Supremum Norm is Norm/Space of Bounded Sequences

Theorem
The supremum norm on the space of bounded sequences is a norm.

Norm Axiom $(\text N 1)$
Let $x \in \ell^\infty$.

By definition of supremum norm:


 * $\ds \norm {\mathbf x}_\infty = \sup_{n \mathop \in \N} \size {x_n}$

The complex modulus of $x_n$ is real and non-negative.

Hence, $\norm {\mathbf x}_\infty \ge 0$.

Suppose $\norm {\mathbf x}_\infty = 0$.

Then:

Thus norm axiom $(\text N 1)$ is satisfied.

Norm Axiom $(\text N 2)$
Suppose $\alpha \in \C$.

Norm Axiom $(\text N 3)$
Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N}, \mathbf y = \sequence {y_n}_{n \mathop \in \N} \in \ell^\infty$.

By Triangle Inequality for Complex Numbers:


 * $\forall n \in \N : \size {x_n + y_n} \le \size {x_n} + \size {y_n}$

Then: