Proof of Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions

Each result cited is universally quantified over the data in its own statement, and is applied here to the data named in the statement of the lemma. Throughout, $S=\operatorname{span}(e)$ denotes the span of the tuple $e$, a linear subspace of $H$ containing every $e_{i}$ by claim 1 of The Span of a Finite Family is the Smallest Subspace Containing It, and $S^{\perp}$ its orthogonal complement. By Orthogonality, Orthogonal Complement and Orthonormal Families in a Real Inner Product Space (iii) the tuple $e$ satisfies $\langle e_{i},e_{j}\rangle=0$ for $i\ne j$ and $|e_{i}|=1$ for every $i$.

The map $P$ of the statement is the map of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace determined by $e$, so that claims 1 to 6 of that lemma are available: $P$ is linear, $Px\in S$ for every $x\in H$, $Px=x$ for every $x\in S$, $x-Px\in S^{\perp}$ for every $x\in H$, $|Px|\le|x|$ and $|x|^{2}=|Px|^{2}+|x-Px|^{2}$; and the map $\kappa:S\to\mathbb{R}^{m}$ with $\kappa(y)=(\langle y,e_{1}\rangle,\dots,\langle y,e_{m}\rangle)$ is a linear bijection with inverse the map sending $\zeta$ to $\sum_{i=1}^{m}\zeta_{i}e_{i}$, and satisfies $\langle y,y'\rangle=\kappa(y)\cdot\kappa(y')$ and $|y|=\lVert\kappa(y)\rVert$ for all $y,y'\in S$.

Claim 1 (clause 1).

The map $\Lambda$ is linear and $\Lambda=\kappa\circ P$. For each $i$ the map $x\mapsto\langle x,e_{i}\rangle$ is additive and homogeneous by claim 1 of Elementary Identities in a Real Inner Product Space together with the symmetry of the inner product (condition (a) of Real Inner Product Spaces (i)), and the vector operations of $\mathbb{R}^{m}$ are coordinatewise by Euclidean Space and Real Matrices: Standing Notation (i); hence $\Lambda$ is linear. For $x\in H$ and each $i$ we have $e_{i}\in S$ and $x-Px\in S^{\perp}$, so $\langle x-Px,e_{i}\rangle=0$ by Orthogonality, Orthogonal Complement and Orthonormal Families in a Real Inner Product Space (ii) and therefore $\langle Px,e_{i}\rangle=\langle x,e_{i}\rangle$ by claim 1 of Elementary Identities in a Real Inner Product Space. Since $Px\in S$, this says exactly $\Lambda x=\kappa(Px)$; in particular $\Lambda y=\kappa(y)$ for $y\in S$.

The identities involving $\Lambda^{\sharp}$. For $\zeta\in\mathbb{R}^{m}$ the vector $\Lambda^{\sharp}\zeta=\sum_{i=1}^{m}\zeta_{i}e_{i}$ lies in $S$ by Span of a Finite Family of Vectors, and $\kappa(\Lambda^{\sharp}\zeta)=\zeta$ because $\Lambda^{\sharp}$ is the inverse of $\kappa$. Consequently

$$\Lambda\Lambda^{\sharp}\zeta=\kappa\bigl(P\Lambda^{\sharp}\zeta\bigr)=\kappa\bigl(\Lambda^{\sharp}\zeta\bigr)=\zeta,$$

using $Py=y$ for $y\in S$; and $|\Lambda^{\sharp}\zeta|=\lVert\kappa(\Lambda^{\sharp}\zeta)\rVert=\lVert\zeta\rVert$. Also $\Lambda Px=\kappa(PPx)=\kappa(Px)=\Lambda x$, again since $Px\in S$. And $\lVert\Lambda x\rVert=\lVert\kappa(Px)\rVert=|Px|\le|x|$. Finally, for $x\in H$ and $\zeta\in\mathbb{R}^{m}$, both $\Lambda^{\sharp}\zeta$ and $Px$ lie in $S$, so

