1 Introduction

Throughout this paper, let $I$ be an index set, and $\{x_i:i\in I$ } be a subset of elements of a vector space $\mathcal{H}$ . For $A\subseteq I$ denote by $x_A$ the sum of vectors $x_i,i\in A$ . i.e. $x_A={\sum }_{i\in A}{x}_i$ . The notation $\overline{n}$ is used to denote the set $\{1,2,\cdots ,n\}$ . For a finite set $A$ , we use $\left|A\right|$ to denote the cardinality of $A$ , and the standard notation for binomial coefficients, $\left(\genfrac{}{}{0.0pt}{}{n}{k}\right)=n!/(k!\left(n-k\right)!)$ is used.

The algorithm given in Theorem 2.3 is meant to test the validity of identities of the form ${\sum }_{A\subseteq \overline{n}}{c_A\Vert x_A\Vert }^2=0$ in inner-product spaces by converting the verification of such an identity to verifying numerical equalities. The algorithm is illustrated in Section 3 by using it to derive several identities. Notable among these results is a generalization of the parallelepiped law, which is deduced, in Corollary 3.5, from a more general result. In Section 4, we prove that all the identities that can be verified by this algorithm only hold in inner-product spaces. Thus, the algorithm can be used to derive characterizations of norms defined by an inner product. This is the application chosen for discussion in the paper.

Investigating norm identities that are satisfied only by norms induced by inner products dates back to the late 19th century (see Amir, 1986, Introduction). Fréchet, 1935 showed that a normed space is an inner product space if and only if

Jordan and von Neumann, 1935 showed that the norm is induced by an inner product if and only if the parallelogram law

holds for all $x,y$ . For a proof of Jordan and von Neumann’s result see Istrăţescu (1987), Theorem 4.3.6. The study of characterizations of inner-product spaces continue to be an active field (see e.g. Adamek 2020; Chelidze 2004; Dadipour and Moslehian 2010; Mendoza and Pakhrou, 2003). The author hopes that the algorithm presented here will lead to new characterization like the ones given in Corollary 4.5.

This paper is self-contained, the only results needed are the Jordan and von Neumann characterization mentioned above and a Lemma due to Fréchet on finite difference (see e.g. Amir 1986 Lemmas 1.1, 1.2 or Reznick, 1978, Lemma 1). We give a proof of this lemma below

let $F_n=\left\{f:\mathrm{ℝ}\to \mathrm{ℝ}:fislocallyintegrableand{\sum }_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right){\left(-1\right)}^{n-k}f\left(r+ks\right)=0\forall \left(r,s\right)\in {\mathrm{ℝ}}^2\right\}$ . The following statements hold:

1. For $n\ge 1$ every function in $F_n$ is infinitely differentiable.

2. If $n>1$ and $f\in F_n$ then $f^{\text{'}}\in F_{n-1}.$

3. For $n\ge 1$ every function in $F_n$ is a polynomial of degree less than $n.$

Take $\varphi$ a $C^{\infty }$ -function on $\mathbb{R}$ with support in $[-1,1]$ such that ${\int }_{\mathbb{R}}\varphi =1$ . By multiplying the above equality by $\varphi \left(s\right)$ and integrating with respect to $s$ , we obtain $f\left(r\right)={\sum }_{k=1}^n\left(\begin{array}{c}n\\ k\end{array}\right){\left(-1\right)}^{1+k}{\psi }_k\left(r\right)$ where ${\psi }_k\left(r\right)={\int }_{\mathbb{R}}f\left(r+ks\right)\varphi \left(s\right)ds$ . Substituting $x=r+ks$ , $s=\frac{x-r}{k}$ yields, for $a>r+k$ $${\psi }_k\left(r\right)={\int }_{\mathbb{R}}f\left(x\right)\varphi \left(\frac{x-r}{k}\right)\frac{\mathit{dx}}{k}={\int }_{\left[-a,a\right]}f\left(x\right)\varphi \left(\frac{x-r}{k}\right)\frac{\mathit{dx}}{k}$$

Repeated differentiation under the integral sign, which is justifiable by the dominated convergence theorem, and the mean value theorem, we have that ${\psi }_k$ is infinitely differentiable with

$\frac{d^m}{d{r}^m}{\psi }_k\left(r\right)={\int }_{\mathbb{R}}f\left(x\right)\frac{d^m}{d{r}^m}\varphi \left(\frac{x-r}{k}\right)\frac{\mathit{dx}}{k}$ for $m\ge 0$ .

Thus, $f$ is infinitely differentiable.

2) By assumption, we have

${\sum }_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right){\left(-1\right)}^{n-k}f\left(r+ks\right)=0$ for all $\left(r,s\right)\in {\mathbb{R}}^2$ .

Differentiating both sides with respect to $s$ yields $$0=\sum _{k=1}^n\left(\begin{array}{c}n\\ k\end{array}\right){\left(-1\right)}^{n-k}kf\text{'}\left(r+ks\right)=n\sum _{k=1}^n\left(\begin{array}{c}n-1\\ k-1\end{array}\right){\left(-1\right)}^{n-k}f\text{'}\left(r+ks\right)$$

Summing over $l=k-1$ and dividing by $n$ , the last expression becomes $$\sum _{l=0}^n\left(\begin{array}{c}n-1\\ l\end{array}\right){\left(-1\right)}^{n-1-l}f\text{'}\left(r+\left(l+1\right)s\right)=0$$

Given $r_1,s\in \mathbb{R}$ , substituting $r=r_1-s$ in the last identity, we obtain $$\sum _{l=0}^n\left(\begin{array}{c}n-1\\ l\end{array}\right){\left(-1\right)}^{n-1-l}f\text{'}\left(r_1+ls\right)$$

Since $r_1,s$ are arbitrary, then $f^{\text{'}}\in F_{n-1}$ .

3) By induction on $n\ge 1$ . For $n=1,$ the identity defining $F_1$ is $-f\left(r\right)+f\left(r+s\right)=0$ for all $r,s$ which clearly implies that $f$ is a constant, i.e., a polynomial of degree less than 1. Suppose $n>1$ , the result is true for $n-1,$ and that $f\in F_n$ . Using 2) we have $f^{\text{'}}$ is a polynomial of degree less than $n-1,$ so by integration, $f$ is a polynomial of degree less than $n.$ ▪

2 A test for a class of norm identities in inner product Spaces.

In this section, a test that can be used to verify the validity of certain identities in inner product spaces is provided. Recall the familiar identity for inner product norms

Or equivalently

holds for all positive integer $n\ge 2$ .

But $$2Re\langle x_1+\cdots +x_n,x_{n+1}\rangle ={\sum }_{i=1}^{n}2Re\langle x_i,x_{n+1}\rangle$$

By induction hypothesis, we have

Substituting (7) and (8) in (6), we obtain $${||x_1+x_2+\cdots +x_{n+1}||}^2=\left(\sum _{\left|A\right|=2,A\subseteq \overline{n}}{||x_A||}^2-\left(n-2\right)\sum _{i=1}^n{||x_i||}^2\right)+{||x_{n+1}||}^2$$ $$+\left(\sum _{\left|A\right|=2,A\subseteq \overline{n+1},n+1\in A}{||x_A||}^2-\sum _{i=1}^n{||x_i||}^2-n{||x_{n+1}||}^2\right)$$ $$=\sum _{\left|A\right|=2,A\subseteq \overline{n+1}}{||x_A||}^2-\left(n-1\right)\sum _{i=1}^{n+1}{||x_i||}^2$$

Thus, the relation (5) is true for $n+1$ . By induction, (5) is true for all $n\in \mathbb{N}$ ,n≥2

By using Theorem 2.1 to substitute for ${\Vert x_A\Vert }^2$ where $\left|A\right|>2$ , the equal expression $$\sum _{\left|B\right|=2,B\subseteq A}{||x_B||}^2-\left(\left|A\right|-2\right)\sum _{i\in A}{||x_i||}^2$$

any identity of the form $\sum _{A\subseteq \overline{n}}{c}_A{||x_A||}^2=0$ can be converted to an identity of the form $$\sum _{B\subseteq \overline{n},1⩽\left|B\right|⩽2}{a}_B{||x_B||}^2=0$$

Thus, testing the validity of the identity $\sum _{A\subseteq \overline{n}}{c}_A{||x_A||}^2=0$ is transformed into testing the validity of the equivalent identity $\sum _{B\subseteq \overline{n},1⩽\left|B\right|⩽2}{a}_B{||x_B||}^2=0$ . The verification of the latter identity can be reduced to verifying that all its coefficients are zero as shown by the following Lemma.

