Proof of Properties of Unitary Operators

Write $0_{V}$ for the zero vector of $V$ and $u-v$ for $u+(-v)$; 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 $V$. By the definition of a unitary operator, $T$ is a linear operator on $V$ that is surjective and satisfies $\langle T(u),T(v)\rangle=\langle u,v\rangle$ for all $u,v\in V$.

Claim 1. Let $u\in V$. By the definition of the induced norm and the inner-product preservation just quoted,

Both $\lVert T(u)\rVert$ and $\lVert u\rVert$ are real numbers that are nonnegative and have equal squares, so they are equal by the uniqueness part of Existence and Uniqueness of the Nonnegative Square Root.

Claim 2. Surjectivity of $T$ is part of the definition of a unitary operator. For injectivity, suppose $u,u'\in V$ satisfy $T(u)=T(u')$. By linearity and Elementary Identities in a Vector Space,

so claim 1 gives $\lVert u-u'\rVert=\lVert T(u-u')\rVert=\lVert 0_{V}\rVert=0$, where $\lVert 0_{V}\rVert=\lVert 0\cdot 0_{V}\rVert=|0|\,\lVert 0_{V}\rVert=0$ by the absolute homogeneity condition of the norm. By the positivity condition of the norm, $u-u'=0_{V}$, that is $u=u'$. Hence $T$ is injective, and being also surjective it is a bijection from $V$ onto $V$.

Claim 3. By claim 1, $\lVert T(u)\rVert=\lVert u\rVert=1\cdot\lVert u\rVert$ for every $u\in V$, and $0\le1$ in the ordered field of real numbers. Hence the number $1$ is a bound for $T$, so $T$ is a bounded linear operator on $V$. Consequently $T$ has exactly one operator norm by Existence and Uniqueness of the Operator Norm, and since $1$ is a bound for $T$, claim 2 of Properties of the Operator Norm gives $\lVert T\rVert_{\mathrm{op}}\le1$.