Is $ L^{\infty} $ a direct limit or inverse limit of the directed system $ (L^p , i_{p}^q )_{p,q \in [1 , + \infty [ } $? 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.
 A: 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.
