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I'd like to show that $C([0,1])$ (that is, the set of functions $\{f:[0,1]\rightarrow \mathbb{R} \, \textrm{ and } \, f \, \textrm{is continuous} \}$ is not a complete mertric space under the $L_1$ distance function:

$$ d(f,g) = \int_0^1 |f(x)-g(x)|dx $$

I can find counter examples (for example, here) but would rather prove it using definitions and principles so that I do not have to rely on committing specific degenerate sequences to memory.

Since all compact metric spaces are complete, I have to figure that the place to start is to show that $C([0,1])$ is not compact and that somehow an infinite cover allows for a divergent Cauchy sequence. However, I don't how to show this (or if it's even the right approach to take).

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AnonSubmitter85
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  • You are trying to show $C[0,1]$ is not complete or not compact? – Mariano Suárez-Álvarez Jun 04 '13 at 04:52
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    In any case, there is no need to «committing specific degenerate sequences to memory». Just remember why the sequences exist, and then come up with one every time you need to. I personally have no specific degenerate sequence commited to memory to show that $C[0,1]$ is not complete under the $L^1$ norm but I am pretty sure that if I sit down a little while I can come up with one... – Mariano Suárez-Álvarez Jun 04 '13 at 04:54
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    Moreover, it is not like you need to train yourself to be able to show that $C[0,1]$ is not complete in $L^1$ in less than 20 seconds... that is a completely useless thing. – Mariano Suárez-Álvarez Jun 04 '13 at 04:58
  • @MarianoSuárez-Alvarez It's for a timed exam, so, no, it would not be useless. – AnonSubmitter85 Jun 04 '13 at 05:21
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    That is an even more absurd reason! – Mariano Suárez-Álvarez Jun 04 '13 at 05:22
  • @MarianoSuárez-Alvarez The question is about completeness. I mentioned compactness as an idea for a starting point since it would have to be noncompact to be noncomplete. – AnonSubmitter85 Jun 04 '13 at 05:23
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    Instead of worrying about this silly counterexample-free thing, you should immediately show that a non-trivial linear subspace of a locally convex space is never compact, complete or not. – Mariano Suárez-Álvarez Jun 04 '13 at 05:24
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    For contrast, here is a question asking for explicit counterexamples: http://math.stackexchange.com/q/21878/ – Jonas Meyer Jun 04 '13 at 05:55

2 Answers2

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I will assume you want to show that $C[0,1]$ is not a complete subspace of $L^1$.

Here is a counterexample-free approach.

Suppose $C[0,1]$ is a complete subspace of $L^1[0,1]$, so that it is in particular closed. Since it is clearly a proper subspace, the Hahn-Banach theorem tells us that there is a non-zero continuous linear functional $\phi:L^1[0,1]\to\mathbb R$ such that $\phi$ vanishes on $C[0,1]$. Now the fact that the dual of $L^1[0,1]$ can be identified with $L^\infty[0,1]$ allows us to translate this: there exists a non-zero function $g\in L^\infty[0,1]$ such that $\int_0^1 fg=0$ for all $f\in C[0,1]$.

That this is not possible is a standard result in measure theory.

Mariano Suárez-Álvarez
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    Of course, this is a rather absurdly complicated argument in comparison to exhibiting a $L^1$-Cauchy sequence in $C[0,1]$ which does not converge to a point in $C[0,1]$! – Mariano Suárez-Álvarez Jun 04 '13 at 05:13
  • @MarianoSuárez-Alvarez I need to be able to do this without resorting to measure theory. Think undergraduate level material. – AnonSubmitter85 Jun 04 '13 at 05:19
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    @Potato Why is this silly? – AnonSubmitter85 Jun 04 '13 at 05:20
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    An undergraduate should learn why the subspace is not closed and be able to construct non-converging sequences. It is infinitely more instructive to pick up the reason why such sequences exist than to come up with strange reasons. – Mariano Suárez-Álvarez Jun 04 '13 at 05:20
  • @AnonSubmitter85 The most elementary way to do this is by constructing a counterexample. You write down continuous functions converging to a step function. It's really not that hard if you remember the intuition behind the construction. – Potato Jun 04 '13 at 05:22
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    @Potato What intuition are you talking about here? The counter example would have to not converge, hence, the non-completeness. – AnonSubmitter85 Jun 04 '13 at 05:34
  • @MarianoSuárez-Alvarez Sorry, but I have no idea what you are talking about. – AnonSubmitter85 Jun 04 '13 at 05:35
  • @AnonSubmitter85 A step function is not in $C[0,1]$, so I'm not sure what you're getting at. Sequences in a non-complete space can converge to things outside the space. – Potato Jun 04 '13 at 05:55
  • @Potato Sorry. That one can simply provide a limit point that lies outside of the space to show that it is not complete was unknown to me. That does make finding a counter example rather trivial. – AnonSubmitter85 Jun 04 '13 at 06:10
  • @AnonSubmitter85 Yes. I would provide further details, but it appears Professor Meyer already has. Now I hope you see why we admonished you to just find a counter-example :) It was not just us being mean! – Potato Jun 04 '13 at 06:13
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$C[0,1]$ can be embedded as a subspace of $L^1[0,1]$. It is dense in $L^1$, but not equal to $L^1$. Therefore it is not closed, and hence not complete.

