# Arzela-Ascoli Theorem in $L^p[0,1]$

I understand the Arzela-Ascoli for $X$, compact metric space. So when $X=L^p[0,1]$, the theorem becomes the following?

If $f_n\in C(L^p[0,1],L^p[0,1])$ that is uniformly bounded and equicontinuous, then there exists a uniformly converging subsequence $f_{n_k}$ with respect to the $L^p$ norm?

• Maybe you have this thm in mind? Jan 30, 2015 at 19:12
• Yes, thats the theorem Jan 31, 2015 at 15:14

This is exactly correct. Let $S$ be a subset of $L^p([0,1])$. Then by Arzela-Ascoli, if the family of functions $f_k$ belonging to $S$ are equicontinuous and uniformly bounded, then the set $S$ is compact. By definition of compactness (sequential) we have a convergent subsequence $f_{n_k}$ converging in $S$. Note that $L^p([0,1])$ itself is not compact.
• Uh...this makes very little sense to me. It seems to me that you are saying that a subset of $\mathcal{L}^p([0,1])$ (note that I used $\mathcal{L}$, because these are functions and not equivalence classes anymore) which is compact as a subset of $C([0,1])$ is compact in $\mathcal{L}^p([0,1])$. I can believe that, but you haven't proven it ($\mathcal{L}^p$ has a messy topology, because it has indistinguishable points).
• At any rate I don't think it's what the OP was asking, which was about functions from $L^p$ to $L^p$. Arzela-Ascoli in this setting would be about functions from $L^p$ to $\mathbb{R}$.
• Is equicontinuity and uniform boundedness enough for a subset in $C(X, Y)$ to be relatively compact? Where $Y$ is Banach and $X$ is compactg metric space? Nov 28 at 22:15