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For $f$ that is integrable on any set of finite measure $|f|_p=sup_gtext simple with left|int fgright|$.
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In Stein & Shakarchi's Functional analysis book, the following lemma is used for the proof of Riesz Thorin interpolation theorem.
Let $pin[1,infty]$ and $p'$ the conjugate of $p$, i.e. so that
$1/p+1/p'=1$. Then for $f$ that is integrable on any set of finitemeasure $|f|_p=sup_gtext simple with left|int fgright|$.
Correct me if I am wrong, it seems like the book only proved the lemma for real-valued functions, however the Riesz Thorin interpolation theorem requires the underlying field to be complex, so we need the lemma to work for complex-valued function. However, the proof for the real version make use of the sign function and it being a simple function, but the complex analog $f/|f|$ is not simple. So I don't know how to generalised the proof to the complex case.
After a lot of effort, I think I have come up with a proof, but it doesn't work for the case $p=1$
If $|f|_p<infty$ and $pnot=1$, then by the Riesz representation theorem, $L^pcong(L^p')'$ isometrically via $fmapstolanglecdot,frangle$, we have
$$|f|_p=sup_left|int fg,dmuright|$$
We can pick $g_n$ simple with $g_nto g$ and $|g_n|leq|g|$, then $|g_n|_p'leq 1$ and by dominated convergence $int fg_n,dmutoint fg,dmu$. Hence the result. If $p=1$, we can just take $g=bar f/|f|$.
Suppose now $|f|_p=infty$ and $pnot=1$, we need to show that the RHS of $(*)$ is not finite. It suffice to show that if the RHS of $(*)$ is finite, then $|f|_p<infty$. Denote the RHS of $(*)$ as $M$. Let $D$ be the space of all simple function that belongs to $L^p'$. Define $T:Dtomathbb C$ by $T(g)=int fg,dmu$, this integral indeed exist since $f$ is locally integrable and $g$ is simple. Now $T$ is a bounded linear map with norm $M$. Since $D$ is dense in $L^p'$, the map $T$ can be extended uniquely to a bounded linear map $L^p'tomathbb C$ with norm $M$. By the Riesz representation theorem, $T$ is represented by some $tildefin L^p$. We know
$$intunderbraceRe(f-tildef)_:=hmathbf 1_Admu+iintIm(f-tildef)mathbf 1_Admu=int(f-tildef)mathbf 1_Admu=0$$
for any set $A$ of finite measure. So $int hmathbf 1_A,dmu=0$ for any set $A$ of finite measure. We claim that $h=0$ a.e. Suppose not, then there exist $epsilon>0$ and a set $A$ of finite non-zero measure such that $epsilonmathbf 1_Aleq|h|$. Then
$$0<epsilonmu(A)leqint|h|dmu=int h(mathbf 1_Acap-mathbf 1_Acap<0)dmu=0.$$
This is a contradiction, so $h=Re(f-tildef)=0$ a.e. Similarly, $Im(f-tildef)=0$ a.e. So $f=tildef$ a.e. So $|f|_p=M$.
Is this proof for the $pnot=1$ case correct? Is there a way to prove the $p=1$ case? Is there a direct way to generlised the proof found in Stein & Shakarchi's book without appealing to the Riesz representation theorem? (The book actaully used this lemma to prove the Riesz representation theorem) Thanks very much in advance!
real-analysis complex-analysis functional-analysis analysis
asked Jul 25 at 15:22
lamkingming
1237
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In Stein & Shakarchi's Functional analysis book, the following lemma is used for the proof of Riesz Thorin interpolation theorem.
Let $pin[1,infty]$ and $p'$ the conjugate of $p$, i.e. so that
$1/p+1/p'=1$. Then for $f$ that is integrable on any set of finitemeasure $|f|_p=sup_gtext simple with left|int fgright|$.
Correct me if I am wrong, it seems like the book only proved the lemma for real-valued functions, however the Riesz Thorin interpolation theorem requires the underlying field to be complex, so we need the lemma to work for complex-valued function. However, the proof for the real version make use of the sign function and it being a simple function, but the complex analog $f/|f|$ is not simple. So I don't know how to generalised the proof to the complex case.
After a lot of effort, I think I have come up with a proof, but it doesn't work for the case $p=1$
If $|f|_p<infty$ and $pnot=1$, then by the Riesz representation theorem, $L^pcong(L^p')'$ isometrically via $fmapstolanglecdot,frangle$, we have
$$|f|_p=sup_left|int fg,dmuright|$$
We can pick $g_n$ simple with $g_nto g$ and $|g_n|leq|g|$, then $|g_n|_p'leq 1$ and by dominated convergence $int fg_n,dmutoint fg,dmu$. Hence the result. If $p=1$, we can just take $g=bar f/|f|$.
