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I sent an e-mail to Erwann Aubry (see Georges Elencwajg's answer) and asked for a scan of this proof from the book "Calcul Différentiel". He replied and send me the proof translated into English. Here is his original paper. And this how this beautiful story goes:

Theorem. Every open star-shaped set $\Omega$ in $\mathbb{R}^n$ is $C^\infty$-diffeomorphic to $\mathbb{R}^n.$

Proof. For convenience assume that $\Omega$ is star-shaped at $0.$

Let $F=\mathbb{R}^n\setminus\Omega$ and $\phi:\mathbb{R}^n\rightarrow\mathbb{R}_+$ (here $\mathbb{R}_+=[0,\infty)$) be a $C^\infty$-function such that $F=\phi^{-1}(\{0\}).$ (such $\phi$ exists due to Whitney extension theorem)

Now we set $f:\Omega\rightarrow\mathbb{R}^n$ by formula: $$f(x)=\overbrace{\left[1+\left(\int_0^1\frac{dv}{\phi(vx)}\right)^2||x||^2\right]}^{\lambda(x)}\cdot x=\left[1+\left(\int_0^{||x||}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot x.$$ Clearly $f$ is smooth on $\Omega.$

We set $A(x)=\sup\{t>0:t\frac{x}{||x||}\in\Omega\}.$ $f$ sends injectively the segment (or ray) $[0,A(x))\frac{x}{||x||}$ to the ray $\mathbb{R_+}\frac{x}{||x||}.$ Moreover $f(0\frac{x}{||X||})=0$ and $$\lim_{r\rightarrow A(x)}||f(r\frac{x}{||x||})||=\lim_{r\rightarrow A(x)}\left[1+\left(\int_0^{r}\frac{dt}{\phi\left(t\cdot\frac{rx}{||x||}\cdot||\frac{||x||}{rx}||\right)}\right)^2\right]\cdot r=\\ \left[1+\left(\int_0^{A(x)}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot A(x)=+\infty.$$ Indeed, if $A(x)=+\infty,$ then it holds for obvious reason. If $A(x)<+\infty,$ then by definitions of $\phi$ and $A(x)$ we get that $\phi(A(x)\frac{x}{||x||})=0.$ Hence by Mean value theorem and the fact that $\phi$ is $C^1$ $$\phi\left(r\frac{x}{||x||}\right)\leqslant M(A(x)-r)$$ for some constant $M$ and every $r.$ As a result $$\int_0^{A(x)}\frac{dt}{\phi\left(t\frac{x}{||x||}\right)}$$ diverges. Hence we infer that $f([0,A(x))\frac{x}{||x||})=\mathbb{R_+}\frac{x}{||x||}$ and so $f(\Omega)=\mathbb{R}^n.$

To end the proof we need to show that $f$ has $C^\infty$-inverse. But as corollary from the Inverse function theorem we get that it is sufficient to show that $df$ vanish nowhere.

Suppose that $d_xf(h)=0$ for some $x\in\Omega$ and $h\neq 0.$ From definition of $f$ we get that $$d_xf(h)=\lambda(x)h+d_x\lambda(h)x.$$ Hence $h=\mu x$ for some $\mu\neq 0$ and from that $x\neq 0.$ As a result $\lambda(x)+d_x\lambda(x)=0.$ But we have that $\lambda(x)\geqslant 1$ and function $g(t):=\lambda(tx)$ is increasing, so $g'(1)=d_x\lambda(x)>0,$ which gives a contradiction.$\square$

I sent an e-mail to Erwann Aubry (see Georges Elencwajg's answer) and asked for a scan of this proof from the book "Calcul Différentiel". He replied and send me the proof translated into English. Here is his original paper. And this is how the proof goes:

Theorem. Every open star-shaped set $\Omega$ in $\mathbb{R}^n$ is $C^\infty$-diffeomorphic to $\mathbb{R}^n.$

Proof. For convenience assume that $\Omega$ is star-shaped at $0.$

Let $F=\mathbb{R}^n\setminus\Omega$ and $\phi:\mathbb{R}^n\rightarrow\mathbb{R}_+$ (here $\mathbb{R}_+=[0,\infty)$) be a $C^\infty$-function such that $F=\phi^{-1}(\{0\}).$ (such $\phi$ exists due to Whitney extension theorem)

