> [!definition] > > Let $(V, ||\cdot||)$ be a [[Normed Vector Space|normed vector space]] over $\mathbb{C}$. $(V, ||\cdot||)$ is a **Banach** space if the induced [[Metric Space|metric]] yields a [[Complete Metric Space|complete]] metric space. > [!theorem] > > Let $X$ be a vector space over $\complex$ and $\norm{\cdot}_1, \norm{\cdot}_2$ be norms on $X$ such that $\norm{\cdot}_1 \le \norm{\cdot}_2$. If $X$ is complete with respect to both of them, then they are equivalent. > > *Proof*. Denote $(X, \norm{\cdot}_2)$ with $\cx$ and $(X, \norm{\cdot}_1)$ with $\cy$, and let $T: \cx \to \cy$ by $x \mapsto x$. Then $T$ is bijective and continuous as > $ > \norm{Tx}_1 \le \norm{x}_2 \quad \forall x \in X > $ > By the [[Open Mapping Theorem]], $T$ is an isomorphism, meaning that there exists $C > 0$ such that > $ > \norm{T^{-1}(x)}_2 \le C\norm{x}_1 \quad \forall x \in X > $ > Chaining the inequality yields that $\norm{\cdot}_1 \sim \norm{\cdot}_2$.