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Post Reopened by Matthew Daws, Yemon Choi, András Bátkai, Will Sawin, Martin Sleziak
Post Closed as "Not suitable for this site" by user44191, Jochen Wengenroth, Ben McKay, abx, YCor
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Gabe Conant
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Let $D$ be an open convex subset of $\mathbf R^d$ with boundary $\partial D$ and closure $\overline D$. For $x\in\mathbf R^d$ and a unit vector $e$ let $L_{x,e}=\{x+\lambda e, \lambda\ge0\}$ be the ray drawn from $x$ in the direction $e$. For $x\in\mathbf R^d$, let $$D(x)=\{y\in \mathbf R^d \colon L_{x,y/|y| \cap \overline D \ne \emptyset}\}.$$$$D(x)=\{y\in \mathbf R^d \colon L_{x,y/|y|} \cap \overline D \ne \emptyset\}.$$ In particular, $D(x)=\overline D$ for all $x\in D$. Furthermore, for $y\in D(x)$, let $$\lambda(x,y/|y|)=\max\{R>0 \colon x+Ry/|y| \in \overline D\}.$$

(From Vassili Kolokoltsov: On fully mixed and multidimensional extensions of the Caputo and Riemann-Liouville derivatives, related Markov processes and fractional differential equations, arXiv:1501.03925, DOI: 10.1515/fca-2015-0060.

enter image description here

Look at the above picture, let us consider a simple case if $d=2$. How to show $D(x)=\bar{D}, x\in D$? This results seems a little bit strange. If $x\in D$, then $L_{x,y/{|y|}}\cap \bar{D}\neq \varnothing$ forever. So $D(x)=R^2,$ why $D(x)=\bar{D}, x\in D$? Is there some misunderstanding? Please correct it.

Let $D$ be an open convex subset of $\mathbf R^d$ with boundary $\partial D$ and closure $\overline D$. For $x\in\mathbf R^d$ and a unit vector $e$ let $L_{x,e}=\{x+\lambda e, \lambda\ge0\}$ be the ray drawn from $x$ in the direction $e$. For $x\in\mathbf R^d$, let $$D(x)=\{y\in \mathbf R^d \colon L_{x,y/|y| \cap \overline D \ne \emptyset}\}.$$ In particular, $D(x)=\overline D$ for all $x\in D$. Furthermore, for $y\in D(x)$, let $$\lambda(x,y/|y|)=\max\{R>0 \colon x+Ry/|y| \in \overline D\}.$$

(From Vassili Kolokoltsov: On fully mixed and multidimensional extensions of the Caputo and Riemann-Liouville derivatives, related Markov processes and fractional differential equations, arXiv:1501.03925, DOI: 10.1515/fca-2015-0060.

enter image description here

Look at the above picture, let us consider a simple case if $d=2$. How to show $D(x)=\bar{D}, x\in D$? This results seems a little bit strange. If $x\in D$, then $L_{x,y/{|y|}}\cap \bar{D}\neq \varnothing$ forever. So $D(x)=R^2,$ why $D(x)=\bar{D}, x\in D$? Is there some misunderstanding? Please correct it.

Let $D$ be an open convex subset of $\mathbf R^d$ with boundary $\partial D$ and closure $\overline D$. For $x\in\mathbf R^d$ and a unit vector $e$ let $L_{x,e}=\{x+\lambda e, \lambda\ge0\}$ be the ray drawn from $x$ in the direction $e$. For $x\in\mathbf R^d$, let $$D(x)=\{y\in \mathbf R^d \colon L_{x,y/|y|} \cap \overline D \ne \emptyset\}.$$ In particular, $D(x)=\overline D$ for all $x\in D$. Furthermore, for $y\in D(x)$, let $$\lambda(x,y/|y|)=\max\{R>0 \colon x+Ry/|y| \in \overline D\}.$$

(From Vassili Kolokoltsov: On fully mixed and multidimensional extensions of the Caputo and Riemann-Liouville derivatives, related Markov processes and fractional differential equations, arXiv:1501.03925, DOI: 10.1515/fca-2015-0060.

