I am trying to understand the following statement from an analysis book:
Two norms are equivalent ($||\cdot ||_2 \sim||\cdot||_1 $) if they induce equivalent metrics.
At first I thought this was straight forward but then the book has the following proposition that leads me to believe that I have a misunderstanding:
Let $||\cdot||_1$ and $||\cdot||_2$ be two norms on a field F. Then $||\cdot ||_2 \sim||\cdot||_1$ iff there exists a positive real number $\alpha$ such that $||x||_2=||x||_1^\alpha , \ \forall x\in F$.
How can it be that $||x||_2=||x||_1^\alpha$ if x is any other value other that 1 if $\alpha\ne$1? Maybe this is related to my misunderstanding... can norms and metrics be thought of as the same thing? Thanks!
They aren't equal only equivalent, which metrics are if they induce the same topology. This means that they have the same open sets, and since a field is also a one-dimensional vector space, this just means they need to have the same unit balls.
But then $$\lVert \cdot \rVert_1\le 1 \iff \lVert \cdot\rVert_2\le 1$$
so choose any $y$ such that $0\ne \lVert y\rVert_1>1$--if this is not possible the proposition is trivial since both give the discrete topology where every non-zero vector has norm $1$--and consider $x\ne 0$ so that $\lVert x\rVert_1=\lVert y\rVert_1^\alpha$, for some $\alpha\in\Bbb R$. Let $m_i/n_i\downarrow \alpha$ be a sequence of rational numbers with $n_i$ positive. Then
$${\lVert x^{n_i}\rVert\over\lVert y^{m_i}\rVert_1}<1\implies {\lVert x^{n_i}\rVert\over \lVert y^{m_i}\rVert_2}<1$$
so that $\lVert x\rVert_2\le \lVert y\rVert_2^{m_i/n_i}$ so that the same holds in the limit giving $\lVert x\rVert_2\le\lVert y\rVert_2^\alpha$. Similarly using a sequence $m_i'/n_i'\uparrow\alpha$ we get the reverse inequality $\lVert x\rVert_2\ge \lVert y\rVert_2^\alpha$.
Then logging both of
$$\begin{cases}\lVert x\rVert_1=\lVert y\rVert_1^\alpha \\ \lVert x\rVert_2=\lVert y\rVert_2^\alpha\end{cases}$$
we get
$${\log \lVert x\rVert_1\over\log \lVert x\rVert_2}={\log \lVert y\rVert_1\over\log \lVert y\rVert_2}=s$$
and using that $y$ is arbitrary and $>0$, we get that $s>0$ and that the equality holds on all of the space.