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Consider the normed space $(C[0,1], \| \cdot\|_1)$ where $C[0,1]=\{f:[0,1] \to \Bbb R : f$ is continuous$\}$ and $\|f\|_1 = \int_0^1|f(t)|dt$. I'm trying to find out if the unit sphere $S=\{f \in C[0,1] : \| f \|_1 = 1\}$ is compact or not.

To prove it's not compact (I don't know if that's true or not) I was thinking of a sequence of functions $\{f_n\}$ whose graphs are triangles of base $1/n$ and height $2n$, and defined as zero elsewhere. So this triangles have constant area $1$, and then the functions have norm $1$. This sequence of functions converges pointwise to the function zero, and not uniformly. But now I'm struggling to prove that this sequence doesn't have a convergent subsequence in the metric induced by the norm (I was trying to prove that by proving that $\{f_n\}$ is not a Cauchy sequence).

Maybe there's an easier counterexample, or maybe the sphere is compact, I don't know. I hope you can give me a counterexample so I can work on it or tell me if it's compact or not in order to know what I actually have to prove. Thanks.

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up vote 1 down vote accepted

The unit ball of a normed space is compact if and only if the space is finite-dimensional.

An explicit way to show non-compactness is as follows: let $f_n$ be the function that is zero outside of $[1/(n+1),1/n]$ and a triangle of height $2n(n+1)$ in there. Then $\|f_n\|_1=1$ for all $n$, and $\|f_n-f_m\|_1=2$ for all $n\ne m$; so not subsequence can be convergent.

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You are right. But it is closed in the same sense that $(0,1)$ is closed it you consider it as a metric space on its own. It is closed, but not complete. – Martin Argerami Feb 17 '14 at 3:41
Again you are right, but it is not a big deal. Formally, you define an equivalence relation where $f\sim g$ iff $f=g$ off a set of measure zero. In practice, everything one says about functions in $L^1$ is "a.e." (almost everywhere), i.e. off sets of measure zero. – Martin Argerami Feb 17 '14 at 4:09
Yes, you are right, thanks. – Martin Argerami Feb 17 '14 at 13:32

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