Mixing
Canonical projection on the set of continuous functions that equal $0$ at $x=0$ to $mathbb R ^d $ is...
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I would like to know if the following is true.
Let $mathcal C_{(0) } $ denote the set of continuous functions from $[0,infty)$ to $mathbb R^d $ that sends $0 $ in $[0,infty)$ to $0 $ in $mathbb R^d $ and equip $mathcal C_{(0) } $ with the metric $$rho(f,g):= sum _{n=1 } ^{infty } 2^{-n }text{min}{1, sup_{0 le t le n } |f(t)-g(t)|} $$ then the canonical projections $pi_t :mathcal C_{(0) } to mathbb R^d ,pi_t(f)=f(t) $ is Lipschitz continuous (with respect to the metric $rho $)?
I would think it isn't since the distance between the constant zero function and the sequence of functions $f_n(t)=(mt,...,mt) $ increases without bound at the point $t $ in $mathbb R^d $, but the book I'm reading, Brownian Motion by R. Schlling and L Partzsch, claims it is:
then later
Have I misunderstood something?
real-analysis
asked Jan 24 at 5:37
MrFranzénMrFranzén
99110
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add a comment |
$begingroup$
I would like to know if the following is true.
Let $mathcal C_{(0) } $ denote the set of continuous functions from $[0,infty)$ to $mathbb R^d $ that sends $0 $ in $[0,infty)$ to $0 $ in $mathbb R^d $ and equip $mathcal C_{(0) } $ with the metric $$rho(f,g):= sum _{n=1 } ^{infty } 2^{-n }text{min}{1, sup_{0 le t le n } |f(t)-g(t)|} $$ then the canonical projections $pi_t :mathcal C_{(0) } to mathbb R^d ,pi_t(f)=f(t) $ is Lipschitz continuous (with respect to the metric $rho $)?
I would think it isn't since the distance between the constant zero function and the sequence of functions $f_n(t)=(mt,...,mt) $ increases without bound at the point $t $ in $mathbb R^d $, but the book I'm reading, Brownian Motion by R. Schlling and L Partzsch, claims it is:
then later
Have I misunderstood something?
real-analysis
asked Jan 24 at 5:37
MrFranzénMrFranzén
99110
$endgroup$
1
$begingroup$
For what it's worth, I agree with you. I think they mean locally Lipschitz, which would then imply continuity. ;)
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– An old man in the sea.
Feb 1 at 12:00
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math.stackexchange.com/questions/1480971/…
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add a comment |
$begingroup$
I would like to know if the following is true.
Let $mathcal C_{(0) } $ denote the set of continuous functions from $[0,infty)$ to $mathbb R^d $ that sends $0 $ in $[0,infty)$ to $0 $ in $mathbb R^d $ and equip $mathcal C_{(0) } $ with the metric $$rho(f,g):= sum _{n=1 } ^{infty } 2^{-n }text{min}{1, sup_{0 le t le n } |f(t)-g(t)|} $$ then the canonical projections $pi_t :mathcal C_{(0) } to mathbb R^d ,pi_t(f)=f(t) $ is Lipschitz continuous (with respect to the metric $rho $)?
I would think it isn't since the distance between the constant zero function and the sequence of functions $f_n(t)=(mt,...,mt) $ increases without bound at the point $t $ in $mathbb R^d $, but the book I'm reading, Brownian Motion by R. Schlling and L Partzsch, claims it is:
then later
Have I misunderstood something?
real-analysis
asked Jan 24 at 5:37
MrFranzénMrFranzén
99110
$endgroup$
I would like to know if the following is true.
Let $mathcal C_{(0) } $ denote the set of continuous functions from $[0,infty)$ to $mathbb R^d $ that sends $0 $ in $[0,infty)$ to $0 $ in $mathbb R^d $ and equip $mathcal C_{(0) } $ with the metric $$rho(f,g):= sum _{n=1 } ^{infty } 2^{-n }text{min}{1, sup_{0 le t le n } |f(t)-g(t)|} $$ then the canonical projections $pi_t :mathcal C_{(0) } to mathbb R^d ,pi_t(f)=f(t) $ is Lipschitz continuous (with respect to the metric $rho $)?
