Universal Upper Bound greater than Supremum Operator Norm
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Theorem
Let $\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $T \in \map {CL} {X, Y}$.
Let $\norm {\, \cdot \,}$ be an operator norm on $\map {CL} {X, Y}$ defined as:
- $\norm T := \sup \set {\norm {Tx}_Y : x \in X : \norm x_X \le 1}$
Suppose:
- $\exists M \in \R_{> 0} : \forall x \in X : \norm {T x}_Y \le M \norm x_X$
Then:
- $\norm {T} \le M$
Proof
Let $x \in X$ be such that $\norm x_X \le 1$.
Then:
\(\ds \norm {T x}_Y\) | \(\le\) | \(\ds M \norm x_X\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds M \cdot 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds M\) |
Hence, $M$ is an upper bound for $S = \set {\norm {T x}_Y : x \in X : \norm x_X \le 1}$.
By definition of the supremum of subset of real numbers:
- $\sup S \le M$.
That is:
- $\norm T \le M$.
$\blacksquare$
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 $\map {CL} {X, Y}$