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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 $B$.

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I can't check it right now, but this paper seems to be addressing related questions: springerlink.com/content/171683521510x846 –  t.b. May 21 '12 at 14:57
    
I'm most likely missing something, but $\| \cdot \|_2 = \frac{1}{2} \| \cdot \|_1 $ seems to be an equivalent norm such that $\| A \|_2 < 1 .$ –  Ragib Zaman May 21 '12 at 15:10
    
@RagibZaman: the norms are given on $X$, so your example gives the same associated operator norm –  Ilya May 21 '12 at 15:12
    
Is $A$ invertible? –  copper.hat May 21 '12 at 15:15
    
@copper.hat not necessary, all the assumptions are in OP –  Ilya May 21 '12 at 15:16
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1 Answer

up vote 7 down vote accepted

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

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That should be true, thanks! –  Ilya May 21 '12 at 15:46
    
Very nice construction. –  copper.hat May 21 '12 at 15:52
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