I would think it isn't since the distance between the constant zero function and the sequence of functions $f_n(t)=(mt,...,mt) $ increases without bound at the point $t $ in $mathbb R^d $, but the book I'm reading, Brownian Motion by R. Schlling and L Partzsch, claims it is:
then later
Have I misunderstood something?
real-analysis
real-analysis
asked Jan 24 at 5:37
MrFranzénMrFranzén
99110
asked Jan 24 at 5:37
MrFranzénMrFranzén
99110
asked Jan 24 at 5:37
MrFranzénMrFranzén
99110
asked Jan 24 at 5:37
MrFranzénMrFranzén
99110
asked Jan 24 at 5:37
MrFranzénMrFranzén
99110
99110
1
$begingroup$
For what it's worth, I agree with you. I think they mean locally Lipschitz, which would then imply continuity. ;)
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– An old man in the sea.
Feb 1 at 12:00
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math.stackexchange.com/questions/1480971/…
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– An old man in the sea.
Feb 1 at 14:39
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math.stackexchange.com/questions/3096130/…
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– An old man in the sea.
Feb 1 at 14:39
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math.stackexchange.com/questions/530605/…
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– An old man in the sea.
Feb 1 at 15:11
add a comment |
1
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For what it's worth, I agree with you. I think they mean locally Lipschitz, which would then imply continuity. ;)
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– An old man in the sea.
Feb 1 at 12:00
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math.stackexchange.com/questions/1480971/…
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– An old man in the sea.
Feb 1 at 14:39
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– An old man in the sea.
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– An old man in the sea.
Feb 1 at 15:11
1
1
$begingroup$
For what it's worth, I agree with you. I think they mean locally Lipschitz, which would then imply continuity. ;)
$endgroup$
– An old man in the sea.
Feb 1 at 12:00
$begingroup$
For what it's worth, I agree with you. I think they mean locally Lipschitz, which would then imply continuity. ;)
$endgroup$
– An old man in the sea.
Feb 1 at 12:00
$begingroup$
math.stackexchange.com/questions/1480971/…
$endgroup$
– An old man in the sea.
Feb 1 at 14:39
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math.stackexchange.com/questions/1480971/…
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– An old man in the sea.
Feb 1 at 14:39
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math.stackexchange.com/questions/3096130/…
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– An old man in the sea.
Feb 1 at 14:39
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math.stackexchange.com/questions/3096130/…
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– An old man in the sea.
Feb 1 at 14:39
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– An old man in the sea.
Feb 1 at 15:11
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I'm reading the same book. I think it's a typo, since the projection is not Lipschitz...However, it's locally lipschitz.
Here's my take on it.
We have that $|w(t)-v(t)|<infty$, for all $t in [0,infty[$, since $w,v in (mathbb{R}^d)^I$.
$sup_{i in [0,n]} |w(i)-v(i)|$ is non-decreasing with $n$, and $1 wedge sup_{i in [0,n]} |w(i)-v(i)|leq 1$. So, $ sup_{i in [0,1]} |w(i)-v(i)|leq rho(w,v)leqsum_{igeq 1} 2^{-i}=1$
The Locally Lipschitz definition applied to our case is:
for all $z in C_{(0)}$, there exists $delta_t$ such that $displaystyle exists_K forall_{w,vin B(z,delta_t)}|pi_t(w)-pi_t(v)|<Krho(w,v)$
Let's fix $t$, with $[t]$ being the smallest integer greater or equal than $t$.
If $sup_{i in [0,[t]]} |w(i)-v(i)|geq 1$, then $displaystyle rho(w,v)geq frac{1}{2^{[t]-1}}$.
So, let's choose $displaystyle delta_tin left]0,frac{1}{2^{[t]-2}}right[$, and we're sure that $displaystyle forall_{w,vin B(z,delta_t)}$ we have
$$rho(w,v)leq rho(w,z)+rho(v,z)<2cdot frac{1}{2^{[t]-2}}=frac{1}{2^{[t]-1}}$$
Now, we know that $|w(t)-v(t)|leq sup_{i in [0,[t]]} |w(i)-v(i)|= frac{2^{[t]}}{2^{[t]}} sup_{i in [0,[t]]} |w(i)-v(i)|< 2^{[t]} rho(w,s)$
Then we just need to choose $K=2^{[t]}$
answered Feb 1 at 14:37
An old man in the sea.An old man in the sea.
