This question is a continuation of this post : Metrization of a topological vector space

Let $C_{lip}(\mathbb R^d)$ be the space of Lipschitz functions on $\mathbb R^d$. We endow $C_{lip}(\mathbb R^d)$ with the following topology: $(f_n)_{n\ge 1} \subset C_{lip}(\mathbb R^d)$ converges to $f\in C_{lip}(\mathbb R^d)$ iff

$$\lim_{n\to\infty} \left\{\left|\int_{\mathbb R^d}(f_n-f)(x)u(x)dx\right| + \left|\int_{\mathbb R^d}\nabla(f_n-f)(x)\cdot w(x)dx\right|\right\} = 0,$$

for all $u:\mathbb R^d\to\mathbb R$ and $w:\mathbb R^d\to\mathbb R^d$ satisfying

$$\int_{\mathbb R^d}|u(x)|(1+|x|)dx<\infty \quad\mbox{and}\quad \int_{\mathbb R^d}|w(x)|dx<\infty.$$

Consider the completion $\overline{C}_{lip}(\mathbb R^d)$ of $C_{lip}(\mathbb R^d)$ w.r.t. this topology. Could we show that any linear continuous function $T: \overline{C}_{lip}(\mathbb R^d)\to\mathbb R$ must be of the form

$$T(f)=\int_{\mathbb R^d}f(x)u(x)dx+\int_{\mathbb R^d}\nabla f(x)w(x)dx?$$


2 Answers 2


This is my understanding of the answer given by Pietro Majer. The reasoning could be divided into three steps:

  1. The dual space of $L^1(\mathbb R^d)^{d+1}$ is identified as $L^{\infty}(\mathbb R^d)^{d+1}$, i.e. $\big(L^1(\mathbb R^d)^{d+1}\big)^*=L^{\infty}(\mathbb R^d)^{d+1}$;

  2. The embedding $E\subset L^{\infty}(\mathbb R^d)^{d+1}$ is weakly* closed. What does "weakly* closed" refer to? Here the definitions of $F^\perp$ and $E_\perp$ are not clear to me either;

  3. The elements $(g,w)$ of $L^1(\mathbb R^d)^{d+1}$ represent all $w^*$-continuous functionals on ${\rm Lip}(\mathbb R^d)$ via $T_{(g,w)}$. Why this is true? We only know $L^{\infty}(\mathbb R^d)^{d+1}$ is the dual space of $L^1(\mathbb R^d)^{d+1}$, but do not know the dual space of $L^{\infty}(\mathbb R^d)^{d+1}$.

Finally, if I only consider the space of linear continuous functionals for the space ${\rm Lip}_0(\mathbb R^d)$, could we simplify the above proof?

  • $\begingroup$ 1) Yes. 2) it refers to the weakly* topology of $L^\infty(\mathbb{R}^d)^{d+1}$ as a dual space, per (1). If $E\subset X^*$, $E_\perp:=\cap_{f\in E}\ker f\subset X$ and if $F\subset X$, $F^\perp:=(j_XF)_\perp\subset X^*$ where $j_X$ is the canonical embedding $j_X:X\to X^{**}$. By Hahn-Banach, $(E_\perp)^\perp$ is the weak* closure of ${\rm span}E$ in $X^*$ and $(F^\perp)_\perp$ is the weak-closure but also norm-closure of ${\rm span}F$ in $X$ . (reference: any textbook of linear functional analysis). $\endgroup$ Nov 25, 2019 at 16:18
  • $\begingroup$ 3) As a general fact, w* continuous linear functionals on $X^*$ are precisely the elements of $X^{**}$ in the image of $j_X$, i.e. the evaluations on the elements of $X$. (reference: ibid.) $\endgroup$ Nov 25, 2019 at 16:18
  • $\begingroup$ Finally: yes. In this case, the embedding is just $\nabla:{\rm Lip}_0(\mathbb{R}^d)\to L^\infty(\mathbb{R}^{d},\mathbb{R}^d)$. But one can simplify everything for ${\rm Lip} (\mathbb{R}^d) $ as well, using the embedding $\eta: {\rm Lip} (\mathbb{R}^d)\to \mathbb{R}\times L^\infty(\mathbb{R}^{d},\mathbb{R}^d)$ defined by $u\mapsto (u(0), \nabla u)$. In the case of $\nabla:{\rm Lip}_0(\mathbb{R}^d)\to L^\infty(\mathbb{R}^{d},\mathbb{R}^d)$ it follows from the definition that the new $E:=\nabla {\rm Lip}_0(\mathbb{R}^d)$ is $F^\perp$ for $F:=\{g\in L^1(\mathbb{R}^d,\mathbb{R}^d): {\rm{div}}g=0\}$ $\endgroup$ Nov 25, 2019 at 16:42
  • $\begingroup$ Whence $\rm{Lip}_0(\mathbb{R}^d),\|\cdot\|_{lip}$ is isometrically isomorphic to the dual space of $L^1(\mathbb{R}^d,\mathbb{R}^d)/F$. The latter is also very standard material about representation of duals of Sobolev spaces (though one usually does it for $W^{k,p}$; more or less any textbook treating Sobolev spaces has it). $\endgroup$ Nov 25, 2019 at 16:42
  • $\begingroup$ @PietroMajer Hi, I met another question (mathoverflow.net/questions/349238/…) under this setting. Could you please take a look? $\endgroup$
    – user128095
    Dec 27, 2019 at 23:25