Since all compact metric spaces are complete, I have to figure that the place to start is to show that $C([0,1])$ is not compact and that somehow an infinite cover allows for a divergent Cauchy sequence.

It is certainly not compact, but the logic is off here. Compact metric spaces are complete, but complete metric spaces need not be compact. For example, think of $\mathbb R$ with its usual metric, or any other nonzero complete normed space.

Jonas Meyer
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  • The meat, of course, lies in showing that the continuous functions are a dense subset (or non-closed), which already would give you a non-convergent Cauchy sequence (let a sequence of continuous functions converge to a non-continuous integrable one). – Henno Brandsma Jun 04 '13 at 05:39
  • Subspace? I am guessing this is a topology argument. I only have the beginning chapters of Rudin's PMA to go off, so I don't know what you are talking about. Thanks though. – AnonSubmitter85 Jun 04 '13 at 05:40
  • @AnonSubmitter85: You're welcome. In the beginning chapters of Rudin's PMA, aren't metric spaces covered, with definitions of such things as open, closed, and dense? A subset of a metric space is also a metric space if you restrict the metric, and in this way can be thought of as a subspace. (Here it is also a vector subspace.) There may be other things in my answer, not fleshed out, that are obstacles for you, but I don't think the word "subspace" should be. – Jonas Meyer Jun 04 '13 at 05:45
  • @Henno: I agree that the meat is in work that was left out, but Cauchy-sequences need not be given to fill it in. One can argue that simple functions are dense, then that step functions approximate simple functions, then that continuous functions approximate step functions, to give an argument for density without seeing any examples. I do agree that to just answer this question it would be easiest to give a simple example of a nonconvergent Cauchy sequence. Also, from AnonSubmitter85's comments I see that filling in this meat is not at an appropriate level for background/interests. – Jonas Meyer Jun 04 '13 at 05:49
  • @JonasMeyer Yeah, they are covered and I understand them. I guess I was trying to read into what you were saying and missed the point entirely. Though I don't recall any material about a metric space having to be closed in order for it to complete. I think, however, that Henno's comment about just choosing a sequence that converges to a non-continuous function shows me how unnecessarily complicated I had made things in my head. Thanks again. – AnonSubmitter85 Jun 04 '13 at 05:52
  • @AnonSubmitter85: I don't have the book handy, but I would guess that showing that a subspace must be closed to be complete is at least featured as an exercise. – Jonas Meyer Jun 04 '13 at 05:54
  • @JonasMeyer If it's not closed, then there is a limit point not in the space, which means what? That you can define a Cauchy sequence that converges to this outside limit point. And then the fact that the limit point is outside the space means it can't be complete? I don't recall the requirments for completeness including that the limit point be within the space. I don't know. – AnonSubmitter85 Jun 04 '13 at 06:00
  • @AnonSubmitter85: Basically, yes. – Jonas Meyer Jun 04 '13 at 06:01
  • @JonasMeyer You have been a tremendous help. Thank you. – AnonSubmitter85 Jun 04 '13 at 06:03
  • This answer is somewhat weird, because it depends on showing that $C$ is dense in $L^1$, which is (since showing it is not equal to it, which is obvious as one can exhibit an example like the characteristic function of an interval as an example) almost precisely what the OP wants to do! – Mariano Suárez-Álvarez Feb 23 '15 at 19:03
  • @Mariano: Fair point. I was vague, and to indicate what I had in mind as an actual approach to proving it is not complete I would have to suggest an approach to proving it is dense. As written it is just a perspective that might be obvious, but wasn't to the OP. What I had in mind was proving it is dense in a counterexample free approach, doing standard approximation arguments from intro measure theory: approximate (in $L^1$ sense) measurable with simple, simple with step, step with continuous. I mentioned this in a comment. Really I think avoiding a direct example is weird, too. – Jonas Meyer Feb 23 '15 at 21:20
  • @Mariano: I like your answer, but it is somewhat weird too, because it depends also on a "standard result in measure theory" for which the proof that comes to mind is showing measurable functions can be approximated by continuous ones, leading to conclude $\int_0^1 g^2=0$ (maybe I'm missing something easier; the fact that $g$ is bounded doesn't make it much easier.) You mention you find your answer "absurdly complicated," in light of what's being asked, and I feel similarly about mine (the intended one). – Jonas Meyer Feb 23 '15 at 21:24