[SOLVED]Closed, Symmetric, Convex Subset Property

Sudharaka

Well-known member
MHB Math Helper
Hi everyone,

Here's one of the questions I had when trying to solve a problem.

Let $$K\subset \mathbb{R}^{n}$$ be a closed symmetric convex subset with non empty interior. Let $$\alpha>\beta$$ where $$\alpha,\,\beta\in \mathbb{R}$$. Then prove that,

$\beta K\subset\alpha K$

I have tried various things but didn't get a breakthrough. Any hint on how to solve this would be greatly appreciated.

Opalg

MHB Oldtimer
Staff member
Hi everyone,

Here's one of the questions I had when trying to solve a problem.

Let $$K\subset \mathbb{R}^{n}$$ be a closed symmetric convex subset with non empty interior. Let $$\alpha>\beta$$ where $$\alpha,\,\beta\in \mathbb{R}$$. Then prove that,

$\beta K\subset\alpha K$

I have tried various things but didn't get a breakthrough. Any hint on how to solve this would be greatly appreciated.
Presumably the condition $$\alpha>\beta$$ should be $$\alpha>\beta>0$$. (Otherwise, you could take $\beta$ to be large and negative, and then the symmetric property would say that $\beta K = (-\beta)K \supset \alpha K$.)

First, use the conditions that $K$ is symmetric, convex and has nonempty interior to show that $0$ is in the interior of $K$. Show that, for each $x\in\mathbb{R}^n$, the set $\{\lambda\in\mathbb{R} : \lambda x\in K\}$ is a symmetric closed interval. The fact that $x\in \alpha K\:\Leftrightarrow\:\alpha^{-1}x \in K$ then implies that $\beta K\subset\alpha K.$

Sudharaka

Well-known member
MHB Math Helper
Presumably the condition $$\alpha>\beta$$ should be $$\alpha>\beta>0$$. (Otherwise, you could take $\beta$ to be large and negative, and then the symmetric property would say that $\beta K = (-\beta)K \supset \alpha K$.)

First, use the conditions that $K$ is symmetric, convex and has nonempty interior to show that $0$ is in the interior of $K$. Show that, for each $x\in\mathbb{R}^n$, the set $\{\lambda\in\mathbb{R} : \lambda x\in K\}$ is a symmetric closed interval. The fact that $x\in \alpha K\:\Leftrightarrow\:\alpha^{-1}x \in K$ then implies that $\beta K\subset\alpha K.$
Thanks so much for answering my question. However there is a little doubt in my mind. How does the fact that $$0$$ is in the interior of $$K$$ connected to the problem. Is it needed to prove the statement that follows?

Opalg

MHB Oldtimer
Staff member
Thanks so much for answering my question. However there is a little doubt in my mind. How does the fact that $$0$$ is in the interior of $$K$$ connected to the problem. Is it needed to prove the statement that follows?
For this problem it is not necessary for $K$ to have nonempty interior, nor for it to be closed. However, this result is possibly part of a more general programme to show that $K$ can be used to define a norm on the space, and then those extra conditions would be relevant.

Sudharaka

Well-known member
MHB Math Helper
For this problem it is not necessary for $K$ to have nonempty interior, nor for it to be closed. However, this result is possibly part of a more general programme to show that $K$ can be used to define a norm on the space, and then those extra conditions would be relevant.
Great. Now this makes perfect sense. Yes, indeed this is part of a more general question. I shall make a separate thread for that and give my answer to the whole question perhaps in the next few days so that I can confirm whether I have done it correctly. Thanks for all your help, I really appreciate it.