${\sum }_{B\subseteq \overline{n},1\le \left|B\right|\le 2}{a_B\Vert x_B\Vert }^2=0$ for all $x_1,\cdots ,x_n\in \mathcal{H}$

( $\Rightarrow$ ) Suppose that $\sum _{B\subseteq \overline{n},1⩽\left|B\right|⩽2}{a}_B{∥x_B∥}^2=0$ for all $x_1,\cdots ,x_n\in \mathcal{H}$ . Pick $u,v\in \mathcal{H}$ orthogonal unit vectors. For $0\le \theta \le 2\pi$ let $u_{\theta }=\mathit{cos}\theta u+\mathit{sin}\theta v$ . For 1 $⩽i\ne j⩽n$ , let $x_i=u,x_j=u_{\theta }$ and $x_k=0$ for $k\in \overline{n}\backslash \{i,j\}$ .

Since ${\Vert u+u_{\theta }\Vert }^2=2+2\mathit{cos}\theta$ we have $$0={\sum }_{B\subseteq \overline{n},1\le \left|B\right|\le 2}{a_B\Vert x_B\Vert }^2$$ $$={\sum }_{k\notin \{i,j\}}{a}_{\{i,k\}}+{\sum }_{k\notin \{i,j\}}{a}_{\{j,k\}}+a_{\{i,j\}}\left(2+2\mathit{cos}\theta \right)+a_{\left\{i\right\}}+a_{\left\{j\right\}}$$

Choose ${\theta }_1,{\theta }_2\in \left[0,2\pi \right]$ with $\mathit{cos}{\theta }_1\ne cos{\theta }_2$ , subtract the above equality at $\theta ={\theta }_2$ from the same equality at $\theta ={\theta }_1$ to obtain $a_{\{i,j\}}=0$ . Since $i\ne j$ were arbitrary elements of $\overline{n}$ , we obtain that $a_B=0$ for all $B\subseteq \overline{n}$ with $\left|B\right|=2$ . Using this and our assumption of the validity of the identity, we obtain $0={\sum }_{B\subseteq \overline{n},1\le \left|B\right|\le 2}{a_B\Vert x_B\Vert }^2={\sum }_{j=1}^n{a}_{\left\{j\right\}}{\Vert x_j\Vert }^2$ . Finally, for each $1\le i\le n$ , take $x_i$ to be a unit vector and $x_j=0$ for $j\ne i$ in the last identity to get $$0={\sum }_{B\subseteq \overline{n},1\le \left|B\right|\le 2}{a_B\Vert x_B\Vert }^2={\sum }_{i=1}^n{a}_{\left\{j\right\}}{\Vert x_j\Vert }^2=a_{\left\{i\right\}}$$

Since $1\le i\le n$ was arbitrary, then $a_B=0$ for all $B\subseteq \overline{n}$ with $\left|B\right|=1$ . ■

The following Theorem uses Theorem 2.1 and a modified version of Lemma 2.2 to test validity of identities of the form $\sum _{A\subseteq \overline{n}}{c}_A{∥x_A∥}^2=0$ . The modification allows us to avoid the need to compute $a_B$ for * with $\left|B\right|=1$ which simplifies the application of the validity test.

The identity ${\sum }_{B\subseteq \overline{n}}{c_B\Vert x_B\Vert }^2=0$ holds if and only if

1. $a_B=0$ for all $B\subseteq \overline{n}$ with $\left|B\right|=2$ , and

2. For each unit vector $u\in \mathcal{H}$ , and for each $1\le i\le n$ , if $x_i=u$ and $x_j=0$ for $j\ne i$ then for this choice of x_k’ s we have $\sum _{A\subseteq \overline{n}}{c}_A{∥x_A∥}^2=0$ .