$$\zeta\cdot\Lambda x=\kappa\bigl(\Lambda^{\sharp}\zeta\bigr)\cdot\kappa(Px)=\bigl\langle\Lambda^{\sharp}\zeta,\,Px\bigr\rangle=\bigl\langle\Lambda^{\sharp}\zeta,\,x\bigr\rangle,$$

the last equality because $\langle\Lambda^{\sharp}\zeta,x-Px\rangle=0$ (as $x-Px\in S^{\perp}$) and claim 1 of Elementary Identities in a Real Inner Product Space. Note also that $P=\Lambda^{\sharp}\Lambda$, since $\Lambda^{\sharp}\Lambda x=\sum_{i=1}^{m}\langle x,e_{i}\rangle e_{i}=Px$.

The map $\Lambda^{\sharp}$ is linear. Let $\zeta,\zeta'\in\mathbb{R}^{m}$ and $a\in\mathbb{R}$, and put $y=\Lambda^{\sharp}(\zeta+\zeta')-\Lambda^{\sharp}\zeta-\Lambda^{\sharp}\zeta'$. For every $x\in H$, the identity just proved and claim 2 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite give

$$\langle y,x\rangle=(\zeta+\zeta')\cdot\Lambda x-\zeta\cdot\Lambda x-\zeta'\cdot\Lambda x=0 ,$$

using claim 1 of Elementary Identities in a Real Inner Product Space on the left. Taking $x=y$ gives $\langle y,y\rangle=0$, so $y=0_{H}$ by the definiteness of the inner product (condition (d) of Real Inner Product Spaces (i)). The same argument with claim 4 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite gives $\Lambda^{\sharp}(a\zeta)=a\,\Lambda^{\sharp}\zeta$. Taking $a=0$ shows that $\Lambda^{\sharp}$ sends the origin of $\mathbb{R}^{m}$ to $0_{H}$, by claim 1 of Elementary Identities in a Vector Space.

The last assertion. If $\Lambda x=\Lambda x'$ then $Px=\Lambda^{\sharp}\Lambda x=\Lambda^{\sharp}\Lambda x'=Px'$; conversely if $Px=Px'$ then $\Lambda x=\Lambda Px=\Lambda Px'=\Lambda x'$. This proves Claim 1.

Claim 2 (clause 2). Let $M,M'\in\mathcal{S}(m)$, let $a\in\mathbb{R}$ and let $z,w,z'\in H$.

$M^{\Lambda}$ is a bounded symmetric bilinear form. Since $M\in\mathcal{S}(m)$ we have $M^{\top}=M$ by Euclidean Space and Real Matrices: Standing Notation (iii), so claim 5 of Elementary Properties of the Transpose of a Real Matrix and claim 1 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite give

$$\Lambda z\cdot(M\,\Lambda w)=\bigl(M^{\top}\Lambda z\bigr)\cdot\Lambda w=\bigl(M\,\Lambda z\bigr)\cdot\Lambda w=\Lambda w\cdot(M\,\Lambda z),$$

which is the symmetry $M^{\Lambda}(z,w)=M^{\Lambda}(w,z)$. Additivity and homogeneity in the first argument follow from the linearity of $\Lambda$ (Claim 1) and claims 2 and 4 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite. By The Cauchy-Schwarz Inequality on $\mathbb{R}^n$, claim 3 of Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix and claim 5 of Elementary Arithmetic in an Ordered Field, together with $\lVert\Lambda z\rVert\le|z|$ and $\lVert\Lambda w\rVert\le|w|$ from Claim 1,

$$\bigl|M^{\Lambda}(z,w)\bigr|\le\lVert\Lambda z\rVert\,\lVert M\,\Lambda w\rVert\le\lVert M\rVert\,\lVert\Lambda z\rVert\,\lVert\Lambda w\rVert\le\lVert M\rVert\,|z|\,|w| .$$

Hence $M^{\Lambda}\in\mathrm{Sym}(H)$ by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (i), and $\lVert M^{\Lambda}\rVert\le\lVert M\rVert$ by the second part of claim 2 of Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity, the number $\lVert M\rVert$ being nonnegative by Euclidean Space and Real Matrices: Standing Notation (v).

