Proof of Properties of the Operator Norm

Throughout, $\lVert\cdot\rVert$ denotes the norm of a vector of $V$ and $\lVert\cdot\rVert_{\mathrm{op}}$ the operator norm of a bounded linear operator on $V$. We use repeatedly the following fact: 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.

Claims 1 and 2. By Existence and Uniqueness of the Operator Norm the number $\lVert T\rVert_{\mathrm{op}}$ is the operator norm of $T$, so by that definition it is a bound for $T$ and satisfies $\lVert T\rVert_{\mathrm{op}}\le C$ for every bound $C$ for $T$. That it is a bound says precisely that $0\le\lVert T\rVert_{\mathrm{op}}$ and $\lVert T(u)\rVert\le\lVert T\rVert_{\mathrm{op}}\,\lVert u\rVert$ for every $u\in V$, which is claim 1; the second property is claim 2.

Claim 3. The maps $S+T$, $\lambda T$ and $ST$ are linear operators on $V$ by Sums, Scalar Multiples, Composites and the Identity are Linear Operators.

Sum. For $u\in V$, by the definition of the sum, the triangle inequality of the norm, claim 1 applied to $S$ and to $T$, and distributivity,

Since $0\le\lVert S\rVert_{\mathrm{op}}$ and $0\le\lVert T\rVert_{\mathrm{op}}$ by claim 1, we get $0\le\lVert S\rVert_{\mathrm{op}}+\lVert T\rVert_{\mathrm{op}}$, so this number is a bound for $S+T$; in particular $S+T$ is bounded, and claim 2 applied to $S+T$ gives $\lVert S+T\rVert_{\mathrm{op}}\le\lVert S\rVert_{\mathrm{op}}+\lVert T\rVert_{\mathrm{op}}$.

Scalar multiple. For $u\in V$, by absolute homogeneity of the norm and claim 1, multiplying by the nonnegative number $|\lambda|$,

and $0\le|\lambda|\,\lVert T\rVert_{\mathrm{op}}$, so this number is a bound for $\lambda T$; hence $\lambda T$ is bounded and $\lVert\lambda T\rVert_{\mathrm{op}}\le|\lambda|\,\lVert T\rVert_{\mathrm{op}}$ by claim 2.

For the reverse inequality, first suppose $\lambda=0$. Then $\lVert(\lambda T)(u)\rVert=|0|\,\lVert T(u)\rVert=0=0\cdot\lVert u\rVert$ for every $u$, so $0$ is a bound for $\lambda T$ and claim 2 gives $\lVert\lambda T\rVert_{\mathrm{op}}\le0$; with $0\le\lVert\lambda T\rVert_{\mathrm{op}}$ from claim 1 and antisymmetry, $\lVert\lambda T\rVert_{\mathrm{op}}=0=|\lambda|\,\lVert T\rVert_{\mathrm{op}}$. Now suppose $\lambda\ne0$. Applying the inequality just proved to the scalar $\lambda^{-1}$ and the bounded operator $\lambda T$, and using $\lambda^{-1}(\lambda T)=T$, which holds because $\lambda^{-1}\bigl(\lambda T(u)\bigr)=(\lambda^{-1}\lambda)T(u)=T(u)$ by the vector space axioms,

Multiplying by the nonnegative number $|\lambda|$ and using $|\lambda|\,|\lambda^{-1}|=|\lambda\lambda^{-1}|=|1|=1$, by claim 4 and claim 8 of Properties of Complex Conjugation and Modulus, gives $|\lambda|\,\lVert T\rVert_{\mathrm{op}}\le\lVert\lambda T\rVert_{\mathrm{op}}$. Antisymmetry now yields $\lVert\lambda T\rVert_{\mathrm{op}}=|\lambda|\,\lVert T\rVert_{\mathrm{op}}$.

Product. For $u\in V$, applying claim 1 to $S$ at the vector $T(u)$, and then multiplying $\lVert T(u)\rVert\le\lVert T\rVert_{\mathrm{op}}\,\lVert u\rVert$ by the nonnegative number $\lVert S\rVert_{\mathrm{op}}$,

Since $0\le\lVert S\rVert_{\mathrm{op}}\,\lVert T\rVert_{\mathrm{op}}$ by claim 1 and the second order axiom, this number is a bound for $ST$; hence $ST$ is bounded and claim 2 gives $\lVert ST\rVert_{\mathrm{op}}\le\lVert S\rVert_{\mathrm{op}}\,\lVert T\rVert_{\mathrm{op}}$.