Is the dilation of a convex body a subset of original convex body

Question

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

Answer 1

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 $x\in\mathcal C$.

Answer 2

It is unclear from your definition, but it seems you're describing a general homography (one of the Affine transformations). If so, the below is a counterexample (assuming the "homothetic center" is at the point at the left).

Answer 3

Counterexample in $\mathbb{R}$: $C=[1,2], \lambda_1=1, \lambda_2=2$. Then $\lambda_1C=C$ and $\lambda_2 C=[2,4]$. But the assertion is true for arbitrary $p$ when $0\in C$.

Edit: Assume that $0\in C$ and that $0<\lambda_1\leq\lambda_2$. Let $x\in \lambda_1 C$. Then there is some $\xi\in C$ such that $x=\lambda_1 \xi$. Put $\mu:=\frac{\lambda_1}{\lambda_2}$. Then $0<\mu\leq 1$ and $\mu \xi=(1-\mu)0+\mu \xi\in C$ since $C$ is convex. This implies that $x=\lambda_1 \xi=\lambda_2\mu\xi\in \lambda_2 C$.