0
$\begingroup$

$\DeclareMathOperator\Lip{Lip}$My problem is slightly different from the title, but I don't have a more straightforward title. Sorry for that.

For $d\ge 1$, denote $\mathbb S^{d-1}:=\{x\in\mathbb R^d: \|x\|=1\}$ and $$\Lip_1(\mathbb R^d):=\{F: \text{$\lvert F(x)-F(y)\rvert \le \|x-y\|$ for all $x,y\in\mathbb R^d$}\}.$$

Define a map $T: \Lip_1(\mathbb R)\times \mathbb S^{d-1}\to \Lip_1(\mathbb R^d)$ as follows: For each $f\in \Lip_1(\mathbb R)$ and $v\in\mathbb S^{d-1}$, $T(f,v)\in \Lip_1(\mathbb R^d)$ is defined by $T(f,v)(x)\mathrel{:=}f(x\cdot v)$. My question is whether there exists a topology for $\Lip_1(\mathbb R^d)$ under which $\lambda T(\Lip_1(\mathbb R)\times \mathbb S^{d-1})^c$ is dense in $\Lip_1(\mathbb R^d)$ for some $\lambda>0$? Here $F\in \lambda T(\Lip_1(\mathbb R)\times \mathbb S^{d-1})^c$ iff $$F(x)=\lambda\sum_{k=1}^n a_k f_k(x\cdot v_k),\quad \text{where $a_k\in\mathbb R_+$, $f_k\in \Lip_1(\mathbb R)$, $v_k\in\mathbb S^{d-1}$, and $\sum_{k=1}^na_k=1$}.$$

Any answer, comment or reference is highly appreciated!

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ It seems unlikely, barring stupid answers like the trivial topology. For any function $g=T(f,v)$ in the image of $T$, we have that $|g(1,0,\ldots,0)-g(0,\ldots,0)|+|g(0,1,0,\ldots,0)-g(0,\ldots,0)|\le |v_1|+|v_2|\le\sqrt{2}$. After taking convex combinations, we get $|F(1,0,\ldots,0)-F(0,\ldots,0)|+|F(0,1,0,\ldots,0)-F(0,\ldots,0)|\le\sqrt{2}$. However, for the function $h(x)=|x|$, this quantity is 2. It's possible you can fix this by only trying approximating, e.g. $Lip_{\alpha(d)}$ functions for some $\alpha(d)$. I'm not sure what the right dependence on $d$ is. $\endgroup$
    – Sam Zbarsky
    Commented Nov 7, 2019 at 3:12
  • $\begingroup$ @SamZbarsky Thanks for pointing out that. $\endgroup$
    – user128095
    Commented Nov 7, 2019 at 3:19

0

Reset to default

You must log in to answer this question.