Supremum Norm

Last updated Nov 6, 2022

Let $(X_{i}, ||\cdot||{i}){i \in I}$ be Normed Vector Spaces over Field $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$ with Index Set $I$ and let $X = \prod\limits_{i \in I}^{} X_{i}$. Then the function $||\cdot||: X \to \overline{\mathbb{R}{\geq 0}}$ defined $$||x|| := \sup\limits{i \in I} ||x_{i}||{i}$$ (where $||x|| = \infty$ if $||x{i}||_{i}$ is unbounded) is an Extended Norm over the Product Vector Space $X$. It is known as the Supremum Norm.

First note that by Completeness of the Real Numbers, the Supremum Norm is well-defined. We check the Axioms of an Extended Norm:

1. Suppose $x \in X$ so that $||x|| = 0$. Then $||x_{i}||{i} \leq 0$ for all $i \in I$. But $||x{i}||{i} \geq 0$ already so $||x{i}||{i} = 0$ for all $i \in I$. Since $||\cdot||{i}$ is a Norm, $x_{i} = 0$ for all $i \in I$. Thus, $x = 0$. $\checkmark$
2. Suppose $\alpha \in \mathbb{K}$. Then $$||\alpha x|| = \sup\limits_{i \in I} ||\alpha x_{i}||{i} = \sup\limits{i \in I} |\alpha| || x_{i}||{i} = |\alpha|\sup\limits{i \in I}||x_{i}||_{i}= |\alpha| ||x||. \checkmark$$
3. Suppose $x, y \in X$. Then $$\begin{align*} ||x + y|| &= \sup\limits_{i \in I} ||x_{i} + y_{i}||{i} \\ &\leq \sup\limits{i \in I} (||x_{i}||{i} + ||y{i}||{i}) \\ &\leq \sup\limits{i \in I} ||x_{i}||{i} + \sup\limits{i \in I} ||y_{i}||_{i} \\ &= ||x + y||. \checkmark \end{align*}$$

$\blacksquare$

• Non-negative Multiplication Commutes with Supremum
• Sum of Supremum bounds Supremum of Sum