Help with the step II in the proof of Riesz Representation Theorem in big Rudin
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In Riesz Representation Theorem, $X$ is a locally compact Hausdorff space, and $Lambda$ is a positive linear functional on $C_c(X)$ which is the set of all continuous functions on $X$ with compact support. One conclusion of the theorem is that $mu(K) < infty$ for all compact set $K$.
This conclusion is proved in step II (page 43, Rudin's Real and Complex Analysis, third edition). In Rudin's proof, he let $K prec f$ ($f$ belongs to $C_c(X)$, take value $1$ on $K$, and $0 leq f leq 1$), then deduce $mu(K) leq Lambda f$ and assert $mu(K) < infty$.
What I don't understand is that can assert $mu(K) < infty$ just from $mu(K) leq Lambda f$? Can $Lambda f$ be infinite?
functional-analysis analysis measure-theory lebesgue-measure riesz-representation-theorem
edited Nov 19 at 17:37
Brahadeesh
5,89342159
asked Sep 13 '17 at 14:37
yangxs
61
add a comment |
up vote
1
down vote
favorite
In Riesz Representation Theorem, $X$ is a locally compact Hausdorff space, and $Lambda$ is a positive linear functional on $C_c(X)$ which is the set of all continuous functions on $X$ with compact support. One conclusion of the theorem is that $mu(K) < infty$ for all compact set $K$.
This conclusion is proved in step II (page 43, Rudin's Real and Complex Analysis, third edition). In Rudin's proof, he let $K prec f$ ($f$ belongs to $C_c(X)$, take value $1$ on $K$, and $0 leq f leq 1$), then deduce $mu(K) leq Lambda f$ and assert $mu(K) < infty$.
What I don't understand is that can assert $mu(K) < infty$ just from $mu(K) leq Lambda f$? Can $Lambda f$ be infinite?
functional-analysis analysis measure-theory lebesgue-measure riesz-representation-theorem
edited Nov 19 at 17:37
Brahadeesh
5,89342159
asked Sep 13 '17 at 14:37
yangxs
61
So he did: he first proved $mu(K)lecdotsle Lambda(2f)$, then, since $Lambda f<infty$, concluded that $mu(K)le Lambda(2f)<infty$.
– Vim
Sep 13 '17 at 14:42
I don't understand why Λf<∞. Can it be ∞?
– yangxs
Sep 13 '17 at 15:29
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In Riesz Representation Theorem, $X$ is a locally compact Hausdorff space, and $Lambda$ is a positive linear functional on $C_c(X)$ which is the set of all continuous functions on $X$ with compact support. One conclusion of the theorem is that $mu(K) < infty$ for all compact set $K$.
This conclusion is proved in step II (page 43, Rudin's Real and Complex Analysis, third edition). In Rudin's proof, he let $K prec f$ ($f$ belongs to $C_c(X)$, take value $1$ on $K$, and $0 leq f leq 1$), then deduce $mu(K) leq Lambda f$ and assert $mu(K) < infty$.
What I don't understand is that can assert $mu(K) < infty$ just from $mu(K) leq Lambda f$? Can $Lambda f$ be infinite?
functional-analysis analysis measure-theory lebesgue-measure riesz-representation-theorem
edited Nov 19 at 17:37
Brahadeesh
5,89342159
asked Sep 13 '17 at 14:37
yangxs
61
In Riesz Representation Theorem, $X$ is a locally compact Hausdorff space, and $Lambda$ is a positive linear functional on $C_c(X)$ which is the set of all continuous functions on $X$ with compact support. One conclusion of the theorem is that $mu(K) < infty$ for all compact set $K$.
This conclusion is proved in step II (page 43, Rudin's Real and Complex Analysis, third edition). In Rudin's proof, he let $K prec f$ ($f$ belongs to $C_c(X)$, take value $1$ on $K$, and $0 leq f leq 1$), then deduce $mu(K) leq Lambda f$ and assert $mu(K) < infty$.
What I don't understand is that can assert $mu(K) < infty$ just from $mu(K) leq Lambda f$? Can $Lambda f$ be infinite?
functional-analysis analysis measure-theory lebesgue-measure riesz-representation-theorem
functional-analysis analysis measure-theory lebesgue-measure riesz-representation-theorem
edited Nov 19 at 17:37
Brahadeesh
5,89342159
asked Sep 13 '17 at 14:37
yangxs
61
edited Nov 19 at 17:37
Brahadeesh
5,89342159
asked Sep 13 '17 at 14:37
yangxs
61
edited Nov 19 at 17:37
Brahadeesh
5,89342159
edited Nov 19 at 17:37
Brahadeesh
5,89342159
edited Nov 19 at 17:37
Brahadeesh
5,89342159
5,89342159
asked Sep 13 '17 at 14:37
yangxs
61
asked Sep 13 '17 at 14:37
yangxs
61
asked Sep 13 '17 at 14:37
yangxs
61
61
So he did: he first proved $mu(K)lecdotsle Lambda(2f)$, then, since $Lambda f<infty$, concluded that $mu(K)le Lambda(2f)<infty$.
– Vim
Sep 13 '17 at 14:42
I don't understand why Λf<∞. Can it be ∞?
– yangxs
Sep 13 '17 at 15:29
add a comment |
So he did: he first proved $mu(K)lecdotsle Lambda(2f)$, then, since $Lambda f<infty$, concluded that $mu(K)le Lambda(2f)<infty$.
– Vim
Sep 13 '17 at 14:42
I don't understand why Λf<∞. Can it be ∞?
– yangxs
Sep 13 '17 at 15:29
So he did: he first proved $mu(K)lecdotsle Lambda(2f)$, then, since $Lambda f<infty$, concluded that $mu(K)le Lambda(2f)<infty$.
– Vim
Sep 13 '17 at 14:42
So he did: he first proved $mu(K)lecdotsle Lambda(2f)$, then, since $Lambda f<infty$, concluded that $mu(K)le Lambda(2f)<infty$.
– Vim
Sep 13 '17 at 14:42
I don't understand why Λf<∞. Can it be ∞?
– yangxs
Sep 13 '17 at 15:29
I don't understand why Λf<∞. Can it be ∞?
– yangxs
Sep 13 '17 at 15:29
add a comment |
2 Answers
2
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oldest
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up vote
1
down vote
Please see the exact words below from Rudin.
Since $fin C_c(X)$ and it is continuous, it is easy to see that $f(X)$ is finite. So, as a result, $Lambda f$ has to be finite.
answered Sep 13 '17 at 15:38
Yuhang
830118
So it can be regarded as a part of definition of the linear functional Λ?
– yangxs
Sep 13 '17 at 16:10
In the last step why did he put the absolute value of $a$ ? Is not it sufficient to assume $f$ is positive?
– Neil hawking
Aug 24 at 21:37
add a comment |
up vote
0
down vote
$Lambda$ is a (positive) linear functional on $C_c(X)$, so it is a linear map from $C_c(X)$ to $mathbb{C}$. In particular, $Lambda(f) neq infty$ for any $f in C_c(X)$, since $infty notin mathbb{C}$.
The other answer by @Yuhang gives the meaning of the adjective "positive" used here to describe $Lambda$. However, that is unnecessary to know why $Lambda(f) neq infty$.
answered Nov 19 at 17:33
Brahadeesh
5,89342159
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Please see the exact words below from Rudin.
Since $fin C_c(X)$ and it is continuous, it is easy to see that $f(X)$ is finite. So, as a result, $Lambda f$ has to be finite.
answered Sep 13 '17 at 15:38
Yuhang
830118
So it can be regarded as a part of definition of the linear functional Λ?
– yangxs
Sep 13 '17 at 16:10
In the last step why did he put the absolute value of $a$ ? Is not it sufficient to assume $f$ is positive?
– Neil hawking
Aug 24 at 21:37
add a comment |
up vote
1
down vote
Please see the exact words below from Rudin.
Since $fin C_c(X)$ and it is continuous, it is easy to see that $f(X)$ is finite. So, as a result, $Lambda f$ has to be finite.
answered Sep 13 '17 at 15:38
Yuhang
830118
So it can be regarded as a part of definition of the linear functional Λ?
– yangxs
Sep 13 '17 at 16:10
In the last step why did he put the absolute value of $a$ ? Is not it sufficient to assume $f$ is positive?
– Neil hawking
Aug 24 at 21:37
add a comment |
up vote
1
down vote
up vote
1
down vote
Please see the exact words below from Rudin.
