Metrization of a topological vector space

Question

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

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

Answer 1

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.

Answer 2

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 $$d(g,f):=\sum_{j=1}^\infty\frac1{2^j}\frac{d_{u_j,w_j}(g,f)}{1+d_{u_j,w_j}(g,f)}. $$