Proof of The Doubling Form on the Product of a Real Hilbert Space with Itself

Write $z=(x_{1},x_{2})$ and $w=(y_{1},y_{2})$ for elements of $H\times H$, and recall from the definition of the product that the operations are componentwise and that $\langle z,w\rangle=\langle x_{1},y_{1}\rangle+\langle x_{2},y_{2}\rangle$.

Claim 1. The map $(x_{1},x_{2})\mapsto x_{1}-x_{2}$ from $H\times H$ to $H$ is linear, by the componentwise operations and the vector space axioms of $H$. Hence $b$ is additive and homogeneous in its first argument by conditions (b) and (c) of Real Inner Product Space §inner-product, and in its second argument by Elementary Identities in a Real Inner Product Space §bilinear; and $b(z,w)=b(w,z)$ by condition (a), the symmetry of $\langle\cdot,\cdot\rangle$.

For the bound, note first that

by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle together with $|-x_{2}|=|x_{2}|$, which holds by Elementary Identities in a Real Inner Product Space §homogeneity. Writing $p=|x_{1}|$ and $q=|x_{2}|$, we have $0\le(p-q)^{2}=p^{2}-pq-pq+q^{2}$, hence $pq+pq\le p^{2}+q^{2}$ and therefore

using claim 2 of Properties of the Product of Two Real Inner Product Spaces in the last step. As $|x_{1}-x_{2}|$ and $p+q$ are nonnegative, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $|x_{1}-x_{2}|^{2}\le(p+q)^{2}\le2|z|^{2}$, and likewise $|y_{1}-y_{2}|^{2}\le2|w|^{2}$. By The Cauchy-Schwarz Inequality in a Real Inner Product Space,

and since $|b(z,w)|$ and $2|z||w|$ are nonnegative, $|b(z,w)|\le2\,|z|\,|w|$. Hence $b$ is a bounded symmetric bilinear form on $H\times H$, and $\lVert b\rVert\le2$ because $\lVert b\rVert$ is the least constant with this property, by the definition of the norm of a form.

For the representing operator, expand using Elementary Identities in a Real Inner Product Space §bilinear:

the last equality by the definition of the product pairing. Since $T_{b}$ is the unique map with $b(z,w)=\langle T_{b}z,w\rangle$ for all $z,w$, by The Bounded Symmetric Operator Represented by a Bounded Symmetric Bilinear Form on a Real Hilbert Space §representation, we conclude $T_{b}z=(x_{1}-x_{2},x_{2}-x_{1})$. Finally, taking $w=z$ gives $b(z,z)=\langle x_{1}-x_{2},x_{1}-x_{2}\rangle=|x_{1}-x_{2}|^{2}$ by the definition of the norm.

Claim 2. The form $\alpha b$ belongs to $\mathrm{Sym}(H\times H)$ by the scalar multiples of forms, and by claim 1,

The space $H\times H$ is a real Hilbert space by Properties of the Product of Two Real Inner Product Spaces §hilbert, so claim 2 of Affine and Quadratic Functions on a Real Hilbert Space are of Class $C^2$, applied to it and to the form $\alpha b$, gives $q\in C^{2}(H\times H)$ with $Dq(z)=T_{\alpha b}z$ and $D^{2}q(z)=\alpha b$ for every $z$. By claim 4 of The Bounded Symmetric Operator Represented by a Bounded Symmetric Bilinear Form on a Real Hilbert Space, $T_{\alpha b}=\alpha T_{b}$, so

Claim 3. This is claim 4 of Affine and Quadratic Functions on a Real Hilbert Space are of Class $C^2$, applied to the real Hilbert space $H\times H$, to the form $\alpha b$ and to the open subset $W$.