Normed Spaces

Definition 18.1.1 (Norm)   Let $ K$ be a field with valuation $ \left\vert \cdot \right\vert$ and let $ V$ be a vector space over $ K$. A real-valued function $ \left\vert \cdot \right\vert$ on $ V$ is called a if
  1. $ \left\Vert v\right\Vert>0$ for all nonzero $ v\in V$ (positivity).
  2. $ \left\Vert v+w\right\Vert \leq \left\Vert v\right\Vert + \left\Vert w\right\Vert$ for all $ v,w\in V$ (triangle inequality).
  3. $ \left\Vert av\right\Vert = \left\vert a\right\vert\left\Vert v\right\Vert$ for all $ a\in K$ and $ v\in V$ (homogeneity).

Note that setting $ \left\Vert v\right\Vert=1$ for all $ v\neq 0$ does not define a norm unless the absolute value on $ K$ is trivial, as $ 1=\left\Vert av\right\Vert =
\left\vert a\right\vert\left\Vert v\right\Vert=\left\vert a\right\vert$. We assume for the rest of this section that $ \left\vert \cdot \right\vert{}$ is not trivial.

Definition 18.1.2 (Equivalent)   Two norms $ \left\vert \cdot \right\vert _1$ and $ \left\vert \cdot \right\vert _2$ on the same vector space $ V$ are if there exists positive real numbers $ c_1$ and $ c_2$ such that for all $ v\in V$

$\displaystyle \left\Vert v\right\Vert _1 \leq c_1 \left\Vert v\right\Vert _2$   and$\displaystyle \qquad
\left\Vert v\right\Vert _2 \leq c_2 \left\Vert v\right\Vert _1.

Lemma 18.1.3   Suppose that $ K$ is a field that is complete with respect to a valuation $ \left\vert \cdot \right\vert{}$ and that $ V$ is a finite dimensional $ K$ vector space. Continue to assume, as mentioned above, that $ K$ is complete with respect to $ \left\vert \cdot \right\vert{}$. Then any two norms on $ V$ are equivalent.

Remark 18.1.4   As we shall see soon (see Theorem 19.1.8), the lemma is usually false if we do not assume that $ K$ is complete. For example, when $ K=\mathbf{Q}$ and $ \left\vert \cdot \right\vert _p$ is the $ p$-adic valuation, and $ V$ is a number field, then there may be several extensions of $ \left\vert \cdot \right\vert _p$ to inequivalent norms on $ V$.

If two norms are equivalent then the corresponding topologies on $ V$ are equal, since very open ball for $ \left\vert \cdot \right\vert _1$ is contained in an open ball for $ \left\vert \cdot \right\vert _2$, and conversely. (The converse is also true, since, as we will show, all norms on $ V$ are equivalent.)

Proof. Let $ v_1,\ldots, v_N$ be a basis for $ V$. Define the max norm $ \left\Vert \cdot \right\Vert _0$ by

$\displaystyle \left\Vert\sum_{n=1}^N a_n v_n\right\Vert _0 = \max \left\{\left\vert a_n\right\vert : n=1,\ldots, N\right\}.

It is enough to show that any norm $ \left\vert \cdot \right\vert$ is equivalent to $ \left\Vert \cdot \right\Vert _0$. We have

$\displaystyle \left\Vert\sum_{n=1}^N a_n v_n\right\Vert$ $\displaystyle \leq \sum_{n=1}^N \left\vert a_n\right\vert \left\Vert v_n\right\Vert$    
  $\displaystyle \leq \sum_{n=1}^N \max{\left\vert a_n\right\vert} \left\Vert v_n\right\Vert$    
  $\displaystyle = c_1 \cdot \left\Vert\sum_{n=1}^N a_n v_n\right\Vert _0,$    

where $ c_1 = \sum_{n=1}^N \left\Vert v_n\right\Vert$.

To finish the proof, we show that there is a $ c_2\in \mathbf{R}$ such that for all $ v\in V$,

$\displaystyle \left\Vert v\right\Vert _0 \leq c_2 \cdot \left\Vert v\right\Vert.

We will only prove this in the case when $ K$ is not just merely complete with respect to $ \left\vert \cdot \right\vert{}$ but also locally compact. This will be the case of primary interest to us. For a proof in the general case, see the original article by Cassels (page 53).

By what we have already shown, the function $ \left\Vert v\right\Vert$ is continuous in the $ \left\Vert \cdot \right\Vert _0$-topology, so by local compactness it attains its lower bound $ \delta$ on the unit circle $ \left\{v\in V :
\left\Vert v\right\Vert _0=1\right\}$. (Why is the unit circle compact? With respect to $ \left\Vert \cdot \right\Vert _0$, the topology on $ V$ is the same as that of a product of copies of $ K$. If the valuation is archimedean then $ K\cong \mathbf{R}$ or $ \mathbf{C}$ with the standard topology and the unit circle is compact. If the valuation is non-archimedean, then we saw (see Remark 17.1.7) that if $ K$ is locally compact, then the valuation is discrete, in which case we showed that the unit disc is compact, hence the unit circle is also compact since it is closed.) Note that $ \delta>0$ by part 1 of Definition 18.1.1. Also, by definition of $ \left\Vert \cdot \right\Vert _0$, for any $ v\in V$ there exists $ a\in K$ such that $ \left\Vert v\right\Vert _0 = \left\vert a\right\vert$ (just take the max coefficient in our basis). Thus we can write any $ v\in V$ as $ a\cdot w$ where $ a\in K$ and $ w\in V$ with $ \left\Vert w\right\Vert _0=1$. We then have

$\displaystyle \frac{\left\Vert v\right\Vert _0}{\left\Vert v\right\Vert} =
...\Vert w\right\Vert}
= \frac{1}{\left\Vert w\right\Vert} \leq \frac{1}{\delta}.

Thus for all $ v$ we have

$\displaystyle \left\Vert v\right\Vert _0\leq c_2\cdot \left\Vert v\right\Vert,$

where $ c_2 = 1/\delta$, which proves the theorem. $ \qedsymbol$

William Stein 2004-05-06