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Prove that $lim_nto infty x_n = 0$ if $lvert f(x) - f(y) rvert leq frac12 lvert x - y rvert$ , $f(0) = 0$ and $x_n = f(x_n-1)$
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Suppose $f: mathbbR to mathbbR$ satisfies $lvert f(x) - f(y) rvert leq frac12 lvert x - y rvert$ for all $x,y in mathbbR$, and $f(0) = 0$. Let $x_0 in mathbbR $ be arbitrary. Define $x_1 = f(x_0)$, $x_2 = f(x_1)$, etc. so that $x_n = f(x_n-1)$ for all $nin mathbbZ^+$. Prove that $lim_nto infty x_n = 0$.
I am aware that one of this means that the function $f$ is Lipschitz which implies that it is uniformly continuous.
This is my attempt at the proof:
Suppose $varepsilon > 0$.
We have
$$lvert x_n - 0 rvert = lvert f(x_n-1) - f(0) rvert leq frac12 lvert x_n-1 rvert.
$$
From uniform continuity, we can choose $delta >0$ such that $lvert x - y rvert < delta$ implies that $lvert f(x) - f(y) rvert < varepsilon$.
I want to somehow use this information to find an $Nin mathbbZ^+$ such that $vert x_n - 0 rvert < varepsilon$ whenever $ngeq N$ but I'm stuck on how to finish.
convergence uniform-continuity
asked Jul 22 at 2:35
user100000000000000
381313
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Suppose $f: mathbbR to mathbbR$ satisfies $lvert f(x) - f(y) rvert leq frac12 lvert x - y rvert$ for all $x,y in mathbbR$, and $f(0) = 0$. Let $x_0 in mathbbR $ be arbitrary. Define $x_1 = f(x_0)$, $x_2 = f(x_1)$, etc. so that $x_n = f(x_n-1)$ for all $nin mathbbZ^+$. Prove that $lim_nto infty x_n = 0$.
I am aware that one of this means that the function $f$ is Lipschitz which implies that it is uniformly continuous.
This is my attempt at the proof:
Suppose $varepsilon > 0$.
We have
$$lvert x_n - 0 rvert = lvert f(x_n-1) - f(0) rvert leq frac12 lvert x_n-1 rvert.
$$
From uniform continuity, we can choose $delta >0$ such that $lvert x - y rvert < delta$ implies that $lvert f(x) - f(y) rvert < varepsilon$.
I want to somehow use this information to find an $Nin mathbbZ^+$ such that $vert x_n - 0 rvert < varepsilon$ whenever $ngeq N$ but I'm stuck on how to finish.
convergence uniform-continuity
asked Jul 22 at 2:35
user100000000000000
381313
add a comment |Â
up vote
0
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up vote
0
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Suppose $f: mathbbR to mathbbR$ satisfies $lvert f(x) - f(y) rvert leq frac12 lvert x - y rvert$ for all $x,y in mathbbR$, and $f(0) = 0$. Let $x_0 in mathbbR $ be arbitrary. Define $x_1 = f(x_0)$, $x_2 = f(x_1)$, etc. so that $x_n = f(x_n-1)$ for all $nin mathbbZ^+$. Prove that $lim_nto infty x_n = 0$.
I am aware that one of this means that the function $f$ is Lipschitz which implies that it is uniformly continuous.
This is my attempt at the proof:
Suppose $varepsilon > 0$.
We have
$$lvert x_n - 0 rvert = lvert f(x_n-1) - f(0) rvert leq frac12 lvert x_n-1 rvert.
$$
From uniform continuity, we can choose $delta >0$ such that $lvert x - y rvert < delta$ implies that $lvert f(x) - f(y) rvert < varepsilon$.
I want to somehow use this information to find an $Nin mathbbZ^+$ such that $vert x_n - 0 rvert < varepsilon$ whenever $ngeq N$ but I'm stuck on how to finish.
convergence uniform-continuity
asked Jul 22 at 2:35
user100000000000000
381313
Suppose $f: mathbbR to mathbbR$ satisfies $lvert f(x) - f(y) rvert leq frac12 lvert x - y rvert$ for all $x,y in mathbbR$, and $f(0) = 0$. Let $x_0 in mathbbR $ be arbitrary. Define $x_1 = f(x_0)$, $x_2 = f(x_1)$, etc. so that $x_n = f(x_n-1)$ for all $nin mathbbZ^+$. Prove that $lim_nto infty x_n = 0$.
