Norm Equivalence is Equivalence

Theorem

Let $X$ be a vector space.

Let $\norm {\, \cdot \,}_a$ and $\norm {\, \cdot \,}_b$ be equivalent norms on $X$.

Denote this relation by $\sim$:

$\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$.

Then $\sim$ is an equivalence relation.

Proof

Reflexivity

Let $\norm {\, \cdot \,}$ be a norm on $X$.

Then for all $x \in X$ we have that:

$\norm x = 1 \cdot \norm {x}$

Therefore:

$1 \cdot \norm x \le \norm x \le 1 \cdot \norm x$

Hence:

$\norm {\, \cdot \,} \sim \norm {\, \cdot \,}$.

$\Box$

Symmetry

Suppose, $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$.

Then:

$\exists m, M \in \R_{> 0} : m \le M : \forall x \in X : m \norm x_b \le \norm x_a \le M \norm x_b$

Consider two inequalities, obtained by division by $M$ and $m$:

$\dfrac m M \norm x_b \le \dfrac 1 M \norm x_a \le \norm x_b$

$\norm x_b \le \dfrac 1 m \norm x_a \le \dfrac M m \norm x_b$

Notice, that:

$\dfrac 1 M \norm x_a \le \norm x_b \le \dfrac 1 m \norm x_a$

We have that $m \le M$ implies $\dfrac 1 m \ge \dfrac 1 M$.

Define $C := \dfrac 1 m$ and $c := \dfrac 1 M$.

Hence, $c \le C$ and:

$\norm {\, \cdot \,}_b \sim \norm {\, \cdot \,}_a$

$\Box$

Transitivity

Suppose, $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$ and $\norm {\, \cdot \,}_b \sim \norm {\, \cdot \,}_c$.

Then:

$\exists m_{a b}, M_{a b} \in \R_{> 0} : m_{a b} \le M_{a b} : \forall x \in X : m_{a b} \norm x_b \le \norm x_a \le M_{a b} \norm x_b$

$\exists m_{b c}, M_{b c} \in \R_{> 0} : m_{b c} \le M_{b c} : \forall x \in X : m_{b c} \norm x_c \le \norm x_b \le M_{b c} \norm x_c$

Generate $2$ more inequalities by multiplying the second inequality by $m_{a b}$ and $M_{a b}$:

$m_{a b} m_{b c} \norm x_c \le m_{a b} \norm x_b \le m_{a b} M_{b c} \norm x_c$

$M_{a b} m_{b c} \norm x_c \le M_{a b} \norm x_b \le M_{a b} M_{b c} \norm x_c$

From above it follows that:

\(\ds m_{a b} m_{b c} \norm x_c\)

\(\le\)

\(\ds m_{a b} \norm x_b\)

as shown above

\(\ds \)

\(\le\)

\(\ds \norm x_a\)

by hypothesis

\(\ds \)

\(\le\)

\(\ds M_{a b} \norm x_b\)

by hypothesis

\(\ds \)

\(\le\)

\(\ds M_{a b} M_{b c} \norm x_c\)

as shown above

Define $k := m_{a b} m_{b c}$ and $K := M_{a b} M_{b c}$.

Then $k \le K$ and:

$k \norm x_c \le \norm x_a \le K \norm x_c$

Therefore:

$\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_c$

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces