0
$\begingroup$

We say that two norm $\|x\|_1$ and $\|x\|_2$ on a vector space $X$ are said to be equivalent if there exists $K>0$ and $M>0$ such that $$ K\|x\|_1\le \|x\|_2\le M\|x\|_1 $$ Prove that on a finite dimensional vector space all norms are equivalent.

Here is the beginning part of a proof I am reading...

Assume that the vector space is $\mathbb{R}^n$. Denote $\|\cdot\|_2$ the Euclidean norm. If $\|\cdot\|$ is another norm on $\mathbb{R}^n.$ One just need to prove that it is equivalent to $\|\cdot\|_2.$ Let $e_1,\cdots, e_n$ be the standard basis for $\mathbb{R}^n.$ If $x=(x_1,\cdots,x_n)=x_1 e_1+\cdots+x_n e_n.$ Then \begin{align} \|x\|&=\|\sum_{i=1}^n x_i e_i\|\\ &\le\sum_{i=1}^n|x_i|\|e_i\|\quad (\text{By triangle inequality})\\ &\le\left(\sum_{i=1}^n\|e_i\|\right)\|x\|_2\quad (\text{I am lost here...}) \end{align}

Could anyone show me the detail from the 2nd line to the 3rd line? All I know is $\left(\sum_{i=1}^n\|e_i\|\right)\|x\|_2=\left(\sum_{i=1}^n\|e_i\|\right)\left(\sum_{i=1}^n|x_i|^2\right)^{1/2}$...but not the rest, please help.

$\endgroup$

1 Answer 1

0
$\begingroup$

This is because $\lvert x_i\rvert=\sqrt{x_i^2}\le\sqrt{\sum_{i=1}^nx_i^2}=\lVert x\rVert_2.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .