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


  • $E$ 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.


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 $|f(.)|$.

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  • $\begingroup$ NB: I'm using $\mathrm{limsup}_{y\to x}(*)$ to mean $\mathrm{lim}_{r\to 0}(\sup_{y\in \mathrm{Ball}(x,r)\setminus\{x\}}(*))$ $\endgroup$ – DCM Jun 6 at 14:09
  • $\begingroup$ Regarding the "representation $(4)$" thing: I think the notation they're using is a bit confusing. What they mean is that $d$ as given in $(3)$ is a metric inducing the Wasserstein distance $(5)$ (Let's denote it by $\operatorname W_d$). Now, and that's true for any complete and separable metric, the Kantorovich-Rubinstein duality theorem yields that $\text W_d(\mu,\nu)=\left\|\mu-\nu\right\|_{\text{Lip}(d)'}$, where we consider $\text{Lip}(d)$ as being equipped with the semi-norm $|\;\cdot\;|_{\text{Lip}(d)}$ defined as in my equation $(5)$ with $\rho$ replaced by $d$. So, this yields $(4)$. $\endgroup$ – 0xbadf00d Jun 6 at 14:23
  • $\begingroup$ I think your last displayed equation is related to their displayed equation in the proof after equation $(26)$, but I don't see why this equation holds either and they don't give an argument. From my $(8)$ we've got $$\frac{|{\rm D}f(x)(x-y)|}{\left\|x-y\right\|_E}\le d(x,y)+\varepsilon\le\left(\frac1\delta+\beta\right)\rho(x,y)+\varepsilon.$$ So, we would only need to show that $\rho(x,y)\le v(x)$. Since $\varepsilon$ was arbitrary, we would obtain the claim. $\endgroup$ – 0xbadf00d Jun 6 at 14:42
  • $\begingroup$ Meanwhile it's at least clear to me that $\limsup_{y\to x}\dfrac{\rho(x,y)}{\Vert x-y\Vert}\le v(x)$; see my equation $(10)$ here: mathoverflow.net/q/362200/91890. $\endgroup$ – 0xbadf00d Jun 7 at 17:46

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