Decomposition of a continuous linear functional on $L^p$ into two positive continuous l.f.
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
add a comment |
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
$1leq p<infty$
– 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
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 Nov 16 at 1:17
Robson
725221
725221
$1leq p<infty$
– 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
add a comment |
$1leq p<infty$
– 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
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
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
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
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
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
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
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 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
add a comment |
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
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000586%2fdecomposition-of-a-continuous-linear-functional-on-lp-into-two-positive-conti%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$1leq p<infty$
– 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