# Is the closure of the ball of $1$-Lipschitz functions still equi-Lipschitz?

$$\DeclareMathOperator\Lip{Lip}$$Let $$\Lip_0(\mathbb R^d)$$ be the space of Lipschitz functions $$f:\mathbb R^d\to\mathbb R$$ vanishing at zero, i.e., $$f(0)=0$$, and equipped with the norm $$\|f\|:=\|\nabla f\|_{\infty}$$. Then $$\big(\Lip_0(\mathbb R^d), \|\cdot\|\big)$$ is a Banach space. Now we endow $$\Lip_0(\mathbb R^d)$$ with an alternative topology, denoted by $$w$$ and generated by the open sets $$\mathcal O_{u}(f;\varepsilon)$$ as below:

$$\mathcal O_{u}(f;\varepsilon) \quad:=\quad \left\{g\in \Lip_0(\mathbb R^d):~ \left|\int_{\mathbb R^d} \big[\nabla(f-g)(x)\cdot u(x)\big]\right| dx <\varepsilon \right\},$$

where $$f\in \Lip_0(\mathbb R^d)$$, $$u\in L^1(\mathbb R^d;\mathbb R^d)$$ and $$\varepsilon>0$$. Let $$\mathcal C:=\{f\in\Lip_0(\mathbb R^d): \|f\|\le 1\}$$ and denote by $$\overline {\mathcal C}$$ its $$w$$-closure. Could we prove

$$\sup_{f \in \overline {\mathcal C}}~ \|f\| ~<~ \infty?$$

• I probably misunderstood the question, but I find one thing unclear: do you ask that the integral $|\int [\dots]|<\epsilon$ should hold for all $u$? Or is the "variable" $u$ also indexing the collection of open sets? (in which case I don't understand why you don't write $\mathcal O_u(f;\epsilon,u)$) May 17, 2020 at 21:07
• @leomonsaingeon Under the topology $w$, $\mathcal O_u(f;\epsilon)$ is an open set indexed by $u$.
– user128095
May 17, 2020 at 21:20
• Ooooops sorry about that, I even copy-pasted some $u$ index twice in my own comment. Darn. Time to go to bed. Leo-"double-indexing"-monsaingeon out. May 17, 2020 at 21:27
• Maybe another stupid question: I can convince myself that conergence in your $w$ topology is not just simply weak-* convergence in $L^\infty$ of the gradients, i-e $f_n\overset{w}{\to} f$ iff. $\int \nabla f_n\cdot u\to \int \nabla f\cdot u$ for all $u\in L^1$. Is this so? May 17, 2020 at 21:42
• If so. then the answer to your question is yes, combining the Banach-Alaoglu theorem with Arzelà-Ascoli. I'll wait for confirmation before I post my answer. Maybe a good night sleep will help me realize my mistake, too... (hope not) May 17, 2020 at 21:46

$$\DeclareMathOperator\Lip{Lip}$$The answer is yes, and in fact I claim that $$C$$ is closed in the $$w$$-topology. This immediately implies $$\sup\limits_{f\in\bar C}\|f\|=\sup\limits_{f \in C}\|f\|\leq 1.$$
Let me record here a preliminary observation, which will be useful in the sequel. As per the OP's comment above, the convergence $$f_n\to f$$ in the $$w$$-topology simply means that the gradients converge weakly-* in $$L^\infty$$, in other words $$f_n\overset{w}{\to} f \qquad \mbox{iff}\qquad \lim\int \nabla f_n\cdot u=\int\nabla f\cdot u \quad \forall\,u\in L^1.$$
The proof goes next as follows: Take $$f\in \overline C$$, meaning that (according to the preliminary observation) there is a sequence $$f_n\in C$$ such that $$\|f_n\|\leq 1$$ and $$\nabla f_n\overset{*}{\to} \nabla f$$. In particular, observe (by definition of the norm $$\|.\|$$ on $$\Lip_0$$) that the sequence $$\{\nabla f_n\}$$ belongs to the unit ball $$B_1^\infty$$ in $$L^\infty(\mathbb R^d)$$, which is weakly-* relatively compact according to the Banach-Alaoglu theorem. By uniqueness of the weak-* limit we conclude that $$\nabla f\in B^\infty_1$$ too, hence $$\|\nabla f\|_\infty\leq 1$$ and in fact $$f\in C$$.