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How is $|f|_p geq bigg( int_A |f|^p d mu bigg)^{1/p}$?
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How is $|f|_p geq bigg( int_A |f|^p d mu bigg)^{1/p}$?
This is presented as "trivial inequality". However since this is essentially the definition of $|f|_p$, then I don't see how $>$ is trivial?
https://ocw.mit.edu/courses/mathematics/18-125-measure-and-integration-fall-2003/lecture-notes/18125_lec17.pdf
real-analysis lp-spaces
asked Jan 30 at 13:34
mavaviljmavavilj
2,84911138
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add a comment |
$begingroup$
How is $|f|_p geq bigg( int_A |f|^p d mu bigg)^{1/p}$?
This is presented as "trivial inequality". However since this is essentially the definition of $|f|_p$, then I don't see how $>$ is trivial?
https://ocw.mit.edu/courses/mathematics/18-125-measure-and-integration-fall-2003/lecture-notes/18125_lec17.pdf
real-analysis lp-spaces
asked Jan 30 at 13:34
mavaviljmavavilj
2,84911138
$endgroup$
add a comment |
$begingroup$
How is $|f|_p geq bigg( int_A |f|^p d mu bigg)^{1/p}$?
This is presented as "trivial inequality". However since this is essentially the definition of $|f|_p$, then I don't see how $>$ is trivial?
https://ocw.mit.edu/courses/mathematics/18-125-measure-and-integration-fall-2003/lecture-notes/18125_lec17.pdf
real-analysis lp-spaces
asked Jan 30 at 13:34
mavaviljmavavilj
2,84911138
$endgroup$
How is $|f|_p geq bigg( int_A |f|^p d mu bigg)^{1/p}$?
This is presented as "trivial inequality". However since this is essentially the definition of $|f|_p$, then I don't see how $>$ is trivial?
https://ocw.mit.edu/courses/mathematics/18-125-measure-and-integration-fall-2003/lecture-notes/18125_lec17.pdf
real-analysis lp-spaces
real-analysis lp-spaces
asked Jan 30 at 13:34
mavaviljmavavilj
2,84911138
asked Jan 30 at 13:34
mavaviljmavavilj
2,84911138
edited Jan 30 at 14:04
user587192
asked Jan 30 at 13:34
mavaviljmavavilj
2,84911138
asked Jan 30 at 13:34
mavaviljmavavilj
2,84911138
asked Jan 30 at 13:34
mavaviljmavavilj
2,84911138
2,84911138
add a comment |
add a comment |
1 Answer
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Because the set $A$ may be a subset of the "whole" space, say $mathbb{R}$, where you define the $L^p$ norm, and the right-hand side of your inequality is
$$
left(int_{mathbb{R}} |f|^p1_A dmuright)^{1/p}.
$$
But
$$
|f|_p=bigg(int_{mathbb{R}}|f|^p dmubigg)^{1/p}.
$$
[Added: On the other hand, note that it is trivially true that $|f|^pge |f|^p1_A$. But Lebesuge integrals have the property that $int |f|leq int |g|$ if $|f|le |g|$.]
Consider for instance $f(x)=e^{-x^2}$, $A=[0,1]$ and $mu$ the Lebesgue measure on $mathbb{R}$. In this case you have the strict inequality.
An even simpler example: let $f(x)=1_{[0,2]}(x)$ and $A=[0,1]$.
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1
$begingroup$
Good answer, but it is probably a good idea to stress that the crucial point is monotonicity of the Lebesgue integral and the fact that $|f|^pge1_A |f|^p$.
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– Mars Plastic
Jan 30 at 13:49
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@MarsPlastic: Good comment :-) I have added a note.
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– user587192
Jan 30 at 13:55
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because the set $A$ may be a subset of the "whole" space, say $mathbb{R}$, where you define the $L^p$ norm, and the right-hand side of your inequality is
$$
left(int_{mathbb{R}} |f|^p1_A dmuright)^{1/p}.
$$
But
$$
|f|_p=bigg(int_{mathbb{R}}|f|^p dmubigg)^{1/p}.
$$
[Added: On the other hand, note that it is trivially true that $|f|^pge |f|^p1_A$. But Lebesuge integrals have the property that $int |f|leq int |g|$ if $|f|le |g|$.]
Consider for instance $f(x)=e^{-x^2}$, $A=[0,1]$ and $mu$ the Lebesgue measure on $mathbb{R}$. In this case you have the strict inequality.
An even simpler example: let $f(x)=1_{[0,2]}(x)$ and $A=[0,1]$.
$endgroup$
1
$begingroup$
Good answer, but it is probably a good idea to stress that the crucial point is monotonicity of the Lebesgue integral and the fact that $|f|^pge1_A |f|^p$.
