Closed ball in L1 not compact

  • #1
joshmccraney
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Hi PF!

When proving a closed ball in ##L_1[0,1]## is not compact, I came across a proof, which states it is enough to prove that the space is not sequentially compact. Counter example: consider the sequence of functions ##g_n:x \mapsto x^n##. The sequence is bounded as for all ##n\in \mathbb N, \|g_n\|=1##. If ##(g_n)## would have a convergent subsequence, the subsequence would converge pointwise to the function equal to 0 on ##[0,1)## and to 1 at 1. As this function is not continuous, ##(g_n)## cannot have a subsequence converging to a map in ##L_1[0,1]##.

Why is it that because ##x=1## does not converge to the same thing as ##x\in[0,1)## this implies there are no subsequences converging to a map in this space?

Original proof here (second proof):
http://www.mathcounterexamples.net/a-non-compact-closed-ball/
 

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  • #2
mathwonk
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I think you may have misread the article you linked. It seems as if the space there is not L1[0,1], but C[0,1]. I.e. this sequence does converge in L1 but not in C, i.e. it has an integrable limit but not a continuous one. You might be interested in the Riesz theorem mentioned there, that the closed unit ball is not compact in any infinite dimensional Banach space. This theorem is proved in an elementary way in chapter V, p. 109, of Foundations of Modern Analysis, by Jean Dieudonne'.

By the way, the first tipoff was that an L1 convergent sequence need not have a pointwise convergent subsequence, but only one that converges almost everywhere. i.e. the integral cannot detect erratic behavior at a single point.
 
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  • #3
joshmccraney
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I think you may have misread the article you linked. It seems as if the space there is not L1[0,1], but C[0,1]. I.e. this sequence does converge in L1 but not in C, i.e. it has an integrable limit but not a continuous one. You might be interested in the Riesz theorem mentioned there, that the closed unit ball is not compact in any infinite dimensional Banach space. This theorem is proved in an elementary way in chapter V, p. 109, of Foundations of Modern Analysis, by Jean Dieudonne'.

By the way, the first tipoff was that an L1 convergent sequence need not have a pointwise convergent subsequence, but only one that converges almost everywhere. i.e. the integral cannot detect erratic behavior at a single point.
Thanks for replying. I actually found the theorem you've mentioned, but it is a little more advanced than I am comfortable with. I did track down Reisz theorem in a more readable form here (see first four lines of the converse):

https://en.wikipedia.org/wiki/Riesz's_lemma
But even this I don't fully track. I did see a counter example ##f_n = 2^n \textbf{1}_{[0,1/2^n]}##. Can you explain to me how this counter example works though?
 
  • #4
wrobel
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But even this I don't fully track. I did see a counter example fn=2n1[0,1/2n]f_n = 2^n \textbf{1}_{[0,1/2^n]}. Can you explain to me how this counter example works though?
this sequence is bounded in ##L^1[0,1]## ,on the other hand ##\|f_i-f_j\|_{L^1[0,1]}\ge const>0,\quad i\ne j## so that this sequence does not contain a convergent subsequence
 
  • #5
joshmccraney
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this sequence is bounded in ##L^1[0,1]## ,on the other hand ##\|f_i-f_j\|_{L^1[0,1]}\ge const>0,\quad i\ne j## so that this sequence does not contain a convergent subsequence
Isn't this sequence unbounded though? ##\lim_{n\to \infty}f_n = \lim_{n\to \infty}2^n \textbf{1}_{[0,1/2^n]}## and since ##2^n \to \infty##, how is this bounded?
 
  • #6
mathwonk
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you must remember again what the notion of size is in L1, namely it is the size of the integral, so the point here is that the integrals are bounded. this is the same issue that arose above, of confusing the norms in C and in L1, i.e. the maximum value of |f|, versus the integral of |f|.
 
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  • #7
joshmccraney
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you must remember again what the notion of size is in L1, namely it is the size of the integral, so the point here is that the integrals are bounded. this is the same issue that arose above, of confusing the norms in C and in L1, i.e. the maximum value of |f|, versus the integral of |f|.
Thanks! So it seems obvious ##\| f_n(x) \|= 1## if ##x \in [0,1/2^n]##. But why does ##\|f_i-f_j \|> 0## imply no subsequence converges in ##[0,1]##?
 
  • #8
mathwonk
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that is not what was said. he said that difference is not just positive but bounded below by a positive constant. since for a sequence no two of whose terms ever get close, the same holds for any subsequence, thus no subsequence is even cauchy.
 
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  • #9
joshmccraney
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that is not what was said. he said that difference is not just positive but bounded below by a positive constant. since for a sequence no two of whose terms ever get close, the same holds for any subsequence, thus no subsequence is even cauchy.
Brilliant! And thanks!

So let me ask, what does the bold 1 denote? I just assumed ##f_n = 2^n## if ##x \in [0,1/2^n]## else ##f_n = 0##.
 
  • #10
WWGD
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Another standard example is the unit ball in Hilbert space with vertices ## e_i = \delta_i^j ## . Then ## || e_i- e_j ||> \sqrt 2 ## and no convergent subsequence. ##||a-b||^2=<a-b,a-b>=2 ## . So no subsequence will be Cauchy. Note that the result on the equivalence with not having a convergent subsequence only applies to metric spaces.
 
  • #11
mathwonk
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I believe you are right. I.e. that the bold 1 denotes the characteristic function of the following subset, i.e. the function that equals 1 on that set and zero elsewhere. To me a more common notation is a "chi" (for characteristic) rather than a "1".
 
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  • #12
WWGD
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Yes, I have seen it used this way too, as the characteristic function.
 

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