I am aware that one of this means that the function $f$ is Lipschitz which implies that it is uniformly continuous.
This is my attempt at the proof:
Suppose $varepsilon > 0$.
We have
$$lvert x_n - 0 rvert = lvert f(x_n-1) - f(0) rvert leq frac12 lvert x_n-1 rvert.
$$
From uniform continuity, we can choose $delta >0$ such that $lvert x - y rvert < delta$ implies that $lvert f(x) - f(y) rvert < varepsilon$.
I want to somehow use this information to find an $Nin mathbbZ^+$ such that $vert x_n - 0 rvert < varepsilon$ whenever $ngeq N$ but I'm stuck on how to finish.
convergence uniform-continuity
asked Jul 22 at 2:35
user100000000000000
381313
asked Jul 22 at 2:35
user100000000000000
381313
asked Jul 22 at 2:35
user100000000000000
381313
asked Jul 22 at 2:35
user100000000000000
381313
381313
add a comment |Â
add a comment |Â
2 Answers
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Why not iterate what you have? $$|x_n| leq frac12|x_n-1| leq frac12^2 |x_n-2| leq dots leq frac12^n |x_0|$$ Note $frac12^n to 0$ as $n to infty.$
edited Jul 22 at 9:12
rtybase
8,85721433
answered Jul 22 at 3:00
Dzoooks
740214
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This is more or less the proof of the Banach Fixed Point Theorem, but I'll include the argument here. Notice that $$lvert x_n+1 - x_n rvert = lvert f(x_n) - f(x_n-1) rvert le frac 1 2 lvert x_n - x_n-1 rvert $$ and since this holds for all $n in mathbb Z^+$, we can use induction to see that $$lvert x_n+1 - x_n rvert le frac12^n lvert f(x_1) - f(x_0) rvert = frac C 2^n$$ where $C = lvert f(x_1) - f(x_0) rvert$ is a constant. Now fix $N in mathbb Z^+$, And note that for $m > n > N$, beginalign* lvert x_m - x_nrvert &= lvert (x_m - x_m-1) - (x_m-1 - x_m-2) - cdots - (x_n+1 - x_n) rvert \ &le sum^m-1_k=n lvert x_k+1 - x_k rvert\
&le sum^m-1_k=n fracC2^k le sum^infty_k=N fracC2^k = fracC(1/2)^N1-(1/2) to 0,,,, text as Nto infty.
endalign* This shows that $x_n$ is a Cauchy sequence and thus converges to some $x^* in mathbb R$. Now taking the limit on both sides of the equation and using continuity of $f$, we see that $$x_n+1 = f(x_n) ,,,,, implies ,,,,, x^* = f(x^*).$$ That is $x^*$ is a fixed point of $f$.
Now suppose that for $x,y in mathbb R$, $$f(x) = x ,,,,,, text and ,,,,,, f(y) = y.$$ Then $$0 le lvert x - y rvert = lvert f(x) - f(y) rvert le frac 1 2 lvert x - y rvert$$ which shows that $lvert x - y rvert = 0$ so $x=y$. Thus $f$ has only one fixed point and so $f(x^*) = x^*$ implies that $x^* = 0$. Thus $x_nto 0.$
answered Jul 22 at 3:01
User8128
10.2k1522
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Why not iterate what you have? $$|x_n| leq frac12|x_n-1| leq frac12^2 |x_n-2| leq dots leq frac12^n |x_0|$$ Note $frac12^n to 0$ as $n to infty.$
edited Jul 22 at 9:12
rtybase
8,85721433
answered Jul 22 at 3:00
Dzoooks
740214
add a comment |Â
up vote
2
down vote
accepted
Why not iterate what you have? $$|x_n| leq frac12|x_n-1| leq frac12^2 |x_n-2| leq dots leq frac12^n |x_0|$$ Note $frac12^n to 0$ as $n to infty.$
edited Jul 22 at 9:12
rtybase
8,85721433
answered Jul 22 at 3:00
Dzoooks
740214
