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 $$|f|_{\operatorname{Lip}(\rho)}:=\sup_{\substack{x,\:y\:\in\:E\\x\:\ne\:y}}\frac{|f(x)-f(y)|}{\rho(x,y)}\;\;\;\text{for }f:E\to\mathbb R$$ and $$\operatorname{Lip}(\rho):=\left\{f:E\to\mathbb R\mid|f|_{\operatorname{Lip}(\rho)}<\infty\right\}.$$ $|\;\cdot\;|_{\operatorname{Lip}(\rho)}$ is a semi-norm on $\operatorname{Lip}(\rho)$. Let $\mu$ be a probability measure on $(E,\mathcal B(E))$ with $$\int\rho(\;\cdot\;,0)\:{\rm d}\mu<\infty\tag2$$ By $(2)$, $$\operatorname{Lip}(\rho)\subseteq\mathcal L^1(\mu)$$ and $$\left\|f\right\|_{\operatorname{Lip}(\rho)}:=\left|\int f\:{\rm d}\mu\right|+|f|_{\operatorname{Lip}(\rho)}\;\;\;\text{for }f\in\operatorname{Lip}(\rho)$$ is a norm.

Let $f\in\operatorname{Lip}(\rho)$ be Fréchet differentiable. How can we show that $$\left\|f\right\|_{\operatorname{Lip}(\rho)}=\sup_{x\in E}\frac{\left\|{\rm D}f(x)\right\|_{E'}}{v^r(x)}+\int f\:{\rm d}\mu?\tag3$$ In particular, how can we show that $$\sup_{\substack{y\:\in\:E\\0\:<\left\|x-y\right\|_E\:<\:\varepsilon}}\frac{|f(x)-f(y)|}{\rho(x,y)}\xrightarrow{\varepsilon\to0}\frac{\left\|{\rm D}f(x)\right\|_{E'}}{v^r(x)}\tag4$$ for all $x\in E$?

EDIT: Let $x\in E$. It's clear to me that $$\sup_{y\in B_\delta(x)\setminus\{x\}}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E}\xrightarrow{\delta\to0+}\left\|{\rm D}f(x)\right\|_{E'}\tag5.$$ So, writing $$\frac{|f(x)-f(y)|}{\rho(x,y)}=\frac{|f(x)-f(y)}{\left\|x-y\right\|_E}\frac{\left\|x-y\right\|_E}{\rho(x,y)}\;\;\;\text{for all }y\in E\tag6,$$ it only remains to show $$\sup_{y\in B_\delta(x)\setminus\{x\}}\frac{\rho(x,y)}{\left\|x-y\right\|_E}\xrightarrow{\delta\to0+}v^r(x)\tag7.$$ Now let $$c(t,y):=(1-t)x+ty\;\;\;\text{for }(t,y)\in[0,1]\times E$$ is clearly continuous and $$\sup_{t\in[0,\:1]}\left\|c(t,y)-x\right\|_E=\left\|x-y\right\|_E\xrightarrow{y\to x}0\tag8.$$ Moreover, \begin{equation}\begin{split}&\sup_{y\in B_\delta(x)}\left|\frac1{\left\|x-y\right\|_E}\int_0^1(v^r\circ c)(t,y)\left\|\frac{{\rm d}c}{{\rm d}t}(t,y)\right\|_E\:{\rm d}t-v^r(x)\right|\\&\;\;\;\;\;\;\;\;\;\;\;\;\le\sup_{y\in B_\delta(x)}\int_0^1\left|(v^r\circ c)(t,y)-(v^r\circ c)(t,x)\right|\:{\rm d}t\\&\;\;\;\;\;\;\;\;\;\;\;\;\le\sup_{(t,\:y)\:\in\:[0,\:1]\times B_\delta(x)}\left|(v^r\circ c)(t,y)-(v^r\circ c)(t,x)\right|\xrightarrow{\delta\to0+}0\end{split}\tag9\end{equation} for all $y\in E$. So, this yields at least \begin{equation}\begin{split}&\sup_{y\in B_\delta(x)\setminus\{x\}}\frac{\rho(x,y)}{\left\|x-y\right\|_E}-v^r(x)\\&\;\;\;\;\;\;\;\;\;\;\;\;\le\sup_{y\in B_\delta(x)\setminus\{x\}}\int_0^1|(v^r\circ c)(t,y)-v^r(x)|\:{\rm d}t\xrightarrow{\delta\to0+}0.\end{split}\tag{10}\end{equation} How can we show the other inequality?


1 Answer 1


To get started: If $c:[0,1]\to E$ is a $C^1$-curve from $x$ to $y$ you have $$ |f(x)-f(y)|=\left|\int_0^1(f\circ c)'(t)dt\right|\le \int_0^1 \|Df(c(t))\|_{E'} \|c'(t)\|dt $$ $$=\int_0^1 \|Df(c(t))\|_{E'} \frac{v^r(c(t))}{v^r(c(t))} \|c'(t)\|dt$$ $$ \le \sup\{\|Df(z)\|_{E'}/v^r(z):z\in E\} \varrho(x,y)(1+\varepsilon)$$ for a suitable curve.

This gives you an inequality for (4) (I guess that $x$ is fixed, there). For the other inequality I would try to take a direction $r\in E$ which (almost) maximizes $|Df(x)(r)|$ and take $y=x+\varepsilon r$. However, I did not try to work this out.

  • $\begingroup$ Thank you for your answer. Please take note of my edit. $\endgroup$
    – 0xbadf00d
    Commented Jun 7, 2020 at 9:44
  • $\begingroup$ I guess you are missing a differential on the right-hand side of your first equality and your $\varrho$ should be a $\rho$. Where does the division by $v^r(z)$ come from in your supremum? $\endgroup$
    – 0xbadf00d
    Commented Jun 7, 2020 at 12:41
  • $\begingroup$ Right, the derivative was missing. I have expanded the calculation. $\endgroup$ Commented Jun 7, 2020 at 12:55
  • $\begingroup$ Thanks for the edit. What do you say to my edit? I've got the feeling we only need to show that $$\frac{\rho(x,y)}{\left\|x-y\right\|_E}\to v(x)$$ and for this it should be enough to show that there is a curve $c$ connecting $x,y$ with $$\left|\frac1{\left\|x-y\right\|_E}\int_0^1(v\circ c)(t)\left\|c'(t)\right\|_E\:{\rm d}t-v(x)\right|=\frac1{\left\|x-y\right\|_E}\left|\int_0^1(v\circ c)(t)\left\|c'(t)\right\|_E\:{\rm d}t-\left\|x-y\right\|_Ev(x)\right|\to0.$$ $\endgroup$
    – 0xbadf00d
    Commented Jun 7, 2020 at 13:02
  • $\begingroup$ I think your last inequality is wrong. It seems like you've thought that $\rho$ is the supremum (not the infimum) over the curves. However, we should still be able to obtain it by letting $\varepsilon>0$ be arbitrary and choosing $c$ with $\int_0^1v^r(c(t))\left\|c'(t)\right\|_E\:{\rm d}t<\rho(x,y)+\varepsilon$. $\endgroup$
    – 0xbadf00d
    Commented Jun 7, 2020 at 14:02

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