equivalent norms

Let $\parallel x\parallel$ and ${\parallel x\parallel }^{\prime }$ be two norms ona vector space $V$. These norms are equivalent norms ifthere exists a number $C>1$ such that

${\displaystyle \frac{1}{C}}\parallel x\parallel \le {\parallel x\parallel }^{\prime }\le C\parallel x\parallel$(1)

for all $x\in V$.

Since equation (1) is equivalent to

${\displaystyle \frac{1}{C}}{\parallel x\parallel }^{\prime }\le \parallel x\parallel \le C{\parallel x\parallel }^{\prime }$(2)

it follows that the definition is well defined. In other words,$\parallel \cdot \parallel$ and $\parallel \cdot {\parallel }^{\prime }$ are equivalent if and only if$\parallel \cdot {\parallel }^{\prime }$ and $\parallel \cdot \parallel$ are equivalent.An alternative condition is that there exist positive realnumbers $c,d$ such that

$$c\parallel x\parallel \le {\parallel x\parallel }^{\prime }\le d\parallel x\parallel .$$

However, this condition is equivalent to the aboveby setting $C=\max \{1/c,d\}$.

Some key results are as follows:

1. If $\gamma >0$ and ${\parallel x\parallel }^{\prime }=\gamma \parallel x\parallel$, then$\parallel \cdot \parallel$ and $\parallel \cdot {\parallel }^{\prime }$ are equivalent. For example,if $\gamma >1$, then condition (1) holds with $C=\gamma$, andfor , condition (2) holds with $C=1/\gamma$.

2. Suppose norms $\parallel \cdot \parallel$ and $\parallel \cdot {\parallel }^{\prime }$ are equivalent normsas in equation (1), and let $B_r(x)$ and $B_r^{\prime }(x)$ be theopen balls with respect to $\parallel \cdot \parallel$ and $\parallel \cdot {\parallel }^{\prime }$, respectively.By this result (http://planetmath.org/ScalingOfTheOpenBallInANormedVectorSpace)it follows that

   $$C{B}_{\epsilon }(x)\subseteq B_{\epsilon }^{\prime }(x)\subseteq \frac{1}{C}{B}_{\epsilon }(x).$$

   It follows that the identity map from $(V,\parallel \cdot \parallel )$ to $(V,\parallel \cdot {\parallel }^{\prime })$is a homeomorphism. Or, alternatively, equivalent norms on $V$ induce the sametopology on $V$.

3. The converse of the last paragraph is also true, i.e. if two norms induce the same topology on $V$ then they are equivalent. This follows from the fact that every continuous linear function between two normed vector spaces is bounded (http://planetmath.org/BoundedOperator) (see this entry (http://planetmath.org/BoundedOperator)).

4. Suppose $⟨\cdot ,\cdot ⟩$ and ${⟨\cdot ,\cdot ⟩}^{\prime }$ are inner product. Suppose further that the induced norms $\parallel \cdot \parallel$ and $\parallel \cdot {\parallel }^{\prime }$ are equivalent as in equation 1. Then, by the polarization identity, the inner products satisfy

   $$\frac{1}{C^2}{⟨v,w⟩}^{\prime }\le ⟨v,w⟩\le C^2⟨v,w⟩.$$

5. On a finite dimensional vector space all norms are equivalent(see this page (http://planetmath.org/ProofThatAllNormsOnFiniteVectorSpaceAreEquivalent)).This is easy to understand as the unit sphere is compact if and only ifa space is finite dimensional.On infinite dimensional spaces this result does not hold (seethis page (http://planetmath.org/AllNormsAreNotEquivalent)).

   It follows that on a finite dimensional vector space,one can check continuity and convergence with respect with any norm.If a sequence converges in one norm, it converges in all norms.In matrix analysis this is particularly useful as one can choose the norm thatis most easily calculated.

6. The concept of equivalent norms also generalize to possibly non-symmetric norms. In this setting, all norms are also equivalent on a finite dimensional vector space. In particular, $\parallel \cdot \parallel$ and $\parallel -\cdot \parallel$ areequivalent, and there exists $C>0$ such that

   $$\parallel -v\parallel \le C\parallel v\parallel ,v\in V.$$