Bound length of a curve in a ball

Question

Let $E$ be a $\mathbb R$-Banach space, $v:E\to[1,\infty)$ be continuous and $v_i:[0,\infty)\to[1,\infty)$ be continuous and nondecreasing with $$v_1(\left\|x\right\|_E)\le v(x)\le v_2(\left\|x\right\|_E)\;\;\;\text{for all }x\in E,\tag1$$ $$v_1(a)\xrightarrow{a\to\infty}\infty\tag2$$ and $$av_2(a)\le C_1v_1^\theta(a)\;\;\;\text{for all }a>0\tag3$$ for some $C_1\ge0$ and $\theta\ge1$. Now, let $r\in(0,1]$ and $$\rho(x,y):=\inf_{\substack{\gamma\:\in\:C^1([0,\:1],\:E)\\ \gamma(0)\:=\:x\\ \gamma(1)\:=\:y}}\int_0^1v^r\left(\gamma(t)\right)\left\|\gamma'(t)\right\|_E\:{\rm d}t\;\;\;\text{for }x,y\in E.$$

Let $k>0$ and $B_k$ denote the open ball around $0\in E$ with radius $k$. Let $x,y\in E$ and $\varepsilon>0$. By definition of the infimum, there is a $\gamma\in C^1([0,1],E)$ with $\gamma(0)=x$, $\gamma(1)=y$ and $$\rho(x,y)\le\int_0^1v^r\left(\gamma(t)\right)\left\|\gamma'(t)\right\|_E\:{\rm d}t<\rho(x,y)+\varepsilon\tag4.$$

Question: Why can we conclude that $$\int_0^11_{B_k}(\gamma(t))\left\|\gamma'(t)\right\|_E\:{\rm d}t\le 2k\left(\frac{v_2(k)}{v_1(0)}\right)^r+\varepsilon?\tag5$$

The argument should be that we could otherwise replace the corresponding piece of curve by a straight line and obtain a value which differed from $\rho(x,y)$ by more than $\varepsilon$, but how can we show this rigorously?

EDIT: I mean, by $(1)$, there is the trivial inequality $$1\le\frac{v_2(\left\|z\right\|_E)}{v_1(\left\|z\right\|_E)}\le\frac{v_2(k)}{v_1(0)}\;\;\;\text{for all }z\in B_k(0)\tag6$$ and I guess a variant of this needs to be used.

EDIT 2: The inequality is clearly trivial, when $\gamma$ never enters $B_k$. Maybe it's useful to consider the entrance and exit times/points: Let $\sigma_0:=\tau_0:=0$, \begin{align}\sigma_n&:=\inf\{t\in(\tau_{n-1},1):\gamma(t)\in B_k\},\\\tau_n&:=\inf\{t\in(\sigma_n,1):\gamma(t)\not\in B_k\}\wedge 1\end{align} for $n\in\mathbb N$, $N:=\{n\in\mathbb N:\sigma_n<\infty\}$ and \begin{align}x_n&:=\gamma(\sigma_n),\\y_n&:=\gamma(\tau_n)\end{align} and $$c_n(t):=\frac{t(y_n-x_n)+\tau_nx_n-\sigma_ny_n}{\tau_n-\sigma_n}\;\;\;\text{for }t\in[\sigma_n,\tau_n]$$ be the straight line connecting $x_n$ and $y_n$ for $n\in N$. The left-hand side of $(5)$ can then be rewritten as $$\int_0^11_{B_k}(\gamma(t))\left\|\gamma'(t)\right\|_E\:{\rm d}t=\sum_{n\in N}\int_{\sigma_n}^{\tau_n}\left\|\gamma'(t)\right\|_E\:{\rm d}t\tag7.$$ Maybe it can be shown that if the desired inequality doesn't hold, we could replace $\gamma$ on $[\sigma_n,\tau_n]$ with $c_n$ and obtain a value of the integral in $(4)$ which differs by more than $\varepsilon$ from $\rho(x,y)$.

Assume, for simplicity, that $N=\{1\}$ and let $$\tilde\gamma(t):=\left.\begin{cases}\gamma(t)&\text{, if }t\in[0,\sigma_1]\\ c_1(t)&\text{, if }t\in[\sigma_1,\tau_1]\\\gamma(t)&\text{, if }t\in[\tau_1,1]\end{cases}\right\}\;\;\;\text{for }t\in[0,1].$$ We may clearly note that, by construction, $$c_1((\sigma_1,\tau_1))\subseteq B_k\tag8$$ and hence $$\int_{\sigma_1}^{\tau_1}v^r(\tilde\gamma(t))\left\|\tilde\gamma'(t)\right\|_E\:{\rm d}t=\frac{\left\|x_1-y_1\right\|_E}{\tau_1-\sigma_1}\int_{\sigma_1}^{\tau_1}v^r(c_1(t))\le 2kv^r_2(k)\tag9$$ by $(1)$. Now, maybe we need to use $(6)$, $v_1\ge1$ and $r\le1$ to obtain $$v^r_2(k)\le\left(\frac{v_2(k)}{v_1(0)}\right)^r\le\left(\frac{v_2(\left\|z\right\|_E)}{v_1(\left\|z\right\|_E)}\right)^r\;\;\;\text{for all }z\in B_k\tag{10}.$$ I think I'm close but still can't complete the puzzle.

EDIT 3: I think we can argue in the following way: Assume $\sigma_1<\infty$ so that the curve enters the ball $B_k$ at time $\sigma_1$. Replace $\gamma$ on $[\sigma_1,\tau_1]$ by $c_1$, which yields the curve $\tilde\gamma$ (as defined above).