(⇐)Assume 1) and 2) hold, and $u$ be any unit vector in $\mathcal{H}$ , for each $1\le i\le n$ , let $x_i=u$ and $x_j=0$ for $j\ne i$ then from $2)$ $$0={\sum }_{B\subseteq \overline{n}}{c_B\Vert x_B\Vert }^2={\sum }_{B\subseteq \overline{n},1\le \left|B\right|\le 2}{a_B\Vert x_B\Vert }^2={\sum }_{B\subseteq \overline{n},1=\left|B\right|}{a_B\Vert x_B\Vert }^2=a_{\left\{i\right\}}$$

where we have used 1) in the second equality. Since $1\le i\le n$ was arbitrary, we have $a_B=0$ for all $B\subseteq \overline{n}$ with $1=\left|B\right|$ . This together with 1) gives us that $a_B=0$ for all $B\subseteq \overline{n},1\le \left|B\right|\le 2$ . Thus ${\sum }_{B\subseteq \overline{n}}{c_B\Vert x_B\Vert }^2={\sum }_{B\subseteq \overline{n},1\le \left|B\right|\le 2}{a_B\Vert x_B\Vert }^2=0$ is an identity. ■

3 Some deduced identities

In this section, we provide some example applications of Theorem 2.3. For future reference we have listed the identities as Lemmas rather than examples.

holds for all $n>k\ge 2$ and all $x_1,\cdots ,x_n\in \mathcal{H}$ .

Using Theorem 2.1 to replace ${\Vert x_A\Vert }^2$ for each $A\subseteq \overline{n}$ with $\left|A\right|=k$ by the equivalent expression ${\sum }_{\left|B\right|=2,B\subseteq A}{\Vert x_B\Vert }^2-\left(k-2\right){\sum }_{i\in A}{\Vert x_i\Vert }^2$ and replace ${\Vert x_1+\cdots +x_n\Vert }^2$ by ${\sum }_{\left|B\right|=2,B\subseteq \overline{n}}{\Vert x_B\Vert }^2-\left(n-2\right){\sum }_{i=1}^n{\Vert x_i\Vert }^2$ converts the left hand side of (10) to

By writing (11) in the form ${\sum }_{B\subseteq \stackrel{-}{n},1\le \left|B\right|\le 2}{a_B\Vert x_B\Vert }^2$ , we get for a set $B$ of size 2, $$a_B=\left(\begin{array}{c}n-2\\ k-2\end{array}\right)-{\sum }_{\left|A\right|=k,B\subseteq A\subseteq \stackrel{-}{n}}1$$

The number of $k$ -subset $A\subseteq \stackrel{-}{n}$ containing $B$ is equal to the number of ( $k-2)$ -subsets C of $\stackrel{-}{n}/B$ so is $\left(\begin{array}{c}n-2\\ k-2\end{array}\right)$ . Thus for $B\subseteq \stackrel{-}{n}$ of cardinality 2, $$a_B=\left(\genfrac{}{}{0.0pt}{}{n-2}{k-2}\right)-\left(\genfrac{}{}{0.0pt}{}{n-2}{k-2}\right)=0$$

giving us that condition (1) of Theorem 2.3. To verify condition (2), let $u$ to be a unit vector, for each $1\le i\le n$ , let $x_i=u$ and $x_j=0$ for $i\ne j\in \stackrel{-}{n}$ , then substituting this choice of $x_i$ ’s in the left hand side of (10) (our original identity), the equation in (10) becomes $$\left(\begin{array}{c}n-2\\ k-2\end{array}\right)-\left(\begin{array}{c}n-1\\ k-1\end{array}\right)+\left(\begin{array}{c}n-2\\ k-1\end{array}\right)=0$$

The middle term in the above equation was computed by noting that ${\Vert x_A\Vert }^2=1$ if $i\in A$ and $0$ otherwise, and that the number of k-subsets $A\subseteq \stackrel{-}{n}$ containing $\left\{i\right\}$ is equal to the number of ( $k-1)$ -subsets of $\overline{n}$ \{i} which is $\left(\begin{array}{c}n-1\\ k-1\end{array}\right)$ . By a well-known recurrence relation for the binomial coefficients (note $k\ge 2$ ) condition (2) also holds. ▪

