[SOLVED] Closed, Symmetric, Convex Subset Property

[Sudharaka]

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]

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]

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]

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]

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.