The remaining identities. That $M^{\Lambda}(z,w)=M^{\Lambda}(Pz,Pw)$ is immediate from $\Lambda Pz=\Lambda z$ and $\Lambda Pw=\Lambda w$ (Claim 1). For $T_{M^{\Lambda}}$: by Claim 1, for every $w\in H$,

$$\bigl\langle\Lambda^{\sharp}\bigl(M\,\Lambda z\bigr),\,w\bigr\rangle=\bigl(M\,\Lambda z\bigr)\cdot\Lambda w=\Lambda z\cdot\bigl(M\,\Lambda w\bigr)=M^{\Lambda}(z,w)=\bigl\langle T_{M^{\Lambda}}z,\,w\bigr\rangle,$$

the middle equality by the symmetry established above and claim 1 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite, and the last by claim 1 of The Bounded Symmetric Operator Represented by a Bounded Symmetric Bilinear Form on a Real Hilbert Space. Subtracting and using definiteness as in Claim 1 gives $T_{M^{\Lambda}}z=\Lambda^{\sharp}(M\,\Lambda z)$.

For the linearity in the matrix: the $i$th coordinate of $(M+M')\eta$ is $\sum_{j=1}^{m}(M_{ij}+M'_{ij})\eta_{j}$, which equals $\sum_{j}M_{ij}\eta_{j}+\sum_{j}M'_{ij}\eta_{j}$ by claim 2 of Properties of Finite Sums, so $(M+M')\eta=M\eta+M'\eta$; likewise $(aM)\eta=a(M\eta)$ by claim 3 of that lemma. With claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite this gives $(M+M')^{\Lambda}=M^{\Lambda}+(M')^{\Lambda}$ and $(aM)^{\Lambda}=a\,M^{\Lambda}$, the operations on forms being those of Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iv).

Finally, suppose $M\preceq M'$, that is $\zeta\cdot(M\zeta)\le\zeta\cdot(M'\zeta)$ for every $\zeta\in\mathbb{R}^{m}$ by Euclidean Space and Real Matrices: Standing Notation (iv). Taking $\zeta=\Lambda z$ gives $M^{\Lambda}(z,z)\le(M')^{\Lambda}(z,z)$ for every $z\in H$, that is $M^{\Lambda}\preceq(M')^{\Lambda}$ by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iii). This proves Claim 2.

Claim 3 (clause 3). Bilinearity and symmetry of $\Pi$ follow from the linearity of $\Lambda$ (Claim 1) and claims 1, 2, 4 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite, and $|\Pi(z,w)|=|\Lambda z\cdot\Lambda w|\le\lVert\Lambda z\rVert\lVert\Lambda w\rVert\le|z||w|$ by The Cauchy-Schwarz Inequality on $\mathbb{R}^n$ and Claim 1; so $\Pi\in\mathrm{Sym}(H)$ with $\lVert\Pi\rVert\le1$, by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (i) and claim 2 of Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity. The identity form $I$ lies in $\mathrm{Sym}(H)$ by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iv), so $N=I-\Pi\in\mathrm{Sym}(H)$ by claim 1 of Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity.

