Let $E$ be a $\mathbb R$-Banach space, $v:E\to[1,\infty)$ be continuous, $$\rho(x,y):=\inf_{\substack{c\:\in\:C^1([0,\:1],\:E)\\c(0)\:=\:x\\c(1)\:=\:y}}\int_0^1v\left(c(t)\right)\left\|c'(t)\right\|_E\:{\rm d}t\;\;\;\text{for }x,y\in E$$ and $$|f|_{\operatorname{Lip}(\rho)}:=\sup_{\substack{x,\:y\:\in\:E\\x\:\ne\:y}}\frac{|f(x)-f(y)|}{\rho(x,y)}\;\;\;\text{for }f:E\to\mathbb R.$$
Let $f:E\to\mathbb R$ be Fréchet differentiable and $\rho$-Lipschitz continuous with Lipschitz constant $1$. How can we show that $$\left\|{\rm D}f(x)\right\|_{E'}\le v(x)\tag1$$ for all $x\in E$?
The idea is clear to me: Let $x\in E$ and assume $(1)$ does not hold. Then there is a $h\in E$ with $\left\|h\right\|_E=1$ and $$|{\rm D}f(x)h|>v(x)\tag2.$$ Let $\varepsilon>0$. Since $$\frac{f(x+th)-f(x)}t\xrightarrow{t\to0}{\rm D}f(x)h\tag3,$$ there is a $\delta>0$ with \begin{equation}\begin{split}&\left|\left|f(x+th)-f(x)\right|-|t|\left|{\rm D}f(x)h\right|\right|\\&\;\;\;\;\;\;\;\;\;\;\;\;\le\left|f(x+th)-f(x)-t{\rm D}f(x)h\right|<|t|\varepsilon\end{split}\tag4\end{equation} and hence $$|f(x+th)-f(x)|>|t|(v(x)-\varepsilon)\tag5$$ for all $0<|t|<\delta$. On the other hand, by considering the straight line connecting $x$ and $x+th$, we see that (since $f$ is $\rho$-Lipschitz with Lipschitz constant $1$) $$|f(x+th)-f(x)|\le\rho(x,x+th)\le|t|\int_0^1v(x+sth)\:{\rm d}s\tag6$$ for all $t\in\mathbb R$. By the mean value theorem (and since $v$ is continuous), $$\forall t\in\mathbb R:\exists s\in[0,1]:v(x+sth)=\int_0^1v(x+rth)\:{\rm d}r\tag7.$$ Combining $(5)$, $(6)$ and $(7)$, we obtain $$\forall t\in(-\delta,\delta)\setminus\{0\}:\exists s\in[0,1]:v(x)-\varepsilon<v(x+sth)\tag8.$$
Is this already a contradiction?
EDIT:
Maybe it's simpler: By continuity of $v$, $$\mathbb R\ni t\mapsto\int_0^1v(x+sth)\:{\rm d}s\tag9$$ should be continuous. So there is a $\tilde\delta>0$ such that $$\left|\int_0^1v(x+sth)\:{\rm d}s-v(x)\right|<\varepsilon\;\;\;\text{for all }|t|<\tilde\delta\tag{10}.$$ Maybe we can obtain a contradiction by combining $(5)$, $(6)$ and $(10)$.