Mixing
Does $|f^prime|<1$ imply that $forall_{x,y}|f(x)-f(y)|<|x-y|$?
$begingroup$
I have a task that I think reduces to proving that $f$ is a contraction mapping. We know that $forall_x|f^prime(x)|<1$. Therefore if I could prove that $|f^prime|<1implies|f(x)-f(y)|<|x-y|$ then I think the task would be solved.
I feel this property does hold and that it is also somewhat obvious. Unfortunately, saying that something is obvious is obviously not a valid proof.
I have feeling proving this belongs to an elementary course on analysis, but it is somehow surprising how much have I forgotten from this course... How to prove this property and does it even hold?
real-analysis derivatives norm
asked Jan 21 at 19:04
gaazkamgaazkam
456314
$endgroup$
|
show 1 more comment
$begingroup$
I have a task that I think reduces to proving that $f$ is a contraction mapping. We know that $forall_x|f^prime(x)|<1$. Therefore if I could prove that $|f^prime|<1implies|f(x)-f(y)|<|x-y|$ then I think the task would be solved.
I feel this property does hold and that it is also somewhat obvious. Unfortunately, saying that something is obvious is obviously not a valid proof.
I have feeling proving this belongs to an elementary course on analysis, but it is somehow surprising how much have I forgotten from this course... How to prove this property and does it even hold?
real-analysis derivatives norm
asked Jan 21 at 19:04
gaazkamgaazkam
456314
$endgroup$
1
$begingroup$
Maybe using the mean value theorem helps?
$endgroup$
– Shashi
Jan 21 at 19:07
$begingroup$
What about $f: Bbb R setminus { 0 } to Bbb R$, $f(x) = x/|x|$ ?
$endgroup$
– Martin R
Jan 21 at 19:07
3
$begingroup$
Ans what about $x=y$?
$endgroup$
– Hagen von Eitzen
Jan 21 at 19:09
$begingroup$
@MartinR Is this supposed to be a counterexample? In each interval on which $f$ is continuous my property holds. Let $x,y>0,xneq y$, then $|f(x)-f(y)|=0$ but $|x-y|>0$
$endgroup$
– gaazkam
Jan 21 at 19:11
1
$begingroup$
@gaazkam: Exactly. So stating where $f$ is defined should be the first step.
$endgroup$
– Martin R
Jan 21 at 19:17
|
show 1 more comment
$begingroup$
I have a task that I think reduces to proving that $f$ is a contraction mapping. We know that $forall_x|f^prime(x)|<1$. Therefore if I could prove that $|f^prime|<1implies|f(x)-f(y)|<|x-y|$ then I think the task would be solved.
I feel this property does hold and that it is also somewhat obvious. Unfortunately, saying that something is obvious is obviously not a valid proof.
I have feeling proving this belongs to an elementary course on analysis, but it is somehow surprising how much have I forgotten from this course... How to prove this property and does it even hold?
real-analysis derivatives norm
asked Jan 21 at 19:04
gaazkamgaazkam
456314
$endgroup$
I have a task that I think reduces to proving that $f$ is a contraction mapping. We know that $forall_x|f^prime(x)|<1$. Therefore if I could prove that $|f^prime|<1implies|f(x)-f(y)|<|x-y|$ then I think the task would be solved.
I feel this property does hold and that it is also somewhat obvious. Unfortunately, saying that something is obvious is obviously not a valid proof.
I have feeling proving this belongs to an elementary course on analysis, but it is somehow surprising how much have I forgotten from this course... How to prove this property and does it even hold?
real-analysis derivatives norm
real-analysis derivatives norm
asked Jan 21 at 19:04
gaazkamgaazkam
456314
asked Jan 21 at 19:04
gaazkamgaazkam
456314
asked Jan 21 at 19:04
gaazkamgaazkam
456314
asked Jan 21 at 19:04
gaazkamgaazkam
456314
asked Jan 21 at 19:04
gaazkamgaazkam
456314
456314
1
$begingroup$
Maybe using the mean value theorem helps?
$endgroup$
– Shashi
Jan 21 at 19:07
$begingroup$
What about $f: Bbb R setminus { 0 } to Bbb R$, $f(x) = x/|x|$ ?
$endgroup$
– Martin R
Jan 21 at 19:07
3
$begingroup$
Ans what about $x=y$?
$endgroup$
– Hagen von Eitzen
Jan 21 at 19:09
$begingroup$
@MartinR Is this supposed to be a counterexample? In each interval on which $f$ is continuous my property holds. Let $x,y>0,xneq y$, then $|f(x)-f(y)|=0$ but $|x-y|>0$
$endgroup$
– gaazkam
Jan 21 at 19:11
1
$begingroup$
@gaazkam: Exactly. So stating where $f$ is defined should be the first step.
