Let $E$ be a $\mathbb R$-Banach space, $d$ be a metric on $E$ and $\mu$ be a probability measure on $(E,\mathcal B(E))$ with $$\int d(\;\cdot\;,0)\:{\rm d}\mu<\infty\tag1.$$ Moreover, let $$|f|_{\operatorname{Lip}(d)}:=\sup_{\substack{x,\:y\:\in\:E\\x\:\ne\:y}}\frac{|f(x)-f(y)|}{d(x,y)}\;\;\;\text{for }f:E\to\mathbb R$$ and $$\operatorname{Lip}(d):=\left\{f:E\to\mathbb R\mid|f|_{\operatorname{Lip}(d)}<\infty\right\}.$$
$|\;\cdot\;|_{\operatorname{Lip}(d)}$ is a semi-norm on $\operatorname{Lip}(d)$. By $(1)$, $$\operatorname{Lip}(d)\subseteq\mathcal L^1(\mu)$$ and $$\left\|f\right\|_{\operatorname{Lip}(d)}:=\left|\int f\:{\rm d}\mu\right|+|f|_{\operatorname{Lip}(d)}\;\;\;\text{for }f\in\operatorname{Lip}(d)$$ is a norm.
Are we able to show that $\operatorname{Lip}(d)\cap C^1(E,\mathbb R)$ is a closed subspace of $\left(\operatorname{Lip}(d),\left\|\;\cdot\;\right\|_{\operatorname{Lip}(d)}\right)$?