Setup:
Let $\mu$ be a measure on a measurable space $(X,\Sigma)$, such that for every $p \in [1,\infty)$$p ,q\in [1,\infty)$, $L^p_{\mu}(\Sigma)\subseteq L^q_{\mu}(\Sigma)$ if $p\leq q$$p\geq q$. Furthermore, the inclusions are continuous. (A precise characterization of when this happens can be found in Rudin's book, for example.)
Background:
Now, let $X^p$ denote the closed unit ball in each $L^p_{\mu}(\Sigma)$; by construction we have that there are continuous inclusions maps $i^{p,q}:X^p\hookrightarrow X^q$. From classical topology, we known that the Top is (co)complete and from the Semadeni-Zidenberg Theorem we also know that Ban$_1$ (the category of unit Balls in Banach spaces and of short linear maps) is (co)complete. Therefore, the limit $\varprojlim X^p$ is well-defined in both Top and in Ban$_1$.
Question:
What is $\varprojlim X^p$ equal to when the limit is taken in Top and in Ban$_1$; and are they the same?
Intuition My intuition is that in $Ban_1$, $\varprojlim_n X^n$ is equal to $$ \left\{ f \in L^{\infty}_{\mu}(\Sigma):\, \|f\|_{\infty}\leq 1 \right\} , $$ with topology given by the norm $\|\cdot\|_1+\|\cdot\|_{\infty}$; but I cannot find a rigorous justification of my hypothesis (I have not shown minimiality in $Ban_1$ of the corresponding coconecone).
Furthermore, I expect that $\varprojlim X^p$ in Top has finer topology than $\varprojlim X^p$ in Ban$_1$. Am I correct in this direction?
