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

Throughout we use that $0\le\lVert b\rVert$ for every $b\in\mathrm{Sym}(H)$ and $0\le\lVert T\rVert$ for every $T\in\mathcal{L}(H)$: by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §norm and Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm §operator-norm these numbers are greatest lower bounds of sets of nonnegative real numbers, of which $0$ is therefore a lower bound. We also use the symmetry $\langle x,y\rangle=\langle y,x\rangle$ of Real Inner Product Space §inner-product without comment.

Claim 1. Let $b\in\mathrm{Sym}(H)$ and fix $x\in H$. The map $\ell_{x}:H\to\mathbb{R}$ given by $\ell_{x}(y)=b(x,y)$ is a linear functional: for $y,y'\in H$ and $\lambda\in\mathbb{R}$, the symmetry, additivity and homogeneity of Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §form give

It is bounded, since $|b(x,y)|\le\lVert b\rVert\,|x|\,|y|$ by Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity §bound, so that the constant $\lVert b\rVert\,|x|$ serves in Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm §functional. By The Riesz Representation Theorem for a Real Hilbert Space §existence there is $z\in H$ with $b(x,y)=\langle y,z\rangle=\langle z,y\rangle$ for every $y\in H$, and by The Riesz Representation Theorem for a Real Hilbert Space §uniqueness such a $z$ is unique. Setting $T_{b}x=z$ for each $x\in H$ defines a map $T_{b}:H\to H$ with $b(x,y)=\langle T_{b}x,y\rangle$ for all $x,y\in H$. If $T:H\to H$ is any map with this property, then for each $x$ we have $\langle y,Tx\rangle=\langle y,T_{b}x\rangle$ for every $y\in H$, so $Tx=T_{b}x$ by The Riesz Representation Theorem for a Real Hilbert Space §uniqueness; the map is therefore unique.

The map $T_{b}$ is linear: for $x,x',y\in H$ and $\lambda\in\mathbb{R}$, Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §form and conditions (b) and (c) of Real Inner Product Space §inner-product give

for every $y\in H$, whence $T_{b}(x+x')=T_{b}x+T_{b}x'$ and $T_{b}(\lambda x)=\lambda\,T_{b}x$ by The Riesz Representation Theorem for a Real Hilbert Space §uniqueness. It is bounded: for $x\in H$,

by Real Inner Product Space §norm, claim 3 of Properties of the Absolute Value in an Ordered Field and Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity §bound. If $T_{b}x=0_{H}$ then $|T_{b}x|=0\le\lVert b\rVert\,|x|$, the last inequality by claim 5 of Elementary Arithmetic in an Ordered Field. Otherwise $|T_{b}x|$ is positive by Elementary Identities in a Real Inner Product Space §vanishing, and multiplying the displayed inequality by the nonnegative number $|T_{b}x|^{-1}$ (claim 5 of Elementary Arithmetic in an Ordered Field, with claim 7 of Elementary Order Arithmetic in an Ordered Field for its positivity) gives $|T_{b}x|\le\lVert b\rVert\,|x|$. In either case $T_{b}\in\mathcal{L}(H)$ by Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm §bounded.

Claim 2. For $x,y\in H$, using claim 1 and the symmetry of Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §form,

Claim 3. The bound $|T_{b}x|\le\lVert b\rVert\,|x|$ established in claim 1 holds for every $x\in H$, and $0\le\lVert b\rVert$, so Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces §bound gives $\lVert T_{b}\rVert\le\lVert b\rVert$. Conversely, for all $x,y\in H$, The Cauchy-Schwarz Inequality in a Real Inner Product Space, Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces §bound and claim 5 of Elementary Arithmetic in an Ordered Field give

so $\lVert b\rVert\le\lVert T_{b}\rVert$ by Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity §bound, since $0\le\lVert T_{b}\rVert$. The order of $\mathbb{R}$ is antisymmetric, being a total order, so $\lVert T_{b}\rVert=\lVert b\rVert$.

Claim 4. By Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §identity and condition (b) of Real Inner Product Space §inner-product, for all $x,y\in H$,

Since $T_{b_{1}}+T_{b_{2}}$ is a map from $H$ to $H$ representing $b_{1}+b_{2}$ in the sense of claim 1, the uniqueness there gives $T_{b_{1}+b_{2}}=T_{b_{1}}+T_{b_{2}}$. The same argument, with condition (c) of Real Inner Product Space §inner-product in place of (b), gives $T_{\lambda b_{1}}=\lambda\,T_{b_{1}}$. Furthermore $I(x,y)=\langle x,y\rangle=\langle\mathrm{id}_{H}x,y\rangle$ and $(cI)(x,y)=c\,\langle x,y\rangle=\langle c\,x,y\rangle$ by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §identity and condition (c), while $0_{\mathrm{Sym}}(x,y)=0=\langle 0_{H},y\rangle$ by Elementary Identities in a Real Inner Product Space §zero; the uniqueness in claim 1 gives $T_{I}=\mathrm{id}_{H}$, $T_{cI}x=c\,x$ and $T_{0_{\mathrm{Sym}}}x=0_{H}$ for every $x\in H$.

Claim 5. The map $b_{T}$ is symmetric, since $b_{T}(x,y)=\langle Tx,y\rangle=\langle x,Ty\rangle=\langle Ty,x\rangle=b_{T}(y,x)$ by hypothesis. It is additive and homogeneous in its first argument, because $T$ is a linear map and the inner product satisfies conditions (b) and (c) of Real Inner Product Space §inner-product. It is bounded, since The Cauchy-Schwarz Inequality in a Real Inner Product Space, Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces §bound and claim 5 of Elementary Arithmetic in an Ordered Field give $|\langle Tx,y\rangle|\le|Tx|\,|y|\le\lVert T\rVert\,|x|\,|y|$ for all $x,y\in H$. Hence $b_{T}\in\mathrm{Sym}(H)$ by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §form. Since $b_{T}(x,y)=\langle Tx,y\rangle$ for all $x,y\in H$, the map $T$ represents $b_{T}$ in the sense of claim 1, so $T_{b_{T}}=T$ by the uniqueness there.