# Equivalence of Norms on Banach Spaces

Recall from The Open Mapping Theorem that if $X$ and $Y$ are Banach spaces and if $T : X \to Y$ is a bounded linear operator then the range $T(X)$ is a closed subspace of $Y$ if and only if $T$ is an open map.

Also recall that two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ defined on a linear space $X$ are said to be equivalence if there exists $C, D > 0$ such that for all $x \in X$ we have that:

(1)(On the Equivalence of Norms in a Finite-Dimensional Linear Space page we proved that any two norms defined on finite-dimensional normed linear spaces must be equivalence, so all interesting cases of equivalence norms on normed linear spaces arise from infinite-dimensional normed linear spaces)

As a consequence of the open mapping theorem we will see that when $X$ is a Banach space with respect to two different norms then determining whether those two norms are equivalent is simpler.

Corollary 1: Let $\| \cdot \|_1$ and $\| \cdot \|_2$ be two norms defined on $X$ and let $X$ be a Banach space with respect to both of these norms. Suppose that there exists a constant $C \in \mathbb{R}$, $C > 0$ such that for all $x \in X$ we have that $\| x \|_2 \leq C \| x \|_1$. Then $\| \cdot \|_1$ and $\| \cdot \|_2$ are equivalent. |

**Proof:**Let $T : X \to X$ be defined for all $x \in X$ by $T(x) = x$. That is, $T$ is the identity function on $X$. It is easy to verify that $T$ is a linear operator and that $T$ is bijective. Furthermore, for all $x \in X$ we have that:

- So $T$ is a bounded linear operator whose domain ($X$) is a Banach space and whose codomain (also $X$) is a Banach space. Also note that $T(X) = X$ is (trivially) a closed subspace of $X$. By the open mapping theorem we have that $T$ is an open map. Therefore $T^{-1}$ is continuous and hence bounded. So for every $x \in X$:

- Hence, for every $x \in X$:

- Finally, for every $x \in X$:

- By setting $\displaystyle{D = \frac{1}{\| T^{-1} \|} > 0}$ we see that $\| \cdot \|_1$ and $\| \cdot \|_2$ are equivalent. $\blacksquare$