Decomposition of a continuous linear functional on $L^p$ into two positive continuous l.f.
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Let $(X,mathcal{F},mu)$ be a $sigma$-finite measure space. Prove that if $I: L^p(mu) to mathbb{R}$ is a continuous linear functional then there exists $I^+$, $I^-$ continuous positive linear functionals such that $I=I^+ + I^-$.
I know the proof for the case that $I$ is a bounded linear functional. We define $I^+(f):=sup{I(g) : g in L^p, 0leq g leq f}$ and $I^-=I^+ - I$ and prove the statement.
But in this case I don't know what to do, and I can't link the proof for the bounded case because, of course, $I^+$ is defined with a supremum, which just make sense if we assume boundness of $I$ ... and I don't think that's a way to fix it.
My attempt was based on: try to define a continuous $I^+$ for simple functions and then use the fact that given $fin L^p$ exist a sequence of simple functions $varphi_n to f$ and use continuity of $I^+$ to give $I^+ (f)$, but I can't do this in a useful way .
Anyone can help me?
integration measure-theory
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Let $(X,mathcal{F},mu)$ be a $sigma$-finite measure space. Prove that if $I: L^p(mu) to mathbb{R}$ is a continuous linear functional then there exists $I^+$, $I^-$ continuous positive linear functionals such that $I=I^+ + I^-$.
I know the proof for the case that $I$ is a bounded linear functional. We define $I^+(f):=sup{I(g) : g in L^p, 0leq g leq f}$ and $I^-=I^+ - I$ and prove the statement.
But in this case I don't know what to do, and I can't link the proof for the bounded case because, of course, $I^+$ is defined with a supremum, which just make sense if we assume boundness of $I$ ... and I don't think that's a way to fix it.
My attempt was based on: try to define a continuous $I^+$ for simple functions and then use the fact that given $fin L^p$ exist a sequence of simple functions $varphi_n to f$ and use continuity of $I^+$ to give $I^+ (f)$, but I can't do this in a useful way .
Anyone can help me?
integration measure-theory
$1leq p<infty$
– Robson
2 days ago
2
Continuity is the same as bounded? $I$ is a linear map between normed spaces, so being continuous is the same as being bounded.
– Lucas
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(X,mathcal{F},mu)$ be a $sigma$-finite measure space. Prove that if $I: L^p(mu) to mathbb{R}$ is a continuous linear functional then there exists $I^+$, $I^-$ continuous positive linear functionals such that $I=I^+ + I^-$.
I know the proof for the case that $I$ is a bounded linear functional. We define $I^+(f):=sup{I(g) : g in L^p, 0leq g leq f}$ and $I^-=I^+ - I$ and prove the statement.
But in this case I don't know what to do, and I can't link the proof for the bounded case because, of course, $I^+$ is defined with a supremum, which just make sense if we assume boundness of $I$ ... and I don't think that's a way to fix it.
My attempt was based on: try to define a continuous $I^+$ for simple functions and then use the fact that given $fin L^p$ exist a sequence of simple functions $varphi_n to f$ and use continuity of $I^+$ to give $I^+ (f)$, but I can't do this in a useful way .
Anyone can help me?
integration measure-theory
Let $(X,mathcal{F},mu)$ be a $sigma$-finite measure space. Prove that if $I: L^p(mu) to mathbb{R}$ is a continuous linear functional then there exists $I^+$, $I^-$ continuous positive linear functionals such that $I=I^+ + I^-$.
I know the proof for the case that $I$ is a bounded linear functional. We define $I^+(f):=sup{I(g) : g in L^p, 0leq g leq f}$ and $I^-=I^+ - I$ and prove the statement.
But in this case I don't know what to do, and I can't link the proof for the bounded case because, of course, $I^+$ is defined with a supremum, which just make sense if we assume boundness of $I$ ... and I don't think that's a way to fix it.
My attempt was based on: try to define a continuous $I^+$ for simple functions and then use the fact that given $fin L^p$ exist a sequence of simple functions $varphi_n to f$ and use continuity of $I^+$ to give $I^+ (f)$, but I can't do this in a useful way .
Anyone can help me?
integration measure-theory
integration measure-theory
asked 2 days ago
Robson
591221
591221
$1leq p<infty$
– Robson
2 days ago
2
Continuity is the same as bounded? $I$ is a linear map between normed spaces, so being continuous is the same as being bounded.
– Lucas
2 days ago
add a comment |
$1leq p<infty$
– Robson
2 days ago
2
Continuity is the same as bounded? $I$ is a linear map between normed spaces, so being continuous is the same as being bounded.
– Lucas
2 days ago
$1leq p<infty$
– Robson
2 days ago
$1leq p<infty$
– Robson
2 days ago
2
2
Continuity is the same as bounded? $I$ is a linear map between normed spaces, so being continuous is the same as being bounded.
