The operator norm as the least bound and as the unit-sphere or unit-ball supremum

Let $X$ and $Y$ be normed spaces over the same scalar field, and let $T:X\to Y$ be bounded in the sense of A bounded linear operator between normed spaces. The operator norm of $T$ is

$\mathrm{\parallel }T\mathrm{\parallel }:=\sup \{\mathrm{\parallel }Tx\mathrm{\parallel }:x\in X,\mathrm{\parallel }x\mathrm{\parallel }\le 1\}\mathrm{.}$

The set on the right contains $0$ and every bound $C$ for $T$ bounds it above. Thus its finite real supremum exists by the proved least-upper-bound property of the library's real field (The Cauchy-sequence reals have the least-upper-bound property). In particular every number strictly below the supremum is exceeded by a member of this set; otherwise it would be a smaller upper bound. No choice principle is used for this assertion.

The same number is the least bound of $T$:

$\mathrm{\parallel }T\mathrm{\parallel }=\inf \{C\ge 0:\mathrm{\parallel }Tx\mathrm{\parallel }\le C\mathrm{\parallel }x\mathrm{\parallel }\text{ for all }x\in X\}\mathrm{.}$

If $X\mathrm{\ne }\{0\}$, positive homogeneity also gives

$\mathrm{\parallel }T\mathrm{\parallel }=\sup \{\mathrm{\parallel }Tx\mathrm{\parallel }:\mathrm{\parallel }x\mathrm{\parallel }=1\}\mathrm{.}$

When $X=\{0\}$, the unit sphere is empty and the unit-ball supremum is $0$, so $\mathrm{\parallel }T\mathrm{\parallel }=0$.

• The inequality $\mathrm{\parallel }Tx\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$ holds for every $x\in X$ by the unit-ball definition and rescaling.
• The unit-ball formula is the one used uniformly below, because it also covers the zero-space case without a separate convention.