Class of Lipschitz Functions on the unit d-dimensional ball

Question

Let $\mathcal{F} = \{f:\mathcal{B}_d \to \mathbb{R}\;:\; \text{f is Lipschitz}\}$, where $\mathcal{B}_d = \{x \in \mathbb{R}^d\;:\: \|x\|_2 \leq 1\}$ is the unit ball in $d$ dimension. Is the class $\mathcal{F}$ finite dimensional?

My approach: For any constant $M$, the set, $\mathcal{F}_M = \{ f \in \mathcal{F}\;:\; \|f\|_\infty \leq M \}$ is compact in the sup norm.
This can be proven the following way
Step 1 : Use Arzela Ascolli to show the $\mathcal{F}_M$ is relatively compact in the uniform norm
Step 2 : Show that this $\mathcal{F}$ is closed in the uniform norm. This means that $\mathcal{F}$ is compact in the unform norm.
The same question have been addressed here, Show a locally bounded Lipschitz function space is compact for sup metric , so I assume it's true. This is also mentioned in the wiki page https://en.wikipedia.org/wiki/Lipschitz_continuity#Properties.

Question Does this show that the space $(\mathcal{F}, \|\;\|_\infty$ is a locally compact vector space?. The wikipedia page suggests that each $\mathcal{F}_M$ is a locally compact convex set of the Banach Space $\mathcal{C}({\mathcal{B}_d})$

If the space $(\mathcal{F}, \|\;\|_\infty)$ is locally compact, then we can deduce that it must be a finite dimensional space from Riesz Lemma (https://en.wikipedia.org/wiki/F._Riesz%27s_theorem#Statement). The question on the forum Space of Lipschitz functions is finite dimensional appears to do this only so I would assume it's true. The only place where it is throwing me off is they have taken $\mathcal{F}$ as a closed subspace of $\mathcal{C}({\mathcal{B}_d})$ such that every $f \in \mathcal{F}$ is Lipschitz. I am not sure how my setup is different? Is my space stritly bigger?

Any help is appreciated.

Answer 1

I think you are confusing yourself about some points. For clarity introduce some notation:

$$\mathcal F_R(L) := \{ f: B\to\Bbb R \mid \forall x,y\in B: |f(x)|\leq R, \ |f(x)-f(y)|\leq L|x-y|\}\\ \mathcal F_R=\bigcup_{L\in\Bbb R}\mathcal F_R(L),\quad \mathcal F(L)=\bigcup_{R\in\Bbb R} \mathcal F_R(L),\quad \mathcal F = \bigcup_{L,R\in\Bbb R}\mathcal F_R(L)$$ So that $\mathcal F$ is the space of all Lipschitz functions from $B$ to $\Bbb R$. Then you have the following:

1. $\mathcal F$ is a vector subspace of $C(B)$. It is not closed.
2. $\mathcal F_R$ is the ball of radius $R$ in $\mathcal F$. It is not pre-compact: Azerla-Ascoli doesn't apply because the Lipschitz constant can be arbitrarily large, and equicontinuity will then fail.
3. $\mathcal F(L)$ is the space of all Lipschitz functions from $B$ to $\Bbb R$ with Lipschitz constant $\leq L$. This is not a vector space, since the implication [$f\in\mathcal F(L)\implies \lambda f\in\mathcal F(L)$ for all $\lambda\in\Bbb R]$ fails.
4. $\mathcal F_R(L)$ is pre-compact by Azerla-Ascoli.

Your strategy of showing finite dimensionality of any of the spaces will then not succeed. For $\mathcal F$ (or its closure) it doesn't work because $\mathcal F$ is not locally compact, because the balls in $\mathcal F$ are not pre-compact. For $\mathcal F(L)$ the "balls" $\mathcal F_R(L)$ are pre-compact, but $\mathcal F(L)$ is not a vector subspace.