$endgroup$
– Mars Plastic
Jan 30 at 13:49
$begingroup$
@MarsPlastic: Good comment :-) I have added a note.
$endgroup$
– user587192
Jan 30 at 13:55
add a comment |
$begingroup$
Because the set $A$ may be a subset of the "whole" space, say $mathbb{R}$, where you define the $L^p$ norm, and the right-hand side of your inequality is
$$
left(int_{mathbb{R}} |f|^p1_A dmuright)^{1/p}.
$$
But
$$
|f|_p=bigg(int_{mathbb{R}}|f|^p dmubigg)^{1/p}.
$$
[Added: On the other hand, note that it is trivially true that $|f|^pge |f|^p1_A$. But Lebesuge integrals have the property that $int |f|leq int |g|$ if $|f|le |g|$.]
Consider for instance $f(x)=e^{-x^2}$, $A=[0,1]$ and $mu$ the Lebesgue measure on $mathbb{R}$. In this case you have the strict inequality.
An even simpler example: let $f(x)=1_{[0,2]}(x)$ and $A=[0,1]$.
$endgroup$
1
$begingroup$
Good answer, but it is probably a good idea to stress that the crucial point is monotonicity of the Lebesgue integral and the fact that $|f|^pge1_A |f|^p$.
$endgroup$
– Mars Plastic
Jan 30 at 13:49
$begingroup$
@MarsPlastic: Good comment :-) I have added a note.
$endgroup$
– user587192
Jan 30 at 13:55
add a comment |
$begingroup$
Because the set $A$ may be a subset of the "whole" space, say $mathbb{R}$, where you define the $L^p$ norm, and the right-hand side of your inequality is
$$
left(int_{mathbb{R}} |f|^p1_A dmuright)^{1/p}.
$$
But
$$
|f|_p=bigg(int_{mathbb{R}}|f|^p dmubigg)^{1/p}.
$$
[Added: On the other hand, note that it is trivially true that $|f|^pge |f|^p1_A$. But Lebesuge integrals have the property that $int |f|leq int |g|$ if $|f|le |g|$.]
Consider for instance $f(x)=e^{-x^2}$, $A=[0,1]$ and $mu$ the Lebesgue measure on $mathbb{R}$. In this case you have the strict inequality.
An even simpler example: let $f(x)=1_{[0,2]}(x)$ and $A=[0,1]$.
$endgroup$
Because the set $A$ may be a subset of the "whole" space, say $mathbb{R}$, where you define the $L^p$ norm, and the right-hand side of your inequality is
$$
left(int_{mathbb{R}} |f|^p1_A dmuright)^{1/p}.
$$
But
$$
|f|_p=bigg(int_{mathbb{R}}|f|^p dmubigg)^{1/p}.
$$
[Added: On the other hand, note that it is trivially true that $|f|^pge |f|^p1_A$. But Lebesuge integrals have the property that $int |f|leq int |g|$ if $|f|le |g|$.]
Consider for instance $f(x)=e^{-x^2}$, $A=[0,1]$ and $mu$ the Lebesgue measure on $mathbb{R}$. In this case you have the strict inequality.
An even simpler example: let $f(x)=1_{[0,2]}(x)$ and $A=[0,1]$.
edited Jan 30 at 13:55
answered Jan 30 at 13:38
user587192
1
$begingroup$
Good answer, but it is probably a good idea to stress that the crucial point is monotonicity of the Lebesgue integral and the fact that $|f|^pge1_A |f|^p$.
$endgroup$
– Mars Plastic
Jan 30 at 13:49
$begingroup$
@MarsPlastic: Good comment :-) I have added a note.
$endgroup$
– user587192
Jan 30 at 13:55
add a comment |
1
$begingroup$
Good answer, but it is probably a good idea to stress that the crucial point is monotonicity of the Lebesgue integral and the fact that $|f|^pge1_A |f|^p$.
$endgroup$
– Mars Plastic
Jan 30 at 13:49
$begingroup$
@MarsPlastic: Good comment :-) I have added a note.
$endgroup$
– user587192
Jan 30 at 13:55
1
1
$begingroup$
Good answer, but it is probably a good idea to stress that the crucial point is monotonicity of the Lebesgue integral and the fact that $|f|^pge1_A |f|^p$.
$endgroup$
– Mars Plastic
Jan 30 at 13:49
$begingroup$
Good answer, but it is probably a good idea to stress that the crucial point is monotonicity of the Lebesgue integral and the fact that $|f|^pge1_A |f|^p$.
$endgroup$
– Mars Plastic
Jan 30 at 13:49
$begingroup$
@MarsPlastic: Good comment :-) I have added a note.
$endgroup$
– user587192
Jan 30 at 13:55
$begingroup$
@MarsPlastic: Good comment :-) I have added a note.
$endgroup$
– user587192
Jan 30 at 13:55
add a comment |
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