how to find adjoint of this operator on the space of C[0,1]? The 2019 Stack Overflow Developer Survey Results Are InFind adjoint operator of an operator TFind the adjoint operatorHow to find adjoint operator?Adjoint operator on Banach spaceFind the adjoint of this non-standard inner product spaceHow to find Adjoint of OperatorThe image of the adjoint operator in a Hilbert spaceadjoint operator of convolution operatorHow do I find adjoint operator in different bases?Uniqueness of the Adjoint operator
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how to find adjoint of this operator on the space of C[0,1]?
The 2019 Stack Overflow Developer Survey Results Are InFind adjoint operator of an operator TFind the adjoint operatorHow to find adjoint operator?Adjoint operator on Banach spaceFind the adjoint of this non-standard inner product spaceHow to find Adjoint of OperatorThe image of the adjoint operator in a Hilbert spaceadjoint operator of convolution operatorHow do I find adjoint operator in different bases?Uniqueness of the Adjoint operator
$begingroup$
We are given $T$ is an operator on $C[0,1]$ as follows
$T(g(x))=sumlimits_k=1^mp_kg(f_k(x)), p_kin [0,1], f_kin C[0,1]$, could anyone tell me how to show adjoint of this operator is as follows?
$T^*mu(cdot)= sumlimits_k=1^mp_kmu(f_k^-1(cdot))$, where $mu$ is a probability measure on $[0,1]$. I am completely blown off, how to relate these. Thanks for helping. I understand there are some gaps, It would be nice if people point out for me. Thanks a lot.
adjoint-operators
asked Mar 30 at 21:31
Ding DongDing Dong
17.4k1160184
$endgroup$
add a comment |
$begingroup$
We are given $T$ is an operator on $C[0,1]$ as follows
$T(g(x))=sumlimits_k=1^mp_kg(f_k(x)), p_kin [0,1], f_kin C[0,1]$, could anyone tell me how to show adjoint of this operator is as follows?
$T^*mu(cdot)= sumlimits_k=1^mp_kmu(f_k^-1(cdot))$, where $mu$ is a probability measure on $[0,1]$. I am completely blown off, how to relate these. Thanks for helping. I understand there are some gaps, It would be nice if people point out for me. Thanks a lot.
adjoint-operators
asked Mar 30 at 21:31
Ding DongDing Dong
17.4k1160184
$endgroup$
add a comment |
$begingroup$
We are given $T$ is an operator on $C[0,1]$ as follows
$T(g(x))=sumlimits_k=1^mp_kg(f_k(x)), p_kin [0,1], f_kin C[0,1]$, could anyone tell me how to show adjoint of this operator is as follows?
$T^*mu(cdot)= sumlimits_k=1^mp_kmu(f_k^-1(cdot))$, where $mu$ is a probability measure on $[0,1]$. I am completely blown off, how to relate these. Thanks for helping. I understand there are some gaps, It would be nice if people point out for me. Thanks a lot.
adjoint-operators
asked Mar 30 at 21:31
Ding DongDing Dong
17.4k1160184
$endgroup$
We are given $T$ is an operator on $C[0,1]$ as follows
$T(g(x))=sumlimits_k=1^mp_kg(f_k(x)), p_kin [0,1], f_kin C[0,1]$, could anyone tell me how to show adjoint of this operator is as follows?
$T^*mu(cdot)= sumlimits_k=1^mp_kmu(f_k^-1(cdot))$, where $mu$ is a probability measure on $[0,1]$. I am completely blown off, how to relate these. Thanks for helping. I understand there are some gaps, It would be nice if people point out for me. Thanks a lot.
adjoint-operators
adjoint-operators
asked Mar 30 at 21:31
Ding DongDing Dong
17.4k1160184
asked Mar 30 at 21:31
Ding DongDing Dong
17.4k1160184
edited Mar 31 at 10:44
Ding Dong
asked Mar 30 at 21:31
Ding DongDing Dong
17.4k1160184
asked Mar 30 at 21:31
Ding DongDing Dong
17.4k1160184
asked Mar 30 at 21:31
Ding DongDing Dong
17.4k1160184
17.4k1160184
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The dual of $C[0,1]$ is the space of all signed/complex measures on $[0,1]$ with total variation norm. we have $T^*(mu)(g)=mu (Tg)=int Tg dmu=int sum p_kg(f_k)dmu=sum p_k int gd(mucirc f_k^-1)$ and hence $T^*=sum p_k mucirc f_k^-1$
edited Mar 31 at 12:29
Ding Dong
17.4k1160184
answered Mar 31 at 0:15
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
So, on which object $T^*$ will act? I am not understanding that. will it act on some subset of $[0,1]$ or the space I am considering and give me a measure of that subset or it will give me a new measure?
