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Show that the closed unit ball $B[0,1]$ in $C[0,1]$ is not compact under the following metrics:

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

$2.d(f,g)=\int _0^1 |f(x)-g(x)| dx$

My try:

In order to show not compact if we can find a sequence which has no convergent subsequence then we are done.

How to approach $1,2$?

Learnmore
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5 Answers5

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Use functions that are $n^2$ on $[0,1/n]$ and zero on $[1/n+\epsilon,1]$ and on $[1/n,1/n+\epsilon]$ connect them to be continuous. Then the integral of any one of them is close to $n$ (choose an appropriate $\epsilon$ that will depend on $n$) and I believe the integral of the difference of any two of them is at least one.

Gregory Grant
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  • I like this answer, but an image would be more than explanatory. However, I know how tedious it can be! +1 – Daniel May 15 '15 at 14:56
  • I don't have good software at my fingertips to make such an image. Can you help? – Gregory Grant May 15 '15 at 14:56
  • Sorry, I don't have software neither. I leaved the comment so that someone could do it, if that someone has the means. – Daniel May 15 '15 at 14:58
  • Thank you, good suggestion, in this case a picture is definitely worth a thousand words. – Gregory Grant May 15 '15 at 14:58
  • what motivated you to think that? – Learnmore May 15 '15 at 15:09
  • @learnmore Are you asking this to me? Are you asking what motivated me to think of this answer? Or are you referring to the idea of making a picture? – Gregory Grant May 15 '15 at 15:10
  • to you sir how you got that – Learnmore May 15 '15 at 15:11
  • It reminds me of an example I saw in graduate school of a sequence of functions that converge to zero pointwise, but not with respect to the $L^1$ metric. In that case you just take $n\cdot 1_{(0,1/n]}$ and you get the integral of every one of them is one but they converge pointwise to zero. For your example though, they need to modified to be continuous and to be so that the difference of any two has integral at least one, so that's why I changed it so they have a height of $n^2$ and not $n$ and I introduced the $\epsilon$ thing to make it continuous. – Gregory Grant May 15 '15 at 15:14
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The uniform case is classic: use $f_n(x)=x^n$. There is a lot to be learned from this example.

That example won't work in the integral case, because this sequence converges to the zero function in the integral sense. Instead, try making a function which is a continuous approximation to a function with a jump: for instance it is $1$ on $[1/4,3/4]$, zero outside $[1/4-1/(4n),3/4+1/(4n)]$, and continuous (say, linear) in between. These look like trapezoids which are approaching a rectangle.

Another possibility is $\sin(nx)$, but I think the proof might be more difficult in this case.

Ian
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For 1. take $f_n$ to be the function whose graph is given by connecting the points $(0,0), ({1 \over n+1}, 0), ({1\over 2} ({1 \over n+1} + {1 \over n}),1),({1 \over n}, 0), (0,0) $. Then $d(0,f_n) = 1$, and $d(f_n,f_m) = 1$ for all $n \neq m$ hence there are no convergent subsequences.

$f_n$ is a 'tooth' of height $1$ with support in $[{1 \over n+1},{1 \over n}]$.

For 2. in a similar fashion, take $f_n$ to be the function whose graph is given by connecting the points $(0,0), ({1 \over n+1}, 0), ({1\over 2} ({1 \over n+1} + {1 \over n}), 2n (n+1)), ({1 \over n}, 0)), (0,0) $. Then , as above, $d(0,f_n) = 1$, and $d(f_n,f_m) = 2$ for all $n \neq m$ hence there are no convergent subsequences.

$f_n$ is a 'tooth' of height $2n(n+1)$ with support in $[{1 \over n+1},{1 \over n}]$.

copper.hat
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  • how you got that – Learnmore May 15 '15 at 15:11
  • How did I get what? – copper.hat May 15 '15 at 15:12
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    If the $f_n$ have (mostly) disjoint support it makes the computations much easier. – copper.hat May 15 '15 at 15:15
  • @learnmore I think the motivation comes from the following: Let $X$ be a metric space, and if $(x_{n})$ is a infinite sequence of points in $X$ such that $d(x_{i},x_{j}) \geq \epsilon$ for all $i,j$ then prove that $X$ cannot be compact. You will see why compactness fails ( think of finite subcover!). – C.S. May 15 '15 at 15:18
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You haven't provided a set of functions for the first sequence. What are your $e_n$'s ?

For the first sequence, consider the set of functions $p_n(x) = x^n$. We have $\sup_{[0,1]} |x^n| = 1$ for all $n$, since $1^n = 1$. However, no subsequence converges uniformly to a continuous function. They converge to the function $$f(x) = \left\{ \begin{array}{cc} 0 & x \neq 1\\ 1 & x=1\end{array}\right.$$

For the second part, I suggest following up on @GregoryGrant's post.

Joel
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As mentioned in previous answers, an animation much aids comprehension. Here's an open cover for the closed unit ball with respect to sup norm metric. Let,

\begin{align} f^n_{\mathrm{upper}}(x) &= \text{the continuous function obtained by connecting the points } \begin{cases} (0,0),\\ (1/n,2),\\(1-1/n,2),\\(1,0) \end{cases}\\ f^n_{\mathrm{lower}}(x) &= \text{the continuous function obtained by connecting the points } \begin{cases} (0,0),\\ (1/n,-2),\\(1-1/n,-2),\\(1,0) \end{cases}. \\ \end{align} And then define each open set, $$ \mathcal{O}_n = \{ g(x) \in C[0,1] : f^n_{\mathrm{lower}}(x) < g(x) < f^n_{\mathrm{upper}}(x) \quad \forall x \in [0,1]\}. $$

And you can check that every function in the unit closed ball lies in at least on of these $ \mathcal{O}_i$s, and each $ \mathcal{O}_i$ is an open set. From the illustation it's clear that the cover does not contain a finite subcover.

The open covers shown in animation

The illustration is created using Desmos.

Kooranifar
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