Since $Pz,Pw\in S$, Claim 1 gives $\Pi(z,w)=\Lambda z\cdot\Lambda w=\kappa(Pz)\cdot\kappa(Pw)=\langle Pz,Pw\rangle$; and $\langle Pz,w\rangle=\langle Pz,Pw\rangle$ because $\langle Pz,w-Pw\rangle=0$, as $Pz\in S$ and $w-Pw\in S^{\perp}$. Taking $w=z$ gives $\Pi(z,z)=|Pz|^{2}=\lVert\Lambda z\rVert^{2}$, the last equality by Claim 1. For $N$: by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iv) and claim 1 of Elementary Identities in a Real Inner Product Space,

$$\langle z-Pz,\,w-Pw\rangle=\langle z,w\rangle-\langle Pz,w\rangle-\langle z,Pw\rangle+\langle Pz,Pw\rangle=\langle z,w\rangle-\langle Pz,Pw\rangle=N(z,w),$$

where $\langle z,Pw\rangle=\langle Pw,z\rangle=\langle Pw,Pz\rangle=\langle Pz,Pw\rangle$ by the previous paragraph and symmetry of the inner product; and $\langle z-Pz,w\rangle=\langle z,w\rangle-\langle Pz,w\rangle=\langle z,w\rangle-\langle Pz,Pw\rangle=N(z,w)$ likewise. Taking $w=z$ gives $N(z,z)=|z-Pz|^{2}$, and claim 2 of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace gives $|z|^{2}=|Pz|^{2}+|z-Pz|^{2}=\Pi(z,z)+N(z,z)$.

That $T_{\Pi}z=Pz$ follows from $\Pi(z,w)=\langle Pz,w\rangle$ for all $w$ and claim 1 of The Bounded Symmetric Operator Represented by a Bounded Symmetric Bilinear Form on a Real Hilbert Space, with definiteness as before. The quadratic forms $\Pi(z,z)=|Pz|^{2}$ and $N(z,z)=|z-Pz|^{2}$ are nonnegative and each is at most $|z|^{2}=I(z,z)$, by the displayed decomposition and Real Inner Product Spaces (ii); so $0_{\mathrm{Sym}}\preceq\Pi\preceq I$ and $0_{\mathrm{Sym}}\preceq N\preceq I$ by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iii), and $\lVert N\rVert\le1$ by claim 5 of Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity. This proves Claim 3.

Claim 4 (clause 4). Write $x_{1}=\Lambda^{\sharp}\zeta_{1}$, so that $\Lambda x_{1}=\zeta_{1}$ by Claim 1, and hence $\Lambda x-\zeta_{1}=\Lambda(x-x_{1})$ for every $x\in H$, by the linearity of $\Lambda$. Consequently, by the definitions of $M^{\Lambda}$ and $\Pi$ and by Claim 1,

$$T(x)=c+\bigl\langle\Lambda^{\sharp}q,\,x-x_{1}\bigr\rangle+\tfrac{1}{2}\,M^{\Lambda}(x-x_{1},x-x_{1})+\eta\,\Pi(x-x_{1},x-x_{1}) =c+\bigl\langle\Lambda^{\sharp}q,\,x-x_{1}\bigr\rangle+\tfrac{1}{2}\,b(x-x_{1},x-x_{1}),$$

where $b=M^{\Lambda}+2\eta\,\Pi$, using $\eta\,\Pi(\cdot,\cdot)=\tfrac{1}{2}(2\eta\,\Pi)(\cdot,\cdot)$ and Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iv). Expanding by bilinearity and symmetry of $b$ and by claim 1 of The Bounded Symmetric Operator Represented by a Bounded Symmetric Bilinear Form on a Real Hilbert Space,

$$\tfrac{1}{2}b(x-x_{1},x-x_{1})=\tfrac{1}{2}b(x,x)-b(x,x_{1})+\tfrac{1}{2}b(x_{1},x_{1})=\tfrac{1}{2}b(x,x)-\bigl\langle T_{b}x_{1},x\bigr\rangle+\tfrac{1}{2}b(x_{1},x_{1}),$$

so that

$$T(x)=\bigl\langle\Lambda^{\sharp}q-T_{b}x_{1},\,x\bigr\rangle+c_{0}+\tfrac{1}{2}b(x,x), \qquad c_{0}=c-\bigl\langle\Lambda^{\sharp}q,x_{1}\bigr\rangle+\tfrac{1}{2}b(x_{1},x_{1}).$$