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Why not iterate what you have? $$|x_n| leq frac12|x_n-1| leq frac12^2 |x_n-2| leq dots leq frac12^n |x_0|$$ Note $frac12^n to 0$ as $n to infty.$
edited Jul 22 at 9:12
rtybase
8,85721433
answered Jul 22 at 3:00
Dzoooks
740214
Why not iterate what you have? $$|x_n| leq frac12|x_n-1| leq frac12^2 |x_n-2| leq dots leq frac12^n |x_0|$$ Note $frac12^n to 0$ as $n to infty.$
edited Jul 22 at 9:12
rtybase
8,85721433
answered Jul 22 at 3:00
Dzoooks
740214
edited Jul 22 at 9:12
rtybase
8,85721433
edited Jul 22 at 9:12
rtybase
8,85721433
edited Jul 22 at 9:12
rtybase
8,85721433
8,85721433
answered Jul 22 at 3:00
Dzoooks
740214
answered Jul 22 at 3:00
Dzoooks
740214
answered Jul 22 at 3:00
Dzoooks
740214
740214
add a comment |Â
add a comment |Â
up vote
-1
down vote
This is more or less the proof of the Banach Fixed Point Theorem, but I'll include the argument here. Notice that $$lvert x_n+1 - x_n rvert = lvert f(x_n) - f(x_n-1) rvert le frac 1 2 lvert x_n - x_n-1 rvert $$ and since this holds for all $n in mathbb Z^+$, we can use induction to see that $$lvert x_n+1 - x_n rvert le frac12^n lvert f(x_1) - f(x_0) rvert = frac C 2^n$$ where $C = lvert f(x_1) - f(x_0) rvert$ is a constant. Now fix $N in mathbb Z^+$, And note that for $m > n > N$, beginalign* lvert x_m - x_nrvert &= lvert (x_m - x_m-1) - (x_m-1 - x_m-2) - cdots - (x_n+1 - x_n) rvert \ &le sum^m-1_k=n lvert x_k+1 - x_k rvert\
&le sum^m-1_k=n fracC2^k le sum^infty_k=N fracC2^k = fracC(1/2)^N1-(1/2) to 0,,,, text as Nto infty.
endalign* This shows that $x_n$ is a Cauchy sequence and thus converges to some $x^* in mathbb R$. Now taking the limit on both sides of the equation and using continuity of $f$, we see that $$x_n+1 = f(x_n) ,,,,, implies ,,,,, x^* = f(x^*).$$ That is $x^*$ is a fixed point of $f$.
Now suppose that for $x,y in mathbb R$, $$f(x) = x ,,,,,, text and ,,,,,, f(y) = y.$$ Then $$0 le lvert x - y rvert = lvert f(x) - f(y) rvert le frac 1 2 lvert x - y rvert$$ which shows that $lvert x - y rvert = 0$ so $x=y$. Thus $f$ has only one fixed point and so $f(x^*) = x^*$ implies that $x^* = 0$. Thus $x_nto 0.$
answered Jul 22 at 3:01
User8128
10.2k1522
add a comment |Â
up vote
-1
down vote
This is more or less the proof of the Banach Fixed Point Theorem, but I'll include the argument here. Notice that $$lvert x_n+1 - x_n rvert = lvert f(x_n) - f(x_n-1) rvert le frac 1 2 lvert x_n - x_n-1 rvert $$ and since this holds for all $n in mathbb Z^+$, we can use induction to see that $$lvert x_n+1 - x_n rvert le frac12^n lvert f(x_1) - f(x_0) rvert = frac C 2^n$$ where $C = lvert f(x_1) - f(x_0) rvert$ is a constant. Now fix $N in mathbb Z^+$, And note that for $m > n > N$, beginalign* lvert x_m - x_nrvert &= lvert (x_m - x_m-1) - (x_m-1 - x_m-2) - cdots - (x_n+1 - x_n) rvert \ &le sum^m-1_k=n lvert x_k+1 - x_k rvert\
&le sum^m-1_k=n fracC2^k le sum^infty_k=N fracC2^k = fracC(1/2)^N1-(1/2) to 0,,,, text as Nto infty.
endalign* This shows that $x_n$ is a Cauchy sequence and thus converges to some $x^* in mathbb R$. Now taking the limit on both sides of the equation and using continuity of $f$, we see that $$x_n+1 = f(x_n) ,,,,, implies ,,,,, x^* = f(x^*).$$ That is $x^*$ is a fixed point of $f$.
