CAL1_L3

Equivalence of norms

We will prove that all norms on a finite dimensional normed linear space are equivalent.

Theorem 1(Lemma).

A norm is a continuous function.

Consider a norm $\Vert \cdot \Vert :V\to \mathbb{R}$. We will prove that $\Vert \cdot \Vert$ is uniformly continuous, which implies that it is also continuous. Let $ϵ>0$. We need to find a $\delta$ such that $\Vert {\mathbf{v}}_1-{\mathbf{v}}_2\Vert <\delta ⟹|\Vert {\mathbf{v}}_1\Vert -\Vert {\mathbf{v}}_2\Vert |<ϵ$. But, we know that $|\Vert {\mathbf{v}}_1\Vert -\Vert {\mathbf{v}}_2\Vert |\le \Vert {\mathbf{v}}_1-{\mathbf{v}}_2\Vert$. Thus, choosing $\delta =ϵ$ will do.

Theorem 2(Lemma).

Suppose $(V,\Vert \cdot \Vert )$ is finite dimensional normed linear space. Consider $S=\{\mathbf{x}\in V:\Vert \mathbf{x}\Vert =1\}$. $S$ is closed in $(V,\Vert \cdot \Vert )$.

Since $\Vert \cdot \Vert$ is continuous, and singleton sets are closed, $\Vert \cdot {\Vert }^{-1}(\{1\})$ is closed in $V$ ($f^{-1}$ of a closed set is closed when $f$ is a continuous function.)

Theorem 3(Theorem).

All norms on a finite dimensional vector space are equivalent.

Let $\Vert \cdot \Vert$ be a norm on a finite dimensional vector space $V$ over $\mathbb{R}$. We will show that $\Vert \cdot \Vert$ is equivalent to $\Vert \cdot {\Vert }_{\infty }$, and the theorem will follow from the transitivity of the equivalence of norms.

Note that the infinity norm is not unique, and depends upon a choice of basis. Fix a basis $\{{\mathbf{e}}_1,{\mathbf{e}}_2,\dots ,{\mathbf{e}}_n\}$ for $V$, and define the infinity norm with respect to this basis. Let $\mathbf{x}\in V$. Then,

$$\begin{array}{rl}\Vert \mathbf{x}\Vert & =\Vert \sum x_i{\mathbf{e}}_i\Vert \\ & \le \sum \Vert x_i{\mathbf{e}}_i\Vert \\ & =\sum |x_i|\Vert {\mathbf{e}}_i\Vert \\ & \le \Vert \mathbf{x}{\Vert }_{\infty }\sum \Vert {\mathbf{e}}_i\Vert \end{array}$$

So, we have found a constant $c_2=\sum \Vert {\mathbf{e}}_i\Vert >0$ such that $\Vert \cdot \Vert \le c_2\Vert \cdot {\Vert }_{\infty }$.

To get $c_1$, consider the unit sphere $S=\{\mathbf{y}\in V|\Vert \mathbf{y}{\Vert }_{\infty }=1\}$ (do not get mislead by the generic name, $S$ is actually a cube). We know that $S$ is closed and bounded in $(V,\Vert \cdot {\Vert }_{\infty })$. We also know that isomorphisms between vector spaces are also homeomorphisms (prove this) (In finite dimensions, any linear map between normed spaces is continuous).

Let $\varphi$ be an isomorphism from ${\mathbb{R}}^n$ to $V$. Note that $\varphi$ is continuous . Thus, ${\varphi }^{-1}(S)$ is closed in ${\mathbb{R}}^n$. Also note that ${\varphi }^{-1}(S)$ is bounded in $({\mathbb{R}}^n,\Vert \cdot {\Vert }_{\infty })$. Thus, from the Heine Borel theorem, ${\varphi }^{-1}(S)$ is compact. Since the image of a compact set under a continuous function is compact, $S$ is compact. From the extreme value theorem, the norm $\Vert \cdot \Vert$ must attain a minimum value on $S$, $c$. Now, if $c=0$, there exists $\mathbf{y}\in S$ such that $\Vert \mathbf{y}\Vert =0⟹\mathbf{y}=0⟹⟹\Vert \mathbf{y}{\Vert }_{\infty }=0$, which is a contradiction. Thus, $c>0$.

Now, we claim that $c\Vert \cdot {\Vert }_{\infty }\le \Vert \cdot \Vert$. Suppose $\mathbf{x}\in V$. If $\mathbf{x}=0$, the claim is trivially true. So, suppose $\mathbf{x}\ne 0$. Then,

$$\frac{\mathbf{x}}{\Vert \mathbf{x}{\Vert }_{\infty }}\in S$$

By the definition of $c$, we have

$$\begin{array}{r}c\le \Vert \frac{\mathbf{x}}{\Vert \mathbf{x}{\Vert }_{\infty }}\Vert \\ ⟹c\Vert \mathbf{x}{\Vert }_{\infty }\le \Vert \mathbf{x}\Vert \end{array}$$

Setting $c_1=c$ proves the theorem. ◻️