1,65211135
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$begingroup$
I'm reading the same book. I think it's a typo, since the projection is not Lipschitz...However, it's locally lipschitz.
Here's my take on it.
We have that $|w(t)-v(t)|<infty$, for all $t in [0,infty[$, since $w,v in (mathbb{R}^d)^I$.
$sup_{i in [0,n]} |w(i)-v(i)|$ is non-decreasing with $n$, and $1 wedge sup_{i in [0,n]} |w(i)-v(i)|leq 1$. So, $ sup_{i in [0,1]} |w(i)-v(i)|leq rho(w,v)leqsum_{igeq 1} 2^{-i}=1$
The Locally Lipschitz definition applied to our case is:
for all $z in C_{(0)}$, there exists $delta_t$ such that $displaystyle exists_K forall_{w,vin B(z,delta_t)}|pi_t(w)-pi_t(v)|<Krho(w,v)$
Let's fix $t$, with $[t]$ being the smallest integer greater or equal than $t$.
If $sup_{i in [0,[t]]} |w(i)-v(i)|geq 1$, then $displaystyle rho(w,v)geq frac{1}{2^{[t]-1}}$.
So, let's choose $displaystyle delta_tin left]0,frac{1}{2^{[t]-2}}right[$, and we're sure that $displaystyle forall_{w,vin B(z,delta_t)}$ we have
$$rho(w,v)leq rho(w,z)+rho(v,z)<2cdot frac{1}{2^{[t]-2}}=frac{1}{2^{[t]-1}}$$
Now, we know that $|w(t)-v(t)|leq sup_{i in [0,[t]]} |w(i)-v(i)|= frac{2^{[t]}}{2^{[t]}} sup_{i in [0,[t]]} |w(i)-v(i)|< 2^{[t]} rho(w,s)$
Then we just need to choose $K=2^{[t]}$
answered Feb 1 at 14:37
An old man in the sea.An old man in the sea.
1,65211135
$endgroup$
add a comment |
$begingroup$
I'm reading the same book. I think it's a typo, since the projection is not Lipschitz...However, it's locally lipschitz.
Here's my take on it.
We have that $|w(t)-v(t)|<infty$, for all $t in [0,infty[$, since $w,v in (mathbb{R}^d)^I$.
$sup_{i in [0,n]} |w(i)-v(i)|$ is non-decreasing with $n$, and $1 wedge sup_{i in [0,n]} |w(i)-v(i)|leq 1$. So, $ sup_{i in [0,1]} |w(i)-v(i)|leq rho(w,v)leqsum_{igeq 1} 2^{-i}=1$
The Locally Lipschitz definition applied to our case is:
for all $z in C_{(0)}$, there exists $delta_t$ such that $displaystyle exists_K forall_{w,vin B(z,delta_t)}|pi_t(w)-pi_t(v)|<Krho(w,v)$
Let's fix $t$, with $[t]$ being the smallest integer greater or equal than $t$.
If $sup_{i in [0,[t]]} |w(i)-v(i)|geq 1$, then $displaystyle rho(w,v)geq frac{1}{2^{[t]-1}}$.
So, let's choose $displaystyle delta_tin left]0,frac{1}{2^{[t]-2}}right[$, and we're sure that $displaystyle forall_{w,vin B(z,delta_t)}$ we have
$$rho(w,v)leq rho(w,z)+rho(v,z)<2cdot frac{1}{2^{[t]-2}}=frac{1}{2^{[t]-1}}$$
Now, we know that $|w(t)-v(t)|leq sup_{i in [0,[t]]} |w(i)-v(i)|= frac{2^{[t]}}{2^{[t]}} sup_{i in [0,[t]]} |w(i)-v(i)|< 2^{[t]} rho(w,s)$
Then we just need to choose $K=2^{[t]}$
answered Feb 1 at 14:37
An old man in the sea.An old man in the sea.
1,65211135
$endgroup$
add a comment |
$begingroup$
I'm reading the same book. I think it's a typo, since the projection is not Lipschitz...However, it's locally lipschitz.
Here's my take on it.