Let us write (13) in the form $${\mathrm{\Sigma }}_{B\subseteq \overline{n},1\le \left|B\right|\le 2}{a}_B{\Vert x_B\Vert }^2$$

For a set $B$ with $\left|B\right|=2$ , we have $a_B={\sum }_{I\supseteq B}{\left(-1\right)}^{\left|I\right|}$ . As the number of sets $I\subseteq \overline{n}$ of cardinality $k$ containing $B$ equals the number of ways of choosing the $k-2$ elements of I\B from the $n-2$ elements of $\stackrel{-}{n}$ \B, this sum is $$a_B={\sum }_{k=2}^n{\left(-1\right)}^k\left(\genfrac{}{}{0.0pt}{}{n-2}{k-2}\right)={\sum }_{l=0}^{n-2}{\left(-1\right)}^l\left(\genfrac{}{}{0.0pt}{}{n-2}{l}\right)={\left(1-1\right)}^{n-2}=0$$

So, condition 1) of Theorem 2.3 is satisfied. To verify condition (2), let $u$ be a unit vector, for each $1\le i\le n$ let $x_i=u$ and $x_j=0$ for $j\ne i$ . By substituting this choice in (12), as the number of sets $I\subseteq \stackrel{-}{n}$ containing $i$ and of cardinality $k$ is $\left(\genfrac{}{}{0.0pt}{}{n-1}{k-1}\right)$ , we have $${\sum }_{I\subseteq \overline{n}}{\left(-1\right)}^{\left|I\right|}{\Vert x_I\Vert }^2={\sum }_{I\subseteq \overline{n},i\in I}{\left(-1\right)}^{\left|I\right|}1={\sum }_{k=1}^n{\left(-1\right)}^k\left(\genfrac{}{}{0.0pt}{}{n-1}{k-1}\right)=-{\left(1-1\right)}^{n-1}=0$$

Thus condition (2) holds.▪

This gives us a test for identities of the form $$\sum _{I=\{i_1,i_2,\cdots ,i_k\}\subseteq \overline{n}}{a}_I\sum _{{∊}_1,\cdots ,{∊}_k\in \{1,-1\}}{\Vert {∊}_1{x}_{i_1}+{∊}_2{x}_{i_2}+\cdots +{∊}_k{x}_{i_k}\Vert }^2=0$$

(where the inner sum is over all possible choices of signs ${∊}_1,\cdots ,{∊}_k$ ) Indeed, the above identity can be transformed to the form $$0={\sum }_{I\subseteq \overline{n}}{a}_I{\sum }_{J\subseteq I}{\Vert x_J-x_{I\backslash J}\Vert }^2={\sum }_{I\subseteq \overline{n}}{a}_I{\sum }_{J\subseteq I}\left(2{\Vert x_J\Vert }^2+2{\Vert x_{I\backslash J}\Vert }^2-{\Vert x_I\Vert }^2\right)$$

which has the form that can be verified using Theorem 2.3.

The test for verifying such identities is given in Theorem 3.4. Corollary 3.5 uses this test to prove the parallelepiped law. Namely

holds for all $x_1,\cdots ,x_n$ in $\mathcal{H}$ if and only if for each $i\in \overline{n}$ , $$0=\sum _{\left\{I\in \mathcal{J}\text{and}i\in I\right\}}{2}^{\left|I\right|}{a}_I$$

For $I$ fixed and $J_1\subseteq I,$ the square norm ${\Vert x_{J_1}\Vert }^2$ occurs (with a factor of 2) twice in the inner sum on the RHS of (16) (once when $J=J_1$ and the other when $J=I\backslash J_1$ ) while the $-{\Vert x_I\Vert }^2$ occurs $2^{\left|I\right|}$ times in the inner sum (once for each $J\subseteq I$ ) Thus, the RHS of (16) is $${\sum }_{J}4{\Vert x_J\Vert }^2{\sum }_{\{I\in \mathcal{J}:I\supseteq J\}}{a}_I-{\sum }_{I\in \mathcal{J}}{a}_I{2}^{\left|I\right|}{\Vert x_I\Vert }^2$$

