Consider a convex body $\mathcal{C} \subset \mathbb{R}^p$, for some $p \in \mathbb{N}$.
That is $\mathcal{C}$ is a convex set in $\mathbb{R}^p$ with a non-empty interior.
For $\lambda \in \mathbb{R}_{>0}$, define the $\lambda$-dilation of the convex body $\mathcal{C}$ as the set:
$$\lambda \mathcal{C} := \{\lambda k \mid k \in \mathcal{C} \} $$
If we fix $\lambda_{1}, \lambda_{2}$ such that $0 < \lambda_{1} \le \lambda_{2}$,
then is $\lambda_{1} \mathcal{C} \subseteq \lambda_{2} \mathcal{C}$?
Or is there a simple counterexample? I expect this to be true since intuitively
the $\lambda$-dilation of the convex body $\mathcal{C}$ just re-scales the
convex body by a constant factor (or stays the same when $\lambda = 1$),
so that the subset should be preserved under the dilations described.
I was not quite able to relate the inequality relation between
$\lambda_{1}, \lambda_{2}$ into the subset relation, when I tried
a "pick an arbitrary set element" type argument.
Could anyone help clarify the above?
Also does this then work under general translation invariance
and scaling i.e. consider a fixed $y \in \mathbb{R}^p$, is the
following true?
$$y – \lambda_{1} \mathcal{C} \subseteq y – \lambda_{2} \mathcal{C}$$
In this case, one is "scaling, flipping, and translating the original"
convex body $\mathcal{C}$ using different scaling factors $0 < \lambda_{1} \le \lambda_{2}$ and translating by the same vector $y$.
Edit: Based on helpful comments by @DavidGStork, @TSF, and @JensSchwaiger, assume $0 \in \text{int}{\mathcal{C}}$, or if required that $\mathcal{C}$ has it's centroid at
the 0 vector.
Any help appreciated
Best Answer
Here's a proof if the origin is in the convex body.
We can without loss of generality assume that $\lambda_2=1$. So, for some $0\leq\lambda\leq 1,$ we need $x\in \mathcal C\implies \lambda x \in\mathcal C$. This is certainly true, since $\lambda x$ is on the line segment connecting $0\in\mathcal C$ and $x\in\mathcal C$.