Now suppose that for $x,y in mathbb R$, $$f(x) = x ,,,,,, text and ,,,,,, f(y) = y.$$ Then $$0 le lvert x - y rvert = lvert f(x) - f(y) rvert le frac 1 2 lvert x - y rvert$$ which shows that $lvert x - y rvert = 0$ so $x=y$. Thus $f$ has only one fixed point and so $f(x^*) = x^*$ implies that $x^* = 0$. Thus $x_nto 0.$
answered Jul 22 at 3:01
User8128
10.2k1522
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
This is more or less the proof of the Banach Fixed Point Theorem, but I'll include the argument here. Notice that $$lvert x_n+1 - x_n rvert = lvert f(x_n) - f(x_n-1) rvert le frac 1 2 lvert x_n - x_n-1 rvert $$ and since this holds for all $n in mathbb Z^+$, we can use induction to see that $$lvert x_n+1 - x_n rvert le frac12^n lvert f(x_1) - f(x_0) rvert = frac C 2^n$$ where $C = lvert f(x_1) - f(x_0) rvert$ is a constant. Now fix $N in mathbb Z^+$, And note that for $m > n > N$, beginalign* lvert x_m - x_nrvert &= lvert (x_m - x_m-1) - (x_m-1 - x_m-2) - cdots - (x_n+1 - x_n) rvert \ &le sum^m-1_k=n lvert x_k+1 - x_k rvert\
&le sum^m-1_k=n fracC2^k le sum^infty_k=N fracC2^k = fracC(1/2)^N1-(1/2) to 0,,,, text as Nto infty.
endalign* This shows that $x_n$ is a Cauchy sequence and thus converges to some $x^* in mathbb R$. Now taking the limit on both sides of the equation and using continuity of $f$, we see that $$x_n+1 = f(x_n) ,,,,, implies ,,,,, x^* = f(x^*).$$ That is $x^*$ is a fixed point of $f$.
Now suppose that for $x,y in mathbb R$, $$f(x) = x ,,,,,, text and ,,,,,, f(y) = y.$$ Then $$0 le lvert x - y rvert = lvert f(x) - f(y) rvert le frac 1 2 lvert x - y rvert$$ which shows that $lvert x - y rvert = 0$ so $x=y$. Thus $f$ has only one fixed point and so $f(x^*) = x^*$ implies that $x^* = 0$. Thus $x_nto 0.$
answered Jul 22 at 3:01
User8128
10.2k1522
This is more or less the proof of the Banach Fixed Point Theorem, but I'll include the argument here. Notice that $$lvert x_n+1 - x_n rvert = lvert f(x_n) - f(x_n-1) rvert le frac 1 2 lvert x_n - x_n-1 rvert $$ and since this holds for all $n in mathbb Z^+$, we can use induction to see that $$lvert x_n+1 - x_n rvert le frac12^n lvert f(x_1) - f(x_0) rvert = frac C 2^n$$ where $C = lvert f(x_1) - f(x_0) rvert$ is a constant. Now fix $N in mathbb Z^+$, And note that for $m > n > N$, beginalign* lvert x_m - x_nrvert &= lvert (x_m - x_m-1) - (x_m-1 - x_m-2) - cdots - (x_n+1 - x_n) rvert \ &le sum^m-1_k=n lvert x_k+1 - x_k rvert\
&le sum^m-1_k=n fracC2^k le sum^infty_k=N fracC2^k = fracC(1/2)^N1-(1/2) to 0,,,, text as Nto infty.
endalign* This shows that $x_n$ is a Cauchy sequence and thus converges to some $x^* in mathbb R$. Now taking the limit on both sides of the equation and using continuity of $f$, we see that $$x_n+1 = f(x_n) ,,,,, implies ,,,,, x^* = f(x^*).$$ That is $x^*$ is a fixed point of $f$.
Now suppose that for $x,y in mathbb R$, $$f(x) = x ,,,,,, text and ,,,,,, f(y) = y.$$ Then $$0 le lvert x - y rvert = lvert f(x) - f(y) rvert le frac 1 2 lvert x - y rvert$$ which shows that $lvert x - y rvert = 0$ so $x=y$. Thus $f$ has only one fixed point and so $f(x^*) = x^*$ implies that $x^* = 0$. Thus $x_nto 0.$
answered Jul 22 at 3:01
User8128
10.2k1522
answered Jul 22 at 3:01
User8128
10.2k1522
answered Jul 22 at 3:01
User8128
10.2k1522
answered Jul 22 at 3:01
User8128
10.2k1522
10.2k1522
add a comment |Â
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