# Is this basic function space compact?

Let $A=L^2(X)$ be the space of square integrable functions on a compact Euclidean space $X$. If we equip $A$ with the usual 2-norm, is $A$ compact?

Edit: And if we restrict AA by adding the assumption that the functions are totally bounded, i.e. the supremum norms of all the functions are bounded by a constant?

• No, for two different reasons: 1) it's a vector space 2) it's not even locally compact.
– user98602
Apr 1, 2016 at 17:10
• And if we restrict $A$ by adding the assumption that the functions are totally bounded, i.e. the supremum norm of all the functions are bounded by a constant? Apr 1, 2016 at 17:22
• Look at my answer. All the functions have sup norm 1. Apr 1, 2016 at 17:26
• I remember something along the lines "A subset is compact iff it is closed and almost finite dimensional.". Apr 1, 2016 at 17:48

No. If $X=[0,1]$, then the sequence $\{e^{2\pi inx}\}_{n\in\mathbb{Z}}$ has no convergent subsequence.
$d(x,y)=\|x-y\|$ is an unbounded metric. Any metric on a compact space is bounded.