$endgroup$
– Martin R
Jan 21 at 19:17
|
show 1 more comment
1
$begingroup$
Maybe using the mean value theorem helps?
$endgroup$
– Shashi
Jan 21 at 19:07
$begingroup$
What about $f: Bbb R setminus { 0 } to Bbb R$, $f(x) = x/|x|$ ?
$endgroup$
– Martin R
Jan 21 at 19:07
3
$begingroup$
Ans what about $x=y$?
$endgroup$
– Hagen von Eitzen
Jan 21 at 19:09
$begingroup$
@MartinR Is this supposed to be a counterexample? In each interval on which $f$ is continuous my property holds. Let $x,y>0,xneq y$, then $|f(x)-f(y)|=0$ but $|x-y|>0$
$endgroup$
– gaazkam
Jan 21 at 19:11
1
$begingroup$
@gaazkam: Exactly. So stating where $f$ is defined should be the first step.
$endgroup$
– Martin R
Jan 21 at 19:17
1
1
$begingroup$
Maybe using the mean value theorem helps?
$endgroup$
– Shashi
Jan 21 at 19:07
$begingroup$
Maybe using the mean value theorem helps?
$endgroup$
– Shashi
Jan 21 at 19:07
$begingroup$
What about $f: Bbb R setminus { 0 } to Bbb R$, $f(x) = x/|x|$ ?
$endgroup$
– Martin R
Jan 21 at 19:07
$begingroup$
What about $f: Bbb R setminus { 0 } to Bbb R$, $f(x) = x/|x|$ ?
$endgroup$
– Martin R
Jan 21 at 19:07
3
3
$begingroup$
Ans what about $x=y$?
$endgroup$
– Hagen von Eitzen
Jan 21 at 19:09
$begingroup$
Ans what about $x=y$?
$endgroup$
– Hagen von Eitzen
Jan 21 at 19:09
$begingroup$
@MartinR Is this supposed to be a counterexample? In each interval on which $f$ is continuous my property holds. Let $x,y>0,xneq y$, then $|f(x)-f(y)|=0$ but $|x-y|>0$
$endgroup$
– gaazkam
Jan 21 at 19:11
$begingroup$
@MartinR Is this supposed to be a counterexample? In each interval on which $f$ is continuous my property holds. Let $x,y>0,xneq y$, then $|f(x)-f(y)|=0$ but $|x-y|>0$
$endgroup$
– gaazkam
Jan 21 at 19:11
1
1
$begingroup$
@gaazkam: Exactly. So stating where $f$ is defined should be the first step.
$endgroup$
– Martin R
Jan 21 at 19:17
$begingroup$
@gaazkam: Exactly. So stating where $f$ is defined should be the first step.
$endgroup$
– Martin R
Jan 21 at 19:17
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Suppose that $|f'(x)|<1 ~forall x in (a,b).$
SIMPLER PROOF (thanks to BigbearZzz)
Consider $y<xin [a, b]$.
By the mean value theorem, there exists a number $c in (y,x)$:
$$f'(c) = frac{f(x)-f(y)}{x-y} Rightarrow \
f(x)-f(y) = (x-y)f'(c) Rightarrow\
|f(x)-f(y)| = |x-y||f'(c)|.$$
Since $|f'(c)| <1 $, then:
$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| Rightarrow \
|f(x)-f(y)| < |x-y|.$$
OLD PROOF
Consider $y<xin [a, b]$. Moreover, consider the function $g(x) = x$ (continuous and differentiable everywhere).
By the Cauchy's mean value theorem, there exists a number $c in (y,x)$:
$$left(f(x)-f(y)right)g'(c) = left(g(x)-g(y)right)f'(c) Rightarrow\
left(f(x)-f(y)right) = left(x-yright)f'(c) Rightarrow \
|f(x)-f(y)| = |x-y||f'(c)|.$$
Since $|f'(c)| <1 $, then:
$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| Rightarrow \
|f(x)-f(y)| < |x-y|.$$
answered Jan 21 at 19:17
the_candymanthe_candyman
8,97832145
$endgroup$
$begingroup$
I wonder why you need to use Cauchy MVT when the normal MVT works just fine?
$endgroup$
– BigbearZzz
Jan 21 at 20:28
$begingroup$
@BigbearZzz you're right! I think I used it because yesterday I was following a seminar about the mathematical heritage of Cauchy. I'm going to fix it.
