Closed unit ball is compact?

Question

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.

Answer 1

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, $\{ g \in Y \mathrel: g(z) \in O_z \text{ and } g(w) \in O_w \}$.