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 $S = \set {\norm {T x} : x \in X : \norm x \le 1}$
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
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$