We say that $\Omega$ is a star-shaped domain (with respect to the origin) of $\mathbb R ^n$ if :

$\Omega = \{x\in \mathbb R ^n : \left \| x \right \| < g(\frac{x}{\left \| x \right \|})\} $ and $\partial \Omega = \{x\in \mathbb R ^n : \left \| x \right \| = g(\frac{x}{\left \| x \right \|})\} $ with $g$ is a continuous, positive function on the unit ball. I have two questions:

1) I know what star-shaped means Geometrically, but it doesn't get linked with the definition given above. Can you help me understand..

2)Is there a map (bijection) between $\Omega$ and the unit sphere $B$?

I appreciate your answers and your help.

EDIT: $g$ is a function on a unit sphere.

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    $\begingroup$ To develop a geometric understanding, take $n = 2$. Then $g$ may be defined as a function of $\theta$ = angle coordinate in polar coordinates. Now draw $\partial \Omega$ for the cases $g(\theta) = 2, \, g(\theta) = 2 + \cos \theta$, and perhaps a few other cases. Finally find $g$ if $\partial \Omega$ is a square with corners at $(\pm1, \pm1)$. Then ask yourself the same question again. $\endgroup$ – Hans Engler Mar 22 '17 at 13:54
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    $\begingroup$ @Mokata: In order for $0 \in \Bbb R^n$ to belong to $\Omega$ (as it is natural to require), the definition of $\Omega$ should be slightly modified to be $\{x \in \Bbb R^n : \dots\} \color{red} {\cup \{0\}}$. $\endgroup$ – Alex M. Mar 22 '17 at 14:24
  • $\begingroup$ @ Alex M: But automatically $0$ is in $\Omega$, since $g$ is a positive function, non!? $\endgroup$ – Motaka Mar 22 '17 at 14:34
  • $\begingroup$ @Mokata $g(\frac{x}{||x||})$ is not defined in $0$ $\endgroup$ – Vincent Mar 22 '17 at 14:36
  • $\begingroup$ @ Vincent : It was a mistake from me: $g$ is a function on the unit sphere. $\endgroup$ – Motaka Mar 22 '17 at 14:47

Your definition is not the most used one. I was introduced to this one https://en.wikipedia.org/wiki/Star_domain . $S$ is called a star domain iff $\exists x_0 \in S, \forall x \in S, [x_0,x] \in S $. This one is quite intuitive : there is at least one point in $S$ from which you "see" every other point in $S$ (the "light" going through $S$).

Those two definitions are not equivalent. I have issues with the hypothesis of $g$ being continuous. In $\mathbb{R}^2$, $A=[-1,1]\times \{0\}$ is a star domain according to my definition, but not according to yours.

Intuitively, from what I see, any domain verifying your definition (except $\{0\}$) has a non-empty interior (to be proven or disproven, I didn't investigate). Are you sure it is not a restriction of the definition of star-shaped domain? (something like non-empty interior star-shaped domain?) Or are you sure that you did not miss something, like a $\mathring \Omega$ somewhere in your definition ? (\mathring)

  • $\begingroup$ I find it like thet in the book "An Introduction to the Mathematical Theory of the Navier-Stokes Equations: Steady-State Problems" Giovanni P. Galdi, I think that he must say "Strongly star-shaped domain" because this one Implies wikipidia's definition $\endgroup$ – Motaka Mar 22 '17 at 14:55
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    $\begingroup$ @Mokata: I believe that the author has in mind a solid $\Omega$ enclosed by a closed hypersurface $\partial \Omega$, like when you want to apply the Gauss-Ostrogradski theorem. $\endgroup$ – Alex M. Mar 22 '17 at 14:57
  • $\begingroup$ @Alex M.: He used a bounded domaine, but after he works with a a domain $\Omega$ with Lipschitz boundary. $\endgroup$ – Motaka Mar 22 '17 at 14:59


As far as I can understand the problem formulation, $g$ is not defined on the $n$–dimensional unit ball $$B^n = \{x\in\mathbb R^n:\|x\| \le 1\}$$ but rather on the ball's surface, that is on the $(n-1)$–dimensional (hyper)sphere $$S^{n-1} = \{x\in\mathbb R^n:\|x\| = 1\}$$


For any $x \ne 0$, $$\frac{x}{\left \| x \right \|}$$ represents a unit vector in the direction of $x$, identifying a single point on a unit sphere.

There is a value $g$ assigned to that point: $g\left(\frac{x}{\left \| x \right \|}\right)$.
The definition says, that $\Omega$ contains for each direction all such points $x$, whose distance from $0$ do not exceed the $g$ defined for that direction.

In other words, the domain is a union of (open-ended) line segments with a common beginning in $0$, whose lengths are defined by $g$.

If $g = r$ is constant, non-zero, then $\Omega$ will be an open ball of radius $r$.
If $g$ is 'close to constant', i.e. $|g - r| \le e$ for $0 \lt e \ll r$, we could say $\Omega$ is a ball of radius $r$ with hills & valeys not exceeding the height or depth of $e$.


The $g$ function is positive. At each point $P$ belonging to a unit (hyper)sphere $S^{n-1}$ (hence $\|P\| = 1$): $$g(P) > 0$$ then all points on a line segment between $0$ and $g(P)\cdot P$ belong to $\Omega$: $$\{x\in\mathbb R^n: x = k\cdot P,\ 0 < k < g(P)\}\subset \Omega$$

So $\Omega$ contains a non-degenerate line segment, which is equipollent with $\mathbb R$, hence $\Omega$ cardinality is at least continuum.
On the other hand $\Omega \subset \mathbb R^n$, so its cardinality is at most continuum; which impiles $|\Omega|=\mathfrak C$.

The cardinality of both the unit sphere $S^{n-1}$ (provided $n\ge 2$) and the unit ball $B^n$ is also continuum, so a bijection between each of them and $\Omega$ certainly exists.

The bijection between the ball and $\Omega$ is straightforward: for any ray (half-line) starting at $0$ and meeting the unit sphere at point $P$, its initial segment of length $g(P)$ is a subset of $\Omega$ – so it's enough to scale the segment by $1/g(P)$ to obtain a unit segment; and a union of all unit segments starting at $0$ is a unit ball.

  • $\begingroup$ :Can you explain more? $\endgroup$ – Motaka Mar 22 '17 at 18:43
  • $\begingroup$ @Mokata I tried to expand my answer. Please see if it's clear enogh now. $\endgroup$ – CiaPan Mar 23 '17 at 6:53
  • $\begingroup$ :Now it's good, I forgot to specify that $g$ is defined just in the unit sphere.Thank's, do you have an idea about how to prove that if $\Omega $ is $\mathcal C ^1$ so g is $\mathcal C ^1$? $\endgroup$ – Motaka Mar 23 '17 at 8:06

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