Proof of Existence and Uniqueness of the Operator Norm

Since $T$ is bounded, fix a real number $C_{0}$ with $0\le C_{0}$ that is a bound for $T$. Write $0_{V}$ for the zero vector. We record three facts. By that lemma and the absolute homogeneity condition of the norm, together with $|0|=0$ from claim 3 of Properties of Complex Conjugation and Modulus, we have $\lVert 0_{V}\rVert=\lVert 0\cdot 0_{V}\rVert=|0|\,\lVert 0_{V}\rVert=0$, and by linearity $T(0_{V})=T(0\cdot 0_{V})=0\,T(0_{V})=0_{V}$. Next, if $x,y,c$ are real numbers with $x\le y$ and $0\le c$, then $0\le c(y-x)=cy-cx$ by the second order axiom of Ordered Field, so $cx\le cy$; we call this multiplying an inequality by a nonnegative number. Finally $0\le1$, since $1=1\cdot1$ and by totality either $0\le1$, or $1\le0$ and then $0\le-1$ so that $0\le(-1)(-1)=1$.

Step 1: $B_{T}$ is nonempty and every bound for $T$ is an upper bound for $B_{T}$. Since $\lVert 0_{V}\rVert=0\le1$, the number $\lVert T(0_{V})\rVert=\lVert 0_{V}\rVert=0$ lies in $B_{T}$, so $B_{T}$ is nonempty. Let $C$ be a bound for $T$ and let $u\in V$ with $\lVert u\rVert\le1$. Multiplying $\lVert u\rVert\le1$ by the nonnegative number $C$ gives $C\lVert u\rVert\le C$, and $\lVert T(u)\rVert\le C\lVert u\rVert$, so $\lVert T(u)\rVert\le C$ by transitivity. Hence $C$ is an upper bound for $B_{T}$. In particular $C_{0}$ is, so $B_{T}$ is nonempty and bounded above and therefore has a least upper bound $\sigma$ by the least upper bound property in The Real Numbers.

Step 2: $0\le\sigma$. The number $0$ lies in $B_{T}$ by step 1 and $\sigma$ is an upper bound for $B_{T}$.

Step 3: $\sigma$ is a bound for $T$. Let $u\in V$. If $u=0_{V}$, both $\lVert T(u)\rVert$ and $\sigma\lVert u\rVert$ are $0$. Suppose $u\ne0_{V}$. By the positivity condition of the norm $\lVert u\rVert\ne0$, so it has an inverse $r=\lVert u\rVert^{-1}$ in the field of real numbers. If $r\le0$ then $0\le-r$, and with $0\le\lVert u\rVert$ the second order axiom gives $0\le(-r)\lVert u\rVert=-1$, hence $1\le0$; together with $0\le1$ this forces $1=0$, which is false. So $0\le r$, and by absolute homogeneity together with $|r|=r$ (claim 8 of Properties of Complex Conjugation and Modulus) we get $\lVert ru\rVert=r\lVert u\rVert=1$. Hence $\lVert T(ru)\rVert\in B_{T}$, so $\lVert T(ru)\rVert\le\sigma$. By linearity and absolute homogeneity $\lVert T(ru)\rVert=r\lVert T(u)\rVert$, so multiplying $r\lVert T(u)\rVert\le\sigma$ by the nonnegative number $\lVert u\rVert$ and using $\lVert u\rVert\,r=1$ gives $\lVert T(u)\rVert\le\sigma\lVert u\rVert$. With step 2 this shows $\sigma$ is a bound for $T$.

Step 4: $\sigma$ is an operator norm of $T$. By step 3 it is a bound for $T$, and by step 1 every bound $C$ for $T$ is an upper bound for $B_{T}$, so $\sigma\le C$ because $\sigma$ is the least upper bound. These are exactly the two conditions of Operator Norm.

Step 5: uniqueness. Let $c$ and $c'$ both be operator norms of $T$. Each is a bound for $T$, and each is at most every bound for $T$, so $c\le c'$ and $c'\le c$; antisymmetry of the order gives $c=c'$. Combined with step 4, $T$ has exactly one operator norm and it is the least upper bound $\sigma$ of $B_{T}$.