We have that $|w(t)-v(t)|<infty$, for all $t in [0,infty[$, since $w,v in (mathbb{R}^d)^I$.
$sup_{i in [0,n]} |w(i)-v(i)|$ is non-decreasing with $n$, and $1 wedge sup_{i in [0,n]} |w(i)-v(i)|leq 1$. So, $ sup_{i in [0,1]} |w(i)-v(i)|leq rho(w,v)leqsum_{igeq 1} 2^{-i}=1$
The Locally Lipschitz definition applied to our case is:
for all $z in C_{(0)}$, there exists $delta_t$ such that $displaystyle exists_K forall_{w,vin B(z,delta_t)}|pi_t(w)-pi_t(v)|<Krho(w,v)$
Let's fix $t$, with $[t]$ being the smallest integer greater or equal than $t$.
If $sup_{i in [0,[t]]} |w(i)-v(i)|geq 1$, then $displaystyle rho(w,v)geq frac{1}{2^{[t]-1}}$.
So, let's choose $displaystyle delta_tin left]0,frac{1}{2^{[t]-2}}right[$, and we're sure that $displaystyle forall_{w,vin B(z,delta_t)}$ we have
$$rho(w,v)leq rho(w,z)+rho(v,z)<2cdot frac{1}{2^{[t]-2}}=frac{1}{2^{[t]-1}}$$
Now, we know that $|w(t)-v(t)|leq sup_{i in [0,[t]]} |w(i)-v(i)|= frac{2^{[t]}}{2^{[t]}} sup_{i in [0,[t]]} |w(i)-v(i)|< 2^{[t]} rho(w,s)$
Then we just need to choose $K=2^{[t]}$
answered Feb 1 at 14:37
An old man in the sea.An old man in the sea.
1,65211135
$endgroup$
I'm reading the same book. I think it's a typo, since the projection is not Lipschitz...However, it's locally lipschitz.
Here's my take on it.
We have that $|w(t)-v(t)|<infty$, for all $t in [0,infty[$, since $w,v in (mathbb{R}^d)^I$.
$sup_{i in [0,n]} |w(i)-v(i)|$ is non-decreasing with $n$, and $1 wedge sup_{i in [0,n]} |w(i)-v(i)|leq 1$. So, $ sup_{i in [0,1]} |w(i)-v(i)|leq rho(w,v)leqsum_{igeq 1} 2^{-i}=1$
The Locally Lipschitz definition applied to our case is:
for all $z in C_{(0)}$, there exists $delta_t$ such that $displaystyle exists_K forall_{w,vin B(z,delta_t)}|pi_t(w)-pi_t(v)|<Krho(w,v)$
Let's fix $t$, with $[t]$ being the smallest integer greater or equal than $t$.
If $sup_{i in [0,[t]]} |w(i)-v(i)|geq 1$, then $displaystyle rho(w,v)geq frac{1}{2^{[t]-1}}$.
So, let's choose $displaystyle delta_tin left]0,frac{1}{2^{[t]-2}}right[$, and we're sure that $displaystyle forall_{w,vin B(z,delta_t)}$ we have
$$rho(w,v)leq rho(w,z)+rho(v,z)<2cdot frac{1}{2^{[t]-2}}=frac{1}{2^{[t]-1}}$$
Now, we know that $|w(t)-v(t)|leq sup_{i in [0,[t]]} |w(i)-v(i)|= frac{2^{[t]}}{2^{[t]}} sup_{i in [0,[t]]} |w(i)-v(i)|< 2^{[t]} rho(w,s)$
Then we just need to choose $K=2^{[t]}$
answered Feb 1 at 14:37
An old man in the sea.An old man in the sea.
1,65211135
edited Feb 1 at 16:11
answered Feb 1 at 14:37
An old man in the sea.An old man in the sea.
1,65211135
answered Feb 1 at 14:37
An old man in the sea.An old man in the sea.
1,65211135
answered Feb 1 at 14:37
An old man in the sea.An old man in the sea.
1,65211135
1,65211135
add a comment |
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For what it's worth, I agree with you. I think they mean locally Lipschitz, which would then imply continuity. ;)
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– An old man in the sea.
Feb 1 at 12:00
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math.stackexchange.com/questions/1480971/…
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Feb 1 at 14:39
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