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:

$\norm T \le M$.

$\blacksquare$


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