
$F'(x)=f'(x)-1 <0$, $F$ is strictly decreasing, has at most one zero.
3) See comment by Jens Schwaiger.Thanks .
Assume $f'(x) >1$ for some $x$, and $f'(y)<1$ for some $y$, say, $x <y$, then by Darboux' Intermediate Value Property, there is a $z in (x,y)$ s.t. $f'(z)=1$. Contradiction.
$endgroup$
$begingroup$
It seems that you have still to exclude the case that $f'(x)<1$ and $f'(y)>1$ for some $x,y$. This can be done by using the fact that the derivative of a function has the Darboux property.
$endgroup$
– Jens Schwaiger
Dec 24 '18 at 23:07
$begingroup$
Jens.Yes, thanks.Assumed f' is continuos(tacitly).
$endgroup$
– Peter Szilas
Dec 25 '18 at 7:21
$begingroup$
Jens.Hopefully fixed.Greetings.
$endgroup$
– Peter Szilas
Dec 25 '18 at 7:55
add a comment |
$begingroup$
$f'(x)not =1$ , $x in mathbb{R}.$
1) $f'(x)>1$, $xin mathbb{R}$, or
2)$f'(x)<1$, $x in mathbb{R}.$
$F(x):=f(x)-x;$
1)Let $f'(x)>1$.
$F'(x)=f'(x)-1>0.$
$F$ is strictly increasing, i.e.
$F$ has at most one zero.
2) Let $f'(x) <1$.
$F'(x)=f'(x)-1 <0$, $F$ is strictly decreasing, has at most one zero.
3) See comment by Jens Schwaiger.Thanks .
Assume $f'(x) >1$ for some $x$, and $f'(y)<1$ for some $y$, say, $x <y$, then by Darboux' Intermediate Value Property, there is a $z in (x,y)$ s.t. $f'(z)=1$. Contradiction.
$endgroup$
$f'(x)not =1$ , $x in mathbb{R}.$
1) $f'(x)>1$, $xin mathbb{R}$, or
2)$f'(x)<1$, $x in mathbb{R}.$
$F(x):=f(x)-x;$
1)Let $f'(x)>1$.
$F'(x)=f'(x)-1>0.$
$F$ is strictly increasing, i.e.
$F$ has at most one zero.
2) Let $f'(x) <1$.
$F'(x)=f'(x)-1 <0$, $F$ is strictly decreasing, has at most one zero.
3) See comment by Jens Schwaiger.Thanks .
Assume $f'(x) >1$ for some $x$, and $f'(y)<1$ for some $y$, say, $x <y$, then by Darboux' Intermediate Value Property, there is a $z in (x,y)$ s.t. $f'(z)=1$. Contradiction.
edited Dec 25 '18 at 7:38
answered Dec 24 '18 at 21:57
Peter SzilasPeter Szilas
11.4k2822
11.4k2822
$begingroup$
It seems that you have still to exclude the case that $f'(x)<1$ and $f'(y)>1$ for some $x,y$. This can be done by using the fact that the derivative of a function has the Darboux property.
$endgroup$
– Jens Schwaiger
Dec 24 '18 at 23:07
$begingroup$
Jens.Yes, thanks.Assumed f' is continuos(tacitly).
$endgroup$
– Peter Szilas
Dec 25 '18 at 7:21
$begingroup$
Jens.Hopefully fixed.Greetings.
$endgroup$
– Peter Szilas
Dec 25 '18 at 7:55
add a comment |
$begingroup$
It seems that you have still to exclude the case that $f'(x)<1$ and $f'(y)>1$ for some $x,y$. This can be done by using the fact that the derivative of a function has the Darboux property.
$endgroup$
– Jens Schwaiger
Dec 24 '18 at 23:07
$begingroup$
Jens.Yes, thanks.Assumed f' is continuos(tacitly).
$endgroup$
– Peter Szilas
Dec 25 '18 at 7:21
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Jens.Hopefully fixed.Greetings.
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– Peter Szilas
Dec 25 '18 at 7:55
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It seems that you have still to exclude the case that $f'(x)<1$ and $f'(y)>1$ for some $x,y$. This can be done by using the fact that the derivative of a function has the Darboux property.
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– Jens Schwaiger
Dec 24 '18 at 23:07
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It seems that you have still to exclude the case that $f'(x)<1$ and $f'(y)>1$ for some $x,y$. This can be done by using the fact that the derivative of a function has the Darboux property.
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– Jens Schwaiger
Dec 24 '18 at 23:07
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Jens.Yes, thanks.Assumed f' is continuos(tacitly).
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– Peter Szilas
Dec 25 '18 at 7:21
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Jens.Yes, thanks.Assumed f' is continuos(tacitly).
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– Peter Szilas
Dec 25 '18 at 7:21
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Jens.Hopefully fixed.Greetings.
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– Peter Szilas
Dec 25 '18 at 7:55
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Jens.Hopefully fixed.Greetings.
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– Peter Szilas
Dec 25 '18 at 7:55
add a comment |
1
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You're right, this isn't true if $f$ isn't differentiable. Continuity isn't even enough - a piecewise function that alternates between slopes $1/2$ and $2$ can intersect the diagonal infinitely often.
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– dbx
Dec 24 '18 at 21:10
2
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Mean Value Theorem
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– Will Jagy
Dec 24 '18 at 21:17