$endgroup$
– the_candyman
Jan 21 at 20:39
add a comment |
$begingroup$
I am making a guess as to what your final problem is. Draw a careful graph of
$$ frac{3x + sqrt{1+x^2 ;}}{4} $$
It has no fixpoint.
answered Jan 21 at 19:31
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Ugh. Is this function a counter-example to my property? If it wasn't, then from Banach fixed-point theorem it would have to have a fixpoint.
$endgroup$
– gaazkam
Jan 21 at 19:45
$begingroup$
@gaazkam your question has a correct answer already. The warning I am making is for non-compact metric spaces. We may then have $d(f(x), f(y)) < d(x,y)$ while still failing to have fixpoints for $f$ There is yet more that could be said...
$endgroup$
– Will Jagy
Jan 21 at 19:57
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose that $|f'(x)|<1 ~forall x in (a,b).$
SIMPLER PROOF (thanks to BigbearZzz)
Consider $y<xin [a, b]$.
By the mean value theorem, there exists a number $c in (y,x)$:
$$f'(c) = frac{f(x)-f(y)}{x-y} Rightarrow \
f(x)-f(y) = (x-y)f'(c) Rightarrow\
|f(x)-f(y)| = |x-y||f'(c)|.$$
Since $|f'(c)| <1 $, then:
$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| Rightarrow \
|f(x)-f(y)| < |x-y|.$$
OLD PROOF
Consider $y<xin [a, b]$. Moreover, consider the function $g(x) = x$ (continuous and differentiable everywhere).
By the Cauchy's mean value theorem, there exists a number $c in (y,x)$:
$$left(f(x)-f(y)right)g'(c) = left(g(x)-g(y)right)f'(c) Rightarrow\
left(f(x)-f(y)right) = left(x-yright)f'(c) Rightarrow \
|f(x)-f(y)| = |x-y||f'(c)|.$$
Since $|f'(c)| <1 $, then:
$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| Rightarrow \
|f(x)-f(y)| < |x-y|.$$
answered Jan 21 at 19:17
the_candymanthe_candyman
8,97832145
$endgroup$
$begingroup$
I wonder why you need to use Cauchy MVT when the normal MVT works just fine?
$endgroup$
– BigbearZzz
Jan 21 at 20:28
$begingroup$
@BigbearZzz you're right! I think I used it because yesterday I was following a seminar about the mathematical heritage of Cauchy. I'm going to fix it.
$endgroup$
– the_candyman
Jan 21 at 20:39
add a comment |
$begingroup$
Suppose that $|f'(x)|<1 ~forall x in (a,b).$
SIMPLER PROOF (thanks to BigbearZzz)
Consider $y<xin [a, b]$.
By the mean value theorem, there exists a number $c in (y,x)$:
$$f'(c) = frac{f(x)-f(y)}{x-y} Rightarrow \
f(x)-f(y) = (x-y)f'(c) Rightarrow\
|f(x)-f(y)| = |x-y||f'(c)|.$$
Since $|f'(c)| <1 $, then:
$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| Rightarrow \
|f(x)-f(y)| < |x-y|.$$
OLD PROOF
Consider $y<xin [a, b]$. Moreover, consider the function $g(x) = x$ (continuous and differentiable everywhere).
By the Cauchy's mean value theorem, there exists a number $c in (y,x)$:
$$left(f(x)-f(y)right)g'(c) = left(g(x)-g(y)right)f'(c) Rightarrow\
left(f(x)-f(y)right) = left(x-yright)f'(c) Rightarrow \
|f(x)-f(y)| = |x-y||f'(c)|.$$
Since $|f'(c)| <1 $, then:
$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| Rightarrow \
|f(x)-f(y)| < |x-y|.$$
answered Jan 21 at 19:17
the_candymanthe_candyman
8,97832145
$endgroup$
$begingroup$
I wonder why you need to use Cauchy MVT when the normal MVT works just fine?
$endgroup$
– BigbearZzz
Jan 21 at 20:28
$begingroup$
@BigbearZzz you're right! I think I used it because yesterday I was following a seminar about the mathematical heritage of Cauchy. I'm going to fix it.
$endgroup$
– the_candyman
Jan 21 at 20:39
add a comment |
$begingroup$
Suppose that $|f'(x)|<1 ~forall x in (a,b).$
SIMPLER PROOF (thanks to BigbearZzz)
Consider $y<xin [a, b]$.
