Size of $L^infty$

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I am confused about the 'size' of $L^infty$:



One the one hand, when we consider $L^infty(Omega)$ and $mu(Omega)$ is finite (and $mu$ some measure), then we have the inclusion $L^infty subset L^p$ for some $p geq 1$. So $L^infty$ is the 'smallest' of the $L^p$ spaces.



On the other hand, we know for arbitrary open $Omega subset mathbbR^n$ that $L^infty$ is not separable, so it is in some sense 'too big' to be approximated by a countable subset, but $L^p$ for $1 leq p < infty$ are separable.



I find this somewhat confusing. What am I missing or getting wrong?



Thanks!




functional-analysis lebesgue-integral




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  • Side remark, if the measure space is atomic like $L^infty(mathbb N) = ell^infty$ then the inclusion is reversed.
    – Calvin Khor
    2 days ago






  • 2




    I'm guessing part of the issue is that $L^infty subset L^p$ as sets, but their separability involves talking about their own distinct norms. $(L^infty,||_p) $ does have a countable $L^p$-dense subset, since it is subset of the separable space $(L^p, ||_p)$.
    – Calvin Khor
    2 days ago














up vote
3
down vote

favorite
1












I am confused about the 'size' of $L^infty$:



One the one hand, when we consider $L^infty(Omega)$ and $mu(Omega)$ is finite (and $mu$ some measure), then we have the inclusion $L^infty subset L^p$ for some $p geq 1$. So $L^infty$ is the 'smallest' of the $L^p$ spaces.



On the other hand, we know for arbitrary open $Omega subset mathbbR^n$ that $L^infty$ is not separable, so it is in some sense 'too big' to be approximated by a countable subset, but $L^p$ for $1 leq p < infty$ are separable.



I find this somewhat confusing. What am I missing or getting wrong?



Thanks!




functional-analysis lebesgue-integral




share|cite|improve this question





















  • Side remark, if the measure space is atomic like $L^infty(mathbb N) = ell^infty$ then the inclusion is reversed.
    – Calvin Khor
    2 days ago






  • 2




    I'm guessing part of the issue is that $L^infty subset L^p$ as sets, but their separability involves talking about their own distinct norms. $(L^infty,||_p) $ does have a countable $L^p$-dense subset, since it is subset of the separable space $(L^p, ||_p)$.
    – Calvin Khor
    2 days ago












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





I am confused about the 'size' of $L^infty$:



One the one hand, when we consider $L^infty(Omega)$ and $mu(Omega)$ is finite (and $mu$ some measure), then we have the inclusion $L^infty subset L^p$ for some $p geq 1$. So $L^infty$ is the 'smallest' of the $L^p$ spaces.



On the other hand, we know for arbitrary open $Omega subset mathbbR^n$ that $L^infty$ is not separable, so it is in some sense 'too big' to be approximated by a countable subset, but $L^p$ for $1 leq p < infty$ are separable.



I find this somewhat confusing. What am I missing or getting wrong?



Thanks!




functional-analysis lebesgue-integral




share|cite|improve this question













I am confused about the 'size' of $L^infty$:



One the one hand, when we consider $L^infty(Omega)$ and $mu(Omega)$ is finite (and $mu$ some measure), then we have the inclusion $L^infty subset L^p$ for some $p geq 1$. So $L^infty$ is the 'smallest' of the $L^p$ spaces.



On the other hand, we know for arbitrary open $Omega subset mathbbR^n$ that $L^infty$ is not separable, so it is in some sense 'too big' to be approximated by a countable subset, but $L^p$ for $1 leq p < infty$ are separable.



I find this somewhat confusing. What am I missing or getting wrong?



Thanks!





functional-analysis lebesgue-integral





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago
























asked 2 days ago









Alex

337




337











  • Side remark, if the measure space is atomic like $L^infty(mathbb N) = ell^infty$ then the inclusion is reversed.
    – Calvin Khor
    2 days ago






  • 2




    I'm guessing part of the issue is that $L^infty subset L^p$ as sets, but their separability involves talking about their own distinct norms. $(L^infty,||_p) $ does have a countable $L^p$-dense subset, since it is subset of the separable space $(L^p, ||_p)$.
    – Calvin Khor
    2 days ago
















  • Side remark, if the measure space is atomic like $L^infty(mathbb N) = ell^infty$ then the inclusion is reversed.
    – Calvin Khor
    2 days ago






  • 2




    I'm guessing part of the issue is that $L^infty subset L^p$ as sets, but their separability involves talking about their own distinct norms. $(L^infty,||_p) $ does have a countable $L^p$-dense subset, since it is subset of the separable space $(L^p, ||_p)$.
    – Calvin Khor
    2 days ago