– Lucas
2 days ago
Continuity is the same as bounded? $I$ is a linear map between normed spaces, so being continuous is the same as being bounded.
– Lucas
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
I am assuming that $1<p<infty$.By a well known result there exists $g in L^{q}$ ($q=frac p {p-1}$) such that $I(f)=int fg$ for all $f in L^{p}$ and all you have to do is define $I^{pm}(f)=int fg^{pm}$. The same idea works for $p=1$ also.
Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
– Robson
2 days ago
1
The usual definition of a positive operator on a function space is $I^{+}f geq 0$ if $f geq 0$. There is no (non-zero) linear operator $T$ on $L^{p}$ for which $Tf geq 0$ for all $f$, so you have to go back to your source and find the correct definitions.
– Kavi Rama Murthy
2 days ago
that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
– Robson
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I am assuming that $1<p<infty$.By a well known result there exists $g in L^{q}$ ($q=frac p {p-1}$) such that $I(f)=int fg$ for all $f in L^{p}$ and all you have to do is define $I^{pm}(f)=int fg^{pm}$. The same idea works for $p=1$ also.
Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
– Robson
2 days ago
1
The usual definition of a positive operator on a function space is $I^{+}f geq 0$ if $f geq 0$. There is no (non-zero) linear operator $T$ on $L^{p}$ for which $Tf geq 0$ for all $f$, so you have to go back to your source and find the correct definitions.
– Kavi Rama Murthy
2 days ago
that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
– Robson
2 days ago
add a comment |
up vote
1
down vote
accepted
I am assuming that $1<p<infty$.By a well known result there exists $g in L^{q}$ ($q=frac p {p-1}$) such that $I(f)=int fg$ for all $f in L^{p}$ and all you have to do is define $I^{pm}(f)=int fg^{pm}$. The same idea works for $p=1$ also.
Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
– Robson
2 days ago
1
The usual definition of a positive operator on a function space is $I^{+}f geq 0$ if $f geq 0$. There is no (non-zero) linear operator $T$ on $L^{p}$ for which $Tf geq 0$ for all $f$, so you have to go back to your source and find the correct definitions.
– Kavi Rama Murthy
2 days ago
that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
– Robson
2 days ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I am assuming that $1<p<infty$.By a well known result there exists $g in L^{q}$ ($q=frac p {p-1}$) such that $I(f)=int fg$ for all $f in L^{p}$ and all you have to do is define $I^{pm}(f)=int fg^{pm}$. The same idea works for $p=1$ also.
I am assuming that $1<p<infty$.By a well known result there exists $g in L^{q}$ ($q=frac p {p-1}$) such that $I(f)=int fg$ for all $f in L^{p}$ and all you have to do is define $I^{pm}(f)=int fg^{pm}$. The same idea works for $p=1$ also.
edited 2 days ago
answered 2 days ago
Kavi Rama Murthy
39.6k31749
39.6k31749
Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
– Robson
2 days ago
1
The usual definition of a positive operator on a function space is $I^{+}f geq 0$ if $f geq 0$. There is no (non-zero) linear operator $T$ on $L^{p}$ for which $Tf geq 0$ for all $f$, so you have to go back to your source and find the correct definitions.
– Kavi Rama Murthy
2 days ago
that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
– Robson
2 days ago
add a comment |
Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
– Robson
2 days ago
1
The usual definition of a positive operator on a function space is $I^{+}f geq 0$ if $f geq 0$. There is no (non-zero) linear operator $T$ on $L^{p}$ for which $Tf geq 0$ for all $f$, so you have to go back to your source and find the correct definitions.
– Kavi Rama Murthy
2 days ago
that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
– Robson
2 days ago
Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
– Robson
2 days ago
Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
– Robson
2 days ago
1
1
The usual definition of a positive operator on a function space is $I^{+}f geq 0$ if $f geq 0$. There is no (non-zero) linear operator $T$ on $L^{p}$ for which $Tf geq 0$ for all $f$, so you have to go back to your source and find the correct definitions.
– Kavi Rama Murthy
2 days ago
The usual definition of a positive operator on a function space is $I^{+}f geq 0$ if $f geq 0$. There is no (non-zero) linear operator $T$ on $L^{p}$ for which $Tf geq 0$ for all $f$, so you have to go back to your source and find the correct definitions.
– Kavi Rama Murthy
2 days ago
that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
– Robson
2 days ago
that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
– Robson
2 days ago
add a comment |
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$1leq p<infty$
– Robson
2 days ago
2
Continuity is the same as bounded? $I$ is a linear map between normed spaces, so being continuous is the same as being bounded.
– Lucas
2 days ago