By claims 1 and 2 of Affine and Quadratic Functions on a Real Hilbert Space are of Class $C^2$ the two summands (the affine one, and $x\mapsto\tfrac12 b(x,x)$) belong to $C^{2}(H)$, with gradients $\Lambda^{\sharp}q-T_{b}x_{1}$ and $T_{b}x$ and Hessians $0_{\mathrm{Sym}}$ and $b$; so by claim 2 of Constants, Sums, Scalar Multiples and Differences of Differentiable Functions on an Open Subset of a Real Inner Product Space, $T\in C^{2}(H)$ with

$$DT(x)=\Lambda^{\sharp}q-T_{b}x_{1}+T_{b}x=\Lambda^{\sharp}q+T_{b}(x-x_{1}), \qquad D^{2}T(x)=b ,$$

the middle step by the linearity of $T_{b}$ (claim 1 of The Bounded Symmetric Operator Represented by a Bounded Symmetric Bilinear Form on a Real Hilbert Space).

By Claims 2 and 3 and claim 4 of The Bounded Symmetric Operator Represented by a Bounded Symmetric Bilinear Form on a Real Hilbert Space, $T_{b}z=\Lambda^{\sharp}(M\,\Lambda z)+2\eta\,Pz=\Lambda^{\sharp}\bigl(M\,\Lambda z+2\eta\,\Lambda z\bigr)$ for $z\in H$, using $Pz=\Lambda^{\sharp}\Lambda z$ and the linearity of $\Lambda^{\sharp}$. Taking $z=x-x_{1}$, so that $\Lambda z=\Lambda x-\zeta_{1}$, and using $|\Lambda^{\sharp}\xi|=\lVert\xi\rVert$, the triangle inequality in $\mathbb{R}^{m}$ and claim 3 of Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix,

$$\bigl|DT(x)-\Lambda^{\sharp}q\bigr|=\bigl\lVert M(\Lambda x-\zeta_{1})+2\eta(\Lambda x-\zeta_{1})\bigr\rVert\le\bigl(\lVert M\rVert+2|\eta|\bigr)\lVert\Lambda x-\zeta_{1}\rVert .$$

In particular $DT(x)=\Lambda^{\sharp}q$ when $\Lambda x=\zeta_{1}$. That $T(x)=T(x')$ whenever $\Lambda x=\Lambda x'$ is immediate from $T=T_{0}\circ\Lambda$. Finally, by The Cauchy-Schwarz Inequality on $\mathbb{R}^n$, claim 3 of Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix and claim 5 of Properties of the Absolute Value in an Ordered Field,

$$\bigl|T_{0}(\zeta)-T_{0}(\zeta_{1})\bigr| \le\bigl|q\cdot(\zeta-\zeta_{1})\bigr|+\tfrac{1}{2}\bigl|(\zeta-\zeta_{1})\cdot\bigl(M(\zeta-\zeta_{1})\bigr)\bigr|+|\eta|\,\lVert\zeta-\zeta_{1}\rVert^{2} \le\lVert q\rVert\lVert\zeta-\zeta_{1}\rVert+\bigl(\tfrac{1}{2}\lVert M\rVert+|\eta|\bigr)\lVert\zeta-\zeta_{1}\rVert^{2}.$$

This proves Claim 4.

Claim 5 (clause 5). That $T_{0}(\zeta_{1})=c=\chi(\zeta_{1})$ is immediate, since $\zeta_{1}-\zeta_{1}$ is the origin of $\mathbb{R}^{m}$ and the remaining three terms vanish there, by claim 1 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product and The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite.