By the mean value theorem, there exists a number $c in (y,x)$:
$$f'(c) = frac{f(x)-f(y)}{x-y} Rightarrow \
f(x)-f(y) = (x-y)f'(c) Rightarrow\
|f(x)-f(y)| = |x-y||f'(c)|.$$
Since $|f'(c)| <1 $, then:
$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| Rightarrow \
|f(x)-f(y)| < |x-y|.$$
OLD PROOF
Consider $y<xin [a, b]$. Moreover, consider the function $g(x) = x$ (continuous and differentiable everywhere).
By the Cauchy's mean value theorem, there exists a number $c in (y,x)$:
$$left(f(x)-f(y)right)g'(c) = left(g(x)-g(y)right)f'(c) Rightarrow\
left(f(x)-f(y)right) = left(x-yright)f'(c) Rightarrow \
|f(x)-f(y)| = |x-y||f'(c)|.$$
Since $|f'(c)| <1 $, then:
$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| Rightarrow \
|f(x)-f(y)| < |x-y|.$$
answered Jan 21 at 19:17
the_candymanthe_candyman
8,97832145
$endgroup$
Suppose that $|f'(x)|<1 ~forall x in (a,b).$
SIMPLER PROOF (thanks to BigbearZzz)
Consider $y<xin [a, b]$.
By the mean value theorem, there exists a number $c in (y,x)$:
$$f'(c) = frac{f(x)-f(y)}{x-y} Rightarrow \
f(x)-f(y) = (x-y)f'(c) Rightarrow\
|f(x)-f(y)| = |x-y||f'(c)|.$$
Since $|f'(c)| <1 $, then:
$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| Rightarrow \
|f(x)-f(y)| < |x-y|.$$
OLD PROOF
Consider $y<xin [a, b]$. Moreover, consider the function $g(x) = x$ (continuous and differentiable everywhere).
By the Cauchy's mean value theorem, there exists a number $c in (y,x)$:
$$left(f(x)-f(y)right)g'(c) = left(g(x)-g(y)right)f'(c) Rightarrow\
left(f(x)-f(y)right) = left(x-yright)f'(c) Rightarrow \
|f(x)-f(y)| = |x-y||f'(c)|.$$
Since $|f'(c)| <1 $, then:
$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| Rightarrow \
|f(x)-f(y)| < |x-y|.$$
answered Jan 21 at 19:17
the_candymanthe_candyman
8,97832145
edited Jan 21 at 20:42
answered Jan 21 at 19:17
the_candymanthe_candyman
8,97832145
answered Jan 21 at 19:17
the_candymanthe_candyman
8,97832145
answered Jan 21 at 19:17
the_candymanthe_candyman
8,97832145
8,97832145
$begingroup$
I wonder why you need to use Cauchy MVT when the normal MVT works just fine?
$endgroup$
– BigbearZzz
Jan 21 at 20:28
$begingroup$
@BigbearZzz you're right! I think I used it because yesterday I was following a seminar about the mathematical heritage of Cauchy. I'm going to fix it.
$endgroup$
– the_candyman
Jan 21 at 20:39
add a comment |
$begingroup$
I wonder why you need to use Cauchy MVT when the normal MVT works just fine?
$endgroup$
– BigbearZzz
Jan 21 at 20:28
$begingroup$
@BigbearZzz you're right! I think I used it because yesterday I was following a seminar about the mathematical heritage of Cauchy. I'm going to fix it.
$endgroup$
– the_candyman
Jan 21 at 20:39
$begingroup$
I wonder why you need to use Cauchy MVT when the normal MVT works just fine?
$endgroup$
– BigbearZzz
Jan 21 at 20:28
$begingroup$
I wonder why you need to use Cauchy MVT when the normal MVT works just fine?
$endgroup$
– BigbearZzz
Jan 21 at 20:28
$begingroup$
@BigbearZzz you're right! I think I used it because yesterday I was following a seminar about the mathematical heritage of Cauchy. I'm going to fix it.
$endgroup$
– the_candyman
Jan 21 at 20:39
$begingroup$
@BigbearZzz you're right! I think I used it because yesterday I was following a seminar about the mathematical heritage of Cauchy. I'm going to fix it.
$endgroup$
– the_candyman
Jan 21 at 20:39
add a comment |
$begingroup$
I am making a guess as to what your final problem is. Draw a careful graph of
$$ frac{3x + sqrt{1+x^2 ;}}{4} $$
It has no fixpoint.
answered Jan 21 at 19:31
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Ugh. Is this function a counter-example to my property? If it wasn't, then from Banach fixed-point theorem it would have to have a fixpoint.
$endgroup$
– gaazkam
Jan 21 at 19:45
$begingroup$
@gaazkam your question has a correct answer already. The warning I am making is for non-compact metric spaces. We may then have $d(f(x), f(y)) < d(x,y)$ while still failing to have fixpoints for $f$ There is yet more that could be said...
