8
$\begingroup$

Let $X$ be a Banach space with a norm $\|\cdot\|_1$ and $A$ be a linear operator on $X$ such that

  1. $\|A\|_1\leq 1$;

  2. $\|A^m\|_1<1$ for some $m\in \mathbb N$.

Is that true that there is an equivalent norm $\|\cdot\|_2$ on $X$ such that $\|A\|_2<1$? If there exists such a norm, how can it be constructed?

Here for operator we use associated (induced norm): given a norm $\|\cdot\|$ on $X,$ $ \|B\| :=\sup\limits_{\|x\|=1}\|Bx\| $ for any linear operator $B$.

  • 0
    @copper.hat not necessary, all the assumptions are in OP2012-05-21

1 Answers 1

7

Let $\|x\|_2 = \|x\|_1+\|Ax\|_1+\dots+\|A^{m-1}x\|_1$, then $\|Ax\|_2 = \|Ax\|_1+\dots+\|A^{m}x\|_1=\|x\|_2+(\|A^{m}x\|_1-\|x\|_1)\le \|x\|_2-\epsilon\|x\|_1\le (1-\epsilon')\|x\|_2$.

  • 0
    Very nice construction.2012-05-21