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Let $C^{0,1}([a,b])$ be the space of all Lipschitz-continuous functions $x\colon [a,b] \to \mathbb{R}$ with the metric $$ d_{0,1}(x,y) := \sup_{a \le t \le b} |x(t) - y(t)| + \sup_{a \le s,t \le b, s\ne t} \left| \frac{x(s) - x(t)}{s-t} - \frac{y(s) - y(t)}{s - t} \right|. $$ So that $x \in C^{0,1}([a,b])$ if $x \in C([a,b])$ and there exists a constant $L$ such that $$ |x(s) - x(t)| \le L|s-t| \qquad \forall s,t \in [a,b]. $$ Show that

a) The sphere $K = \{ x \in C^{0,1}([a,b]) : d_{0,1}(x,0) \le 1 \}$ is a compact subset of $(C([a,b]), \Delta)$.

b) $C^{0,1}([a,b])$ is not separable.

I knew the definitions, but I have no idea how the attack these problems, do you have any hints?

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What is $\Delta$? –  Davide Giraudo Oct 26 '12 at 21:24
    
a) Hint: Ascoli - Arzelà –  Giuseppe Negro Oct 26 '12 at 21:44
    
I am not sure, but i guess the $\Delta$ means the metric. –  Stefan Oct 26 '12 at 21:47
    
In case you need some further hints than what is in Rudolf L.'s answer you should probably say what is missing. Note that $K$ is uniformly bounded ($\Delta(0,k) \leq 1$ for all $k \in K$) and equicontinuous ($L \leq 1$), so, using Arzela-Ascoli, you only have to check that $K$ is closed in $(C([a,b]),\Delta)$, but that's relatively easy. Also, Rudolf L. shows how to exhibit a discrete subset of size $\#\mathbb{R}$ in $C^{0,1}([a,b])$ so that it cannot be separable. –  commenter Oct 29 '12 at 9:54

1 Answer 1

Presumably $\Delta(x,y) = \sup_{a \leq t \leq b} |x(t) - y(t)|$.

For a): as Giuseppe said: use the Arzelà-Ascoli theorem (for $f \in K$ you have $L \leq 1$).

For b): For $p \in [a,b]$ consider the function $f_p(t) = |p-t|$ and show that for $p \neq q$ $$ d_{0,1}(f_p,f_q) = \underbrace{\sup_{t} \cdots}_{= d(p,q)} + \underbrace{\sup_{s,t} \cdots}_{\geq 2} \geq d(p,q) + 2 $$ (plug $p$ and $q$ for $s$ and $t$ into the definition of $d_{0,1}$ to get those estimates).

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Arzel`a-Ascoli allows you to show that the set is relatively compact. To prove compactness, you need to show additionally that the metric space is complete. How would you do that? –  jens Apr 21 at 11:59

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