Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.