Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm

Let $\mathbb{R}$ be the ordered field of real numbers, with the notation of that item, and let $E$ and $F$ be real inner product spaces with norms $|\cdot|_{E}$ and $|\cdot|_{F}$ in the ambient notation, let $0_{F}$ be the zero vector of $F$, and for a real number $s$ let $|s|$ be its absolute value.

1. (Bounded linear maps)(i) A linear map $T:E\to F$ is bounded if there is a real number $C$ such that $|Tx|_{F}\le C\,|x|_{E}$ for every $x\in E$. We write $\mathcal{L}(E,F)$ for the set of all bounded linear maps from $E$ to $F$ and $\mathcal{L}(E)$ for $\mathcal{L}(E,E)$. We write $\mathrm{id}_{E}$ for the identity map $E\to E$, $x\mapsto x$.

2. (Operator norm)(ii) Let $T\in\mathcal{L}(E,F)$ and let $B_{T}$ be the set of real numbers $C$ with $0\le C$ such that $|Tx|_{F}\le C\,|x|_{E}$ for every $x\in E$. The set $B_{T}$ is nonempty: if $C$ is as in clause 1 and $0\le C$ then $C\in B_{T}$, while if $C<0$ then, by claim 5 of Elementary Arithmetic in an Ordered Field applied to $C\le 0$ with a nonnegative multiplier, $|Tx|_{F}\le C|x|_{E}\le 0\le 0\cdot|x|_{E}$ for every $x$, so $0\in B_{T}$. It is bounded below by $0$, so by Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below it has a greatest lower bound. The operator norm of $T$ is

$$\lVert T\rVert=\inf B_{T}.$$

3. (Bounded linear functionals)(iii) A linear functional on $E$ is a map $\ell:E\to\mathbb{R}$ with $\ell(x+y)=\ell(x)+\ell(y)$ and $\ell(\lambda x)=\lambda\,\ell(x)$ for all $x,y\in E$ and $\lambda\in\mathbb{R}$. It is bounded if there is a real number $C$ with $|\ell(x)|\le C\,|x|_{E}$ for every $x\in E$. For a bounded linear functional $\ell$ the set $B_{\ell}$ of real numbers $C$ with $0\le C$ and $|\ell(x)|\le C\,|x|_{E}$ for every $x\in E$ is nonempty and bounded below by $0$, by the argument of clause 2 with $|\ell(x)|$ in place of $|Tx|_{F}$, and the norm of $\ell$ is $\lVert\ell\rVert=\inf B_{\ell}$.