The space of Lipschitz functions $\text{Lip}(\mathbb{R}^d)$ embeds via $f\mapsto (\frac{f}{1+|x|},\partial_1 f,\dots,\partial_d f)$ into the space $ L^\infty(\mathbb{R}^d)^{d+1}$, which we isometrically identify with the dual space of $ L^1(\mathbb{R}^d)^{d+1}$.

The image of this embedding of $\text{Lip}(\mathbb{R}^d)$ is a weakly* closed subspace $E$ of $ L^\infty(\mathbb{R}^d)^{d+1}$, in fact, presented by weak equations

$$E=\{ (g ,w_1,\dots,w_d)\in L^\infty(\mathbb{R}^d)^{d+1}: \partial_i\big((1+|x|)g \big)= w_i, \; \partial_iw_j=\partial_jw_i,\; \text{for all } i,j \}$$ as the annihilator $E=F^\perp$ of the $\|\cdot\|_1$-closed linear span $$F:=\overline{\text{span}}\big\{(1+|x|)\phi e_0+\partial_i\phi e_i,\;\partial_i\phi e_j-\partial_j\phi e_i:\;\phi\in C^\infty_0(\mathbb{R}^d) ,\, 1\le i\le j\le d\big\}.$$ (For our needs it is not necessary to characterize better the space $F$: it sufficient to know that $E$ is weakly* closed, thus $E=F^\perp$ for $F:=E_\perp\subset L^1(\mathbb{R}^d)^{d+1}$).

Therefore $E$ is a dual space, as shown by the isometry $$E= F^\perp\sim \bigg({L^1(\mathbb{R}^d)^{d+1}\over F }\bigg)^*$$ and induces its weak* topology $\tau^*$ on $\text{Lip}(\mathbb{R}^d)$, and the convergence structure of it. In particular, the elements $(g,w)$ of $L^1(\mathbb{R}^d)^{d+1}$ represents all $w^*$-continuous functionals on $\text{Lip}(\mathbb{R}^d)$ via $$T_{(g,w)}:\text{Lip}(\mathbb{R}^d)\ni f \mapsto \int_{\mathbb{R}^d}g(x){f(x)\over 1+|x|}dx+\int_{\mathbb{R}^d}w\cdot \nabla f(x)dx.$$

So the convergence you are considering is exactly the convergence associated with this weak* topology, which is not metrizable, of course; but as a general fact, a linear functional on a dual of a separable Banach $X$ is $w^*$-continuous iff it is $w^*$-sequentially continuous (iff is in the image of the embedding $E\to E^{**}$), so at the end the answer is, yes.

  • $\begingroup$ In fact I think you can even replace the first integral in the definition of $T(f)$ with $\lambda f(0)$ $\endgroup$ Nov 23, 2019 at 4:42
  • $\begingroup$ In fact the whole description can be made simpler; I'll edit and simplify. $\endgroup$ Nov 23, 2019 at 11:49
  • $\begingroup$ Many thanks for the answer which helps me really a lot! $\endgroup$
    – user128095
    Nov 23, 2019 at 12:21
  • $\begingroup$ Is everything clear to you? $\endgroup$ Nov 23, 2019 at 12:22
  • $\begingroup$ It is simpler to factor out the constants, and consider $\text{Lip}(\mathbb{R}^d)=\mathbb{R}\oplus\text{Lip}_0(\mathbb{R}^d)$, where $\text{Lip}_0(\mathbb{R}^d)$ is the closed hyperplane consisting of the functions vanishing at $0$; $\text{Lip}_0(\mathbb{R}^d)$ has a simpler embedding in $L^\infty$, just $f\mapsto \nabla f$; it is a norm- isometry by definition of the Lipschitz norm. $\endgroup$ Nov 23, 2019 at 12:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.