Proof of Vanishing of Uniformly Small Linear, Quadratic and Bilinear Terms in a Real Inner Product Space

Throughout we use the following facts without further comment: products and multiplicative inverses of positive real numbers are positive (claims 5 and 7 of Elementary Order Arithmetic in an Ordered Field); $|x|$ is positive for $x\ne 0_{E}$ (Elementary Identities in a Real Inner Product Space §vanishing); and $0\le\lVert b\rVert$ for every $b\in\mathrm{Sym}(E)$, because by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §norm the number $\lVert b\rVert$ is the greatest lower bound of a set of nonnegative real numbers, of which $0$ is therefore a lower bound.

Claim 1. If $p=0_{E}$ there is nothing to prove, so suppose $p\ne 0_{E}$; then $0<|p|$. Let $\varepsilon\in\mathbb{R}$ be positive and let $\delta$ be a radius as in the hypothesis for this $\varepsilon$. Put

Here $\delta\cdot 2^{-1}$ is positive and smaller than $\delta$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, so $t$ is positive. By Elementary Identities in a Real Inner Product Space §homogeneity and claim 1 of Properties of the Absolute Value in an Ordered Field, $|z|=|t|\,|p|=t\,|p|=\delta\cdot 2^{-1}<\delta$, so the hypothesis applies to $z$. Since $\langle p,z\rangle=t\,\langle p,p\rangle=t\,|p|^{2}$ by Elementary Identities in a Real Inner Product Space §bilinear and Real Inner Product Space §norm, and $t|p|^{2}$ is nonnegative, claim 1 of Properties of the Absolute Value in an Ordered Field gives

Multiplying both sides by the nonnegative number $t^{-1}|p|^{-1}$ (claim 5 of Elementary Arithmetic in an Ordered Field) yields $|p|\le\varepsilon$. As $\varepsilon$ was an arbitrary positive real number and $0\le|p|$, claim 3 of Comparison of Real Numbers with Arbitrary Positive Slack gives $|p|=0$ and hence $p=0_{E}$ by Elementary Identities in a Real Inner Product Space §vanishing, contradicting $p\ne 0_{E}$. Therefore $p=0_{E}$.

Claim 2. Let $\varepsilon\in\mathbb{R}$ be positive and let $\delta$ be a radius as in the hypothesis. We first show that

If $x=0_{E}$, then $0_{E}=0\cdot 0_{E}$ by claim 3 of Elementary Identities in a Vector Space, so homogeneity in the first argument (Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §form) gives $b(0_{E},0_{E})=0\cdot b(0_{E},0_{E})=0$, while $|x|=0$ by Elementary Identities in a Real Inner Product Space §zero; both sides are then $0$. If $x\ne 0_{E}$, put $s=(\delta\cdot 2^{-1})|x|^{-1}$, a positive real number, and $z=s\,x$; exactly as in claim 1, $|z|=s\,|x|=\delta\cdot 2^{-1}<\delta$. By the homogeneity and symmetry of Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction §form, $b(z,z)=b(sx,sx)=s\,b(x,sx)=s\,b(sx,x)=s^{2}\,b(x,x)$, and $|z|^{2}=s^{2}|x|^{2}$ by Elementary Identities in a Real Inner Product Space §homogeneity. Since $0\le s^{2}$, claims 1 and 4 of Properties of the Absolute Value in an Ordered Field give $|b(z,z)|=s^{2}|b(x,x)|$, so the hypothesis reads

and multiplying by the nonnegative number $(s^{2})^{-1}$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives the asserted bound.

Since $0\le\varepsilon$, Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity §quadratic-norm now yields $\lVert b\rVert\le\varepsilon$. As $\varepsilon$ was an arbitrary positive real number and $0\le\lVert b\rVert$, claim 3 of Comparison of Real Numbers with Arbitrary Positive Slack gives $\lVert b\rVert=0$, hence $b=0_{\mathrm{Sym}}$ by Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity §norm-axioms.

Claim 3. Let $\varepsilon\in\mathbb{R}$ be positive and let $\delta$ be a radius as in the hypothesis of this claim. Taking $y=w$ there shows that every $z\in E$ with $|z|<\delta$ satisfies $|b(z,z)|\le\varepsilon\,|z|\,|z|=\varepsilon\,|z|^{2}$. Since $\varepsilon$ was arbitrary, this is precisely the hypothesis of claim 2, which therefore gives $b=0_{\mathrm{Sym}}$.

Claim 4. Write $c=\lVert b\rVert\cdot 2^{-1}$, which is nonnegative by claim 5 of Elementary Arithmetic in an Ordered Field applied to $0\le\lVert b\rVert$ with the nonnegative multiplier $2^{-1}$, and put $M=c+1$; then $1\le M$ by claim 3 of Elementary Arithmetic in an Ordered Field, so $0<M$ by claim 6 of Elementary Order Arithmetic in an Ordered Field and claim 2 there.

Let $\delta_{1}$ be a radius as in the hypothesis for the positive number $1$. For $z\in E$ with $|z|<\delta_{1}$, claims 2 and 5 of Properties of the Absolute Value in an Ordered Field together with Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity §bound give

Now let $\varepsilon\in\mathbb{R}$ be positive and let $\delta$ be the lesser of $\delta_{1}$ and $\varepsilon M^{-1}$ (claim 9 of Elementary Order Arithmetic in an Ordered Field), which is positive because it is one of them. For $z$ with $|z|<\delta$ we have $|z|\le\delta$ and $\delta\le\varepsilon M^{-1}$, so two applications of claim 5 of Elementary Arithmetic in an Ordered Field, with the nonnegative multipliers $M$ and $|z|$, give

whence $|\langle p,z\rangle|\le\varepsilon|z|$. Since $\varepsilon$ was an arbitrary positive real number, claim 1 gives $p=0_{E}$.

With $p=0_{E}$ we have $\langle p,z\rangle=0$ for every $z$ by Elementary Identities in a Real Inner Product Space §zero, so the hypothesis reads: for every positive $\varepsilon$ there is a positive $\delta$ with $|\tfrac{1}{2}b(z,z)|\le\varepsilon|z|^{2}$ whenever $|z|<\delta$. Let $\eta\in\mathbb{R}$ be positive and apply this with $\varepsilon=\eta\cdot 2^{-1}$, positive by claim 8 of Elementary Order Arithmetic in an Ordered Field, obtaining a positive $\delta$ with $2^{-1}|b(z,z)|\le\eta\cdot 2^{-1}|z|^{2}$ for $|z|<\delta$, by claims 1 and 4 of Properties of the Absolute Value in an Ordered Field. Multiplying by the nonnegative number $2$ gives $|b(z,z)|\le\eta|z|^{2}$ for every such $z$. As $\eta$ was arbitrary, claim 2 gives $b=0_{\mathrm{Sym}}$.