If $g circ f=g$, prove that $g$ is constant Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)If $f circ g = f$, prove that $f$ is a constant function.Prove that if $f:mathbbRtomathbbR$ is decreasing and continuous, then there is a unique point $c in mathbbR$ such that $,f(c)=c$.The set of all fixed points of a continuous function $f:[0,1] to [0,1]$ , satisfying $f circ f=f$ , is a non-empty interval?Prove that $fcirc f = mboxid_x$Integrable function whose Fourier transform is constantLet $f$ be a differentiable function on $mathbbR$ then prove the followingProve that this function is constantIf $fin mathcalC^1(mathbbR^2)$, then there exists $g:[0,1]to mathbbR^2$ such that $fcirc g $ is constant.Show that $Phi$ is not a contraction, but $Phi circ Phi$ is a contractionProve that the function is bijective
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If $g circ f=g$, prove that $g$ is constant
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)If $f circ g = f$, prove that $f$ is a constant function.Prove that if $f:mathbbRtomathbbR$ is decreasing and continuous, then there is a unique point $c in mathbbR$ such that $,f(c)=c$.The set of all fixed points of a continuous function $f:[0,1] to [0,1]$ , satisfying $f circ f=f$ , is a non-empty interval?Prove that $fcirc f = mboxid_x$Integrable function whose Fourier transform is constantLet $f$ be a differentiable function on $mathbbR$ then prove the followingProve that this function is constantIf $fin mathcalC^1(mathbbR^2)$, then there exists $g:[0,1]to mathbbR^2$ such that $fcirc g $ is constant.Show that $Phi$ is not a contraction, but $Phi circ Phi$ is a contractionProve that the function is bijective
$begingroup$
Let $f:mathbbR to [0,1]$ be a monotonic function so that $|f(x) - f(y) | <|x-y|$, $forall x, y in mathbbR $, $xneq y$. If $g:mathbbR to mathbbR $ is a continuous function and $g circ f=g$, prove that $g$ is constant.
This problem also previously asked to prove that $f$ has a unique fixed point and I could show this by proving that $f$ is continuous and considering the function $h(x) =f(x) - x$. Yet, I don't know how to use this to prove that $g$ is constant.
real-analysis functions
edited Mar 31 at 18:50
TonyK
44.1k358137
asked Mar 31 at 18:45
MathEnthusiastMathEnthusiast
47913
$endgroup$
add a comment |
$begingroup$
Let $f:mathbbR to [0,1]$ be a monotonic function so that $|f(x) - f(y) | <|x-y|$, $forall x, y in mathbbR $, $xneq y$. If $g:mathbbR to mathbbR $ is a continuous function and $g circ f=g$, prove that $g$ is constant.
This problem also previously asked to prove that $f$ has a unique fixed point and I could show this by proving that $f$ is continuous and considering the function $h(x) =f(x) - x$. Yet, I don't know how to use this to prove that $g$ is constant.
real-analysis functions
edited Mar 31 at 18:50
TonyK
44.1k358137
asked Mar 31 at 18:45
MathEnthusiastMathEnthusiast
47913
$endgroup$
add a comment |
$begingroup$
Let $f:mathbbR to [0,1]$ be a monotonic function so that $|f(x) - f(y) | <|x-y|$, $forall x, y in mathbbR $, $xneq y$. If $g:mathbbR to mathbbR $ is a continuous function and $g circ f=g$, prove that $g$ is constant.
This problem also previously asked to prove that $f$ has a unique fixed point and I could show this by proving that $f$ is continuous and considering the function $h(x) =f(x) - x$. Yet, I don't know how to use this to prove that $g$ is constant.
real-analysis functions
edited Mar 31 at 18:50
TonyK
44.1k358137
asked Mar 31 at 18:45
MathEnthusiastMathEnthusiast
47913
$endgroup$
Let $f:mathbbR to [0,1]$ be a monotonic function so that $|f(x) - f(y) | <|x-y|$, $forall x, y in mathbbR $, $xneq y$. If $g:mathbbR to mathbbR $ is a continuous function and $g circ f=g$, prove that $g$ is constant.
