Mixing
Total variation of a measure in a topological dual of the space of continuous functions
$begingroup$
I read the definition of total variation of a measure in the following link
https://www.encyclopediaofmath.org/index.php/Signed_measure
And then I have a question:
If $mu$ is a vector measure in $C^*([0,T];mathbb{R}^{m})$ which is a topological dual of the space of continuous functions. Let $mu(t)=mu^i$ for $tin[t^i,t^{i+1})$, $i=0,ldots,n-1$, where $t^0=0$, $t^n=T$. Each partition $[t^i,t^{i+1}]$ has a length of $h_n$ which tends to zero as $ntoinfty$.
Then we have the total variation of $mu$
$$|mu|([0,T])=supleft{ sum_{i=0}^{infty}|mu([t^i,t^{i+1}])|: {[t^i,t^{i+1}]} textrm{ is a countable partition of } [0,T] right}.$$
Can we obtain the following?
begin{equation}label{1}
|mu|([0,T])=supleft{ sum_{i=0}^{infty}|mu^i|right}.
end{equation}
So the uniform boundedness of $mu$ implies $sum_{i=0}^{infty}|mu^i|$ is uniformly bounded with respect to (w.r.t.) $n$? Is that correct?
Thank you so much for your time and help.
functional-analysis measure-theory norm bounded-variation total-variation
asked Jan 26 at 17:13
RossRoss
1237
$endgroup$
|
show 3 more comments
$begingroup$
I read the definition of total variation of a measure in the following link
https://www.encyclopediaofmath.org/index.php/Signed_measure
And then I have a question:
If $mu$ is a vector measure in $C^*([0,T];mathbb{R}^{m})$ which is a topological dual of the space of continuous functions. Let $mu(t)=mu^i$ for $tin[t^i,t^{i+1})$, $i=0,ldots,n-1$, where $t^0=0$, $t^n=T$. Each partition $[t^i,t^{i+1}]$ has a length of $h_n$ which tends to zero as $ntoinfty$.
Then we have the total variation of $mu$
$$|mu|([0,T])=supleft{ sum_{i=0}^{infty}|mu([t^i,t^{i+1}])|: {[t^i,t^{i+1}]} textrm{ is a countable partition of } [0,T] right}.$$
Can we obtain the following?
begin{equation}label{1}
|mu|([0,T])=supleft{ sum_{i=0}^{infty}|mu^i|right}.
end{equation}
So the uniform boundedness of $mu$ implies $sum_{i=0}^{infty}|mu^i|$ is uniformly bounded with respect to (w.r.t.) $n$? Is that correct?
Thank you so much for your time and help.
functional-analysis measure-theory norm bounded-variation total-variation
asked Jan 26 at 17:13
RossRoss
1237
$endgroup$
$begingroup$
Some things are not understandable. You should edit your question.
$endgroup$
– amsmath
Jan 26 at 17:29
$begingroup$
I have just edited it. Is that okay now?
$endgroup$
– Ross
Jan 26 at 17:34
$begingroup$
No. I neither understand the definition of $mu$ nor is it clear what you take the supremum over.
$endgroup$
– amsmath
Jan 26 at 17:36
$begingroup$
We have a countable partition ${[t^i,t^{i+1}]}$ of $[0,T]$. Then we define $mu(t)=mu^i$ on each interval $[t^i,t^{i+1}]$. And we take the supremum over all the partitions of $[0,T]$.
$endgroup$
– Ross
Jan 26 at 17:49
$begingroup$
And what is $mu^i$?
$endgroup$
– amsmath
Jan 26 at 18:08
|
show 3 more comments
$begingroup$
I read the definition of total variation of a measure in the following link
https://www.encyclopediaofmath.org/index.php/Signed_measure
And then I have a question:
If $mu$ is a vector measure in $C^*([0,T];mathbb{R}^{m})$ which is a topological dual of the space of continuous functions. Let $mu(t)=mu^i$ for $tin[t^i,t^{i+1})$, $i=0,ldots,n-1$, where $t^0=0$, $t^n=T$. Each partition $[t^i,t^{i+1}]$ has a length of $h_n$ which tends to zero as $ntoinfty$.
Then we have the total variation of $mu$
$$|mu|([0,T])=supleft{ sum_{i=0}^{infty}|mu([t^i,t^{i+1}])|: {[t^i,t^{i+1}]} textrm{ is a countable partition of } [0,T] right}.$$
Can we obtain the following?
begin{equation}label{1}
|mu|([0,T])=supleft{ sum_{i=0}^{infty}|mu^i|right}.
end{equation}
So the uniform boundedness of $mu$ implies $sum_{i=0}^{infty}|mu^i|$ is uniformly bounded with respect to (w.r.t.) $n$? Is that correct?
Thank you so much for your time and help.
functional-analysis measure-theory norm bounded-variation total-variation
asked Jan 26 at 17:13
RossRoss
1237
$endgroup$
I read the definition of total variation of a measure in the following link
https://www.encyclopediaofmath.org/index.php/Signed_measure
And then I have a question:
If $mu$ is a vector measure in $C^*([0,T];mathbb{R}^{m})$ which is a topological dual of the space of continuous functions. Let $mu(t)=mu^i$ for $tin[t^i,t^{i+1})$, $i=0,ldots,n-1$, where $t^0=0$, $t^n=T$. Each partition $[t^i,t^{i+1}]$ has a length of $h_n$ which tends to zero as $ntoinfty$.
