Truncated wave cones: convexity, piecewise C1 presentation and outward normals

The function $\phi$ is convex: for $p=(x,t)$, $q=(y,s)\in {\mathbb{R}}^{n+1}$ and $\lambda \in [0,1]$, writing $u:=(1-\lambda )(x-x_0)$ and $v:=\lambda (y-x_0)$, the triangle inequality and absolute homogeneity of the Euclidean norm [F3] give $\mathrm{\mid }u+v\mathrm{\mid }\le (1-\lambda )\mathrm{\mid }x-x_0\mathrm{\mid }+\lambda \mathrm{\mid }y-x_0\mathrm{\mid }$, while $c((1-\lambda )t+\lambda s)=(1-\lambda )ct+\lambda cs$ by linearity, so $\phi ((1-\lambda )p+\lambda q)\le (1-\lambda )\phi (p)+\lambda \phi (q)$; consequently $K(t_1,t_2)=\{\phi <c{t}_0\}\cap \{t_1<t<t_2\}$ is convex, because both sets are convex: for $\{\phi <c{t}_0\}$ this is the inequality just proved applied to two points with values below $c{t}_0$, and the slab is defined by two affine conditions [F9].

Basic topological properties: $K(t_1,t_2)$ is open since [F3] gives $\mathrm{\mid }\mathrm{\mid }x-x_0\mathrm{\mid }-\mathrm{\mid }y-x_0\mathrm{\mid }\mathrm{\mid }\le \mathrm{\mid }x-y\mathrm{\mid }$, so $\phi$ and the coordinate $t$ are continuous and the half-lines $(t_1,t_2)$ and $(-\mathrm{\infty },c{t}_0)$ are open; it is nonempty because $(x_0,(t_1+t_2)\mathrm{/}2)$ has $\phi =0+c(t_1+t_2)\mathrm{/}2<c{t}_0$ and lies in the slab; it is bounded because every point has $t_1<t<t_2$ and $\mathrm{\mid }x-x_0\mathrm{\mid }<c(t_0-t)<c{t}_0$. This is (i).

The boundary decomposition: $\mathrm{\partial }K(t_1,t_2)=D_b\cup D_t\cup L$ with $D_b:={\bar{B}}_{c(t_0-t_1)}(x_0)\times \{t_1\}$, $D_t:={\bar{B}}_{c(t_0-t_2)}(x_0)\times \{t_2\}$ and $L$ as in the statement; the only overlaps are $D_b\cap L=S_{c(t_0-t_1)}(x_0)\times \{t_1\}$ and $D_t\cap L=S_{c(t_0-t_2)}(x_0)\times \{t_2\}$, and $D_b\cap D_t=\mathrm{\varnothing }$. Indeed $\mathrm{\partial }\{\phi <c{t}_0\}\subseteq \{\phi =c{t}_0\}$ and $\mathrm{\partial }\{t_1<t<t_2\}\subseteq \{t=t_1\}\cup \{t=t_2\}$, and $K$ is the intersection of these three open sets, so a boundary point of $K$ lies in one of the three level sets; conversely a point $p=(x,t)$ with $\phi (p)=c{t}_0$ and $t\in (t_1,t_2)$ (a point of $L$) has $p+\delta (x-x_0,c)\mathrm{\notin }K$ and $p-\delta (x-x_0,c)\in K$ for small $\delta >0$, while a point with $t=t_1$ and $\mathrm{\mid }x-x_0\mathrm{\mid }<c(t_0-t_1)$ has $p-\delta e_t\mathrm{\notin }K$ and $p+\delta e_t\in K$, and at a rim point $\mathrm{\mid }x-x_0\mathrm{\mid }=c(t_0-t_1)$, $t=t_1$, the points $(x_0+\frac{t_0-t_1-2\epsilon }{t_0-t_1}(x-x_0),t_1+\epsilon )$ lie in $K$ for $0<\epsilon <\min ((t_0-t_1)\mathrm{/}2,t_2-t_1)$, since their spatial radius is $c(t_0-t_1-2\epsilon )<c(t_0-t_1-\epsilon )$, and tend to $p$; at a top rim point $(x,t_2)$ the points $(x,t_2-\epsilon )$ lie in $K$ for $0<\epsilon <t_2-t_1$ and tend to it, while the interior of the top disk is approached vertically.

Each face is a compact Borel subset of a regular $C^1$ hypersurface patch: $D_b$ and $D_t$ are closed balls in the hyperplanes $\{t=t_i\}$, which are graphs of the constant (hence $C^1$) functions over ${\mathbb{R}}^n$ with nonvanishing gradient of $(x,t)\mapsto t$; the lateral face lies in the graphic hypersurface $\{(y,h(y)):y\in O\}$ where $O:={\mathbb{R}}^n\setminus \{x_0\}$ and $h(y):=t_0-\mathrm{\mid }y-x_0\mathrm{\mid }\mathrm{/}c$, which is $C^1$ because $y\mapsto ⟨y-x_0,y-x_0⟩$ is a finite sum of products of the $C^1$ coordinate functions [F6], the square root is differentiable on $(0,\mathrm{\infty })$ with derivative $\frac{1}{2}{s}^{-1\mathrm{/}2}$ [F4], and the chain rule [F5] applies on the open set where the inner value is positive, namely $O$.

