Smooth convex set intersects any vector non-coplanar with supporting hyperplane?

Question

Suppose I have a convex, compact, and $n$-dimensional set $K \subseteq \mathbb{R}^n$, and the origin $o$ is a smooth point on the boundary of $K$.

Let $b_1, ..., b_{n-1}$ be a basis for the unique supporting hyperplane of $K$ at the origin. Is it the case that if vector $v \notin span(b_1, ..., b_{n-1})$, then $span(v) \cap (K \setminus \{o\})$ is non-empty?

It seems it must be the case to me, but the one-line proof I imagine exists eludes me.

Answer 1

Since $C$ has dimension $n$ it has interior points. Since $K$ is smooth at the origin, its normal cone is a ray, i.e., $N_K(0)= \mathbb R^+ u$ for some non-zero normal vector $u$, where $u\perp b_i$ for all $i=1\dots n$.

Assume $span(v) \cap K = \{0\}$. Then the convex sets $span (v)$ and $int(K)$ are disjoint, and can be separated by a hyperplane. So there is $a\ne 0$ such that $$ a^T(sv) \le a^Tz \quad \forall s\in \mathbb R, z\in int (K). $$ It follows $a^Tv=0$, $a^Tz \ge 0$ for all $z \in cl(int(K))=K$. This implies $-a$ is a normal direction, $v$ is perpendicular to it, so $v$ is in the span of $b_1\dots b_{n-1}$.

Answer 2

This is correct if your boundary is of class $C^2$ locally at the origin, because in that case your domain has the interior ball property: there is a ball such that $0$ is on its boundary and is entirely contained in $K$. Next, observe that given the plane $x_n=0$ in $\mathbb{R}^n$ and a ball $$ x_1^2+x_2^2+\dots +(x_n-r)^2=r^2;\qquad r>0 $$ tangent to that plane at the origin, then any line $l$ in the direction of $v$ with $v_n\neq 0$ ($v$ not in the plane) intersects the ball along a segment (very simple computation). Replacing the plane $x_n=0$ by the tangent plane to $K$ at the origin and the ball by the interior ball (guaranteed to exist) at the origin, the desired property is proved.

Answer 3

Suppose the surface is just locally $C^1$ and has non-empty interior. For simplicity, I will take the supporting plane to be $x_n=0$ and I will call the rest of coordinates $x'$, so a generic point has coordinates $(x',x_n)$. Let $Q$ be in the interior. Without loss of generality we can assume that $Q$ is on the $x_n$-axis, $Q=(0,x_{0n})$ with $x_{0n}>0$ (Otherwise we can perform a shear transformation fixing the $x_n=0$ hyper plane). The boundary of $K$ is locally given by a $C^1$-function, $x_n=f(x')\ge 0$, defined on some small disk $D=\{x':\|x'\|<\epsilon\}$. Since $x'=0$ is a local minimum, $\nabla f(0)=0$.

Pick a line $l$ not contained in $\{x_n=0\}$. Parametrically, it is given by $x'=v't,\, x_n=tv_n$ with $v_n>0$. Since the directional derivative of $f$ along $v'$ is zero, we have $f(tv')=o(t)$ as $t\to 0$. In particular, $$ f(tv')<tv_n/2\qquad (*) $$ for small $t\le\epsilon_0$.

By convexity, the segments joining $Q$ with the points $(tv',f(tv'))$ with $t\le\epsilon_0$ on the boundary are contained in $K$. Thanks to (*), Those segments intersect the line $l$ and, therefore, the points on $l$ with $t\le\epsilon_0$ belong to $K$.