Closed unit ball is compact?

Let $X$ be a Banach space and let $\operatorname{Lip}_{0}(X)$ be the space of all real-valued Lipschitz functions which vanish at $0$. The space $\operatorname{Lip}_{0}(X)$ is a Banach space when it is equipped with the Lipschitz norm, defined by:

$$L(f)=\|f\|_{\operatorname{Lip}}=\sup\left\{\frac{f(x)-f(y)}{\|x-y\|}:\,x,y\in X,\,x\neq y\right\}$$

My goal is to show that the closed unit ball of the space $\operatorname{Lip}_{0}(X)$ is compact for the topology of pointwise convergence. I have try with no success to use Tychonoff theorem or Banach-Alaoglu theorem. I failed because the Banach-Alaoglu theorem concerned the closed unit ball of the dual space with the weak-star topology.

Thank for any help.

• I'm confused. If you're interested in the topology of pointwise convergence, then what do you need the $||.||_{\textrm{Lip}}$-norm for?
– fgp
Commented May 10, 2013 at 22:58
• @fgp: to define the unit ball? Commented May 10, 2013 at 23:02
• Uh, right. Note to self: Don't write comments when tired! Sorry for the noise
– fgp
Commented May 10, 2013 at 23:04
• I don't have a counter-example yet, but I'm very suspicious of this result unless you assume that $X$ is the dual of another Banach space or perhaps if you use the topology of uniform convergence. In general, elementary compactness theorems usually follow from either Banach-Alaoglu or Arzela-Ascoli, or they aren't correct. Commented May 10, 2013 at 23:12
• Well, standard Cantor diagonal method implies the result when $X$ is separable. Commented May 10, 2013 at 23:24

Tychonoff's theorem does give it to you — the closed unit ball $B$ of $\def\Lip{\operatorname{Lip}}\Lip_0(X)$ with the topology of pointwise convergence is naturally realized as a subspace of the product
$Y = \prod_{x \in X} I_x$
where $I_x = [\hspace{.1 cm}-||x||, ||x||\hspace{.1 cm}]$. The intervals are chosen by considering the Lipschitz expression with $y = 0$. Now, the product space $Y$ is compact by Tychonoff's Theorem, so we must only show that $B$ is closed in this space, or that $Y \setminus B$ is open.
Suppose $f \in Y \setminus B$. Then there are some $z, w \in X$ such that $\cfrac{f(z) - f(w)}{||z - w||} = 1 + \varepsilon > 1$. Choose open subsets $O_z$ and $O_w$ of $I_z$ and $I_w$ respectively such that $f(z) \in O_z$, $f(w) \in O_w$, and if $a \in O_z, b \in O_w$, then $\cfrac{a - b}{||z - w||} > 1 + \frac{\varepsilon}{2}$. This allows us to form a neighborhood of $f$ disjoint from $B$, namely, $\{ g \in Y \mathrel: g(z) \in O_z \text{ and } g(w) \in O_w \}$.
• $\operatorname{Lip}_0(X)$ isn't the unit ball, it's the whole vector space. I think your answer could be further improved by introducing the notation $I_x$ for the closed interval from $-\lVert x \rVert$ to $\lVert x \rVert$. Commented May 29, 2013 at 5:57
• I asked you to introduce $I_x$ because it oculd be used several times in the proof, as per my latest edit. Additionally, it's required that $O_z$ be open in $I_z$. Is this relatively open, or open as a subset of the real line? And does it matter? Commented May 29, 2013 at 6:09