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I asked my question in overflow, so far with no answers.
Perhaps here, I'll get an answer.
https://mathoverflow.net/questions/282048/a-lemma-on-convex-domain-which-is-a-lipschitz-domain
[admin edit: Below is the actual question posted, so our community doesn't have to follow multiple links:]
[box=yellow]I am reading the following paper:
Whitney Estimates for Convex Domains with Applications to Multivariate Piecewise Polynomial Approximation
I don't see why does the property (i) of Lipschitz domain is satisfied? (it's listed on the bottom of page 5), what is $\delta:=\delta(d,R)$ explicitly?
I don't understand the last paragraph:
[box=gray]Now, for any $x\in \partial \Omega \cap U_j$, the cone defined by the convex closure of $x\cup B(0,1)$ is contained in the closure of $\Omega$. Any head angle $\alpha$ of this cone satisfies $\sin(\alpha/2)\ge 1/R$ and thus the boundary is $Lip 1$ function with $M:=M(d,R)$.[/box]
Can you elaborate on the two sentences in the quote above? and with the specifics, cause I don't see why the convex closure of $x\cup B(0,1)$ cannot exceed the closure of $\Omega$; I also don't see how did they arrive at the inequality in the last sentence of $\sin(\alpha/2)$, perhaps I am missing something from elementary geometry. why is the boundary $Lip1$?
Thanks.[/box]
Cheers!
Perhaps here, I'll get an answer.
https://mathoverflow.net/questions/282048/a-lemma-on-convex-domain-which-is-a-lipschitz-domain
[admin edit: Below is the actual question posted, so our community doesn't have to follow multiple links:]
[box=yellow]I am reading the following paper:
Whitney Estimates for Convex Domains with Applications to Multivariate Piecewise Polynomial Approximation
I don't see why does the property (i) of Lipschitz domain is satisfied? (it's listed on the bottom of page 5), what is $\delta:=\delta(d,R)$ explicitly?
I don't understand the last paragraph:
[box=gray]Now, for any $x\in \partial \Omega \cap U_j$, the cone defined by the convex closure of $x\cup B(0,1)$ is contained in the closure of $\Omega$. Any head angle $\alpha$ of this cone satisfies $\sin(\alpha/2)\ge 1/R$ and thus the boundary is $Lip 1$ function with $M:=M(d,R)$.[/box]
Can you elaborate on the two sentences in the quote above? and with the specifics, cause I don't see why the convex closure of $x\cup B(0,1)$ cannot exceed the closure of $\Omega$; I also don't see how did they arrive at the inequality in the last sentence of $\sin(\alpha/2)$, perhaps I am missing something from elementary geometry. why is the boundary $Lip1$?
Thanks.[/box]
Cheers!