Suppose now $|f|_p=infty$ and $pnot=1$, we need to show that the RHS of $(*)$ is not finite. It suffice to show that if the RHS of $(*)$ is finite, then $|f|_p<infty$. Denote the RHS of $(*)$ as $M$. Let $D$ be the space of all simple function that belongs to $L^p'$. Define $T:Dtomathbb C$ by $T(g)=int fg,dmu$, this integral indeed exist since $f$ is locally integrable and $g$ is simple. Now $T$ is a bounded linear map with norm $M$. Since $D$ is dense in $L^p'$, the map $T$ can be extended uniquely to a bounded linear map $L^p'tomathbb C$ with norm $M$. By the Riesz representation theorem, $T$ is represented by some $tildefin L^p$. We know
$$intunderbraceRe(f-tildef)_:=hmathbf 1_Admu+iintIm(f-tildef)mathbf 1_Admu=int(f-tildef)mathbf 1_Admu=0$$
for any set $A$ of finite measure. So $int hmathbf 1_A,dmu=0$ for any set $A$ of finite measure. We claim that $h=0$ a.e. Suppose not, then there exist $epsilon>0$ and a set $A$ of finite non-zero measure such that $epsilonmathbf 1_Aleq|h|$. Then
$$0<epsilonmu(A)leqint|h|dmu=int h(mathbf 1_Acap-mathbf 1_Acap<0)dmu=0.$$
This is a contradiction, so $h=Re(f-tildef)=0$ a.e. Similarly, $Im(f-tildef)=0$ a.e. So $f=tildef$ a.e. So $|f|_p=M$.
Is this proof for the $pnot=1$ case correct? Is there a way to prove the $p=1$ case? Is there a direct way to generlised the proof found in Stein & Shakarchi's book without appealing to the Riesz representation theorem? (The book actaully used this lemma to prove the Riesz representation theorem) Thanks very much in advance!
real-analysis complex-analysis functional-analysis analysis
asked Jul 25 at 15:22
lamkingming
1237
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
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In Stein & Shakarchi's Functional analysis book, the following lemma is used for the proof of Riesz Thorin interpolation theorem.
Let $pin[1,infty]$ and $p'$ the conjugate of $p$, i.e. so that
$1/p+1/p'=1$. Then for $f$ that is integrable on any set of finitemeasure $|f|_p=sup_gtext simple with left|int fgright|$.
Correct me if I am wrong, it seems like the book only proved the lemma for real-valued functions, however the Riesz Thorin interpolation theorem requires the underlying field to be complex, so we need the lemma to work for complex-valued function. However, the proof for the real version make use of the sign function and it being a simple function, but the complex analog $f/|f|$ is not simple. So I don't know how to generalised the proof to the complex case.
After a lot of effort, I think I have come up with a proof, but it doesn't work for the case $p=1$
If $|f|_p<infty$ and $pnot=1$, then by the Riesz representation theorem, $L^pcong(L^p')'$ isometrically via $fmapstolanglecdot,frangle$, we have
$$|f|_p=sup_left|int fg,dmuright|$$
We can pick $g_n$ simple with $g_nto g$ and $|g_n|leq|g|$, then $|g_n|_p'leq 1$ and by dominated convergence $int fg_n,dmutoint fg,dmu$. Hence the result. If $p=1$, we can just take $g=bar f/|f|$.
Suppose now $|f|_p=infty$ and $pnot=1$, we need to show that the RHS of $(*)$ is not finite. It suffice to show that if the RHS of $(*)$ is finite, then $|f|_p<infty$. Denote the RHS of $(*)$ as $M$. Let $D$ be the space of all simple function that belongs to $L^p'$. Define $T:Dtomathbb C$ by $T(g)=int fg,dmu$, this integral indeed exist since $f$ is locally integrable and $g$ is simple. Now $T$ is a bounded linear map with norm $M$. Since $D$ is dense in $L^p'$, the map $T$ can be extended uniquely to a bounded linear map $L^p'tomathbb C$ with norm $M$. By the Riesz representation theorem, $T$ is represented by some $tildefin L^p$. We know
$$intunderbraceRe(f-tildef)_:=hmathbf 1_Admu+iintIm(f-tildef)mathbf 1_Admu=int(f-tildef)mathbf 1_Admu=0$$
for any set $A$ of finite measure. So $int hmathbf 1_A,dmu=0$ for any set $A$ of finite measure. We claim that $h=0$ a.e. Suppose not, then there exist $epsilon>0$ and a set $A$ of finite non-zero measure such that $epsilonmathbf 1_Aleq|h|$. Then
$$0<epsilonmu(A)leqint|h|dmu=int h(mathbf 1_Acap-mathbf 1_Acap<0)dmu=0.$$
This is a contradiction, so $h=Re(f-tildef)=0$ a.e. Similarly, $Im(f-tildef)=0$ a.e. So $f=tildef$ a.e. So $|f|_p=M$.