Now we set $f:\Omega\rightarrow\mathbb{R}^n$ by formula: $$f(x)=\overbrace{\left[1+\left(\int_0^1\frac{dv}{\phi(vx)}\right)^2||x||^2\right]}^{\lambda(x)}\cdot x=\left[1+\left(\int_0^{||x||}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot x.$$ Clearly $f$ is smooth on $\Omega.$

We set $A(x)=\sup\{t>0:t\frac{x}{||x||}\in\Omega\}.$ $f$ sends injectively the segment (or ray) $[0,A(x))\frac{x}{||x||}$ to the ray $\mathbb{R_+}\frac{x}{||x||}.$ Moreover $f(0\frac{x}{||X||})=0$ and $$\lim_{r\rightarrow A(x)}||f(r\frac{x}{||x||})||=\lim_{r\rightarrow A(x)}\left[1+\left(\int_0^{r}\frac{dt}{\phi\left(t\cdot\frac{rx}{||x||}\cdot||\frac{||x||}{rx}||\right)}\right)^2\right]\cdot r=\\ \left[1+\left(\int_0^{A(x)}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot A(x)=+\infty.$$ Indeed, if $A(x)=+\infty,$ then it holds for obvious reason. If $A(x)<+\infty,$ then by definitions of $\phi$ and $A(x)$ we get that $\phi(A(x)\frac{x}{||x||})=0.$ Hence by Mean value theorem and the fact that $\phi$ is $C^1$ $$\phi\left(r\frac{x}{||x||}\right)\leqslant M(A(x)-r)$$ for some constant $M$ and every $r.$ As a result $$\int_0^{A(x)}\frac{dt}{\phi\left(t\frac{x}{||x||}\right)}$$ diverges. Hence we infer that $f([0,A(x))\frac{x}{||x||})=\mathbb{R_+}\frac{x}{||x||}$ and so $f(\Omega)=\mathbb{R}^n.$

To end the proof we need to show that $f$ has $C^\infty$-inverse. But as corollary from the Inverse function theorem we get that it is sufficient to show that $df$ vanish nowhere.

Suppose that $d_xf(h)=0$ for some $x\in\Omega$ and $h\neq 0.$ From definition of $f$ we get that $$d_xf(h)=\lambda(x)h+d_x\lambda(h)x.$$ Hence $h=\mu x$ for some $\mu\neq 0$ and from that $x\neq 0.$ As a result $\lambda(x)+d_x\lambda(x)=0.$ But we have that $\lambda(x)\geqslant 1$ and function $g(t):=\lambda(tx)$ is increasing, so $g'(1)=d_x\lambda(x)>0,$ which gives a contradiction.$\square$

I sent an e-mail to Erwann Aubry (see Georges Elencwajg's answer) and asked for a scan of this proof from the book "Calcul Différentiel". He replied and send me the proof translated into English. Here is his original paper. And this how this beautiful story goes:

Theorem. Every open star-shaped set $\Omega$ in $\mathbb{R}^n$ is $C^\infty$-diffeomorphic to $\mathbb{R}^n.$

Proof. For convenience assume that $\Omega$ is star-shaped at $0.$

Let $F=\mathbb{R}^n\setminus\Omega$ and $\phi:\mathbb{R}^n\rightarrow\mathbb{R}_+$ (here $\mathbb{R}_+=[0,\infty)$) be a $C^\infty$-function such that $F=\phi^{-1}(\{0\}).$ (such $\phi$ exists due to Whitney extension theorem)

Now we set $f:\Omega\rightarrow\mathbb{R}^n$ by formula: $$f(x)=\overbrace{\left[1+\left(\int_0^1\frac{dv}{\phi(vx)}\right)^2||x||^2\right]}^{\lambda(x)}\cdot x=\left[1+\left(\int_0^{||x||}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot x.$$ Clearly $f$ is smooth on $\Omega.$

