Show that $(x,y)\mapsto\inf_{\text{curve connecting }x,\:y}\int_0^1v^r\left(c(t)\right)\left\|c'(t)\right\|\:{\rm d}t$ is a metric 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$^1$ $$\rho(x,y):=\inf_{\substack{c\:\in\:C^1([0,\:1],\:E)\\ c(0)=x\\ c(1)=y}}\int_0^1v^r\left(c(t)\right)\left\|c'(t)\right\|_E\:{\rm d}t\;\;\;\text{for }x,y\in E.$$

Question 1: How can we show that $\rho$ is a metric on $E$?

We will need to show


*

*$\rho(x,y)=0\Leftrightarrow x=y$;

*$\rho(x,y)=\rho(y,x)$;

*$\rho(x,z)\le\rho(x,y)+\rho(y,z)$
for all $x,y,z\in E$.
1.: If $x=y$, then $$c(t):=x\;\;\;\text{for }t\in[0,1]$$ is in $C^1([0,1],E)$ with $c(0)=x$, $c(1)=y$ and $c'(t)=0$. So, $\rho(x,y)=0$. But how can we conclude $x=y$ from $\rho(x,y)=0$?
2.: I'd argue that if $c\in C^1([0,1],E)$ with $c(0)=x$ and $c(1)=y$, then $$\tilde c(t):=c(1-t)\;\;\;\text{for }t\in[0,1]$$ is in $C^1([0,1],E)$ with $\tilde c(0)=y$ and $\tilde c(1)=x$. But what's the rigorous reason why this yields the claim?
3.: If $c_1,c_2\in C^1([0,1],E)$ with $c_1(0)=x$, $c_1(1)=y=c_2(0)$ and $c_2(1)=z$, then $$c(t):=(1-t)c_1(t)+tc_2(t)\;\;\;\text{for }t\in[0,1]$$ is in $C^1([0,1],E)$ with $c(0)=c_1(0)=x$ and $c(1)=c_2(1)=z$. Again: What's the rigorous reason why this yields the claim?

Question 2: What can we say about the metric $\rho$? Is it complete or separable? And how is it related to the canonical metric on $E$?


$^1$ As usual, $C^1([0,1],E):=\{c\in C^0([0,1]):\left.c\right|_{(0,\:1)}\in C^1((0,1),E)\}$.
 A: To solve such a problem you have to think in terms of Finsler geometry, except you will work with the (potentially) infinite-dimensional Finsler manifold $E$. If hope you do not mind if I denote your function $v^r(x)$ by $f(x)$, since $v$ is normally used to denote vectors.
The relevant Finsler metric $F=|\cdot|_F$ on $E$, is a fiberwise norm on the tangent bundle $TE$, given by
$$
|v|_F= f(x)||v||, v\in T_xE
$$
is conformal to the flat metric on $E$ given by the original norm of $E$ as the Banach space.  I will use the notation $M$ for the Finsler manifold $(E, F)$. For most of the arguments, it does not matter at all that your Finsler metric is conformal to the flat metric or that $E$ was a Banach space, any Banach manifold would do.
The standard reference for Finsler geometry is
Bao, D.; Chern, S.-S.; Shen, Z., An introduction to Riemann-Finsler geometry, Graduate Texts in Mathematics. 200. New York, NY: Springer. xx, 431 p. (2000). ZBL0954.53001.
They work with finite-dimensional manifolds but for most arguments this does not matter at all. The relevant sections to read are 6.1, 6.2, even more specifically, Lemma 6.2.1.
With this in mind, the Finsler length of a path $c:[a,b]\to M$ is defined as the integral
$$
L(c)=\int_a^b F(c'(t))dt. 
$$
(Keep in mind that I am thinking of $c'(t)$ as an element of the tangent bundle $TM$.) This is exactly your formula. It is convenient to allow continuous paths $c$ which are only piecewise-smooth (smooth here and below means $C^1$).
Define a function $d_F$ on $M\times M$ by
$$
d_F(p,q)= \inf_c L(c)
$$
where the infimum is taken over all piecewise-smooth paths connecting $p$ to $q$. We will prove that $d_F$ is a distance function on $M$ locally equivalent to the distance function $d_E$ given by the Finsler norm on $E$.
First, one uses the change of variables formula (for real-valued functions of one variable) to verify that the length does not depend on parameterization in the sense that if
$$
h: I_1=[a_1,b_1]\to I_2=[a_2,b_2]
$$
is a piecewise-smooth homeomorphism then for each piecewise-smooth path $c_2: I_2\to M$ and $c_1= c_2\circ h$,
$$
L(c_1)=L(c_2). 
$$
This proves that $d(p,q)=d(q,p)$ since we can use an orientation-reversing linear homeomorphism $h: [a,b]\to [b,a]$, just as in your question.
Another thing that this will prove is that the infimum taken over all piecewise-smooth paths is the same as the infimum taken over all smooth paths: If $c:I\to M$ has the set $T$ of non-smoothness points $T=\{t_1,...,t_n\}\subset  I$, then choose a  smooth homeomorphism $h: I\to I$ fixing all the points $t_i$, smooth on the complementary open intervals $I\setminus T$ and has vanishing derivatives at all points $t_i$. The composition $c_1=c\circ h$ will then be smooth everywhere and, of course, $L(c_1)=L(c)$ as before.
As for the triangle inequality, the proof is the one that you started to write, except I again use independence of parameterization: Consider two paths
$$
c_1: I_1=[a_1,b_1]\to M, c_2: I_2=[a_2,b_2]\to M, 
$$
such that $c_1(b_1)=c_2(a_2)$.
WLOG, $b_1=b_2$. Set $I=[a_1,b_2]$.
Then we have the concatenation path $c: I\to M$:
$$
c|_{I_{1}}=c_1, c|_{I_2}=c_2. 
$$
Then additivity of the integral (again, functions of one real variable) implies:
$$
L(c)= \int_{a_1}^{b_1} F(c_1'(t))dt + \int_{b_1}^{b_2} F(c_2'(t))dt =L(c_1)+ L(c_2). 
$$
From this you get the triangle inequality for  $d_F$, hence, verifying that it is a pseudo-metric.
Lastly, let's prove that $d_F$ is a metric. I will now use the extra assumption that we have, namely, that $M=(E,F)$ and $F$ is conformal to the Banach norm on $E$ with the conformality factor $f$. Then fact that $f$ is assumed in your question to bounded away from $0$ on the entire $E$ simplifies matter quite a bit (for a general proof see Lemma 6.2.1 mentioned earlier):
$$
f(x)\ge \epsilon >0, \forall x\in E. 
$$
Then for every path $c: [a,b]\to E$
$$
L(c)\ge \epsilon \int_a^b ||c'(t)||dt= \epsilon L_E(c),
$$
where $L_E$ is the length of paths with respect to the original Banach norm on $E$.
Then, as explained in many places, e.g. this Wikipedia article (I am sure you can find it also on MSE):
For $c: [a,b]\to E$,
$$
L_E(c)= \lim_{N\to\infty} \sum_{i=1}^N ||c(t_i)- c(t_{i-1})|| 
$$
where we use regular partitions of $[a,n]$ into $N$ equal length intervals. By the triangle inequality for the Banach norm, for each $N$:
$$
\sum_{i=1}^N ||c(t_i)- c(t_{i-1})||\ge ||c(b)-c(a)||.  
$$
Therefore, taking the limit, we obtain
$$
L_E(c)\ge ||c(b)- c(a)||> 0, 
$$
unless $c(a)=c(b)$. Thus, $d_F(p,q)\ge \epsilon ||p-q||>0$ whenever $p\ne q$.
Thus, $d_F$ is a metric. By continuity, the function $f: E\to {\mathbb R}_+$ is locally bounded:
$$
\forall p\in E ~~\exists \delta_p, C_p: ~~||x-p||\le \delta_p \Rightarrow f(x)\le C_p.   
$$
Thus, for each $p\in E$ the restriction of $d_F$ to the Banach ball $B(p,\delta_p)$ satisfies:
$$
\epsilon d_E(x,y)\le d_F(x,y)\le C_p d_E(x,y). 
$$
In particular, the topology on $E$ defined by $d_F$ is the same as the original Banach space topology.
A: When $c_{xy},\ c_{yz}$ are curves s.t. $$\rho(x,y) +\epsilon
> \int_0^1\ v^r(c_{xy}(t)) \| c_{xy}'(t)\| ,\ \rho(y,z) +\epsilon
> \int_0^1\ v^r(c_{yz}(t)) \| c_{yz}'(t)\|,$$ then consider their union $c$ so that \begin{align*}
\rho (x,z) &\leq \int_0^2\ v^r(c(t) )\| c'(t)\| = \int_0^1\
v^r(c_{xy}(t)) \| c_{xy}'(t)\| + \int_0^1\ v^r(c_{yz}(t)) \|
c_{yz}'(t)\| \\&\leq  \rho(x,y) + \rho(y,z) + 2 \epsilon
\end{align*}
Since $\epsilon$ is arbitrary, then $\rho $ is a metric.
