# Metrization of a topological vector space

Let $$C(\mathbb R^d)$$ be the space of continuous functions on $$\mathbb R^d$$, and $$C_{lip}(\mathbb R^d)\subset C(\mathbb R^d)$$ be the subspace of Lipschitz functions. 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

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

Is this topology metrizable?

PS: The motivation is to apply Baire's theorem. Any answers or comments are highly appreciated.

PS2: Thank Iosif Pinelis for the answer. I wish to ask a further question: Consider the completion $$\overline{C}_{lip}(\mathbb R^d)$$ of $$C_{lip}(\mathbb R^d)$$ w.r.t. this metric. Could we show that any linear continuous function $$T: \overline{C}_{lip}(\mathbb R^d)\to\mathbb R$$ must be of the form

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

I think the answer is no, no topology with this convergence is metrizable, for the following reason.

The space of the Lipschitz functions on $$\mathbb{R}^r$$, with the norm $$\|f\|_{\text{Lip}}:=\big\|{f\over1+|x|}\big\|_\infty+\|\nabla f\|_\infty$$ has a pre-dual Banach space $$X$$ (see Dual space of the completion of the space of Lipschitz functions). As said there, the convergence you are considering is the corresponding weak* convergence, that is, the point-wise convergence of functionals on $$X$$. edit here is a nice online reference.

It is well known that the weak* topology of the dual of an infinite dimensional Banach space $$X$$ is never metrizable (although its trace on bounded sets is). But is there any distance $$d$$ on $$X^*$$, whose metric topology $$\tau_d$$ has the same convergence of the weak* topology $$\tau_{w^*}$$?

The answer is no: assume by contradiction $$d$$ is such a distance. Then $$(X^*,\tau_d)\to (X^*,\tau_{w^*})$$ is sequentially continuous, hence continuous, that is $$\tau_d\subset \tau_{w^*}$$. On the other hand, the evaluation on elements of $$X$$ are continuous functionals on $$(X,\tau_d)$$ so $$\tau_d\supset \tau_{w^*}$$ because $$\tau_{w^*}$$ is the weaker topology that makes them continuous. So $$\tau_d=\tau_{w^*}$$, a contradiction, because as said $$\tau_{w^*}$$ is not metrizable.

• What i'm saying is that the op defines a topologic, non metrizable convergence. It is the convergence of a w* topology and of no metric topology. Nov 23, 2019 at 4:19
• I think not really "iff": the convergence $d(f_n,f)\to0$ is equivalent to the OP's condition only on bounded sets, in general it is weaker and depends on the particolar chosen dense set. Nov 23, 2019 at 4:27
• And yes, of course the question should have been more precisely phrased: a) Is this convergence associated to some topology? And b) Is this convergence associated to some metric topology?. Nov 23, 2019 at 4:36
• Thanks for the answer. Indeed, what I am saying is that, this convergence defines the collection of open sets, and then the collection of open sets defines a topology.
– user128095
Nov 23, 2019 at 12:23
• A convergence does not identify a topology; though here a natural choice is the w* topology. It is not metrizable, but it is metrizable on bounded sets. Nov 23, 2019 at 12:28

According to the answer by Pietro Majer, the convergence in question is incompatible with a metrizable topology.

On the hand, as Pietro Majer commented, the $$w^*$$ topology, which is compatible with this convergence, is metrizable on bounded sets. Let us provide details on this comment.
Let $$\||u\||:=\int_{\mathbb R^d}|u(x)|(1+|x|)dx\quad\quad\mbox{and}\quad \|w\|:=\int_{\mathbb R^d}|w(x)|dx<\infty.$$ Let $$U$$ be the normed space of all $$u$$ with $$\||u\||<\infty$$, and let $$W$$ be defined similarly, so that $$W=L^1$$. These two normed spaces are separable. Let $$\{u_j\colon j\in\mathbb N\}$$ and $$\{w_j\colon j\in\mathbb N\}$$ be corresponding countable dense sets in $$U$$ and $$W$$. Let $$d_{u,w}(g,f):=\left|\int_{\mathbb R^d}(g-f)(x)u(x)dx\right| + \left|\int_{\mathbb R^d}\nabla(g-f)(x)\cdot w(x)dx\right|.$$

Then it is easy to see that, for any sequence $$(f_n)$$ in $$C_{lip}(\mathbb R^d)$$ bounded with respect to the norm $$C_{lip}(\mathbb R^d)\ni g\mapsto\sup_x|g(x)|+\sup_{x\ne y}\frac{|g(x)-g(y)|}{|x-y|}$$, we have $$f_n\to f$$ iff $$d_{u_j,w_j}(f_n,f)\to0$$ for each natural $$j$$ iff $$d(f_n,f)\to0$$, where $$d$$ is the metric defined by $$d(g,f):=\sum_{j=1}^\infty\frac1{2^j}\frac{d_{u_j,w_j}(g,f)}{1+d_{u_j,w_j}(g,f)}.$$

• I was thinking about this, but I thought I saw a flaw. What is the approximation argument that would ensure $f_n$ goes to $f$ if the distances $d_j(f_n,f)$ go to zero? Nov 22, 2019 at 16:47
• Thanks for the reply. I've a related question (detailed above). Do you think the answer is also yes?
– user128095
Nov 22, 2019 at 17:43
• @PierrePC : I have added a detail concerning your question. Nov 22, 2019 at 19:07
• @IosifPinelis Thank you, but I still don't see it... To do it the way I think of it, I need the estimate $|f_n-f|=O(1+|x|)$ uniformly in $n$. I get that $\langle f,u\rangle$ is uniquely determined by $\langle f,u_j\rangle$, but I cannot seem to see why it would suffice. Is there a hidden Banach-Steinhaus somewhere? Nov 22, 2019 at 19:54
• @IosifPinelis I think the question is clarified a bit if we put it in a more abstract setting: For a dual of a separable infinite dimensional Banach space $X$, is there a distance on $X^*$ that induces the point-wise convergence of bounded functionals ---that is, the same convergence of the w*-topology? I think the answer is always no (see my answer) Nov 24, 2019 at 10:01