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I am trying to show that $C[0,1],$ the space of all real - valued continuous functions with the sup metric is not sequentially compact with the sup metric by showing that the sequence $f_n = x^n$ has no convergent subsequence. The sup metric $\|\cdot\|$ is defined as

$$\|f - g \| = \sup_{x \in [0,1]} |f(x) - g(x)|$$

where $|\cdot|$ is the ordinary Euclidean metric. Now I know that $f_n \rightarrow f$ pointwise, where

$$f = \begin{cases} 0, & 0 \leq x < 1 \\ 1, & x = 1.\end{cases}$$

However $f \notin C[0,1]$ so this means by theorem 7.12 of Baby Rudin that $f_n$ cannot converge to $f$ uniformly. However how does this tell me that no subsequence of $f_n$ can converge to something in $C[0,1]$?

Thanks.

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I think you can use the same argument for any subsequence: $f_{n_k}$ converges to $f$ pointwise. –  Martin Sleziak Jun 17 '12 at 7:16
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@AmiteshDatta I think I get it, when you say "Therefore $g = f$ because we know that $f_n \rightarrow f$ pointwise (this sentence and the last... $C[0,1]$) don't we somehow need to "enlarge" our space and say just consider the space of all functions on $[0,1]$? Namely because to talk about limits being unique,etc don't we need to live in one big ambient space? –  user38268 Jun 17 '12 at 7:59
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@BenjaminLim You can also consider the space $B[0,1]$ of all real bounded functions on $[0,1]$ along with the sup metric and observe that $f \in B[0,1]$. –  Sayantan Jun 17 '12 at 8:19
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Why don't you just compute (or estimate) the maximum of $x^n - x^m$ on $[0,1]$ if $n \neq m$ and show that it is larger than $1/2$? –  t.b. Jun 17 '12 at 9:05
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@t.b. That's exactly what I did. We want to show that there exists $\epsilon > 0$ such that for all $N$, there exist $m,n \geq N$ such that $||x^m - x^n ||<\epsilon$. Assume wlog that $n > m$, we then have by taking the derivative and stuff that the maxium value of the difference is achieved at the point $x$ that you calculated, and we get that $||x^m - x^n || = (m/n)^{m/(n-m)}\left(1 - \left(\frac{m}{n}\right)^{n-m}\right)$. However, how can I just choose an $\epsilon $ such that this expression is greater than $\epsilon$ for $m,n $ sufficiently large? I tried to fix $n$ and try taking $m$ –  user38268 Jun 17 '12 at 9:41

4 Answers 4

up vote 4 down vote accepted

This is (more-or-less) a modification of the hint from t.b's comment.

It suffices to show that any subsequence of $(f_n)$ is not Cauchy. (Since every convergent sequence is a Cauchy sequence.)

To do this, notice that $$\lVert f_{2n}-f_n \rVert = \sup_{x\in[0,1]} (x^n - x^{2n}) = \sup_{x\in[0,1]} (x^n - (x^{n})^2) = \sup_{t\in[0,1]} (t-t^2)=\frac14.$$ (It is easy to find maximum of the quadratic function $f(t)=t-t^2=t(1-t)$.)

In addition, if we use the monotonicity of the sequence $(f_n)$, we see that $$k\ge 2n \qquad \Rightarrow \qquad \lVert f_{k}-f_n \rVert \ge \frac14.$$

Now, if we have any subsequence of $(f_n)$ then the above estimate shows that this subsequence is not Cauchy. For any given $k_0$ we can find $k'>k_0$ such that $n_{k'}>2n_{k_0}$ and $\lVert f_{n_{k'}}-f_{n_{k_0}} \rVert \ge \frac 14$.

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Regarding the motivation for your question (sequential compactness):

First, commenting on your specific question: a topological vector space over the real or complex numbers is never sequentially compact. Only bounded closed subset are possibly (sequentially) compact. Your attempt of proof is, therefore, only relevant if you restrict to bounded sequences (which applies to your example but should be made an explicit assumption).

As a general remark: it is known that compactness and sequential compactness are equivalent in metric spaces, hence in subsets of normed vector spaces (like $C^0$).

It is also known (though a bit more involved when it comes to proving it) that a normed (real or complex) vector space has the so called 'Heine Borel' property (which says that a subset is compact if and only if it is closed and bounded) if and only if it is of finite dimension.

Edit: the following sentence is obviouly not correct (see the comment of t.b. -- thanks for pointing this out), but is a nice illustration of a premature und uncautios conclusion, so I don't delete it ;-): Since $C^0$ is not finite dimenional, it follows that bounded closed sets of $C^0$ are (sequentially) compact only if they are contained in a finite dimensional subspace.

