# If two normed spaces are Lipschitz equivalent, then one if complete iff the other is

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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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.
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