Bound the operator norm of the Fréchet derivative of a Lipschitz function in this setting

Question

I want to find a bound for the operator norm of the Fréchet derivative of a Lipschitz continuous function in the following setting:

Let

• $E$ be a $\mathbb R$-Banach space;
• $v:E\to[1,\infty)$ be continuous;
• $v_i:[0,\infty)\to[1,\infty)$ be continuous and nondecreasing with $$v_1(\left\|x\right\|_E)\le v(x)\le v_2(\left\|x\right\|_E)\;\;\;\text{for all }x\in E,\tag1$$ $$v_1(r)\xrightarrow{r\to\infty}\infty\tag2$$ 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 $$\rho(x,y):=\inf_{\substack{c\:\in\:C^1([0,\:1],\:E)\\ c(0)=x\\ c(1)=y}}\int_0^1v\left(c(t)\right)\left\|c'(t)\right\|_E\:{\rm d}t\;\;\;\text{for }x,y\in E.$$ Moreover, let $(\delta,\beta)\in(0,\infty)\times[0,\infty)$ and note that $$d:=1\wedge\frac\rho\delta+\beta\rho\le\left(\frac1\delta+\beta\right)\rho\tag4$$ is a metric equivalent to $\rho$. Let $f:E\to\mathbb R$ be Fréchet differentiable with $f(0)=0$ $$|f|_{\operatorname{Lip}(\rho)}:=\sup_{\substack{x,\:y\:\in\:E\\x\:\ne\:y}}\frac{|f(x)-f(y)|}{\rho(x,y)}\le1\tag5.$$

I want to show that $$\left\|{\rm D}f(x)\right\|_{E'}\le\left(\frac1\delta+\beta\right)v(x)\tag6.$$

Unfortunately, I'm struggling to see how we obtain $(6)$. Let $x\in E$. Clearly, if $\varepsilon>0$, then the Fréchet differentiability of $f$ at $x$ implies $$|f(x)-f(y)-{\rm D}f(x)(x-y)|\le\varepsilon\left\|x-y\right\|_E\;\;\;\text{for all }y\in B_\delta(x)\tag7$$ for some $\delta>0$. From $(5)$ we infer that $$|{\rm D}f(x)(x-y)|\le d(x,y)+\varepsilon\left\|x-y\right\|_E\tag8\;\;\;\text{for all }y\in B_\delta(x).$$ We may use this inequality for arbitrary $y\in E\setminus\{x\}$ by applying it for $\tilde y:=(1-t)x+ty$ with some $t\in\left(0,\delta^{-1}\left\|x-y\right\|_E\right)$, but that doesn't seem to help.

I guess we need to use $(4)$ and observe that for the straight line connecting $x$ and $y$ we obtain $$\rho(x,y)\le\left\|x-y\right\|_E\int_0^1v((1-t)x+ty)\:{\rm d}t\tag9$$ for all $y\in E$. Let $B:=\{y\in E:\left\|x-y\right\|_E\le1\}$. Then $$f:[0,1]\times B\to\mathbb R\;,\;\;\;(t,y)\mapsto v((1-t)x+ty)$$ is bounded and continuous; hence $$F:B\to\mathbb R\;,\;\;\;y\mapsto\int_0^1f(t,y)\:{\rm d}t$$ is bounded and continuous. So, $$\lim_{y\to0}F(y)=\int v((1-t)x)\:{\rm d}t\tag{10},$$ for whatever this is useful to know.

Remark: The claim can be found in equation $(24)$ in https://arxiv.org/abs/math/0602479.

EDIT: I guess something like $(10)$ is needed and is what the authors used in the displayed equation in the proof after equation $(26)$; cf. my related question.

Answer 1

The fact that the authors of https://arxiv.org/pdf/math/0602479.pdf refer to using "representation (4) for the distance" makes me think that the estimate you want is supposed to come from something along the lines of:

$$ \begin{array}{lllll} \Vert Df(x)\Vert &=& \limsup_{y\to x}\dfrac{|f(x)-f(y)|}{\Vert x-y\Vert}\\ &\leq & \limsup_{y\to x} \dfrac{d(x,y)}{\Vert x-y\Vert} \\ &\leq & (\delta^{-1}+\beta)\limsup_{y\to x} \dfrac{\rho(x,y)}{\Vert x-y\Vert} \\ &=& (\delta^{-1}+\beta)v(x) \end{array} $$

using $d(x,y) = \sup\{|\phi(x)-\phi(y)|:\mathrm{Lip}_d(\phi)\leq 1\}$ and $\mathrm{Lip}_d(f)\leq 1$ to get from the first line to the second, the second display on page 16 to get from the second to the third, and then waving your hands a bit to argue that

$$ \limsup_{y\to x}\dfrac{\rho(x,y)}{\Vert x-y\Vert}= v(x) $$

I might be missing something, but I don't think your (1), (2) and (3) are relevant for the derivative estimate, only the one for $|f(.)|$.