$endgroup$
– Ding Dong
Mar 31 at 10:45
$begingroup$
$T:C([0,1]) to C([0,1])$ so $T^*:(C([0,1])^* to C([0,1])^*$. $C([0,1])^*$ ois the space of measures on Borel subsets of $[0,1]$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 11:28
$begingroup$
Then why did you write $T^*(mu)(g)$? Would it be $T^*(mu)(A)$? where $A$ is some Borel subset of $[0,1]$?
$endgroup$
– Ding Dong
Mar 31 at 12:19
$begingroup$
When you think of the dual of $C[0,1]$ as the space of measures you make an identification: a measure $mu$ is also regarded as a continuous linear functional on $C[0,1]$ defined by $mu (f)=int_0^1f dmu$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 12:21
1
$begingroup$
@DingDong $int g(f) dmu=int g d(mucirc f^-1)$. This is the change of variables formula.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 23:13
|
show 2 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The dual of $C[0,1]$ is the space of all signed/complex measures on $[0,1]$ with total variation norm. we have $T^*(mu)(g)=mu (Tg)=int Tg dmu=int sum p_kg(f_k)dmu=sum p_k int gd(mucirc f_k^-1)$ and hence $T^*=sum p_k mucirc f_k^-1$
edited Mar 31 at 12:29
Ding Dong
17.4k1160184
answered Mar 31 at 0:15
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
So, on which object $T^*$ will act? I am not understanding that. will it act on some subset of $[0,1]$ or the space I am considering and give me a measure of that subset or it will give me a new measure?
$endgroup$
– Ding Dong
Mar 31 at 10:45
$begingroup$
$T:C([0,1]) to C([0,1])$ so $T^*:(C([0,1])^* to C([0,1])^*$. $C([0,1])^*$ ois the space of measures on Borel subsets of $[0,1]$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 11:28
$begingroup$
Then why did you write $T^*(mu)(g)$? Would it be $T^*(mu)(A)$? where $A$ is some Borel subset of $[0,1]$?
$endgroup$
– Ding Dong
Mar 31 at 12:19
$begingroup$
When you think of the dual of $C[0,1]$ as the space of measures you make an identification: a measure $mu$ is also regarded as a continuous linear functional on $C[0,1]$ defined by $mu (f)=int_0^1f dmu$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 12:21
1
$begingroup$
@DingDong $int g(f) dmu=int g d(mucirc f^-1)$. This is the change of variables formula.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 23:13
|
show 2 more comments
$begingroup$
The dual of $C[0,1]$ is the space of all signed/complex measures on $[0,1]$ with total variation norm. we have $T^*(mu)(g)=mu (Tg)=int Tg dmu=int sum p_kg(f_k)dmu=sum p_k int gd(mucirc f_k^-1)$ and hence $T^*=sum p_k mucirc f_k^-1$
edited Mar 31 at 12:29
Ding Dong
17.4k1160184
answered Mar 31 at 0:15
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
So, on which object $T^*$ will act? I am not understanding that. will it act on some subset of $[0,1]$ or the space I am considering and give me a measure of that subset or it will give me a new measure?
$endgroup$
– Ding Dong
Mar 31 at 10:45
$begingroup$
$T:C([0,1]) to C([0,1])$ so $T^*:(C([0,1])^* to C([0,1])^*$. $C([0,1])^*$ ois the space of measures on Borel subsets of $[0,1]$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 11:28
$begingroup$
Then why did you write $T^*(mu)(g)$? Would it be $T^*(mu)(A)$? where $A$ is some Borel subset of $[0,1]$?
$endgroup$
– Ding Dong
Mar 31 at 12:19
$begingroup$
When you think of the dual of $C[0,1]$ as the space of measures you make an identification: a measure $mu$ is also regarded as a continuous linear functional on $C[0,1]$ defined by $mu (f)=int_0^1f dmu$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 12:21
1
$begingroup$
@DingDong $int g(f) dmu=int g d(mucirc f^-1)$. This is the change of variables formula.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 23:13
|
show 2 more comments
$begingroup$
The dual of $C[0,1]$ is the space of all signed/complex measures on $[0,1]$ with total variation norm. we have $T^*(mu)(g)=mu (Tg)=int Tg dmu=int sum p_kg(f_k)dmu=sum p_k int gd(mucirc f_k^-1)$ and hence $T^*=sum p_k mucirc f_k^-1$
edited Mar 31 at 12:29
Ding Dong
17.4k1160184
answered Mar 31 at 0:15
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
The dual of $C[0,1]$ is the space of all signed/complex measures on $[0,1]$ with total variation norm. we have $T^*(mu)(g)=mu (Tg)=int Tg dmu=int sum p_kg(f_k)dmu=sum p_k int gd(mucirc f_k^-1)$ and hence $T^*=sum p_k mucirc f_k^-1$
edited Mar 31 at 12:29
Ding Dong
17.4k1160184
answered Mar 31 at 0:15
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
edited Mar 31 at 12:29
Ding Dong
17.4k1160184
edited Mar 31 at 12:29
Ding Dong
17.4k1160184
edited Mar 31 at 12:29
Ding Dong
17.4k1160184
17.4k1160184
answered Mar 31 at 0:15
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered Mar 31 at 0:15
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered Mar 31 at 0:15
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
74.1k53270
$begingroup$
So, on which object $T^*$ will act? I am not understanding that. will it act on some subset of $[0,1]$ or the space I am considering and give me a measure of that subset or it will give me a new measure?
