The operator norm is a norm on the space of bounded linear operators

By [L1], $\mathrm{\parallel }T\mathrm{\parallel }\ge 0$ for every $T\in \mathcal{B}(X,Y)$. If $\mathrm{\parallel }T\mathrm{\parallel }=0$, then $\mathrm{\parallel }Tx\mathrm{\parallel }\le 0$ for every $x$ with $\mathrm{\parallel }x\mathrm{\parallel }\le 1$, so $Tx=0$ there. If $x\mathrm{\ne }0$, apply this to $u:=x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }$ to get $Tu=0$, hence $Tx=\mathrm{\parallel }x\mathrm{\parallel }Tu=0$. Thus $\mathrm{\parallel }T\mathrm{\parallel }=0$ implies $T=0$.

Conversely, if $T=0$ then $\mathrm{\parallel }Tx\mathrm{\parallel }=0$ for every $x$ in the unit ball, so the supremum in [L1] is $0$. This proves definiteness.

For every $x$ with $\mathrm{\parallel }x\mathrm{\parallel }\le 1$, [L2] and [L1] give $\mathrm{\parallel }(\lambda T)x\mathrm{\parallel }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }Tx\mathrm{\parallel }\le \mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }T\mathrm{\parallel }$. Taking the supremum over the unit ball yields $\mathrm{\parallel }\lambda T\mathrm{\parallel }\le \mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }T\mathrm{\parallel }$. The reverse inequality is immediate when $\lambda =0$, and for $\lambda \mathrm{\ne }0$ the same estimate applied to $T={\lambda }^{-1}(\lambda T)$ gives $\mathrm{\parallel }T\mathrm{\parallel }\le \mathrm{\mid }\lambda {\mathrm{\mid }}^{-1}\mathrm{\parallel }\lambda T\mathrm{\parallel }$. Hence $\mathrm{\parallel }\lambda T\mathrm{\parallel }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }T\mathrm{\parallel }$.

For every $x$ with $\mathrm{\parallel }x\mathrm{\parallel }\le 1$, [L2] and [L1] give $\mathrm{\parallel }(S+T)x\mathrm{\parallel }\le \mathrm{\parallel }Sx\mathrm{\parallel }+\mathrm{\parallel }Tx\mathrm{\parallel }\le \mathrm{\parallel }S\mathrm{\parallel }+\mathrm{\parallel }T\mathrm{\parallel }$. Taking the supremum over the unit ball gives $\mathrm{\parallel }S+T\mathrm{\parallel }\le \mathrm{\parallel }S\mathrm{\parallel }+\mathrm{\parallel }T\mathrm{\parallel }$.

Steps 1.1, 1.2, 1.3, and 1.4 are the norm axioms, so the operator norm is a norm on $\mathcal{B}(X,Y)$.