$sigma$-finite measure $mu$ so that $L^p(mu) subsetneq L^q(mu)$ (proper subset)
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I'm looking for a $sigma$-finite measure $mu$ and a measure space so that for
$1 le p <q le infty$
$$L^p(mu) subsetneq L^q(mu)$$
I tried the following:
Let $1 le p <q le infty$ and $lambda$ the Lebesgue measure on $(1,infty)$ which is $sigma$-finite.
$x^alpha$ is integrable on $(1,infty) Leftrightarrow alpha <-1$.
Choose $b$ so that $1/q<b<1/p Leftrightarrow -bq<-1, -bp>-1$.
Then $x^{-b}chi_{(1,infty)}$$in L^q$ but $notin L^p$ because $x^{-bq}$ is integrable because the exponent $-bq<-1$ and $x^{-bp}$ isn't integrable because the exponent $-bp>-1$. Now I found a function that is in $L^p$ but not in $L^q$. But that doesn't really show that $L^p subsetneq L^q$, meaning $L^p$ is a proper subset of $L^q$, right (because I don't know if every element of $L^p$ is also an element of $L^q$)?
Thanks in advance!
real-analysis analysis measure-theory
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up vote
0
down vote
favorite
I'm looking for a $sigma$-finite measure $mu$ and a measure space so that for
$1 le p <q le infty$
$$L^p(mu) subsetneq L^q(mu)$$
I tried the following:
Let $1 le p <q le infty$ and $lambda$ the Lebesgue measure on $(1,infty)$ which is $sigma$-finite.
$x^alpha$ is integrable on $(1,infty) Leftrightarrow alpha <-1$.
Choose $b$ so that $1/q<b<1/p Leftrightarrow -bq<-1, -bp>-1$.
Then $x^{-b}chi_{(1,infty)}$$in L^q$ but $notin L^p$ because $x^{-bq}$ is integrable because the exponent $-bq<-1$ and $x^{-bp}$ isn't integrable because the exponent $-bp>-1$. Now I found a function that is in $L^p$ but not in $L^q$. But that doesn't really show that $L^p subsetneq L^q$, meaning $L^p$ is a proper subset of $L^q$, right (because I don't know if every element of $L^p$ is also an element of $L^q$)?
Thanks in advance!
real-analysis analysis measure-theory
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm looking for a $sigma$-finite measure $mu$ and a measure space so that for
$1 le p <q le infty$
$$L^p(mu) subsetneq L^q(mu)$$
I tried the following:
Let $1 le p <q le infty$ and $lambda$ the Lebesgue measure on $(1,infty)$ which is $sigma$-finite.
$x^alpha$ is integrable on $(1,infty) Leftrightarrow alpha <-1$.
Choose $b$ so that $1/q<b<1/p Leftrightarrow -bq<-1, -bp>-1$.
Then $x^{-b}chi_{(1,infty)}$$in L^q$ but $notin L^p$ because $x^{-bq}$ is integrable because the exponent $-bq<-1$ and $x^{-bp}$ isn't integrable because the exponent $-bp>-1$. Now I found a function that is in $L^p$ but not in $L^q$. But that doesn't really show that $L^p subsetneq L^q$, meaning $L^p$ is a proper subset of $L^q$, right (because I don't know if every element of $L^p$ is also an element of $L^q$)?
Thanks in advance!
real-analysis analysis measure-theory
I'm looking for a $sigma$-finite measure $mu$ and a measure space so that for
$1 le p <q le infty$
$$L^p(mu) subsetneq L^q(mu)$$
I tried the following:
Let $1 le p <q le infty$ and $lambda$ the Lebesgue measure on $(1,infty)$ which is $sigma$-finite.
$x^alpha$ is integrable on $(1,infty) Leftrightarrow alpha <-1$.
Choose $b$ so that $1/q<b<1/p Leftrightarrow -bq<-1, -bp>-1$.
Then $x^{-b}chi_{(1,infty)}$$in L^q$ but $notin L^p$ because $x^{-bq}$ is integrable because the exponent $-bq<-1$ and $x^{-bp}$ isn't integrable because the exponent $-bp>-1$. Now I found a function that is in $L^p$ but not in $L^q$. But that doesn't really show that $L^p subsetneq L^q$, meaning $L^p$ is a proper subset of $L^q$, right (because I don't know if every element of $L^p$ is also an element of $L^q$)?
