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Prove that if two normed spaces are Lipschitz equivalent, then one if complete iff the other is.

My thoughts:

Let $ (V_1, \Vert\cdot\Vert_1) $ and $ (V_2, \Vert\cdot\Vert_2) $ be Lipschitz equivalent normed vector spaces. Then there exists $f : V_1 \to V_2 $, and constants $h, k > 0 $, such that $ h\Vert f(x) - f(y)\Vert_2 \leq \Vert x-y\Vert_1 \leq k\Vert f(x) - f(y)\Vert_2 $ for all $ x,y, \in V_1 $. Suppose $(V_2, \Vert\cdot\Vert_2) $ is complete.

Clearly everything is symmetrical, so we only really need to prove this in one direction. I can see that if $ (x_n) $ is a Cauchy sequence in $V_1$, then $(f(x_n))$ is Cauchy in $ V_2 $. I can also see that $ f $ is uniformly continuous. How can I turn this into a proof?

Thanks

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1 Answer 1

If you can see that having $x_n$ Cauchy in $V_1$ implies $f(x_n)$ is Cauchy in $V_2$, then you are almost there. Completeness of $V_2$ gives that $f(x_n)$ converges to some $y \in V_2$. Now your conditions on $f$ guarantee that $f$ has a continuous (indeed, Lipschitz) inverse $f^{-1} : V_2 \to V_1$ (verify this), so we have $x_n = f^{-1}(f(x_n)) \to f^{-1}(y)$ and thus $x_n$ converges.

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One could add that the fact that $V_1$ and $V_2$ are vector spaces doesn't matter at all, the fact that they are metric spaces and one of them is complete is the only thing used in the proof. –  t.b. May 10 '11 at 13:48

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