Let $E$ be a $\mathbb R$-Banach space, $v:E\to(0,\infty)$ be continuous with $$\inf_{x\in E}v(x)>0\tag1,$$ $r\in(0,1]$ and$^1$ $$\rho(x,y):=\inf_{\substack{c\:\in\:C^1([0,\:1],\:E)\\ c(0)=x\\ c(1)=y}}\int_0^1v^r\left(c(t)\right)\left\|c'(t)\right\|_E\:{\rm d}t\;\;\;\text{for }x,y\in E.$$ Note that $\rho$ is a well-defined metric on $E$.

Let $\mu$ be a probability measure on $(E,\mathcal B(E))$ and $f:E\to\mathbb R$ be Fréchet differentiable. How can we show that \begin{equation}\begin{split}\left\|f\right\|&:=\sup_{\substack{x,\:y\:\in\:E\\ x\:\ne\:y}}\frac{|f(x)-f(y)|}{\rho(x,y)}+\left|\int f\:{\rm d}\mu\right|\\&=\sup_{x\in E}\frac{\left\|{\rm D}f(x)\right\|_{E'}}{v^r(x)}+\int f\:{\rm d}\mu?\end{split}\tag2\end{equation} In particular, how can we show that $$\sup_{\substack{x,\:y\:\in\:E\\0\:<\left\|x-y\right\|_E\:<\:\varepsilon}}\frac{|f(x)-f(y)|}{\rho(x,y)}\xrightarrow{\varepsilon\to0}\frac{\left\|{\rm D}f(x)\right\|_{E'}}{v^r(x)}\tag3?$$