enter image description here

Look at the above picture, let us consider a simple case if $d=2$. How to show $D(x)=\bar{D}, x\in D$? This results seems a little bit strange. If $x\in D$, then $L_{x,y/{|y|}}\cap \bar{D}\neq \varnothing$ forever. So $D(x)=R^2,$ why $D(x)=\bar{D}, x\in D$? Is there some misunderstanding? Please correct it.

retyped the text from the picture
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Martin Sleziak
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Let $D$ be an open convex subset of $\mathbf R^d$ with boundary $\partial D$ and closure $\overline D$. For $x\in\mathbf R^d$ and a unit vector $e$ let $L_{x,e}=\{x+\lambda e, \lambda\ge0\}$ be the ray drawn from $x$ in the direction $e$. For $x\in\mathbf R^d$, let $$D(x)=\{y\in \mathbf R^d \colon L_{x,y/|y| \cap \overline D \ne \emptyset}\}.$$ In particular, $D(x)=\overline D$ for all $x\in D$. Furthermore, for $y\in D(x)$, let $$\lambda(x,y/|y|)=\max\{R>0 \colon x+Ry/|y| \in \overline D\}.$$

(From Vassili Kolokoltsov: On fully mixed and multidimensional extensions of the Caputo and Riemann-Liouville derivatives, related Markov processes and fractional differential equations, arXiv:1501.03925, DOI: 10.1515/fca-2015-0060.

enter image description here

Look at the above picture, let us consider a simple case if $d=2$. How to show $D(x)=\bar{D}, x\in D$? This results seems a little bit strange. If $x\in D$, then $L_{x,y/{|y|}}\cap \bar{D}\neq \varnothing$ forever. So $D(x)=R^2,$ why $D(x)=\bar{D}, x\in D$? Is there some misunderstanding? Please correct it.

enter image description here

Look at the above picture, let us consider a simple case if $d=2$. How to show $D(x)=\bar{D}, x\in D$? This results seems a little bit strange. If $x\in D$, then $L_{x,y/{|y|}}\cap \bar{D}\neq \varnothing$ forever. So $D(x)=R^2,$ why $D(x)=\bar{D}, x\in D$? Is there some misunderstanding? Please correct it.

Let $D$ be an open convex subset of $\mathbf R^d$ with boundary $\partial D$ and closure $\overline D$. For $x\in\mathbf R^d$ and a unit vector $e$ let $L_{x,e}=\{x+\lambda e, \lambda\ge0\}$ be the ray drawn from $x$ in the direction $e$. For $x\in\mathbf R^d$, let $$D(x)=\{y\in \mathbf R^d \colon L_{x,y/|y| \cap \overline D \ne \emptyset}\}.$$ In particular, $D(x)=\overline D$ for all $x\in D$. Furthermore, for $y\in D(x)$, let $$\lambda(x,y/|y|)=\max\{R>0 \colon x+Ry/|y| \in \overline D\}.$$

(From Vassili Kolokoltsov: On fully mixed and multidimensional extensions of the Caputo and Riemann-Liouville derivatives, related Markov processes and fractional differential equations, arXiv:1501.03925, DOI: 10.1515/fca-2015-0060.

enter image description here

Look at the above picture, let us consider a simple case if $d=2$. How to show $D(x)=\bar{D}, x\in D$? This results seems a little bit strange. If $x\in D$, then $L_{x,y/{|y|}}\cap \bar{D}\neq \varnothing$ forever. So $D(x)=R^2,$ why $D(x)=\bar{D}, x\in D$? Is there some misunderstanding? Please correct it.

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How to prove the following the set are equal

enter image description here

Look at the above picture, let us consider a simple case if $d=2$. How to show $D(x)=\bar{D}, x\in D$? This results seems a little bit strange. If $x\in D$, then $L_{x,y/{|y|}}\cap \bar{D}\neq \varnothing$ forever. So $D(x)=R^2,$ why $D(x)=\bar{D}, x\in D$? Is there some misunderstanding? Please correct it.