$\DeclareMathOperator\Lip{Lip}$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

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

  • $\begingroup$ 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 $\mathcal O_u(f;\epsilon,u)$) $\endgroup$ May 17, 2020 at 21:07
  • $\begingroup$ @leomonsaingeon Under the topology $w$, $\mathcal O_u(f;\epsilon)$ is an open set indexed by $u$. $\endgroup$
    – user128095
    May 17, 2020 at 21:20
  • $\begingroup$ 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. $\endgroup$ May 17, 2020 at 21:27
  • $\begingroup$ 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? $\endgroup$ May 17, 2020 at 21:42
  • $\begingroup$ 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) $\endgroup$ May 17, 2020 at 21:46

1 Answer 1


$\DeclareMathOperator\Lip{Lip}$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 $f\in C$.


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