Supremum Norm is Norm/Space of Bounded Sequences

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

By Triangle Inequality for Complex Numbers:

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

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.

$\blacksquare$


Sources

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