For a convex compact set $K$ in $R^n$, if $x \in \partial K,\ \lambda x $ is not in $K$ if $\lambda>1.$

Question

Is it true that for a convex compact set $K$ in $\mathbb{R}^n$ that contains $0$ in its interior, if $x \in \partial K$ , then $\lambda x\not\in K$ for $\lambda > 1$? This is something that feels intuitively true but I am having some trouble proving it. If $\lambda x \in K$ for $\lambda >0$ then the fact that $K$ is convex means it must contain all line segments of all points in $K$ with $\lambda x$. But then for an $\varepsilon$ ball small enough, $x$ would not be a boundary point. This is clear for cases where $K$ is a closed disk, but this is much harder to visualize for general convex compact sets.

Answer 1

Let $K$ be a bounded convex set in $\mathbb{R}^n,$ suppose $x\in\partial K,\ 0\in\text{int}(K),$ and $\lambda x\in K$ where $\lambda>1.$

Sketch of proof that this leads to a contradiction

Since $0\in\text{int}(K),\ \exists \delta>0$ such that $x\not\in B(0,\delta)\subset K.$

Let $y_1, y_2,\ldots y_n$ be points in $B(0,\delta)$ such that $\vec{O y_j} \perp \vec{O x}$ for all $j\in \{1,\ldots,n\}$ and $0$ is an interior point of the convex hull of the $(n-1)$-dimensional shape formed by the vertices $y_1 y_2 \ldots y_n.$

Since $K$ is convex and $y_1, \ldots, y_n, \lambda x \in K,$ the convex hull of the $n$- dimensional shape formed by the vertices $y_1 y_2 \ldots y_n (\lambda x)$ is a subset of $K$ and has $x$ as an interior point, contradicting $x\in\partial K=\overline{K}\setminus \text{int}(K).$

Answer 2

Hint if $y = \lambda x\in K$ and $\lambda > 1$, the homothety centered at $y$ which sends $0$ on $x$ sends a ball around $0$ on a ball centered at $x$. If this ball is contained in $K$, the image of the ball is a neighborhood of $x$ contained in $K$.