This problem also previously asked to prove that $f$ has a unique fixed point and I could show this by proving that $f$ is continuous and considering the function $h(x) =f(x) - x$. Yet, I don't know how to use this to prove that $g$ is constant.
real-analysis functions
real-analysis functions
edited Mar 31 at 18:50
TonyK
44.1k358137
asked Mar 31 at 18:45
MathEnthusiastMathEnthusiast
47913
edited Mar 31 at 18:50
TonyK
44.1k358137
asked Mar 31 at 18:45
MathEnthusiastMathEnthusiast
47913
edited Mar 31 at 18:50
TonyK
44.1k358137
edited Mar 31 at 18:50
TonyK
44.1k358137
edited Mar 31 at 18:50
TonyK
44.1k358137
44.1k358137
asked Mar 31 at 18:45
MathEnthusiastMathEnthusiast
47913
asked Mar 31 at 18:45
MathEnthusiastMathEnthusiast
47913
asked Mar 31 at 18:45
MathEnthusiastMathEnthusiast
47913
47913
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Denote by $c$ the fixed point of $f$.
Let $a in mathbb R$ be arbitrary. Then
$$g(a)=g(f(a))=g(f^2(a))=....=g(f^n(a))=...$$
The sequence $x_n =f^n(a)$ converges to the fixed point $c$ of $f$. Therefore by continuity of $g$
$$g(a)=g(x_n) to g(c)$$
This shows that $g(a)=g(c)$ for all $a in mathbb R$.
answered Mar 31 at 18:49
N. S.N. S.
105k7115210
$endgroup$
$begingroup$
Wow, nice proof! Could you explain to me why $x_n$ converges to $c$?
$endgroup$
– MathEnthusiast
Mar 31 at 18:52
$begingroup$
@MathEnthusiast Try Googling the "Banach fixed point theorem" (or the "contraction mapping principle").
$endgroup$
– Xander Henderson
Mar 31 at 18:57
1
$begingroup$
@MathEnthusiast Didn't you prove it in part 1? If not, First show that there exists some $D<1$ such that $|f(x)-f(y)| <D |x-y|$ for all $x,y in mathbb R$... Then $$|x_n+1-c|=|f(x_n)-f(c)| leq D |x_n -c|$$ and hence, by induction $$|x_n-c| < D^n-1 |x_1-c|$$ Since $D<1$, it follows immediately that $x_n to c$.
$endgroup$
– N. S.
Mar 31 at 19:09
$begingroup$
@Xander Henderson You are right, $f$ is a contraction from the initial constraint. Yet, I don't know how the fact that $f$ is monotonous helps.
$endgroup$
– MathEnthusiast
Mar 31 at 19:09
$begingroup$
@N. S. Now I understand, I could have used Banach's fixed point theorem directly. However, this doesn't use the fact that $f$ is monotonous. Is that condition redundant?
$endgroup$
– MathEnthusiast
Mar 31 at 19:12
|
show 4 more comments
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Denote by $c$ the fixed point of $f$.
Let $a in mathbb R$ be arbitrary. Then
$$g(a)=g(f(a))=g(f^2(a))=....=g(f^n(a))=...$$
The sequence $x_n =f^n(a)$ converges to the fixed point $c$ of $f$. Therefore by continuity of $g$
$$g(a)=g(x_n) to g(c)$$
This shows that $g(a)=g(c)$ for all $a in mathbb R$.
answered Mar 31 at 18:49
N. S.N. S.
105k7115210
$endgroup$
$begingroup$
Wow, nice proof! Could you explain to me why $x_n$ converges to $c$?
$endgroup$
– MathEnthusiast
Mar 31 at 18:52
$begingroup$
@MathEnthusiast Try Googling the "Banach fixed point theorem" (or the "contraction mapping principle").
$endgroup$
– Xander Henderson
Mar 31 at 18:57
1
$begingroup$
@MathEnthusiast Didn't you prove it in part 1? If not, First show that there exists some $D<1$ such that $|f(x)-f(y)| <D |x-y|$ for all $x,y in mathbb R$... Then $$|x_n+1-c|=|f(x_n)-f(c)| leq D |x_n -c|$$ and hence, by induction $$|x_n-c| < D^n-1 |x_1-c|$$ Since $D<1$, it follows immediately that $x_n to c$.
$endgroup$
– N. S.
Mar 31 at 19:09
$begingroup$
@Xander Henderson You are right, $f$ is a contraction from the initial constraint. Yet, I don't know how the fact that $f$ is monotonous helps.
$endgroup$
– MathEnthusiast
Mar 31 at 19:09
$begingroup$
@N. S. Now I understand, I could have used Banach's fixed point theorem directly. However, this doesn't use the fact that $f$ is monotonous. Is that condition redundant?
