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Let $(X, ||\cdot||)$ be a complete normed space. Let $||\cdot||$ be a norm on $X$, and assume that there are constants $c_{1}$, $c_{2} \in (0,\infty)$ such that:

$c_{1}||x-y||\le||x-y||_{0}\le c_{2}||x-y||$

for all $x,y\in X.$ Show that $(X,||\cdot||_{0})$ complete.

To be honest I have absolutely no idea where to start.

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2 Answers

up vote 1 down vote accepted

Let $\{x_n\}$ be a sequence of points in $X$ which is $\lVert\cdot\rVert_0$-Cauchy; that is, for every $\epsilon\gt 0$ there exists $N(\epsilon)\gt 0$ such that if $n,m\geq N$, then $\lVert x_n-x_m\rVert_0\lt\epsilon$. You want to show that the sequence converges in the $\lVert\ \rVert_0$ norm.

Use the fact that $\lVert x-y\rVert\leq \frac{1}{c_1}\lVert x-y\rVert $ to show the sequence is also a Cauchy sequence relative to the original norm. In particular, it must converge using the original norm.

Now you have a candidate for limit of the sequence; use your assumption to see whether it is also the limit under the new norm.

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Why can you say that $||x-y||\le \frac{1}{c_{1}}$? –  dplanet Feb 29 '12 at 20:03
    
@PeterR: You can't; some LaTeX got eaten up; it's been added back. (Did you notice there was an errant $\lVert$ in that formula?) –  Arturo Magidin Feb 29 '12 at 20:15
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Hint: suppose $x_n$ is a Cauchy sequence in $(X, \|\cdot\|_0)$. Is it also a Cauchy sequence in the norm $\|\cdot\|$?

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