Universal Upper Bound greater than Supremum Operator Norm

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:

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:


 * $\norm T \le M$.