$endgroup$
– MathEnthusiast
Mar 31 at 19:12
|
show 4 more comments
$begingroup$
Denote by $c$ the fixed point of $f$.
Let $a in mathbb R$ be arbitrary. Then
$$g(a)=g(f(a))=g(f^2(a))=....=g(f^n(a))=...$$
The sequence $x_n =f^n(a)$ converges to the fixed point $c$ of $f$. Therefore by continuity of $g$
$$g(a)=g(x_n) to g(c)$$
This shows that $g(a)=g(c)$ for all $a in mathbb R$.
answered Mar 31 at 18:49
N. S.N. S.
105k7115210
$endgroup$
$begingroup$
Wow, nice proof! Could you explain to me why $x_n$ converges to $c$?
$endgroup$
– MathEnthusiast
Mar 31 at 18:52
$begingroup$
@MathEnthusiast Try Googling the "Banach fixed point theorem" (or the "contraction mapping principle").
$endgroup$
– Xander Henderson
Mar 31 at 18:57
1
$begingroup$
@MathEnthusiast Didn't you prove it in part 1? If not, First show that there exists some $D<1$ such that $|f(x)-f(y)| <D |x-y|$ for all $x,y in mathbb R$... Then $$|x_n+1-c|=|f(x_n)-f(c)| leq D |x_n -c|$$ and hence, by induction $$|x_n-c| < D^n-1 |x_1-c|$$ Since $D<1$, it follows immediately that $x_n to c$.
$endgroup$
– N. S.
Mar 31 at 19:09
$begingroup$
@Xander Henderson You are right, $f$ is a contraction from the initial constraint. Yet, I don't know how the fact that $f$ is monotonous helps.
$endgroup$
– MathEnthusiast
Mar 31 at 19:09
$begingroup$
@N. S. Now I understand, I could have used Banach's fixed point theorem directly. However, this doesn't use the fact that $f$ is monotonous. Is that condition redundant?
$endgroup$
– MathEnthusiast
Mar 31 at 19:12
|
show 4 more comments
$begingroup$
Denote by $c$ the fixed point of $f$.
Let $a in mathbb R$ be arbitrary. Then
$$g(a)=g(f(a))=g(f^2(a))=....=g(f^n(a))=...$$
The sequence $x_n =f^n(a)$ converges to the fixed point $c$ of $f$. Therefore by continuity of $g$
$$g(a)=g(x_n) to g(c)$$
This shows that $g(a)=g(c)$ for all $a in mathbb R$.
answered Mar 31 at 18:49
N. S.N. S.
105k7115210
$endgroup$
Denote by $c$ the fixed point of $f$.
Let $a in mathbb R$ be arbitrary. Then
$$g(a)=g(f(a))=g(f^2(a))=....=g(f^n(a))=...$$
The sequence $x_n =f^n(a)$ converges to the fixed point $c$ of $f$. Therefore by continuity of $g$
$$g(a)=g(x_n) to g(c)$$
This shows that $g(a)=g(c)$ for all $a in mathbb R$.
answered Mar 31 at 18:49
N. S.N. S.
105k7115210
answered Mar 31 at 18:49
N. S.N. S.
105k7115210
answered Mar 31 at 18:49
N. S.N. S.
105k7115210
answered Mar 31 at 18:49
N. S.N. S.
105k7115210
105k7115210
$begingroup$
Wow, nice proof! Could you explain to me why $x_n$ converges to $c$?
$endgroup$
– MathEnthusiast
Mar 31 at 18:52
$begingroup$
@MathEnthusiast Try Googling the "Banach fixed point theorem" (or the "contraction mapping principle").
$endgroup$
– Xander Henderson
Mar 31 at 18:57
1
$begingroup$
@MathEnthusiast Didn't you prove it in part 1? If not, First show that there exists some $D<1$ such that $|f(x)-f(y)| <D |x-y|$ for all $x,y in mathbb R$... Then $$|x_n+1-c|=|f(x_n)-f(c)| leq D |x_n -c|$$ and hence, by induction $$|x_n-c| < D^n-1 |x_1-c|$$ Since $D<1$, it follows immediately that $x_n to c$.
$endgroup$
– N. S.
Mar 31 at 19:09
$begingroup$
@Xander Henderson You are right, $f$ is a contraction from the initial constraint. Yet, I don't know how the fact that $f$ is monotonous helps.
