Star-shaped multiplier signs

First establish the geometry under the stated $C^1$ hypotheses. Because $U$ is open, $\mathrm{\partial }U=\bar{U}\setminus U$. At any $p\in \mathrm{\partial }U$, shrink a one-sided chart: $U$ is locally $s<h(y)$, and its closure is locally $s\le h(y)$. Indeed, points strictly above the graph have an open neighborhood disjoint from $U$ by continuity of $h$, while graph points are approached from below and points below are in $U$; the neighborhood characterization of closure gives both inclusions. Every graph point also has points strictly above arbitrarily close, so it is not interior to $\mathrm{\Theta }$. Consequently $\mathrm{int}\mathrm{\Theta }=U$ and $\mathrm{\partial }\mathrm{\Theta }=\mathrm{\partial }U$, and $\mathrm{\Theta }$ is closed. Locally its complement is $s>h(y)$; reversing the last coordinate writes this side as a subgraph and gives its outward normal $-n_U$, where $n_U$ is the outward normal of $U$.

The open set ${\mathrm{\Omega }}_0=B_{R_0}(0)\setminus \mathrm{\Theta }$ has boundary exactly ${\mathrm{\Gamma }}_1\cup {\mathrm{\Gamma }}_2$. To see this, the Euclidean triangle inequality and the points $(1\pm \epsilon )x$ for $\mathrm{\mid }x\mathrm{\mid }=R_0$ give $\mathrm{\partial }{B}_{R_0}(0)=\{x:\mathrm{\mid }x\mathrm{\mid }=R_0\}$. At a point of this sphere, closedness of $\mathrm{\Theta }$ and $\mathrm{\Theta }\subset B_{R_0}(0)$ give a neighborhood disjoint from $\mathrm{\Theta }$, so ${\mathrm{\Omega }}_0$ locally agrees with the ball and both sides meet every neighborhood. At a point of $\mathrm{\partial }\mathrm{\Theta }$, a small neighborhood lies inside the ball and the two chart sides from step 1.1 meet every neighborhood, so this too is a boundary point of ${\mathrm{\Omega }}_0$. A point on neither boundary has a neighborhood wholly on one side of each of the ball and $\mathrm{\Theta }$; intersecting these neighborhoods shows that it is either interior to ${\mathrm{\Omega }}_0$ or exterior to its closure. No other boundary points occur. The pieces are disjoint since $\mathrm{\partial }\mathrm{\Theta }\subset \mathrm{\Theta }\subset B_{R_0}(0)$. Thus ${\mathrm{\Omega }}_0$ has one-sided $C^1$ boundary, and on ${\mathrm{\Gamma }}_2$ its outward normal is $\nu =-n_U$.

At a point $x\in {\mathrm{\Gamma }}_1$, rotate coordinates so that $x=R_0{e}_n$. The ball is locally the subgraph of $H(y)=\sqrt{R_0^2-\mathrm{\mid }y{\mathrm{\mid }}^2}$ near $y=0$. Here $\mathrm{\mid }H(y)-R_0\mathrm{\mid }=\mathrm{\mid }y{\mathrm{\mid }}^2\mathrm{/}(R_0+\sqrt{R_0^2-\mathrm{\mid }y{\mathrm{\mid }}^2})\le \mathrm{\mid }y{\mathrm{\mid }}^2\mathrm{/}{R}_0$, so the total derivative of $H$ at zero is zero by its remainder criterion. The graph-normal formula therefore gives $e_n$ at this point; transporting back yields $\nu =x\mathrm{/}{R}_0$ on ${\mathrm{\Gamma }}_1$. The local agreement with the ball in step 2.1 makes this also the outward normal of ${\mathrm{\Omega }}_0$.

Fix $x\in {\mathrm{\Gamma }}_2$ and use the chart of step 1.1, with $x$ represented by $(y,s)$ and $s=h(y)$. Put $d=x_0-x$, with chart vector components $(d\mathrm{\prime },d_n)$. The segment hypothesis applied to $x\in \mathrm{\Theta }$ puts $x+td$ in $\mathrm{\Theta }$ for $0\le t\le 1$; for sufficiently small positive $t$ this point remains in the chart, hence $s+t{d}_n\le h(y+td\mathrm{\prime })$. Subtract $s=h(y)$ and divide by $t>0$. Since $h$ is $C^1$ on an open parameter set, the directional-derivative theorem gives $(h(y+td\mathrm{\prime })-h(y))\mathrm{/}t\to Dh(y)d\mathrm{\prime }$ as $t\downarrow 0$, also when $d\mathrm{\prime }=0$. Therefore $d_n\le Dh(y)d\mathrm{\prime }$, and $d\cdot n_U=(d_n-Dh(y)d\mathrm{\prime })\mathrm{/}\sqrt{1+\mathrm{\mid }Dh(y){\mathrm{\mid }}^2}\le 0$. Scalar products are preserved by the rigid chart, and $q=-d$ while $\nu =-n_U$, so $q\cdot \nu =d\cdot n_U\le 0$ on ${\mathrm{\Gamma }}_2$.

Since $x_0\in \mathrm{\Theta }\subset B_{R_0}(0)$ and $\mathrm{\mid }x\mathrm{\mid }=R_0$ on ${\mathrm{\Gamma }}_1$, the Euclidean Cauchy–Schwarz inequality gives $q\cdot \nu =R_0-x_0\cdot x\mathrm{/}{R}_0\ge R_0-\mathrm{\mid }{x}_0\mathrm{\mid }>0$ there.

The disjoint boundary decomposition in step 2.1, the positive sign on ${\mathrm{\Gamma }}_1$ in step 4.1 and the nonpositive sign on ${\mathrm{\Gamma }}_2$ in step 3.2 give ${\mathrm{\Gamma }}_1=\{x\in \mathrm{\partial }{\mathrm{\Omega }}_0:q(x)\cdot \nu (x)>0\}$.