We set $A(x)=\sup\{t>0:t\frac{x}{||x||}\in\Omega\}.$ $f$ sends injectively segment $[0,A(x))\frac{x}{||x||}$ to ray $\mathbb{R_+}\frac{x}{||x||}.$ Moreover $f(0\frac{x}{||X||})=0$ and $$\lim_{r\rightarrow A(x)}||f(r\frac{x}{||x||})||=\lim_{r\rightarrow A(x)}\left[1+\left(\int_0^{r}\frac{dt}{\phi\left(t\cdot\frac{rx}{||x||}\cdot||\frac{||x||}{rx}||\right)}\right)^2\right]\cdot r=\\ \left[1+\left(\int_0^{A(x)}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot A(x)=+\infty.$$ Indeed, if $A(x)=+\infty,$ then it holds for obvious reason. If $A(x)<+\infty,$ then by definitions of $\phi$ and $A(x)$ we get that $\phi(A(x)\frac{x}{||x||})=0.$ $\color{red}{\text{Hence}}$ $$\color{red}{\phi\left(r\frac{x}{||x||}\right)=O(rA(x))\hspace{7pt}\text{as}\hspace{7pt}r\rightarrow A(x).}$$ $\color{red}{\text{Since }\phi\text{ is smooth, we get that}}$ $$\color{red}{\int_0^{A(x)}\frac{dt}{\phi\left(t\frac{x}{||x||}\right)}}$$ $\color{red}{\text{diverges.}}$ Hence we infer that $f([0,A(x))\frac{x}{||x||})=\mathbb{R_+}\frac{x}{||x||}$ and so $f(\Omega)=\mathbb{R}^n.$

To end the proof we need to show that $f$ has $C^\infty$-inverse. But as corollary from the Inverse function theorem we get that it is sufficient to show that $df$ vanish nowhere.

Suppose that $d_xf(h)=0$ for some $x\in\Omega$ and $h\neq 0.$ From definition of $f$ we get that $$d_xf(h)=\lambda(x)h+d_x\lambda(h)x.$$ Hence $h=\mu x$ for some $\mu\neq 0$ and from that $x\neq 0.$ As a result $\lambda(x)+d_x\lambda(x)=0.$ But we have that $\lambda(x)\geqslant 1$ and function $g(t):=\lambda(tx)$ is increasing, so $g'(1)=d_x\lambda(x)>0,$ which gives a contradiction.$\square$

$\color{red}{\text{I not fully understand the red part.}}$

I sent an e-mail to Erwann Aubry (see Georges Elencwajg's answer) and asked for a scan of this proof from the book "Calcul Différentiel". He replied and send me the proof translated into English. Here is his original paper. And this how this beautiful story goes:

Theorem. Every open star-shaped set $\Omega$ in $\mathbb{R}^n$ is $C^\infty$-diffeomorphic to $\mathbb{R}^n.$

Proof. For convenience assume that $\Omega$ is star-shaped at $0.$

Let $F=\mathbb{R}^n\setminus\Omega$ and $\phi:\mathbb{R}^n\rightarrow\mathbb{R}_+$ (here $\mathbb{R}_+=[0,\infty)$) be a $C^\infty$-function such that $F=\phi^{-1}(\{0\}).$ (such $\phi$ exists due to Whitney extension theorem)

Now we set $f:\Omega\rightarrow\mathbb{R}^n$ by formula: $$f(x)=\overbrace{\left[1+\left(\int_0^1\frac{dv}{\phi(vx)}\right)^2||x||^2\right]}^{\lambda(x)}\cdot x=\left[1+\left(\int_0^{||x||}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot x.$$ Clearly $f$ is smooth on $\Omega.$

We set $A(x)=\sup\{t>0:t\frac{x}{||x||}\in\Omega\}.$ $f$ sends injectively the segment (or ray) $[0,A(x))\frac{x}{||x||}$ to the ray $\mathbb{R_+}\frac{x}{||x||}.$ Moreover $f(0\frac{x}{||X||})=0$ and $$\lim_{r\rightarrow A(x)}||f(r\frac{x}{||x||})||=\lim_{r\rightarrow A(x)}\left[1+\left(\int_0^{r}\frac{dt}{\phi\left(t\cdot\frac{rx}{||x||}\cdot||\frac{||x||}{rx}||\right)}\right)^2\right]\cdot r=\\ \left[1+\left(\int_0^{A(x)}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot A(x)=+\infty.$$ Indeed, if $A(x)=+\infty,$ then it holds for obvious reason. If $A(x)<+\infty,$ then by definitions of $\phi$ and $A(x)$ we get that $\phi(A(x)\frac{x}{||x||})=0.$ Hence by Mean value theorem and the fact that $\phi$ is $C^1$ $$\phi\left(r\frac{x}{||x||}\right)\leqslant M(A(x)-r)$$ for some constant $M$ and every $r.$ As a result $$\int_0^{A(x)}\frac{dt}{\phi\left(t\frac{x}{||x||}\right)}$$ diverges. Hence we infer that $f([0,A(x))\frac{x}{||x||})=\mathbb{R_+}\frac{x}{||x||}$ and so $f(\Omega)=\mathbb{R}^n.$

