The Bounded Symmetric Operator Represented by a Bounded Symmetric Bilinear Form on a Real Hilbert Space

In the setting of Real Hilbert Spaces: Standing Notation and Background, let $H$ be a real Hilbert space, with its inner product $\langle\cdot,\cdot\rangle$, norm $|\cdot|$ and zero vector $0_{H}$ as fixed there. Let $\mathrm{Sym}(H)$ be the set of bounded symmetric bilinear forms on $H$, with its norm $\lVert\cdot\rVert$, its sums and scalar multiples, and the identity form $I=I_{H}$, its multiples $cI$ and the zero form $0_{\mathrm{Sym}}$, as fixed there. Let $\mathcal{L}(H)$ be the set of bounded linear maps from $H$ to itself, with the operator norm $\lVert\cdot\rVert$ and the identity map $\mathrm{id}_{H}$ as fixed there; for $S,T\in\mathcal{L}(H)$ and $\lambda\in\mathbb{R}$, $S+T$ and $\lambda T$ denote the maps $x\mapsto Sx+Tx$ and $x\mapsto\lambda\,Tx$, which lie in $\mathcal{L}(H)$ by Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces (iv).

Then the following hold.

1. (Representation)(i) Let $b\in\mathrm{Sym}(H)$. There is exactly one map $T_{b}:H\to H$ such that

$$b(x,y)=\langle T_{b}x,y\rangle\qquad\text{for all }x,y\in H,$$

and this map belongs to $\mathcal{L}(H)$.

2. (Symmetry)(ii) For $b\in\mathrm{Sym}(H)$ and all $x,y\in H$, $\langle T_{b}x,y\rangle=\langle x,T_{b}y\rangle$.

3. (Norm)(iii) For $b\in\mathrm{Sym}(H)$, $\lVert T_{b}\rVert=\lVert b\rVert$.

4. (Linearity of the correspondence)(iv) For all $b_{1},b_{2}\in\mathrm{Sym}(H)$ and all $\lambda,c\in\mathbb{R}$:

$$T_{b_{1}+b_{2}}=T_{b_{1}}+T_{b_{2}},\qquad T_{\lambda b_{1}}=\lambda\,T_{b_{1}},\qquad T_{I}=\mathrm{id}_{H},$$

the map $T_{0_{\mathrm{Sym}}}$ sends every $x\in H$ to $0_{H}$, and $T_{cI}$ sends every $x\in H$ to $c\,x$.

5. (Symmetric bounded operators arise this way)(v) Let $T\in\mathcal{L}(H)$ satisfy $\langle Tx,y\rangle=\langle x,Ty\rangle$ for all $x,y\in H$. Then the map $b_{T}$ assigning the real number $\langle Tx,y\rangle$ to each pair $x,y\in H$ belongs to $\mathrm{Sym}(H)$, and $T_{b_{T}}=T$.