Side remark, if the measure space is atomic like $L^infty(mathbb N) = ell^infty$ then the inclusion is reversed.
– Calvin Khor
2 days ago




Side remark, if the measure space is atomic like $L^infty(mathbb N) = ell^infty$ then the inclusion is reversed.
– Calvin Khor
2 days ago




2




2




I'm guessing part of the issue is that $L^infty subset L^p$ as sets, but their separability involves talking about their own distinct norms. $(L^infty,||_p) $ does have a countable $L^p$-dense subset, since it is subset of the separable space $(L^p, ||_p)$.
– Calvin Khor
2 days ago




I'm guessing part of the issue is that $L^infty subset L^p$ as sets, but their separability involves talking about their own distinct norms. $(L^infty,||_p) $ does have a countable $L^p$-dense subset, since it is subset of the separable space $(L^p, ||_p)$.
– Calvin Khor
2 days ago










1 Answer
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active

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up vote
3
down vote



accepted










What's going on here, is that the separability doesn't just tell you about the size of sets in the sense of set theoretic inclusions but also tells you something about the topology on that set. On a finite measure space, $L^infty(Omega)$ is continuously embedded in $L^p(Omega)$ but it gets a very different topology than the one we would give it if we just thought of it as a subset of $L^p(Omega)$. Now let me give some details of this.



For simplicity lets fix a finite measure space $(Omega, mathcalF, mu)$ and consider $L^p(Omega)$.



In this case, when considered as sets, $L^infty(Omega) subset L^p(Omega)$. So we could consider $L^infty(Omega)$ with the subspace topology induced by the $L^p(Omega)$-norm. For this topology, $L^infty(Omega)$ would be separable (as a subspace of a separable space). Of course, we don't want to think of $L^infty(Omega)$ with the subspace topology because it comes with its own natural topology, the one induced by its usual norm!



In fact, the embedding $L^infty(Omega) hookrightarrow L^p(Omega)$ is even continuous since if $f in L^infty(Omega)$ then $$int_Omega |f|^p dmu leq mu(Omega) |f|_infty^p. $$



However, we don't have such an inequality going the other way. So whilst the embedding is continuous, it is not a homeomorphic embedding.



This means that the norm topology on $L^infty(Omega)$ is stronger than the one induced on $L^infty(Omega)$ by $L^p(Omega)$ and so it is harder to approximate functions in this topology. The lack of separability of $L^infty(Omega)$ is a symptom of the fact that its inherent notion of approximation is much stronger, rather than of its 'size'.






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  • Great, thank you!
    – Alex
    2 days ago










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










What's going on here, is that the separability doesn't just tell you about the size of sets in the sense of set theoretic inclusions but also tells you something about the topology on that set. On a finite measure space, $L^infty(Omega)$ is continuously embedded in $L^p(Omega)$ but it gets a very different topology than the one we would give it if we just thought of it as a subset of $L^p(Omega)$. Now let me give some details of this.



For simplicity lets fix a finite measure space $(Omega, mathcalF, mu)$ and consider $L^p(Omega)$.



In this case, when considered as sets, $L^infty(Omega) subset L^p(Omega)$. So we could consider $L^infty(Omega)$ with the subspace topology induced by the $L^p(Omega)$-norm. For this topology, $L^infty(Omega)$ would be separable (as a subspace of a separable space). Of course, we don't want to think of $L^infty(Omega)$ with the subspace topology because it comes with its own natural topology, the one induced by its usual norm!



In fact, the embedding $L^infty(Omega) hookrightarrow L^p(Omega)$ is even continuous since if $f in L^infty(Omega)$ then $$int_Omega |f|^p dmu leq mu(Omega) |f|_infty^p. $$



However, we don't have such an inequality going the other way. So whilst the embedding is continuous, it is not a homeomorphic embedding.



This means that the norm topology on $L^infty(Omega)$ is stronger than the one induced on $L^infty(Omega)$ by $L^p(Omega)$ and so it is harder to approximate functions in this topology. The lack of separability of $L^infty(Omega)$ is a symptom of the fact that its inherent notion of approximation is much stronger, rather than of its 'size'.






share|cite|improve this answer





















  • Great, thank you!
    – Alex
    2 days ago














up vote
3
down vote



accepted










What's going on here, is that the separability doesn't just tell you about the size of sets in the sense of set theoretic inclusions but also tells you something about the topology on that set. On a finite measure space, $L^infty(Omega)$ is continuously embedded in $L^p(Omega)$ but it gets a very different topology than the one we would give it if we just thought of it as a subset of $L^p(Omega)$. Now let me give some details of this.



For simplicity lets fix a finite measure space $(Omega, mathcalF, mu)$ and consider $L^p(Omega)$.



