Proof of Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity

We use the notation and claims of Elementary Identities in a Real Inner Product Space and The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity, and Elementary Order Arithmetic in an Ordered Field, Elementary Arithmetic in an Ordered Field, Properties of the Absolute Value in an Ordered Field for real numbers. For $b\in\mathrm{Sym}(E)$ let $B_{b}$ be the set of nonnegative real numbers $C$ with $|b(x,y)|\le C|x||y|$ for all $x,y\in E$, as in Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (ii), so $\lVert b\rVert=\inf B_{b}$, and note $0\le\lVert b\rVert$ since $0$ is a lower bound of $B_{b}$. Let $b\in\mathrm{Sym}(E)$ and $x,y,z\in E$, $\lambda\in\mathbb{R}$. From the symmetry, additivity and homogeneity conditions of Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (i): $b(x,y+z)=b(y+z,x)=b(y,x)+b(z,x)=b(x,y)+b(x,z)$ and $b(x,\lambda y)=b(\lambda y,x)=\lambda b(y,x)=\lambda b(x,y)$ (additivity and homogeneity in the second argument); $b(-x,y)=b((-1)x,y)=-b(x,y)$ by claim 5 of Elementary Identities in a Vector Space, and likewise $b(x,-y)=-b(x,y)$; hence $b(x-y,z)=b(x,z)-b(y,z)$ and $b(x,y-z)=b(x,y)-b(x,z)$; and $b(0_{E},y)=b(0\,y,y)=0\cdot b(y,y)=0$ by claim 3 of Elementary Identities in a Vector Space. Expanding $b(x\pm y,x\pm y)$ by additivity in both arguments and symmetry gives the expansions

$$b(x\pm y,x\pm y)=b(x,x)\pm 2b(x,y)+b(y,y),\qquad\text{hence}\qquad 4\,b(x,y)=b(x+y,x+y)-b(x-y,x-y). \tag{P}$$

Claim 1. $0_{\mathrm{Sym}}$ is symmetric and bilinear and satisfies $|0|\le 0\cdot|x||y|$, so $0_{\mathrm{Sym}}\in\mathrm{Sym}(E)$. The operations $b_{1}+b_{2}$ and $\lambda b$ stay in $\mathrm{Sym}(E)$ by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iv), and the eight conditions of Vector Space over a Field hold because they hold pointwise for real numbers; the zero vector is $0_{\mathrm{Sym}}$ and the additive inverse of $b$ is $(-1)b$.

Claim 2. Let $x,y\in E$. If $|x|=0$ or $|y|=0$, then $x=0_{E}$ or $y=0_{E}$ (Elementary Identities in a Real Inner Product Space (iii)), so $b(x,y)=0$ and the inequality holds. Otherwise $|x||y|>0$ (claim 5 of Elementary Order Arithmetic in an Ordered Field), and for every $C\in B_{b}$, $|b(x,y)|\,(|x||y|)^{-1}\le C$ (claim 5 of Elementary Arithmetic in an Ordered Field); thus $|b(x,y)|(|x||y|)^{-1}$ is a lower bound of $B_{b}$, hence at most $\lVert b\rVert$, and multiplying by $|x||y|$ gives the inequality. If $0\le C$ and $|b(x',y')|\le C|x'||y'|$ for all $x',y'$, then $C\in B_{b}$ and $\lVert b\rVert\le C$.

Claim 3. By claim 5 of Properties of the Absolute Value in an Ordered Field and claim 2, $|(b_{1}+b_{2})(x,y)|\le|b_{1}(x,y)|+|b_{2}(x,y)|\le(\lVert b_{1}\rVert+\lVert b_{2}\rVert)|x||y|$, and the constant is nonnegative, so $\lVert b_{1}+b_{2}\rVert\le\lVert b_{1}\rVert+\lVert b_{2}\rVert$ by claim 2. By claim 4 of Properties of the Absolute Value in an Ordered Field, $|(\lambda b)(x,y)|=|\lambda||b(x,y)|\le|\lambda|\lVert b\rVert|x||y|$, so $\lVert\lambda b\rVert\le|\lambda|\lVert b\rVert$; if $\lambda=0$ both sides are $0$ ($0_{\mathrm{Sym}}$ has norm $0$ since $0\in B_{0_{\mathrm{Sym}}}$), and if $\lambda\ne 0$ then $b=\lambda^{-1}(\lambda b)$ gives $\lVert b\rVert\le|\lambda^{-1}|\,\lVert\lambda b\rVert=|\lambda|^{-1}\lVert\lambda b\rVert$ (claim 4 of Properties of the Absolute Value in an Ordered Field applied to $\lambda\lambda^{-1}=1$), so equality holds. If $\lVert b\rVert=0$ then $|b(x,y)|\le 0$ for all $x,y$ by claim 2, so $b=0_{\mathrm{Sym}}$; the converse was noted. For the identity form, $|\langle x,y\rangle|\le 1\cdot|x||y|$ by The Cauchy–Schwarz Inequality in a Real Inner Product Space gives $\lVert I\rVert\le 1$, and if $x\ne 0_{E}$ then $|x|^{2}=|\langle x,x\rangle|\le\lVert I\rVert|x|^{2}$ by claim 2, so $1\le\lVert I\rVert$ after dividing by $|x|^{2}>0$.