$endgroup$
– Ding Dong
Mar 31 at 10:45
$begingroup$
$T:C([0,1]) to C([0,1])$ so $T^*:(C([0,1])^* to C([0,1])^*$. $C([0,1])^*$ ois the space of measures on Borel subsets of $[0,1]$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 11:28
$begingroup$
Then why did you write $T^*(mu)(g)$? Would it be $T^*(mu)(A)$? where $A$ is some Borel subset of $[0,1]$?
$endgroup$
– Ding Dong
Mar 31 at 12:19
$begingroup$
When you think of the dual of $C[0,1]$ as the space of measures you make an identification: a measure $mu$ is also regarded as a continuous linear functional on $C[0,1]$ defined by $mu (f)=int_0^1f dmu$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 12:21
1
$begingroup$
@DingDong $int g(f) dmu=int g d(mucirc f^-1)$. This is the change of variables formula.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 23:13
|
show 2 more comments
$begingroup$
So, on which object $T^*$ will act? I am not understanding that. will it act on some subset of $[0,1]$ or the space I am considering and give me a measure of that subset or it will give me a new measure?
$endgroup$
– Ding Dong
Mar 31 at 10:45
$begingroup$
$T:C([0,1]) to C([0,1])$ so $T^*:(C([0,1])^* to C([0,1])^*$. $C([0,1])^*$ ois the space of measures on Borel subsets of $[0,1]$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 11:28
$begingroup$
Then why did you write $T^*(mu)(g)$? Would it be $T^*(mu)(A)$? where $A$ is some Borel subset of $[0,1]$?
$endgroup$
– Ding Dong
Mar 31 at 12:19
$begingroup$
When you think of the dual of $C[0,1]$ as the space of measures you make an identification: a measure $mu$ is also regarded as a continuous linear functional on $C[0,1]$ defined by $mu (f)=int_0^1f dmu$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 12:21
1
$begingroup$
@DingDong $int g(f) dmu=int g d(mucirc f^-1)$. This is the change of variables formula.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 23:13
$begingroup$
So, on which object $T^*$ will act? I am not understanding that. will it act on some subset of $[0,1]$ or the space I am considering and give me a measure of that subset or it will give me a new measure?
$endgroup$
– Ding Dong
Mar 31 at 10:45
$begingroup$
So, on which object $T^*$ will act? I am not understanding that. will it act on some subset of $[0,1]$ or the space I am considering and give me a measure of that subset or it will give me a new measure?
$endgroup$
– Ding Dong
Mar 31 at 10:45
$begingroup$
$T:C([0,1]) to C([0,1])$ so $T^*:(C([0,1])^* to C([0,1])^*$. $C([0,1])^*$ ois the space of measures on Borel subsets of $[0,1]$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 11:28
$begingroup$
$T:C([0,1]) to C([0,1])$ so $T^*:(C([0,1])^* to C([0,1])^*$. $C([0,1])^*$ ois the space of measures on Borel subsets of $[0,1]$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 11:28
$begingroup$
Then why did you write $T^*(mu)(g)$? Would it be $T^*(mu)(A)$? where $A$ is some Borel subset of $[0,1]$?
$endgroup$
– Ding Dong
Mar 31 at 12:19
$begingroup$
Then why did you write $T^*(mu)(g)$? Would it be $T^*(mu)(A)$? where $A$ is some Borel subset of $[0,1]$?
$endgroup$
– Ding Dong
Mar 31 at 12:19
$begingroup$
When you think of the dual of $C[0,1]$ as the space of measures you make an identification: a measure $mu$ is also regarded as a continuous linear functional on $C[0,1]$ defined by $mu (f)=int_0^1f dmu$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 12:21
$begingroup$
When you think of the dual of $C[0,1]$ as the space of measures you make an identification: a measure $mu$ is also regarded as a continuous linear functional on $C[0,1]$ defined by $mu (f)=int_0^1f dmu$.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 12:21
1
1
$begingroup$
@DingDong $int g(f) dmu=int g d(mucirc f^-1)$. This is the change of variables formula.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 23:13
$begingroup$
@DingDong $int g(f) dmu=int g d(mucirc f^-1)$. This is the change of variables formula.
$endgroup$
– Kavi Rama Murthy
Mar 31 at 23:13
|
show 2 more comments
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