1. ## Equivalent Norms

Dear Colleagues,

Let $X$ be the space of all $x$ such that $x=(x_{1}, x_{2}, ..., x_{n})$, and $x_{1}, x_{2}, ..., x_{n}$ are numbers, and $||.||$ be any norm on $X$, and define the norm $||.||_{2}$ on $X$ to be $||x||_{2}=(|x_{1}|+|x_{2}|+...+|x_{n}|)^{1/2}$. We want to show directly that there is $a>0$ such that $a||x||_{2}\leqslant ||x||$ for any $x\in X$, but without using the fact that all norms on a finite dimensional vector space are equivalent. Instead we want to use the fact that " A continuous mapping of a compact subset $M$ of a normed space $X$ into the real numbers $R$ assumes a maximum and a minimum at some points of $M$ ".

I think the continuous mapping in my problem will be the norm but which norms will be used and what is the compact set.

Best Regards.

Raed.

2. Hint: It's enough to show this for $x$ such that $||x||=1.$

3. Originally Posted by ymar
Hint: It's enough to show this for $x$ such that $||x||=1.$
Thank you for your reply. In fact, $||x||=1$ is not closed so not compact.

Regards.

4. Originally Posted by raed
In fact, $||x||=1$ is not closed so not compact.
No, the set $\{x\in X:\|x\|=1\}$ is closed, and ymar's hint was a good one.

It's probably best to start with the 2-norm, and to look at the set $\{x\in X:\|x\|_2=1\}$, which is closed and bounded (as a subset of n-dimensional euclidean space) and therefore compact. You need to show that the function $x\mapsto \|x\|$ is continuous on this space and hence attains its upper and lower bounds. The tricky part is to do this without implicitly assuming the result that you are trying to prove.

5. raed, why do you think that that set isn't closed?