Then we have the total variation of $mu$
$$|mu|([0,T])=supleft{ sum_{i=0}^{infty}|mu([t^i,t^{i+1}])|: {[t^i,t^{i+1}]} textrm{ is a countable partition of } [0,T] right}.$$
Can we obtain the following?
begin{equation}label{1}
|mu|([0,T])=supleft{ sum_{i=0}^{infty}|mu^i|right}.
end{equation}
So the uniform boundedness of $mu$ implies $sum_{i=0}^{infty}|mu^i|$ is uniformly bounded with respect to (w.r.t.) $n$? Is that correct?
Thank you so much for your time and help.
functional-analysis measure-theory norm bounded-variation total-variation
functional-analysis measure-theory norm bounded-variation total-variation
asked Jan 26 at 17:13
RossRoss
1237
asked Jan 26 at 17:13
RossRoss
1237
edited Jan 26 at 20:49
Ross
asked Jan 26 at 17:13
RossRoss
1237
asked Jan 26 at 17:13
RossRoss
1237
asked Jan 26 at 17:13
RossRoss
1237
1237
$begingroup$
Some things are not understandable. You should edit your question.
$endgroup$
– amsmath
Jan 26 at 17:29
$begingroup$
I have just edited it. Is that okay now?
$endgroup$
– Ross
Jan 26 at 17:34
$begingroup$
No. I neither understand the definition of $mu$ nor is it clear what you take the supremum over.
$endgroup$
– amsmath
Jan 26 at 17:36
$begingroup$
We have a countable partition ${[t^i,t^{i+1}]}$ of $[0,T]$. Then we define $mu(t)=mu^i$ on each interval $[t^i,t^{i+1}]$. And we take the supremum over all the partitions of $[0,T]$.
$endgroup$
– Ross
Jan 26 at 17:49
$begingroup$
And what is $mu^i$?
$endgroup$
– amsmath
Jan 26 at 18:08
|
show 3 more comments
$begingroup$
Some things are not understandable. You should edit your question.
$endgroup$
– amsmath
Jan 26 at 17:29
$begingroup$
I have just edited it. Is that okay now?
$endgroup$
– Ross
Jan 26 at 17:34
$begingroup$
No. I neither understand the definition of $mu$ nor is it clear what you take the supremum over.
$endgroup$
– amsmath
Jan 26 at 17:36
$begingroup$
We have a countable partition ${[t^i,t^{i+1}]}$ of $[0,T]$. Then we define $mu(t)=mu^i$ on each interval $[t^i,t^{i+1}]$. And we take the supremum over all the partitions of $[0,T]$.
$endgroup$
– Ross
Jan 26 at 17:49
$begingroup$
And what is $mu^i$?
$endgroup$
– amsmath
Jan 26 at 18:08
$begingroup$
Some things are not understandable. You should edit your question.
$endgroup$
– amsmath
Jan 26 at 17:29
$begingroup$
Some things are not understandable. You should edit your question.
$endgroup$
– amsmath
Jan 26 at 17:29
$begingroup$
I have just edited it. Is that okay now?
$endgroup$
– Ross
Jan 26 at 17:34
$begingroup$
I have just edited it. Is that okay now?
$endgroup$
– Ross
Jan 26 at 17:34
$begingroup$
No. I neither understand the definition of $mu$ nor is it clear what you take the supremum over.
$endgroup$
– amsmath
Jan 26 at 17:36
$begingroup$
No. I neither understand the definition of $mu$ nor is it clear what you take the supremum over.
$endgroup$
– amsmath
Jan 26 at 17:36
$begingroup$
We have a countable partition ${[t^i,t^{i+1}]}$ of $[0,T]$. Then we define $mu(t)=mu^i$ on each interval $[t^i,t^{i+1}]$. And we take the supremum over all the partitions of $[0,T]$.
$endgroup$
– Ross
Jan 26 at 17:49
$begingroup$
We have a countable partition ${[t^i,t^{i+1}]}$ of $[0,T]$. Then we define $mu(t)=mu^i$ on each interval $[t^i,t^{i+1}]$. And we take the supremum over all the partitions of $[0,T]$.
$endgroup$
– Ross
Jan 26 at 17:49
$begingroup$
And what is $mu^i$?
$endgroup$
– amsmath
Jan 26 at 18:08
$begingroup$
And what is $mu^i$?
$endgroup$
– amsmath
Jan 26 at 18:08
|
show 3 more comments
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$begingroup$
Some things are not understandable. You should edit your question.
$endgroup$
– amsmath
Jan 26 at 17:29
$begingroup$
I have just edited it. Is that okay now?
$endgroup$
– Ross
Jan 26 at 17:34
$begingroup$
No. I neither understand the definition of $mu$ nor is it clear what you take the supremum over.
$endgroup$
– amsmath
Jan 26 at 17:36
$begingroup$
We have a countable partition ${[t^i,t^{i+1}]}$ of $[0,T]$. Then we define $mu(t)=mu^i$ on each interval $[t^i,t^{i+1}]$. And we take the supremum over all the partitions of $[0,T]$.
$endgroup$
– Ross
Jan 26 at 17:49
$begingroup$
And what is $mu^i$?
$endgroup$
– amsmath
Jan 26 at 18:08