Timeline for Bound the operator norm of the Fréchet derivative of a Lipschitz function in this setting
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 7, 2020 at 17:46 | comment | added | 0xbadf00d | 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. | |
| Jun 6, 2020 at 22:30 | history | edited | DCM | CC BY-SA 4.0 |
deleted 132 characters in body
|
| Jun 6, 2020 at 15:21 | history | edited | DCM | CC BY-SA 4.0 |
added 60 characters in body
|
| Jun 6, 2020 at 14:42 | comment | added | 0xbadf00d | 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. | |
| Jun 6, 2020 at 14:24 | history | edited | DCM | CC BY-SA 4.0 |
added 132 characters in body
|
| Jun 6, 2020 at 14:23 | comment | added | 0xbadf00d | 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 $(4)$. | |
| Jun 6, 2020 at 14:11 | history | edited | DCM | CC BY-SA 4.0 |
added 27 characters in body
|
| Jun 6, 2020 at 14:09 | comment | added | DCM | NB: I'm using $\mathrm{limsup}_{y\to x}(*)$ to mean $\mathrm{lim}_{r\to 0}(\sup_{y\in \mathrm{Ball}(x,r)\setminus\{x\}}(*))$ | |
| Jun 6, 2020 at 14:06 | history | answered | DCM | CC BY-SA 4.0 |