Since $fin C_c(X)$ and it is continuous, it is easy to see that $f(X)$ is finite. So, as a result, $Lambda f$ has to be finite.
answered Sep 13 '17 at 15:38
Yuhang
830118
Please see the exact words below from Rudin.
Since $fin C_c(X)$ and it is continuous, it is easy to see that $f(X)$ is finite. So, as a result, $Lambda f$ has to be finite.
answered Sep 13 '17 at 15:38
Yuhang
830118
answered Sep 13 '17 at 15:38
Yuhang
830118
answered Sep 13 '17 at 15:38
Yuhang
830118
answered Sep 13 '17 at 15:38
Yuhang
830118
830118
So it can be regarded as a part of definition of the linear functional Λ?
– yangxs
Sep 13 '17 at 16:10
In the last step why did he put the absolute value of $a$ ? Is not it sufficient to assume $f$ is positive?
– Neil hawking
Aug 24 at 21:37
add a comment |
So it can be regarded as a part of definition of the linear functional Λ?
– yangxs
Sep 13 '17 at 16:10
In the last step why did he put the absolute value of $a$ ? Is not it sufficient to assume $f$ is positive?
– Neil hawking
Aug 24 at 21:37
So it can be regarded as a part of definition of the linear functional Λ?
– yangxs
Sep 13 '17 at 16:10
So it can be regarded as a part of definition of the linear functional Λ?
– yangxs
Sep 13 '17 at 16:10
In the last step why did he put the absolute value of $a$ ? Is not it sufficient to assume $f$ is positive?
– Neil hawking
Aug 24 at 21:37
In the last step why did he put the absolute value of $a$ ? Is not it sufficient to assume $f$ is positive?
– Neil hawking
Aug 24 at 21:37
add a comment |
up vote
0
down vote
$Lambda$ is a (positive) linear functional on $C_c(X)$, so it is a linear map from $C_c(X)$ to $mathbb{C}$. In particular, $Lambda(f) neq infty$ for any $f in C_c(X)$, since $infty notin mathbb{C}$.
The other answer by @Yuhang gives the meaning of the adjective "positive" used here to describe $Lambda$. However, that is unnecessary to know why $Lambda(f) neq infty$.
answered Nov 19 at 17:33
Brahadeesh
5,89342159
add a comment |
up vote
0
down vote
$Lambda$ is a (positive) linear functional on $C_c(X)$, so it is a linear map from $C_c(X)$ to $mathbb{C}$. In particular, $Lambda(f) neq infty$ for any $f in C_c(X)$, since $infty notin mathbb{C}$.
The other answer by @Yuhang gives the meaning of the adjective "positive" used here to describe $Lambda$. However, that is unnecessary to know why $Lambda(f) neq infty$.
answered Nov 19 at 17:33
Brahadeesh
5,89342159
add a comment |
up vote
0
down vote
up vote
0
down vote
$Lambda$ is a (positive) linear functional on $C_c(X)$, so it is a linear map from $C_c(X)$ to $mathbb{C}$. In particular, $Lambda(f) neq infty$ for any $f in C_c(X)$, since $infty notin mathbb{C}$.
The other answer by @Yuhang gives the meaning of the adjective "positive" used here to describe $Lambda$. However, that is unnecessary to know why $Lambda(f) neq infty$.
answered Nov 19 at 17:33
Brahadeesh
5,89342159
$Lambda$ is a (positive) linear functional on $C_c(X)$, so it is a linear map from $C_c(X)$ to $mathbb{C}$. In particular, $Lambda(f) neq infty$ for any $f in C_c(X)$, since $infty notin mathbb{C}$.
The other answer by @Yuhang gives the meaning of the adjective "positive" used here to describe $Lambda$. However, that is unnecessary to know why $Lambda(f) neq infty$.
answered Nov 19 at 17:33
Brahadeesh
5,89342159
answered Nov 19 at 17:33
Brahadeesh
5,89342159
answered Nov 19 at 17:33
Brahadeesh
5,89342159
answered Nov 19 at 17:33
Brahadeesh
5,89342159
5,89342159
add a comment |
add a comment |
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So he did: he first proved $mu(K)lecdotsle Lambda(2f)$, then, since $Lambda f<infty$, concluded that $mu(K)le Lambda(2f)<infty$.
– Vim
Sep 13 '17 at 14:42
I don't understand why Λf<∞. Can it be ∞?
– yangxs
Sep 13 '17 at 15:29