Let $X$ be a finite measure space. Then, for any $ 1≤p<q≤+∞ $ : $ L^q(X,B,m)⊂L^p(X,B,m) $. I would like to know if the space $ L^{\infty} ( X , B , m ) $ is the direct limit or the inverse limit of the direct system or the inverse system $ ( L^p ( X , B , m ), i_{p}^q )_{p \in [1 , + \infty [ } $ with $ i_{p}^{q} : L^q ( X , B , m ) \to L^p ( X , B , m ) $ an embedding.

Thanks a lot for your help.

  • $\begingroup$ Have you tried to check if it satisfies the UP? $\endgroup$
    – Pedro
    Jun 5, 2014 at 1:45
  • $\begingroup$ What is UP please ? $\endgroup$
    – Bryan261
    Jun 5, 2014 at 1:47
  • $\begingroup$ The universal property of the direct limit. $\endgroup$
    – Pedro
    Jun 5, 2014 at 1:52
  • $\begingroup$ In which category are you considering the $L^p$ spaces? Are you considering them only as vector spaces or as Banach spaces, or maybe something else? $\endgroup$
    – user123641
    Jun 5, 2014 at 11:13
  • $\begingroup$ Also, the direction in which your $i^q_p$ are going hints at the answer. Inclusion is fine from $L^q$ to $L^p$, but it doesn't work the other way. So $i^q_p$ isn't going to create an upward directed system, thus a direct limit seems out of the question. To consider a direct limit, you need to consider a different set of morphisms (like send all $L^q$-integerable functions to themselves or else to $0$ when we consider a morphism $f:L^p\rightarrow L^q$). $\endgroup$
    – user123641
    Jun 5, 2014 at 11:37

1 Answer 1


Since you didn't tell us what the maps associated with $j^\infty_p:L^\infty(X,B,m)\rightarrow L^p(X,B,m)$ were supposed to be, I'm assuming that $j^\infty_p$ maps integrable, essentially bounded functions to themselves and non-integrable essentially bounded functions to $0$.

This set of maps with your $i^q_p$ creates a commuting diagram (there is a slight worry that an essentially bounded integrable function may not be in all $L^p$, but this question addresses that). Now whether $L^\infty$ satisfies the universal property is another matter. The prospect that we have to worry about is whether or not there is an $f$ such that $f\in L^p$ for all $p$ but $f\not\in L^\infty$. If this happens, we can consider the span of $f$ as a Banach space (or vector space) in itself (call it $V$) and the collection of inclusion maps from this subspace into each of the $L^p$ (let's call these inclusion maps $k_p$). This will create another commutative diagram. Thus if $L^\infty$ were the inverse limit, there would be a unique map $u:V\rightarrow L^\infty$ such that $k_p=j^\infty_p\circ u$ for all $p$. However $u$ will have to send $f$ to an essentially bounded function. And then $j^\infty_p$ will send this image to another essentially bounded function in each of the $L^p$. But $k_p$ doesn't do that. Thus $k_p\neq j^\infty_p\circ u$ for any $u:V\rightarrow L^\infty$. Thus $L^\infty$ is not an inverse limit of the system.

Now does there exist such an $f$ for when $X$ has finite measure? See this question for an affirmative. Thus $L^\infty$ need not be an inverse limit.

  • $\begingroup$ Thank you very much. So, in all cases, $ L^{\infty} \neq \bigcap_p L^p $. Isn't it ? :-) $\endgroup$
    – Bryan261
    Jun 5, 2014 at 13:35
  • $\begingroup$ I'm not sure about that. I'd be surprised if there weren't an example for some measure space especially if $X$ is finite. I'm thinking about it now. $\endgroup$
    – user123641
    Jun 5, 2014 at 13:49
  • $\begingroup$ Yeah, if $X$ is finite equipped with the counting measure then all the $L^p$ spaces are equal and $L^\infty$ is trivially the inverse limit. $\endgroup$
    – user123641
    Jun 5, 2014 at 13:51
  • $\begingroup$ So, why do you say, that $ L^{\infty} $ "need not" be an inverse limit ? I don't understand, i'm not good at english, i'm from a foreign country, i'm sorry ... Please, could you explain to me more about this sentence that you say ? Thanks a lot. $\endgroup$
    – Bryan261
    Jun 5, 2014 at 14:00
  • $\begingroup$ That means it is not true in general. It doesn't have to be true. The link I posted gave an example where it isn't true. But for finite cases (with counting measure), it is true. $\endgroup$
    – user123641
    Jun 5, 2014 at 14:20

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