Being of class $C^{2}$ on $\Omega$, the function $\chi$ is, by claim 2 of Basic Properties of Twice Differentiability at a Point, twice differentiable at $\zeta_{1}$ with first-order coefficient $D\chi(\zeta_{1})=q$ and Hessian $D^{2}\chi(\zeta_{1})=M$. Suppose first $0<\eta$. Applying Twice Differentiability at a Point (i) with $\eta$ in the role of the accuracy, there is a positive $\rho\in\mathbb{R}$ such that every $h\in\mathbb{R}^{m}$ with $\lVert h\rVert<\rho$ satisfies $\zeta_{1}+h\in\Omega$ and

$$\Bigl|\chi(\zeta_{1}+h)-\chi(\zeta_{1})-q\cdot h-\tfrac{1}{2}\,h\cdot(Mh)\Bigr|\le\eta\,\lVert h\rVert^{2}.$$

Let $\zeta\in\mathbb{R}^{m}$ satisfy $\lVert\zeta-\zeta_{1}\rVert<\rho$ and put $h=\zeta-\zeta_{1}$, so that $\zeta_{1}+h=\zeta$. Then $\zeta\in\Omega$ and, by claim 6 of Properties of the Absolute Value in an Ordered Field,

$$\chi(\zeta)\le\chi(\zeta_{1})+q\cdot(\zeta-\zeta_{1})+\tfrac{1}{2}(\zeta-\zeta_{1})\cdot\bigl(M(\zeta-\zeta_{1})\bigr)+\eta\,\lVert\zeta-\zeta_{1}\rVert^{2}=T_{0}(\zeta).$$

Suppose next $\eta<0$, so that $-\eta$ is positive by claim 4 of Elementary Order Arithmetic in an Ordered Field. Applying Twice Differentiability at a Point (i) with $-\eta$ in the role of the accuracy gives a positive $\rho$ such that every $\zeta$ with $\lVert\zeta-\zeta_{1}\rVert<\rho$ lies in $\Omega$ and satisfies

$$T_{0}(\zeta)=\chi(\zeta_{1})+q\cdot(\zeta-\zeta_{1})+\tfrac{1}{2}(\zeta-\zeta_{1})\cdot\bigl(M(\zeta-\zeta_{1})\bigr)-(-\eta)\lVert\zeta-\zeta_{1}\rVert^{2}\le\chi(\zeta),$$

again by claim 6 of Properties of the Absolute Value in an Ordered Field. This proves Claim 5.

Claim 6 (clause 6). By $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence the map $d_{E}$ is a metric on $\mathbb{R}^{m}$ with $d_{E}(\xi,\xi')=\lVert\xi-\xi'\rVert$, so the triangle inequality gives, first,

$$\lVert\zeta'-\omega'\rVert=d_{E}(\zeta',\omega')\le d_{E}(\zeta',\zeta)+d_{E}(\zeta,\omega)+d_{E}(\omega,\omega')=\lVert\zeta-\omega\rVert+\lVert\zeta-\zeta'\rVert+\lVert\omega-\omega'\rVert,$$

using the symmetry of a metric; this is the first assertion. Exchanging the roles of the two pairs gives also $\lVert\zeta-\omega\rVert\le\lVert\zeta'-\omega'\rVert+\lVert\zeta-\zeta'\rVert+\lVert\omega-\omega'\rVert$, so, writing $s=\lVert\zeta-\omega\rVert$, $t=\lVert\zeta'-\omega'\rVert$ and $c=\lVert\zeta-\zeta'\rVert+\lVert\omega-\omega'\rVert$, we have $-c\le s-t\le c$ and hence $|s-t|\le c$ by claim 6 of Properties of the Absolute Value in an Ordered Field. Since $s^{2}-t^{2}=(s-t)(s+t)$ and $0\le s+t$, claim 4 of Properties of the Absolute Value in an Ordered Field and claim 5 of Elementary Arithmetic in an Ordered Field give

$$\bigl|s^{2}-t^{2}\bigr|=|s-t|\,(s+t)\le c\,(s+t),$$

which is the second assertion. This proves Claim 6 and completes the proof of the lemma.