Claim 4. By claim 3: $d_{\mathrm{Sym}}(b_{1},b_{2})=\lVert b_{1}-b_{2}\rVert\ge 0$; it vanishes if and only if $b_{1}-b_{2}=0_{\mathrm{Sym}}$, that is $b_{1}=b_{2}$; $\lVert b_{2}-b_{1}\rVert=\lVert(-1)(b_{1}-b_{2})\rVert=\lVert b_{1}-b_{2}\rVert$; and $\lVert b_{1}-b_{3}\rVert=\lVert(b_{1}-b_{2})+(b_{2}-b_{3})\rVert\le\lVert b_{1}-b_{2}\rVert+\lVert b_{2}-b_{3}\rVert$. These are the four conditions of Metric Space.

Claim 5. If $\lVert b\rVert\le c$, then $|b(x',x')|\le\lVert b\rVert|x'|^{2}\le c|x'|^{2}$ by claim 2 and claim 5 of Elementary Arithmetic in an Ordered Field. Conversely suppose $|b(x',x')|\le c|x'|^{2}$ for every $x'\in E$, and let $x,y\in E$. By (P), claim 5 of Properties of the Absolute Value in an Ordered Field and Elementary Identities in a Real Inner Product Space (vi),

$$4|b(x,y)|\le|b(x+y,x+y)|+|b(x-y,x-y)|\le c\bigl(|x+y|^{2}+|x-y|^{2}\bigr)=2c\bigl(|x|^{2}+|y|^{2}\bigr).$$

If $|x|=1=|y|$ this gives $|b(x,y)|\le c$. For general $x,y$ with $x\ne 0_{E}\ne y$, put $x'=|x|^{-1}x$ and $y'=|y|^{-1}y$, unit vectors by Elementary Identities in a Real Inner Product Space (iv); bilinearity gives $b(x,y)=|x||y|\,b(x',y')$, so $|b(x,y)|=|x||y|\,|b(x',y')|\le c|x||y|$. If $x=0_{E}$ or $y=0_{E}$ the inequality $|b(x,y)|\le c|x||y|$ is trivial. Hence $c\in B_{b}$ and $\lVert b\rVert\le c$.

Claim 6. By Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iii) and Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iv), $-cI\preceq b\preceq cI$ means $-c|x|^{2}\le b(x,x)\le c|x|^{2}$ for every $x\in E$, which by claim 6 of Properties of the Absolute Value in an Ordered Field (with $c|x|^{2}$ in the role of the bound) is equivalent to $|b(x,x)|\le c|x|^{2}$ for every $x$, hence to $\lVert b\rVert\le c$ by claim 5. The particular case is $c=\lVert b\rVert$.

Claim 7. Reflexivity and transitivity follow pointwise from those of $\le$ on $\mathbb{R}$ (axioms of Total Order on a Set). If $b_{1}\preceq b_{2}$ and $b_{2}\preceq b_{1}$, then $b_{1}(x,x)=b_{2}(x,x)$ for every $x$ by antisymmetry of $\le$, so $b=b_{1}-b_{2}$ satisfies $b(x,x)=0$ for every $x$, and (P) gives $4b(x,y)=0$, hence $b(x,y)=0$, for all $x,y$; thus $b_{1}=b_{2}$.

