Supremum Norm is Norm/Space of Bounded Sequences

Theorem

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

Proof

Norm Axiom $\text N 1$: Positive Definiteness

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:

 $\ds \norm {\mathbf x}_\infty$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \sup_{n \mathop \in \N} \size {x_n}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds x_n$ $=$ $\ds 0$ Complex Modulus equals Zero iff Zero $\ds \leadsto \ \$ $\ds \bf x$ $=$ $\ds \sequence 0_{n \mathop \in \N}$

Thus Norm Axiom $\text N 1$: Positive Definiteness is satisfied.

$\Box$

Norm Axiom $\text N 2$: Positive Homogeneity

Suppose $\alpha \in \C$.

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

Thus Norm Axiom $\text N 2$: Positive Homogeneity is satisfied.

$\Box$

Norm Axiom $\text N 3$: Triangle Inequality

Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N}, \mathbf y = \sequence {y_n}_{n \mathop \in \N} \in \ell^\infty$.

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

Then:

 $\ds \norm {\mathbf x + \mathbf y}_\infty$ $=$ $\ds \sup_{n \mathop \in \N} \size {x_n + y_n}$ Definition of Supremum Norm $\ds$ $\le$ $\ds \sup_{n \mathop \in \N} \size {x_n} + \sup_{n \mathop \in \N} \size {y_n}$ $\ds$ $=$ $\ds \norm {\mathbf x}_\infty + \norm {\mathbf y}_\infty$ Definition of Supremum Norm

Thus Norm Axiom $\text N 3$: Triangle Inequality is satisfied.

$\blacksquare$