If $(5)$ would not hold, then \begin{equation}\begin{split}\int_{\sigma_1}^{\tau_1}v^r(\tilde\gamma(t))\left\|\tilde\gamma'(t)\right\|_E\:{\rm d}t&\le 2kv_2^r(k)\le2ke\left(\frac{v_2(k)}{v_1(0)}\right)^r\\&<2ke\left(\frac{v_2(k)}{v_1(0)}\right)^r+\varepsilon<\int_{\sigma_1}^{\tau_1}1_{B_k}(\gamma(t))\left\|\gamma'(t)\right\|_E\:{\rm d}t\\&\le\int_{\sigma_1}^{\tau_1}\underbrace{v^r(\gamma(t))}_{\ge\:1}\left\|\gamma'(t)\right\|_E\:{\rm d}t\end{split}\tag{11}\end{equation} by $(9)$ and $(10)$.

Is this a contradiction to $(4)$?

Answer 1

1. For $k>0$, $v_2(k)\geq v_2(0)\geq v_1(0)\geq 1$ so that $\frac{v_2(k)}{v_1(0)} \geq 1$

2. Since $v(x)\geq 1$, then $\int_0^1\ v^r(\gamma (t)) \|\gamma'(t)\| \geq \int_0^1\ \|\gamma '(t)\|$

3. When $\rho(x,y) =\int\ v^r(\gamma(t))\|\gamma'(t)\|$, then we consider the simple case where $\gamma $ is union of two curves $\gamma_1$ and $\gamma_2$ s.t. $\gamma_1$ is not in $B_k$ and $\gamma_2$ in $B_k$ up to finite points.

Then $$ \int \ 1_{B_k}(\gamma(t))\|\gamma'(t)\| = \int \ \|\gamma_2'(t)\| =\ast$$

When $\gamma_2$ starts at a point $x'\in \partial B_k$ and escape at a point $y'\in \partial B_k$, then $\ast \leq \rho(x',y')$, since $\gamma$ is minimizing so that so is its part $\gamma_2$.

Hence $$ \ast \leq \rho (x',y')\leq \rho (0,x') +\rho (0,y')\leq 2k \leq 2k (\frac{v_2(k)}{v_1(0)} )^r $$

[Add] For the convenience we call $\int\ v^r(c(t))\| c'(t)\|$ $\rho$-length for a curve $c$. When $c$ is a minimizing curve and $t_1$ is the first time going into the ball $B_k$ and $t_2$ is last time escaping the ball, then $\rho$-length of $c|[t_1,t_2]$ is $\leq 2k$ : Note that there are curves from $0$ to $c(t_i)$ in the ball s.t. their $\rho$-length is $k$. Clearly their union has a $\rho$-length larger than that of $c|[t_1,t_2]$

Answer 2

Partial Answer:

I'll assume that we can assume that $\gamma$ enters $B_k$ at most once. So, hopefully someone can provide a proof for this claim.

Everything is already established in the question, but for clarity, I'll repeat some of the arguments:

Let \begin{align}\sigma&:=\inf\{t\in(0,1):\gamma(t)\in B_k\},\\\tau&:=\inf\{t\in(\sigma,1):\gamma(t)\not\in B_k\}\wedge 1.\end{align} Assume $\sigma<\infty$ (otherwise the claim is trivial). Let $x_\sigma:=\gamma(\sigma)$, $x_\tau:=\gamma(\tau)$, $$c(t):=\frac{t(x_\tau-x_\sigma)+\tau x_\sigma-\sigma x_\tau}{\tau-\sigma}\;\;\;\text{for }t\in[\sigma,\tau]$$ and $$\tilde\gamma(t):=\left.\begin{cases}\gamma(t)&\text{, if }t\in[0,\sigma]\\ c(t)&\text{, if }t\in[\sigma,\tau]\\\gamma(t)&\text{, if }t\in[\tau,1]\end{cases}\right\}\;\;\;\text{for }t\in[0,1].$$ By construction, $c((\sigma,\tau))\subseteq B_k$ and hence (since $v_1\ge1$ and $r\le1$), \begin{equation}\begin{split}\int_\sigma^\tau v^r(c(t))\left\|c'(t)\right\|_E\:{\rm d}&=\frac{\overbrace{\left\|x_\sigma-x_\tau\right\|_E}^{\le\:2k}}{\tau-\sigma}\int_\sigma^\tau\underbrace{v^r(c(t))}_{\le\:v_2^r(k)}\:{\rm d}t\\&\le2kv_2^r(k)\le2k\left(\frac{v_2(k)}{v_1(0)}\right)^r\end{split}\tag{12}\end{equation} Now assume $(5)$ does not hold. Then, \begin{equation}\begin{split}\int_\sigma^\tau v^r(\tilde\gamma(t))\left\|\tilde\gamma'(t)\right\|_E\:{\rm d}t+\varepsilon&\le2k\left(\frac{v_2(k)}{v_1(0)}\right)^r+\varepsilon\\&<\int_0^11_{B_k}(\gamma(t))\left\|\gamma'(t)\right\|_E=\int_\sigma^\tau\left\|\gamma'(t)\right\|_E\:{\rm d}t\\&\le\int_\sigma^\tau v^r(\gamma(t))\left\|\gamma'(t)\right\|_E\:{\rm d}t,\end{split}\tag{13}\end{equation} where I've used the assumption that $\gamma$ enters $B_k$ only in the time interval $(\sigma,\tau)$ in the equality in the second line. $(12)$ is clearly a contradiction to $(4)$ and hence we are done.