The presentation is verified with $E:=(S_{c(t_0-t_1)}(x_0)\times \{t_1\})\cup (S_{c(t_0-t_2)}(x_0)\times \{t_2\})$: the three faces are closed bounded Borel subsets, hence compact by [F10], of regular $C^1$ patches by step 4.1 and cover $\mathrm{\partial }K$ by step 3.1; $E\subset \mathrm{\partial }K$ is compact and contains the relative boundaries of the faces in their patches (the rim circles of the two disks and the two boundary circles of the annulus $\{c(t_0-t_2)\le \mathrm{\mid }y-x_0\mathrm{\mid }\le c(t_0-t_1)\}$ parametrizing $L$) and all pairwise overlaps, which by step 3.1 are exactly the two rim circles; off $E$ the boundary is locally a single $C^1$ graph with $K$ on one side, namely $t=t_i$ over a small ball in the interior of each disk with $K$ on the side $t>t_1$, respectively $t<t_2$, and $t=h(y)$ over a small ball in $O$ for interior points of $L$, with $K$ locally $\{t<h(y)\}$ by the definition of $K$; and $S_j\cap E$ is surface-null in each face: on a disk the surface measure is $n$-dimensional Lebesgue measure transported by the graph chart [F2], whose rim is a sphere of positive radius, null by [F7] and [F8] applied to the homothety $z\mapsto x_0+rz$ (and for $n=1$ a two-point set), while on $L$ the graph density is the constant $\sqrt{1+1\mathrm{/}{c}^2}$ because $\mathrm{\mid }Dh\mathrm{\mid }=1\mathrm{/}c$ on $O$, so a Borel subset of $L$ is surface-null exactly when its $y$-projection is ${\lambda }_n$-null [F2], and the projection of $E\cap L$ is the union of two positive-radius spheres, null by [F7] and [F8]. Thus $K(t_1,t_2)$ has the specified finite piecewise $C^1$ presentation with faces $D_b,D_t,L$ and edge set $E$: this is (ii).

The outward normals: on the bottom disk the region lies locally in $\{t>t_1\}$, so the outward unit normal is $(0,-1)$; on the top disk it is $(0,1)$; on $L$ the field $\mathrm{\nabla }\phi =((x-x_0)\mathrm{/}\mathrm{\mid }x-x_0\mathrm{\mid },c)$ is continuous and nonvanishing near $L$ because $\mathrm{\mid }x-x_0\mathrm{\mid }=c(t_0-t)\ge c(t_0-t_2)>0$ there, $K$ is locally the side $\{\phi <c{t}_0\}$ and $L\subseteq \{\phi =c{t}_0\}$, so the outward unit normal is $\mathrm{\nabla }\phi \mathrm{/}\mathrm{\mid }\mathrm{\nabla }\phi \mathrm{\mid }=((x-x_0)\mathrm{/}\mathrm{\mid }x-x_0\mathrm{\mid },c)\mathrm{/}\sqrt{1+c^2}$, using the outward-normal convention for one-sided graph boundaries [F2] and the gradient of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case. This is (iii).

The closed cone: $K^{-}=\{\phi \le c{t}_0\}\cap \{t\ge 0\}\cap \{t\le t_0\}$ is closed because $\phi$ is continuous, bounded because $0\le t\le t_0$ and $\mathrm{\mid }x-x_0\mathrm{\mid }\le c{t}_0$, and thus compact by [F10], and convex because the sublevel set is convex by the inequality of step 1.1 and the two half-spaces are convex [F9]; its interior is $\{0<t<t_0,\phi <c{t}_0\}$: the inclusion $\supseteq$ is openness of the right-hand set inside $K^{-}$, and conversely a point with $\phi =c{t}_0$ is not interior, since for $z:=(x_0,0)$ with $\phi (z)=0<c{t}_0$ the points $r(\sigma ):=z+\sigma (p-z)$, $\sigma >1$, satisfy $p=\frac{1}{\sigma }r(\sigma )+(1-\frac{1}{\sigma })z$, so convexity gives $\phi (p)\le \frac{1}{\sigma }\phi (r(\sigma ))+(1-\frac{1}{\sigma })\phi (z)$ and hence $\phi (r(\sigma ))\ge c{t}_0+(\sigma -1)(c{t}_0-\phi (z))>c{t}_0$, with $r(\sigma )\to p$ as $\sigma \downarrow 1$; a point with $t=0$ or $t=t_0$ is not interior because $p-\delta e_t$, respectively $p+\delta e_t$, lies outside $K^{-}$ for every $\delta >0$. Intersecting $\mathrm{int}{K}^{-}$ with the slab $\{t_1<t<t_2\}\subseteq \{0<t<t_0\}$ gives exactly $K(t_1,t_2)$, which is (iv).