$endgroup$
– MathEnthusiast
Mar 31 at 19:09
$begingroup$
@N. S. Now I understand, I could have used Banach's fixed point theorem directly. However, this doesn't use the fact that $f$ is monotonous. Is that condition redundant?
$endgroup$
– MathEnthusiast
Mar 31 at 19:12
|
show 4 more comments
$begingroup$
Wow, nice proof! Could you explain to me why $x_n$ converges to $c$?
$endgroup$
– MathEnthusiast
Mar 31 at 18:52
$begingroup$
@MathEnthusiast Try Googling the "Banach fixed point theorem" (or the "contraction mapping principle").
$endgroup$
– Xander Henderson
Mar 31 at 18:57
1
$begingroup$
@MathEnthusiast Didn't you prove it in part 1? If not, First show that there exists some $D<1$ such that $|f(x)-f(y)| <D |x-y|$ for all $x,y in mathbb R$... Then $$|x_n+1-c|=|f(x_n)-f(c)| leq D |x_n -c|$$ and hence, by induction $$|x_n-c| < D^n-1 |x_1-c|$$ Since $D<1$, it follows immediately that $x_n to c$.
$endgroup$
– N. S.
Mar 31 at 19:09
$begingroup$
@Xander Henderson You are right, $f$ is a contraction from the initial constraint. Yet, I don't know how the fact that $f$ is monotonous helps.
$endgroup$
– MathEnthusiast
Mar 31 at 19:09
$begingroup$
@N. S. Now I understand, I could have used Banach's fixed point theorem directly. However, this doesn't use the fact that $f$ is monotonous. Is that condition redundant?
$endgroup$
– MathEnthusiast
Mar 31 at 19:12
$begingroup$
Wow, nice proof! Could you explain to me why $x_n$ converges to $c$?
$endgroup$
– MathEnthusiast
Mar 31 at 18:52
$begingroup$
Wow, nice proof! Could you explain to me why $x_n$ converges to $c$?
$endgroup$
– MathEnthusiast
Mar 31 at 18:52
$begingroup$
@MathEnthusiast Try Googling the "Banach fixed point theorem" (or the "contraction mapping principle").
$endgroup$
– Xander Henderson
Mar 31 at 18:57
$begingroup$
@MathEnthusiast Try Googling the "Banach fixed point theorem" (or the "contraction mapping principle").
$endgroup$
– Xander Henderson
Mar 31 at 18:57
1
1
$begingroup$
@MathEnthusiast Didn't you prove it in part 1? If not, First show that there exists some $D<1$ such that $|f(x)-f(y)| <D |x-y|$ for all $x,y in mathbb R$... Then $$|x_n+1-c|=|f(x_n)-f(c)| leq D |x_n -c|$$ and hence, by induction $$|x_n-c| < D^n-1 |x_1-c|$$ Since $D<1$, it follows immediately that $x_n to c$.
$endgroup$
– N. S.
Mar 31 at 19:09
$begingroup$
@MathEnthusiast Didn't you prove it in part 1? If not, First show that there exists some $D<1$ such that $|f(x)-f(y)| <D |x-y|$ for all $x,y in mathbb R$... Then $$|x_n+1-c|=|f(x_n)-f(c)| leq D |x_n -c|$$ and hence, by induction $$|x_n-c| < D^n-1 |x_1-c|$$ Since $D<1$, it follows immediately that $x_n to c$.
$endgroup$
– N. S.
Mar 31 at 19:09
$begingroup$
@Xander Henderson You are right, $f$ is a contraction from the initial constraint. Yet, I don't know how the fact that $f$ is monotonous helps.
$endgroup$
– MathEnthusiast
Mar 31 at 19:09
$begingroup$
@Xander Henderson You are right, $f$ is a contraction from the initial constraint. Yet, I don't know how the fact that $f$ is monotonous helps.
$endgroup$
– MathEnthusiast
Mar 31 at 19:09
$begingroup$
@N. S. Now I understand, I could have used Banach's fixed point theorem directly. However, this doesn't use the fact that $f$ is monotonous. Is that condition redundant?
$endgroup$
– MathEnthusiast
Mar 31 at 19:12
$begingroup$
@N. S. Now I understand, I could have used Banach's fixed point theorem directly. However, this doesn't use the fact that $f$ is monotonous. Is that condition redundant?
$endgroup$
– MathEnthusiast
Mar 31 at 19:12
|
show 4 more comments
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