Supremum Operator Norm is Well-Defined

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

By definition of the norm:


 * $S \subseteq \R$

$S$ is non-empty
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:

and:

Thus, by definition of $S$,


 * $0 \in S$

and


 * $S \ne \empty$

$S$ is bounded above
Let $\map {CL} {X, Y}$.

Then:

Since $\exists 0_X \in X: \norm 0_X \le 1$, let $x \in X : \norm x \le 1$.

By continuity of linear transformations on normed vector spaces:

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

Hence:


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