Norm Equivalence

Definition: Let $X$ be a linear space. Two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ defined on $X$ are said to be Equivalent if there exists constants $C, D > 0$ such that $C \| x \|_1 \leq \| x \|_2 \leq D \| x \|_1$ for all $x \in X$.

When two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on $X$ are equivalent, the resulting normed spaces $(X, \| \cdot \|_1)$ and $(X, \| \cdot \|_2)$ have the same topology as noted in the following proposition.

Lemma 1: Let $(X, d_1)$ and $(X, d_2)$ be metric spaces and let $B_1(x, r)$ be the open ball centered at $x$ with radius $r$ (with respect to the metric $d_1$) and let $B_2(x, r)$ be the open ball centered at $x$ with radius $r$ (with respect to the metric $d_2$). If for all $\epsilon > 0$ there exists a $\delta > 0$ such that $B_1(x, \delta) \subseteq B_2(x, \epsilon)$ and for all $\epsilon > 0$ there exists a $\delta > 0$ such that $B_2(x, \delta) \subseteq B_1(x, \epsilon)$ then $U$ is open in $(X, d_1)$ if and only if $U$ is open in $(X, d_2)$. That is, $(X, d_1)$ and $(X, d_2)$ have the same topologies.

• Proof: Let $U$ be open in $(X, d_1)$. For $x \in U$ there exists an $\epsilon > 0$ such that $B_1(x, \epsilon) \subseteq U$. But then by the hypothesis there exists a $\delta > 0$ such that $B_2(x, \delta) \subseteq B_1(x, \epsilon) \subseteq U$. So $U$ is open in $(X, d_2)$.

• Similarly, if $U$ is open in $(X, d_2)$. For $x \in U$ there exists an $\epsilon > 0$ such that $B_2(x, \epsilon) \subseteq U$. But then by the hypothesis there exists a $\delta > 0$ such that $B_1(x, \delta) \subseteq B_2(x, \epsilon) \subseteq U$. So $U$ is open in $(X, d_1)$. $\blacksquare$

Proposition 2: Let $X$ be a linear space and let $\| \cdot \|_1$ and $\| \cdot \|_2$ be equivalent norms on $X$. Then the topologies on $(X, \| \cdot \|_1)$ and $(X, \| \cdot \|_2)$ are the same.

• Proof: Let $B_1(x, r) = \{ y \in X : \| x - y \|_1 < r \}$ and let $B_2(x, r) = \{ y \in X : \| x - y \|_2 < r \}$. Note that $B_1(x, r)$ is simply the open ball (with respect to the $1$-norm) centered at $x$ with radius $r$ and $B_2(x, r)$ is simply the open ball (with respect to the $2$-norm) centered at $x$ with radius $r$.

• We will show two things.

• First we will show that for all $\epsilon > 0$ there exists a $\delta > 0$ such that:

(1)

\begin{align} \quad B_1(x, \delta) \subseteq B_2(x, \epsilon) \end{align}

• And secondly we will show that for all $\epsilon > 0$ there exists a $\delta > 0$ such that:

(2)

\begin{align} \quad B_2(x, \delta) \subseteq B_1(x, \epsilon) \end{align}

• Now since $\| \cdot \|_1$ and $\| \cdot \|_2$ are equivalent norms, there exists positive numbers $C, D > 0$ such that for all $x \in X$ we have that:

(3)

\begin{align} \quad C \| x \|_1 \leq \| x \|_2 \leq D \| x \|_1 \end{align}

• Let $\epsilon > 0$ be given. Let $\delta = \frac{\epsilon}{D}$. Then:

(4)

\begin{align} \quad B_1(x, \delta) \subseteq B_2(x, \epsilon) \end{align}

• To see this, let $y \in B_1(x, \delta)$. Then $\| x - y \|_1 < \delta = \frac{\epsilon}{D}$. So $D \| x - y \|_1 < \epsilon$. But $\| x - y \|_2 \leq D \| x - y \|_1$. So $\| x - y \|_2 < \epsilon$. So $y \in B_2(x, \epsilon)$.

• Again, let $\epsilon > 0$ be given. Let $\delta = C \epsilon$. Then:

(5)

\begin{align} \quad B_2(x, \delta) \subseteq B_1(x, \epsilon) \end{align}

• To see this, let $y \in B_2(x, \delta)$. Then $\| x - y \|_2 < \delta = C \epsilon$. Since $C \| x - y \|_1 \leq \| x - y \|_2$ we have that $C \| x - y \|_1 < C \epsilon$. So $\| x - y \|_1 < \epsilon$. So $y \in B_1(x, \epsilon)$.

• Therefore the topologies on $(X, \| \cdot \|_1)$ and $(X, \| \cdot \|_2)$ are the same. $\blacksquare$