Compactness of a Closed Ball in C([0,1])

[Lee33]

Homework Statement

Show that the closed ball in $C([0,1])$ of center $0$ and radius $1$ is not compact.2. The attempt at a solution

I was given a hint, to look at the sequence of continuous functions $f_n(x) = x^n$ on the closed ball in $C([0,1])$. Why is that sequence continuous? Isn't it discontinuous in the closed ball $[0,1]$?

 

[pasmith]

Homework Statement

Show that the closed ball in $C([0,1])$ of center $0$ and radius $1$ is not compact.


2. The attempt at a solution

I was given a hint, to look at the sequence of continuous functions $f_n(x) = x^n$ on the closed ball in $C([0,1])$. Why is that sequence continuous? Isn't it discontinuous in the closed ball $[0,1]$?

Continuity is not a notion that applies to sequences. Convergence is.

I assume you are using the sup metric, d(f,g) = \sup \{|f(x) - g(x)| : x \in [0,1]\}.

The function f_n : x \mapsto x^n is continuous (indeed, differentiable) on [0,1] for any n \in \mathbb{N}. Hence f_n \in C([0,1]). It is straightforward to show that d(f_n,0) = 1, so f_n is in the closed unit ball for all n.

The question is: Does this sequence converge to a limit in C([0,1]) or not? Does it have any subsequences which do? And what does this tell you about the compactness of the closed unit ball in C([0,1])?

 

[Lee33]

But I thought $f_n : x \mapsto x^n$ is not continuous on $[0,1]$ since $f(x) = \begin{cases} 1 & \quad \text{if }x= 1 \\ 0 & \quad \text{if }x\in [0,1) \\ \end{cases}$

 

[LCKurtz]

But I thought $f_n : x \mapsto x^n$ is not continuous on $[0,1]$ since $f(x) = \begin{cases} 1 & \quad \text{if }x= 1 \\ 0 & \quad \text{if }x\in [0,1) \\ \end{cases}$

$f_n$ is a polynomial and is continuous for all $x$. Surely you know that. All polynomials are.

 

[Lee33]

LCKurtz - Wow, you're right! I should go run outside and clear my mind. Very dumb mistake by me. Thank you both!

 

[WannabeNewton]

What you wrote down is the pointwise limit of $(f_n)$ and indeed the pointwise limit is discontinuous. However $f_n$ is continuous for all $n$.

Regardless, does the uniform limit of $(f_n)$ exist? Furthermore, does any subsequence $(f_{n_k})$ of $(f_n)$ have a uniform limit? Recall that convergence in $C([0,1])$ is uniform convergence in $[0,1]$. And if no $(f_{n_k})$ has a uniform limit then what does that tell you about sequential compactness of the closed unit ball in $C([0,1])$?

 

[Lee33]

Thank you, WannabeNewton! I know what to do now. Since it is not sequentially compact, therefore it is not compact.