In this case, when considered as sets, $L^infty(Omega) subset L^p(Omega)$. So we could consider $L^infty(Omega)$ with the subspace topology induced by the $L^p(Omega)$-norm. For this topology, $L^infty(Omega)$ would be separable (as a subspace of a separable space). Of course, we don't want to think of $L^infty(Omega)$ with the subspace topology because it comes with its own natural topology, the one induced by its usual norm!



In fact, the embedding $L^infty(Omega) hookrightarrow L^p(Omega)$ is even continuous since if $f in L^infty(Omega)$ then $$int_Omega |f|^p dmu leq mu(Omega) |f|_infty^p. $$



However, we don't have such an inequality going the other way. So whilst the embedding is continuous, it is not a homeomorphic embedding.



This means that the norm topology on $L^infty(Omega)$ is stronger than the one induced on $L^infty(Omega)$ by $L^p(Omega)$ and so it is harder to approximate functions in this topology. The lack of separability of $L^infty(Omega)$ is a symptom of the fact that its inherent notion of approximation is much stronger, rather than of its 'size'.






share|cite|improve this answer





















  • Great, thank you!
    – Alex
    2 days ago












up vote
3
down vote



accepted







up vote
3
down vote



accepted






What's going on here, is that the separability doesn't just tell you about the size of sets in the sense of set theoretic inclusions but also tells you something about the topology on that set. On a finite measure space, $L^infty(Omega)$ is continuously embedded in $L^p(Omega)$ but it gets a very different topology than the one we would give it if we just thought of it as a subset of $L^p(Omega)$. Now let me give some details of this.



For simplicity lets fix a finite measure space $(Omega, mathcalF, mu)$ and consider $L^p(Omega)$.



In this case, when considered as sets, $L^infty(Omega) subset L^p(Omega)$. So we could consider $L^infty(Omega)$ with the subspace topology induced by the $L^p(Omega)$-norm. For this topology, $L^infty(Omega)$ would be separable (as a subspace of a separable space). Of course, we don't want to think of $L^infty(Omega)$ with the subspace topology because it comes with its own natural topology, the one induced by its usual norm!



In fact, the embedding $L^infty(Omega) hookrightarrow L^p(Omega)$ is even continuous since if $f in L^infty(Omega)$ then $$int_Omega |f|^p dmu leq mu(Omega) |f|_infty^p. $$



However, we don't have such an inequality going the other way. So whilst the embedding is continuous, it is not a homeomorphic embedding.



This means that the norm topology on $L^infty(Omega)$ is stronger than the one induced on $L^infty(Omega)$ by $L^p(Omega)$ and so it is harder to approximate functions in this topology. The lack of separability of $L^infty(Omega)$ is a symptom of the fact that its inherent notion of approximation is much stronger, rather than of its 'size'.






share|cite|improve this answer













What's going on here, is that the separability doesn't just tell you about the size of sets in the sense of set theoretic inclusions but also tells you something about the topology on that set. On a finite measure space, $L^infty(Omega)$ is continuously embedded in $L^p(Omega)$ but it gets a very different topology than the one we would give it if we just thought of it as a subset of $L^p(Omega)$. Now let me give some details of this.



For simplicity lets fix a finite measure space $(Omega, mathcalF, mu)$ and consider $L^p(Omega)$.



In this case, when considered as sets, $L^infty(Omega) subset L^p(Omega)$. So we could consider $L^infty(Omega)$ with the subspace topology induced by the $L^p(Omega)$-norm. For this topology, $L^infty(Omega)$ would be separable (as a subspace of a separable space). Of course, we don't want to think of $L^infty(Omega)$ with the subspace topology because it comes with its own natural topology, the one induced by its usual norm!



In fact, the embedding $L^infty(Omega) hookrightarrow L^p(Omega)$ is even continuous since if $f in L^infty(Omega)$ then $$int_Omega |f|^p dmu leq mu(Omega) |f|_infty^p. $$



However, we don't have such an inequality going the other way. So whilst the embedding is continuous, it is not a homeomorphic embedding.



This means that the norm topology on $L^infty(Omega)$ is stronger than the one induced on $L^infty(Omega)$ by $L^p(Omega)$ and so it is harder to approximate functions in this topology. The lack of separability of $L^infty(Omega)$ is a symptom of the fact that its inherent notion of approximation is much stronger, rather than of its 'size'.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered 2 days ago









Rhys Steele

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  • Great, thank you!
    – Alex
    2 days ago
















  • Great, thank you!
    – Alex
    2 days ago















Great, thank you!
– Alex
2 days ago




Great, thank you!
– Alex
2 days ago












 

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