Lipschitz-free space of countable uniformly discrete metric space

I assume here that the reader is familiar with the concept of Lipschitz-free space $$\mathcal{F}(X)$$ of a metric space $$X$$. I will follow the definition of $$\mathcal{F}(X)$$ as the completion of the spaces of molecules on $$X$$. See the book "Lipschitz Algebras" by Nik Weaver, for details.

Let $$X$$ be a metric space. Given $$x,y\in X$$, we denote by $$m_{x,y}$$ the elementary molecule $$m_{x,y}=1_{\{x\}}-1_{\{y\}}$$ If $$X$$ is a pointed metric space with distinguished pointed denoted by $$0$$, we write $$m_{x}:=m_{x,0}$$

From now on, we assume that $$X$$ is a pointed metric space, with distinguished point $$0$$.

It is easy to see that $$\{m_{x}:x\in X\setminus\{0\}\}$$ is an algebraic basis of the space of molecules on $$X$$. Concretely, if $$m:X\longrightarrow \mathbb{K}$$ is a molecule on $$X$$, then $$m=\sum_{x\neq 0}m(x)\cdot m_{x},$$ the sum being finite because $$m$$ is finitely supported.

In particular, if $$X$$ is a finite metric space, then the space of molecules (and therefore $$\mathcal{F}(X)$$) is a finite-dimensional normed space with dimension $$\# X-1$$.

Suppose now that $$X$$ is a countable uniformly discrete (pointed) metric space (uniformly discrete means that $$\theta:=\inf_{x\neq y}d(x,y)>0$$). Let $$\mu$$ be the measure on $$X\setminus\{0\}$$ defined by $$\mu(\{x\})=d(x,0)$$. Consider the linear map $$S:L_{1}(\mu)\longrightarrow \mathcal{F}(X)$$ defined by

$$S((\alpha_{x})_{x\neq 0})=\sum_{x\neq 0}\alpha_{x}m_{x}$$ This map is well-defined because the sum on the right-hand side is absolutely convergent. We have

$$\|S(\alpha)\|_{\mathcal{F}(X)}\leq \|\alpha\|_{L_{1}(\mu)}$$ for all $$\alpha\in L_{1}(\mu)$$. It is clear also that the range of $$S$$ contains the space of molecules on $$X$$.

It is not hard to see that $$S$$ is injective. To see that, we will use that usual isometric duality $$\mathcal{F}(X)^{*}=\text{Lip}_{0}(X)$$. Suppose that $$\alpha\in L_{1}(\mu)$$ is such that $$S(\alpha)=0$$. Fix $$y\in X\setminus\{0\}$$, and let $$f:X\longrightarrow \mathbb{K}$$ be the map defined as $$f(x)=1_{\{x=y\}}$$. It is immediate that $$f\in \text{Lip}_{0}(X)$$, with $$\text{Lip}(f)=\frac{1}{d(y,X\setminus\{y\})}$$. We have

$$0=\langle f, S(\alpha)\rangle=\langle f, \sum_{x\neq 0}\alpha_{x}m_{x}\rangle=\sum_{x\neq 0}\alpha_{x}f(x)=\alpha_{y}$$

Since $$y\in X\setminus\{0\}$$ was arbitrary, we conclude that $$\alpha=0$$. (this argument shows that $$S$$ is injective also if we have the weaker condition that $$X$$ is a countable discrete metric space)

My main question is: ¿is the range of $$S$$ closed?

If that is so, then (since it contains the space of molecules) it must be surjective, and then, by the Open Mapping Theorem, it must be an isomorphism between Banach spaces.

Here is my try: suppose that $$(\alpha^{(n)})_{n\in\mathbb{N}}$$ is a sequence in $$L_{1}(\mu)$$ such that $$(S(\alpha^{(n)}))_{n\in\mathbb{N}}$$ is a Cauchy sequence in $$\mathcal{F}(X)$$. Given $$\varepsilon>0$$, there exists $$N\in\mathbb{N}$$ such that for all $$m,n\geq N$$, we have

$$\varepsilon>\|S(\alpha^{(n)})-S(\alpha^{(m)})\|_{\mathcal{F}(X)}=\sup\limits_{f\in B_{\text{Lip}_{0}(X)}}|\langle f, S(\alpha^{(n)})-S(\alpha^{(m)})\rangle|=\sup\limits_{f\in B_{\text{Lip}_{0}(X)}}|\sum_{x\neq 0}(\alpha_{x}^{(n)}-\alpha_{x}^{(m)})f(x)|$$

By considering the Lipschitz maps $$f^{y}(x)=d(y,X\setminus\{y\})\cdot 1_{\{y=x\}}$$, we see that

$$|\alpha_{x}^{(n)}-\alpha_{x}^{(m)}|d(x,X\setminus\{x\})<\varepsilon\;\forall m,n\geq N\text{ and }x\in X\setminus\{0\}$$

Therefore,

$$|\alpha_{x}^{(n)}-\alpha_{x}^{(m)}|<\frac{\varepsilon}{\theta}\;\forall\; n,m\geq N\text{ and }x\in X\setminus\{0\}$$, where $$\theta:=\inf\limits_{x\neq y}d(x,y)>0$$

Consequently, for each $$x\in X\setminus\{0\}$$, the limit $$\beta_{x}:=\lim\limits_{n\rightarrow \infty}\alpha_{x}^{(n)}$$ exists in $$\mathbb{K}$$, and the convergence of $$(\alpha^{(n)})_{n\in\mathbb{N}}$$ to $$\beta$$ is uniform.

But all this does not seem enough to infer that the limit of $$(S(\alpha^{(n)}))_{n\in\mathbb{N}}$$ belongs to the range of $$S$$

This works if $$X$$ has finite diameter, but not in general. An easy way to see why not is to look at the elementary molecules $$m_{xy}$$, since $$\|m_{xy}\|_{\rm AE} = d(x,y)$$. The norm of the corresponding element of $$L^1(X)$$ is $$d(x,0) + d(y,0)$$, which can be arbitrarily larger than $$d(x,y)$$. So your map couldn't be a linear homeomorphism because the two norms aren't comparable.