To end the proof we need to show that $f$ has $C^\infty$-inverse. But as corollary from the Inverse function theorem we get that it is sufficient to show that $df$ vanish nowhere.

Suppose that $d_xf(h)=0$ for some $x\in\Omega$ and $h\neq 0.$ From definition of $f$ we get that $$d_xf(h)=\lambda(x)h+d_x\lambda(h)x.$$ Hence $h=\mu x$ for some $\mu\neq 0$ and from that $x\neq 0.$ As a result $\lambda(x)+d_x\lambda(x)=0.$ But we have that $\lambda(x)\geqslant 1$ and function $g(t):=\lambda(tx)$ is increasing, so $g'(1)=d_x\lambda(x)>0,$ which gives a contradiction.$\square$

I sent an e-mail to Erwann Aubry (see Georges Elencwajg's answer) and asked for a scan of this proof from the book "Calcul Différentiel". He replied and send me the proof translated into English. Here is his original paper. And this how this beautiful story goes:

Theorem. Every open star-shaped set $\Omega$ in $\mathbb{R}^n$ is $C^\infty$-diffeomorphic to $\mathbb{R}^n.$

Proof. For convenience assume that $\Omega$ is star-shaped at $0.$

Let $F=\mathbb{R}^n\setminus\Omega$ and $\phi:\mathbb{R}^n\rightarrow\mathbb{R}_+$ (here $\mathbb{R}_+=[0,\infty)$) be a $C^\infty$-function such that $F=\phi^{-1}(\{0\}).$ (such $\phi$ exists due to Whitney extension theorem)

Now we set $f:\Omega\rightarrow\mathbb{R}^n$ by formula: $$f(x)=\overbrace{\left[1+\left(\int_0^1\frac{dv}{\phi(vx)}\right)^2||x||^2\right]}^{\lambda(x)}\cdot x=\left[1+\left(\int_0^{||x||}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot x.$$ Clearly f is smooth on $\Omega.$

We set $A(x)=\sup\{t>0:t\frac{x}{||x||}\in\Omega\}.$ f sends injectively segment $[0,A(x))\frac{x}{||x||}$ to ray $\mathbb{R_+}\frac{x}{||x||}.$ Moreover $f(0\frac{x}{||X||})=0$ and $$\lim_{r\rightarrow A(x)}||f(r\frac{x}{||x||})||=\lim_{r\rightarrow A(x)}\left[1+\left(\int_0^{r}\frac{dt}{\phi\left(t\cdot\frac{rx}{||x||}\cdot||\frac{||x||}{rx}||\right)}\right)^2\right]\cdot r=\\ \left[1+\left(\int_0^{A(x)}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot A(x)=+\infty.$$ Indeed, if $A(x)=+\infty,$ then it holds for obvious reason. If $A(x)<+\infty,$ then by definitions of $\phi$ and $A(x)$ we get that $\phi(A(x)\frac{x}{||x||})=0.$ $\color{red}{\text{Hence}}$ $$\color{red}{\phi\left(r\frac{x}{||x||}\right)=O(rA(x))\hspace{7pt}\text{as}\hspace{7pt}r\rightarrow A(x).}$$ $\color{red}{\text{Since }\phi\text{ is smooth, we get that}}$ $$\color{red}{\int_0^{A(x)}\frac{dt}{\phi\left(t\frac{x}{||x||}\right)}}$$ $\color{red}{\text{diverges.}}$ Hence we infer that $f([0,A(x))\frac{x}{||x||})=\mathbb{R_+}\frac{x}{||x||}$ and so $f(\Omega)=\mathbb{R}^n.$

To end the proof we need to show that $f$ has $C^\infty$-inverse. But as corollary from the Inverse function theorem we get that it is sufficient to show that $df$ vanish nowhere.

