Supremum Operator Norm is Well-Defined

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Theorem

Let $K$ be a field.

Let $X, Y$ be normed vector spaces over $K$.

Let $\map {CL} {X, Y}$ be the continuous linear transformation space.

Let $\norm {\, \cdot \,}$ be an operator norm on $\map {CL} {X, Y}$ defined as:

$\norm T := \map \sup {\norm {Tx} : x \in X : \norm x \le 1}$

Then $\norm {\, \cdot \,}$ is well-defined.


Proof

Let $S = \set {\norm {T x} : x \in X : \norm x \le 1}$

By definition of the norm:

$S \subseteq \R$

$S$ is non-empty


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By definition, $X$ is a normed vector space, and thus a vector space.

Hence, by Zero Vector is Unique, the zero vector exists in $X$, and $X$ is non-empty.

Let $x = \mathbf 0_X \in X$ be the zero vector.

Then:


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\(\ds \norm x\) \(=\) \(\ds \norm {\mathbf 0_X}\) by hypothesis
\(\ds \) \(=\) \(\ds 0\) Norm Axioms (Vector Space)
\(\ds \) \(\le\) \(\ds 1\) Definition of Usual Ordering

and:

\(\ds \norm {T x}\) \(=\) \(\ds \norm {T \mathbf 0_X}\) by hypothesis
\(\ds \) \(=\) \(\ds \norm {\mathbf 0_Y}\) Linear Transformation Maps Zero Vector to Zero Vector
\(\ds \) \(=\) \(\ds 0\) Norm Axioms (Vector Space)

Thus, by definition of $S$:

$0 \in S$

and:

$S \ne \O$

$\Box$


$S$ is bounded above

Let $\map {CL} {X, Y}$.

Because:

$\exists 0_X \in X: \norm 0_X \le 1$


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let $x \in X : \norm x \le 1$.

Then:

\(\ds \exists M \in \R_{>0} : \forall x \in X: \, \) \(\ds \norm {T x}\) \(\le\) \(\ds M \norm x\) Continuity of Linear Transformation between Normed Vector Spaces
\(\ds \) \(\le\) \(\ds M \cdot 1\) by hypothesis
\(\ds \) \(=\) \(\ds M\) Definition of Multiplicative Identity

By the least upper bound property of $\R$, a supremum of $S$ exists.

$\Box$

Hence:

$\forall T \in \map {CL} {X, Y} : \norm T := \sup \set {\norm {Tx} : x \in X : \norm x \le 1} < \infty$

$\blacksquare$


Sources

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