Is this proof for the $pnot=1$ case correct? Is there a way to prove the $p=1$ case? Is there a direct way to generlised the proof found in Stein & Shakarchi's book without appealing to the Riesz representation theorem? (The book actaully used this lemma to prove the Riesz representation theorem) Thanks very much in advance!
real-analysis complex-analysis functional-analysis analysis
asked Jul 25 at 15:22
lamkingming
1237
In Stein & Shakarchi's Functional analysis book, the following lemma is used for the proof of Riesz Thorin interpolation theorem.
Let $pin[1,infty]$ and $p'$ the conjugate of $p$, i.e. so that
$1/p+1/p'=1$. Then for $f$ that is integrable on any set of finitemeasure $|f|_p=sup_gtext simple with left|int fgright|$.
Correct me if I am wrong, it seems like the book only proved the lemma for real-valued functions, however the Riesz Thorin interpolation theorem requires the underlying field to be complex, so we need the lemma to work for complex-valued function. However, the proof for the real version make use of the sign function and it being a simple function, but the complex analog $f/|f|$ is not simple. So I don't know how to generalised the proof to the complex case.
After a lot of effort, I think I have come up with a proof, but it doesn't work for the case $p=1$
If $|f|_p<infty$ and $pnot=1$, then by the Riesz representation theorem, $L^pcong(L^p')'$ isometrically via $fmapstolanglecdot,frangle$, we have
$$|f|_p=sup_left|int fg,dmuright|$$
We can pick $g_n$ simple with $g_nto g$ and $|g_n|leq|g|$, then $|g_n|_p'leq 1$ and by dominated convergence $int fg_n,dmutoint fg,dmu$. Hence the result. If $p=1$, we can just take $g=bar f/|f|$.
Suppose now $|f|_p=infty$ and $pnot=1$, we need to show that the RHS of $(*)$ is not finite. It suffice to show that if the RHS of $(*)$ is finite, then $|f|_p<infty$. Denote the RHS of $(*)$ as $M$. Let $D$ be the space of all simple function that belongs to $L^p'$. Define $T:Dtomathbb C$ by $T(g)=int fg,dmu$, this integral indeed exist since $f$ is locally integrable and $g$ is simple. Now $T$ is a bounded linear map with norm $M$. Since $D$ is dense in $L^p'$, the map $T$ can be extended uniquely to a bounded linear map $L^p'tomathbb C$ with norm $M$. By the Riesz representation theorem, $T$ is represented by some $tildefin L^p$. We know
$$intunderbraceRe(f-tildef)_:=hmathbf 1_Admu+iintIm(f-tildef)mathbf 1_Admu=int(f-tildef)mathbf 1_Admu=0$$
for any set $A$ of finite measure. So $int hmathbf 1_A,dmu=0$ for any set $A$ of finite measure. We claim that $h=0$ a.e. Suppose not, then there exist $epsilon>0$ and a set $A$ of finite non-zero measure such that $epsilonmathbf 1_Aleq|h|$. Then
$$0<epsilonmu(A)leqint|h|dmu=int h(mathbf 1_Acap-mathbf 1_Acap<0)dmu=0.$$
This is a contradiction, so $h=Re(f-tildef)=0$ a.e. Similarly, $Im(f-tildef)=0$ a.e. So $f=tildef$ a.e. So $|f|_p=M$.
Is this proof for the $pnot=1$ case correct? Is there a way to prove the $p=1$ case? Is there a direct way to generlised the proof found in Stein & Shakarchi's book without appealing to the Riesz representation theorem? (The book actaully used this lemma to prove the Riesz representation theorem) Thanks very much in advance!
real-analysis complex-analysis functional-analysis analysis
asked Jul 25 at 15:22
lamkingming
1237
asked Jul 25 at 15:22
lamkingming
1237
asked Jul 25 at 15:22
lamkingming
1237
asked Jul 25 at 15:22
lamkingming
1237
1237
add a comment |Â
add a comment |Â
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