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?










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  • $1leq p<infty$
    – Robson
    Nov 16 at 1:17






  • 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
    Nov 16 at 3:55















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?










share|cite|improve this question






















  • $1leq p<infty$
    – Robson
    Nov 16 at 1:17






  • 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
    Nov 16 at 3:55













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?










share|cite|improve this question














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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asked Nov 16 at 1:17









Robson

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  • $1leq p<infty$
    – Robson
    Nov 16 at 1:17






  • 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
    Nov 16 at 3:55


















  • $1leq p<infty$
    – Robson
    Nov 16 at 1:17






  • 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
    Nov 16 at 3:55
















$1leq p<infty$
– Robson
Nov 16 at 1:17




$1leq p<infty$
– Robson
Nov 16 at 1:17




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
Nov 16 at 3:55




Continuity is the same as bounded? $I$ is a linear map between normed spaces, so being continuous is the same as being bounded.
– Lucas
Nov 16 at 3:55










1 Answer
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1
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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.






share|cite|improve this answer























  • Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
    – Robson
    Nov 16 at 6:19






  • 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
    Nov 16 at 6:21












  • that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
    – Robson
    Nov 16 at 6:35













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1 Answer
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1 Answer
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active

oldest

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active

oldest

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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.






share|cite|improve this answer























  • Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
    – Robson
    Nov 16 at 6:19






  • 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
    Nov 16 at 6:21












  • that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
    – Robson
    Nov 16 at 6:35

















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.






share|cite|improve this answer























  • Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
    – Robson
    Nov 16 at 6:19






  • 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
    Nov 16 at 6:21












  • that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
    – Robson
    Nov 16 at 6:35















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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 16 at 6:23

























answered Nov 16 at 6:04









Kavi Rama Murthy

42.1k31751




42.1k31751












  • Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
    – Robson
    Nov 16 at 6:19






  • 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
    Nov 16 at 6:21












  • that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
    – Robson
    Nov 16 at 6:35




















  • Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
    – Robson
    Nov 16 at 6:19






  • 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
    Nov 16 at 6:21












  • that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
    – Robson
    Nov 16 at 6:35


















Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
– Robson
Nov 16 at 6:19




Hmm... but if $f$ is negative can I conclude that $I^+(f) geq 0$ ?
– Robson
Nov 16 at 6:19




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
Nov 16 at 6:21






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
Nov 16 at 6:21














that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
– Robson
Nov 16 at 6:35






that's right, because $T(f)=T(-(-f))=-T(-f)$ and can't be the case that both of them are positive! Thanks
– Robson
Nov 16 at 6:35




















 

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