$endgroup$
– Will Jagy
Jan 21 at 19:57
add a comment |
$begingroup$
I am making a guess as to what your final problem is. Draw a careful graph of
$$ frac{3x + sqrt{1+x^2 ;}}{4} $$
It has no fixpoint.
answered Jan 21 at 19:31
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Ugh. Is this function a counter-example to my property? If it wasn't, then from Banach fixed-point theorem it would have to have a fixpoint.
$endgroup$
– gaazkam
Jan 21 at 19:45
$begingroup$
@gaazkam your question has a correct answer already. The warning I am making is for non-compact metric spaces. We may then have $d(f(x), f(y)) < d(x,y)$ while still failing to have fixpoints for $f$ There is yet more that could be said...
$endgroup$
– Will Jagy
Jan 21 at 19:57
add a comment |
$begingroup$
I am making a guess as to what your final problem is. Draw a careful graph of
$$ frac{3x + sqrt{1+x^2 ;}}{4} $$
It has no fixpoint.
answered Jan 21 at 19:31
Will JagyWill Jagy
104k5102201
$endgroup$
I am making a guess as to what your final problem is. Draw a careful graph of
$$ frac{3x + sqrt{1+x^2 ;}}{4} $$
It has no fixpoint.
answered Jan 21 at 19:31
Will JagyWill Jagy
104k5102201
answered Jan 21 at 19:31
Will JagyWill Jagy
104k5102201
answered Jan 21 at 19:31
Will JagyWill Jagy
104k5102201
answered Jan 21 at 19:31
Will JagyWill Jagy
104k5102201
104k5102201
$begingroup$
Ugh. Is this function a counter-example to my property? If it wasn't, then from Banach fixed-point theorem it would have to have a fixpoint.
$endgroup$
– gaazkam
Jan 21 at 19:45
$begingroup$
@gaazkam your question has a correct answer already. The warning I am making is for non-compact metric spaces. We may then have $d(f(x), f(y)) < d(x,y)$ while still failing to have fixpoints for $f$ There is yet more that could be said...
$endgroup$
– Will Jagy
Jan 21 at 19:57
add a comment |
$begingroup$
Ugh. Is this function a counter-example to my property? If it wasn't, then from Banach fixed-point theorem it would have to have a fixpoint.
$endgroup$
– gaazkam
Jan 21 at 19:45
$begingroup$
@gaazkam your question has a correct answer already. The warning I am making is for non-compact metric spaces. We may then have $d(f(x), f(y)) < d(x,y)$ while still failing to have fixpoints for $f$ There is yet more that could be said...
$endgroup$
– Will Jagy
Jan 21 at 19:57
$begingroup$
Ugh. Is this function a counter-example to my property? If it wasn't, then from Banach fixed-point theorem it would have to have a fixpoint.
$endgroup$
– gaazkam
Jan 21 at 19:45
$begingroup$
Ugh. Is this function a counter-example to my property? If it wasn't, then from Banach fixed-point theorem it would have to have a fixpoint.
$endgroup$
– gaazkam
Jan 21 at 19:45
$begingroup$
@gaazkam your question has a correct answer already. The warning I am making is for non-compact metric spaces. We may then have $d(f(x), f(y)) < d(x,y)$ while still failing to have fixpoints for $f$ There is yet more that could be said...
$endgroup$
– Will Jagy
Jan 21 at 19:57
$begingroup$
@gaazkam your question has a correct answer already. The warning I am making is for non-compact metric spaces. We may then have $d(f(x), f(y)) < d(x,y)$ while still failing to have fixpoints for $f$ There is yet more that could be said...
$endgroup$
– Will Jagy
Jan 21 at 19:57
add a comment |
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1
$begingroup$
Maybe using the mean value theorem helps?
$endgroup$
– Shashi
Jan 21 at 19:07
$begingroup$
What about $f: Bbb R setminus { 0 } to Bbb R$, $f(x) = x/|x|$ ?
$endgroup$
– Martin R
Jan 21 at 19:07
3
$begingroup$
Ans what about $x=y$?
$endgroup$
– Hagen von Eitzen
Jan 21 at 19:09
$begingroup$
@MartinR Is this supposed to be a counterexample? In each interval on which $f$ is continuous my property holds. Let $x,y>0,xneq y$, then $|f(x)-f(y)|=0$ but $|x-y|>0$
$endgroup$
– gaazkam
Jan 21 at 19:11
1
$begingroup$
@gaazkam: Exactly. So stating where $f$ is defined should be the first step.
$endgroup$
– Martin R
Jan 21 at 19:17