Claim 8. Pointwise: if $b_{1}(x,x)\le b_{2}(x,x)$ then $b_{1}(x,x)+b(x,x)\le b_{2}(x,x)+b(x,x)$ (order axiom of Ordered Field), $\lambda b_{1}(x,x)\le\lambda b_{2}(x,x)$ for $0\le\lambda$ (claim 5 of Elementary Arithmetic in an Ordered Field) and $\lambda b_{2}(x,x)\le\lambda b_{1}(x,x)$ for $\lambda\le 0$ (apply the previous case to $-\lambda\ge 0$ and claim 4 of Elementary Order Arithmetic in an Ordered Field); and two inequalities add by the compatibility of the order with addition (an axiom of Ordered Field), applied twice, and transitivity.

Claim 9. By bilinearity, $b(x,x)-b(y,y)=b(x-y,x)+b(y,x-y)$, so by claims 5 and 4 of Properties of the Absolute Value in an Ordered Field and claim 2, $|b(x,x)-b(y,y)|\le\lVert b\rVert|x-y||x|+\lVert b\rVert|y||x-y|=\lVert b\rVert(|x|+|y|)|x-y|$. Next $b_{1}(x,y)-b_{2}(x,y)=(b_{1}-b_{2})(x,y)$, so claim 2 gives the second inequality. Third, $b(x',y')-b(x,y)=b(x'-x,y')+b(x,y'-y)$, so $|b(x',y')-b(x,y)|\le\lVert b\rVert(|x'-x||y'|+|x||y'-y|)$ by claim 2. For the consequence, choose $N_{0}$ with $|x_{m}-x|<1$ and $|y_{m}-y|<1$ for $m\ge N_{0}$, so that $|x_{m}|\le|x|+1$ and $|y_{m}|\le|y|+1$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity (i). For $m\ge N_{0}$, by the second and third inequalities,

$$|b_{m}(x_{m},y_{m})-b(x,y)|\le|(b_{m}-b)(x_{m},y_{m})|+|b(x_{m},y_{m})-b(x,y)|\le\lVert b_{m}-b\rVert(|x|+1)(|y|+1)+\lVert b\rVert\bigl(|x_{m}-x|(|y|+1)+|x||y_{m}-y|\bigr).$$

Given $\varepsilon>0$, choose $N\ge N_{0}$ such that for $m\ge N$: $\lVert b_{m}-b\rVert<\varepsilon/(3(|x|+1)(|y|+1))$, $|x_{m}-x|<\varepsilon/(3(\lVert b\rVert+1)(|y|+1))$ and $|y_{m}-y|<\varepsilon/(3(\lVert b\rVert+1)(|x|+1))$, using Convergent Sequence in a Metric Space in $(E,d)$ and in $(\mathrm{Sym}(E),d_{\mathrm{Sym}})$ and claim 1 of Elementary Properties of the Maximum of Two Elements, applied twice, to combine the three indices. Then each of the three terms is less than $\varepsilon/3$, so $|b_{m}(x_{m},y_{m})-b(x,y)|<\varepsilon$ for $m\ge N$, which is convergence in the sense of Limit of a Sequence of Real Numbers.

Claim 10. For $x,y\in V$, claim 2 and claim 5 of Elementary Arithmetic in an Ordered Field (twice, as in Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (v)) give $|b(x,y)|\le\lVert b\rVert|x||y|\le\lVert b\rVert|x|_{V}|y|_{V}$, so $\lVert b\rVert$ belongs to the set whose infimum is $\lVert b|_{V}\rVert_{V}$, whence $\lVert b|_{V}\rVert_{V}\le\lVert b\rVert$. The identities $(b_{1}+b_{2})|_{V}=b_{1}|_{V}+b_{2}|_{V}$ and $(\lambda b)|_{V}=\lambda(b|_{V})$ hold pointwise on $V\times V$. If $b_{1}\preceq b_{2}$, then $b_{1}(x,x)\le b_{2}(x,x)$ for all $x\in E$, in particular for $x\in V$, so $b_{1}|_{V}\preceq_{V}b_{2}|_{V}$. Finally $I_{E}|_{V}(x,x)=\langle x,x\rangle=|x|^{2}\le|x|_{V}^{2}=\langle x,x\rangle_{V}=I_{V}(x,x)$ for $x\in V$, by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field applied to $0\le|x|\le|x|_{V}$; so $I_{E}|_{V}\preceq_{V}I_{V}$.