Thanks in advance!
real-analysis analysis measure-theory
real-analysis analysis measure-theory
asked Nov 13 at 17:04
user610431
627
627
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The inclusion $L^p(mu) subset L^q(mu)$ for $1leq p < q leq +infty$ holds if and only if
$$
inf{mu(A): Ainmathcal{M}, mu(A)>0} > 0.
$$
Hence your $mu$ must necessarily be an atomic measure.
For example, you can consider the spaces $ell^p$. More precisely, you equip $mathbb{N}$ with the counting measure, so that
$$
ell^p := {x = (x_1, x_2, ldots): sum_{j=1}^infty |x_j|^p < infty}.
$$
Thanks, but unfortunately I don't really understand what exactly you mean. Can you give me an example of such an measure?
– user610431
Nov 13 at 17:57
Done. (Edit in the main text.)
– Rigel
Nov 14 at 6:41
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The inclusion $L^p(mu) subset L^q(mu)$ for $1leq p < q leq +infty$ holds if and only if
$$
inf{mu(A): Ainmathcal{M}, mu(A)>0} > 0.
$$
Hence your $mu$ must necessarily be an atomic measure.
For example, you can consider the spaces $ell^p$. More precisely, you equip $mathbb{N}$ with the counting measure, so that
$$
ell^p := {x = (x_1, x_2, ldots): sum_{j=1}^infty |x_j|^p < infty}.
$$
Thanks, but unfortunately I don't really understand what exactly you mean. Can you give me an example of such an measure?
– user610431
Nov 13 at 17:57
Done. (Edit in the main text.)
– Rigel
Nov 14 at 6:41
add a comment |
up vote
0
down vote
The inclusion $L^p(mu) subset L^q(mu)$ for $1leq p < q leq +infty$ holds if and only if
$$
inf{mu(A): Ainmathcal{M}, mu(A)>0} > 0.
$$
Hence your $mu$ must necessarily be an atomic measure.
For example, you can consider the spaces $ell^p$. More precisely, you equip $mathbb{N}$ with the counting measure, so that
$$
ell^p := {x = (x_1, x_2, ldots): sum_{j=1}^infty |x_j|^p < infty}.
$$
Thanks, but unfortunately I don't really understand what exactly you mean. Can you give me an example of such an measure?
– user610431
Nov 13 at 17:57
Done. (Edit in the main text.)
– Rigel
Nov 14 at 6:41
add a comment |
up vote
0
down vote
up vote
0
down vote
The inclusion $L^p(mu) subset L^q(mu)$ for $1leq p < q leq +infty$ holds if and only if
$$
inf{mu(A): Ainmathcal{M}, mu(A)>0} > 0.
$$
Hence your $mu$ must necessarily be an atomic measure.
For example, you can consider the spaces $ell^p$. More precisely, you equip $mathbb{N}$ with the counting measure, so that
$$
ell^p := {x = (x_1, x_2, ldots): sum_{j=1}^infty |x_j|^p < infty}.
$$
The inclusion $L^p(mu) subset L^q(mu)$ for $1leq p < q leq +infty$ holds if and only if
$$
inf{mu(A): Ainmathcal{M}, mu(A)>0} > 0.
$$
Hence your $mu$ must necessarily be an atomic measure.
For example, you can consider the spaces $ell^p$. More precisely, you equip $mathbb{N}$ with the counting measure, so that
$$
ell^p := {x = (x_1, x_2, ldots): sum_{j=1}^infty |x_j|^p < infty}.
$$
edited Nov 14 at 6:40
answered Nov 13 at 17:30
Rigel
10.6k11319
10.6k11319
Thanks, but unfortunately I don't really understand what exactly you mean. Can you give me an example of such an measure?
– user610431
Nov 13 at 17:57
Done. (Edit in the main text.)
– Rigel
Nov 14 at 6:41
add a comment |
Thanks, but unfortunately I don't really understand what exactly you mean. Can you give me an example of such an measure?
– user610431
Nov 13 at 17:57
Done. (Edit in the main text.)
– Rigel
Nov 14 at 6:41
Thanks, but unfortunately I don't really understand what exactly you mean. Can you give me an example of such an measure?
– user610431
Nov 13 at 17:57
Thanks, but unfortunately I don't really understand what exactly you mean. Can you give me an example of such an measure?
– user610431
Nov 13 at 17:57
Done. (Edit in the main text.)
– Rigel
Nov 14 at 6:41
Done. (Edit in the main text.)
– Rigel
Nov 14 at 6:41
add a comment |
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