The following remains true, though: sets like the closed unit ball (and therefore any set containing an open ball) are not (sequentially) compact in $C^0$ and, in general, bounded sequences need not have a converquent subsequence, and this applies to every infinite dimensional normed vector space.

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The set $\{0\} \cup \{\frac{1}{n}x^n\,:\,n \geq 1\}$ is compact in $C[0,1]$, hence bounded and closed but it is not contained in a finite-dimensional subspace... –  t.b. Jun 17 '12 at 9:05
    
@t.b.: thank you for pointing this out. I edited my posting. –  user20266 Jun 17 '12 at 9:21

After all of this discussion, here is my understanding of how the problem concludes: We already know that $f_n \rightarrow f$ pointwise, where $f$ is the discontinuous function that I defined on $C[0,1]$. Now go up to a bigger metric space like $B[0,1]$ that contains $C[0,1]$ and equip it with the sup metric. In there we know that $f_n \rightarrow f$ pointwise. This means that for each $x \in [0,1]$, the ordinary sequence of real numbers $f_n(x) \rightarrow f(x)$ and since $\Bbb{R}$ with the Euclidean topology is Hausdorff it follows that the pointwise limit of $f(x)$ is unique. Viz, there is no other function on $B[0,1]$ that converges to $f$ pointwise.

Now I claim that there is no function $g \in C[0,1]$ for which $f_n$ converges to uniformly. If there were, then $f_n \rightarrow g$ pointwise on $C[0,1]$. Furthermore, a continuous real valued function on a compact set is bounded and so $g$ lives in $B[0,1]$. By what I said about limits being unique, this means that $g = f$. However this is a contradiction because $f \notin C[0,1]$.

It follows that there is no continuous function on $[0,1]$ that $f_n$ converges to. Q.E.D.

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I have a couple of corrections/comments. Firstly, as I said before, you don't need to equip $B[0,1]$ with the supremum metric because the metric structure of $B[0,1]$ is irrelevant to the problem at hand. Indeed, we are studying pointwise convergence and we only need to know that $C[0,1]\subseteq B[0,1]$ as sets but not as metric spaces; indeed, I emphasize that we don't care about any metric space structure on $B[0,1]$ for the purposes of this problem. We only need $B[0,1]$ as a convenient set in which the function $f$ lives ... –  Amitesh Datta Jun 17 '12 at 10:28
    
... Secondly, I think your sentence "... there is no other function on $B[0,1]$ that converges to $f$ pointwise ..." doesn't make sense; a more meaningful sentence might be "... if $g\in B[0,1]$ is such that $f_n\to g$ pointwise, then $f=g$ ...". I emphasize that there is nothing "special" about $B[0,1]$ in this regard; the only relevant property is that $f\in B[0,1]$ and it doesn't matter that "$B[0,1]$" even means as long as it is a set, $C[0,1]\subseteq B[0,1]$, and $f\in B[0,1]$ ... –  Amitesh Datta Jun 17 '12 at 10:32
    
... In other words, my point is that you're making the problem more complicated by introducing $B[0,1]$. A simpler argument is to note the uniqueness of pointwise limits; this involves the metric space structure of $\mathbb{R}$ (notice $B[0,1]$ is not relevant here). You can then finish the argument by noting that the pointwise limit of any subsequence of $\{f_n\}_{n\in\mathbb{N}}$ is equal to $f$ (notice $B[0,1]$ is not relevant here) ... –  Amitesh Datta Jun 17 '12 at 10:37
    
... Of course, this leads to a contradiction because uniform convergence implies pointwise convergence and $f$ is not continuous (notice $B[0,1]$ is not relevant here). You can introduce $B[0,1]$ if you want but even if you do, you shouldn't need to care about any metric space structure on $B[0,1]$. However, the solution is much simpler if you forget about $B[0,1]$. –  Amitesh Datta Jun 17 '12 at 10:37
    
My "ultimate" point is that you can speak of pointwise convergence of any sequence of functions wherever they live provided that their (common) range is a metric space; this doesn't involve any metric space structure (especially any metric space structure on a $B[0,1]$ (even if we care what that means which we don't)) except the metric space structure on the (common) range of the sequence of functions. –  Amitesh Datta Jun 17 '12 at 10:46

Since $f_n$ converges to $f$ pointwise, all subsequences $f_{n_k}$ also converge pointwise to $f$. So $f_{n_k}$ can not converge uniformly to continuous $g$, because uniform convergence implies pointwise convergence as well, and $f$ is not continuous.

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