Suppose that $d_xf(h)=0$ for some $x\in\Omega$ and $h\neq 0.$ From definition of $f$ we get that $$d_xf(h)=\lambda(x)h+d_x\lambda(h)x.$$ Hence $h=\mu x$ for some $\mu\neq 0$ and from that $x\neq 0.$ As a result $\lambda(x)+d_x\lambda(x)=0.$ But we have that $\lambda(x)\geqslant 1$ and function $g(t):=\lambda(tx)$ is increasing, so $g'(1)=d_x\lambda(x)>0,$ which gives a contradiction.$\square$

$\color{red}{\text{I not fully understand the red part.}}$

I sent an e-mail to Erwann Aubry (see Georges Elencwajg's answer) and asked for a scan of this proof from the book "Calcul Différentiel". He replied and send me the proof translated into English. Here is his original paper. And this how this beautiful story goes:

Theorem. Every open star-shaped set $\Omega$ in $\mathbb{R}^n$ is $C^\infty$-diffeomorphic to $\mathbb{R}^n.$

Proof. For convenience assume that $\Omega$ is star-shaped at $0.$

Let $F=\mathbb{R}^n\setminus\Omega$ and $\phi:\mathbb{R}^n\rightarrow\mathbb{R}_+$ (here $\mathbb{R}_+=[0,\infty)$) be a $C^\infty$-function such that $F=\phi^{-1}(\{0\}).$ (such $\phi$ exists due to Whitney extension theorem)

Now we set $f:\Omega\rightarrow\mathbb{R}^n$ by formula: $$f(x)=\overbrace{\left[1+\left(\int_0^1\frac{dv}{\phi(vx)}\right)^2||x||^2\right]}^{\lambda(x)}\cdot x=\left[1+\left(\int_0^{||x||}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot x.$$ Clearly $f$ is smooth on $\Omega.$

We set $A(x)=\sup\{t>0:t\frac{x}{||x||}\in\Omega\}.$ $f$ sends injectively segment $[0,A(x))\frac{x}{||x||}$ to ray $\mathbb{R_+}\frac{x}{||x||}.$ Moreover $f(0\frac{x}{||X||})=0$ and $$\lim_{r\rightarrow A(x)}||f(r\frac{x}{||x||})||=\lim_{r\rightarrow A(x)}\left[1+\left(\int_0^{r}\frac{dt}{\phi\left(t\cdot\frac{rx}{||x||}\cdot||\frac{||x||}{rx}||\right)}\right)^2\right]\cdot r=\\ \left[1+\left(\int_0^{A(x)}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot A(x)=+\infty.$$ Indeed, if $A(x)=+\infty,$ then it holds for obvious reason. If $A(x)<+\infty,$ then by definitions of $\phi$ and $A(x)$ we get that $\phi(A(x)\frac{x}{||x||})=0.$ $\color{red}{\text{Hence}}$ $$\color{red}{\phi\left(r\frac{x}{||x||}\right)=O(rA(x))\hspace{7pt}\text{as}\hspace{7pt}r\rightarrow A(x).}$$ $\color{red}{\text{Since }\phi\text{ is smooth, we get that}}$ $$\color{red}{\int_0^{A(x)}\frac{dt}{\phi\left(t\frac{x}{||x||}\right)}}$$ $\color{red}{\text{diverges.}}$ Hence we infer that $f([0,A(x))\frac{x}{||x||})=\mathbb{R_+}\frac{x}{||x||}$ and so $f(\Omega)=\mathbb{R}^n.$

To end the proof we need to show that $f$ has $C^\infty$-inverse. But as corollary from the Inverse function theorem we get that it is sufficient to show that $df$ vanish nowhere.

Suppose that $d_xf(h)=0$ for some $x\in\Omega$ and $h\neq 0.$ From definition of $f$ we get that $$d_xf(h)=\lambda(x)h+d_x\lambda(h)x.$$ Hence $h=\mu x$ for some $\mu\neq 0$ and from that $x\neq 0.$ As a result $\lambda(x)+d_x\lambda(x)=0.$ But we have that $\lambda(x)\geqslant 1$ and function $g(t):=\lambda(tx)$ is increasing, so $g'(1)=d_x\lambda(x)>0,$ which gives a contradiction.$\square$

$\color{red}{\text{I not fully understand the red part.}}$