So, the equality in (16) becomes

Using Theorem 2.1 to substitute for ${\Vert x_J\Vert }^2$ and ${\Vert x_I\Vert }^2$ in (17), we get that for each $B$ a subset of $\stackrel{-}{n}$ of cardinality 2, $$a_B={\sum }_{B\subseteq J\subseteq \overline{n}}4{\sum }_{\{I\in \mathcal{J}:I\supseteq J\}}{a}_I-{\sum }_{\{I\in \mathcal{J}:I\supseteq B\}}{a}_I{2}^{\left|I\right|}$$

The first sum is $${\sum }_{B\subseteq J\subseteq I\in \mathcal{J}}4a_I={\sum }_{B\subseteq I\in \mathcal{J}}{4a}_I{\sum }_{B\subseteq J\subseteq I}1$$

As the sets $J$ satisfying $B\subseteq J\subseteq I$ are in bijective correspondence with the subsets of $I\backslash B$ the number of the former is $2^{\left|I\backslash B\right|}=2^{\left|I\right|-2}$ , thus $$a_B={\sum }_{B\subseteq I\in \mathcal{J}}{2}^{\left|I\right|-2}4a_I-{\sum }_{B\subseteq I\in \mathcal{J}}{a}_I{2}^{\left|I\right|}=0$$

So, condition (1) of Theorem 2.3 is satisfied. Therefore (16) holds if and only if condition (1) of Theorem 2.3 holds. Hence, it suffices to show that in this case condition (2) of Theorem 2.3 is equivalent to the condition in Theorem 3.4. Let $u$ be a unit vector. For $1\le i\le n$ let $x_i=u$ and $x_j=0$ for $j\ne i$ . With this choice of values for the $x_k$ ,s

For $J\subseteq I$ that ${\Vert x_J-x_{I\backslash J}\Vert }^2=0$ if $i\notin I$ , and if $i\in I$ , we have $${\Vert x_J-x_{I\backslash J}\Vert }^2=\left\{\begin{array}{c}{\Vert x_i\Vert }^2\text{i}\text{f}i\in J\\ {\Vert {-x}_i\Vert }^2\text{i}\text{f}i\notin J\end{array}\right.$$

in either case ${\Vert x_J-x_{I\backslash J}\Vert }^2={\Vert x_i\Vert }^2=1$ . Thus $${\sum }_{I\in \mathcal{J}}{a}_I{\sum }_{J\subseteq I}{\Vert x_J-x_{I\backslash J}\Vert }^2={\sum }_{\{I\in \mathcal{J}:i\in I\}}{\sum }_{J\subseteq I}1={\sum }_{\{I\in \mathcal{J}:i\in I\}}{2}^{\left|I\right|}{a}_I$$

proving the desired equivalence. ▯

for each $1\le i\le n$ , $${\sum }_{i\in I\subseteq \overline{n}}{c}_I=c_i{\sum }_{I\subseteq \overline{n}\backslash \left\{i\right\}}{c}_I=0$$

For a nonempty index set $J$ , by an easy induction on the cardinality of $J$ , we get that $${\sum }_{I\subseteq J}{c}_I=\prod _{l\in J}\left(1+c_l\right)$$

Thus, the condition of Theorem 3.4 becomes, for each $i\in \overline{n}$ $$2a_i{\prod }_{i\ne j\in J}\left(1+2a_j\right)=0$$

Which always hold if there are two distinct indices $i\ne j$ so that $a_i=a_j=-\frac{1}{2}$ ▪

Note that factoring out $2^n$ from (14), it becomes a special case of Corollary 3.4 where $a_1=a_2=\cdots =a_n=-1/2$ .

4 Sufficient conditions for an inner product

In the previous sections we examined identities that follow from the norm being derived from an inner product. In this section, we show that any such identity (any identity of the form ${\sum }_{A\subseteq \overline{n}}{c_A\Vert x_A\Vert }^2=0$ ) implies that the norm is derived from an inner product. The proof is divided into two lemmas.

1. If some identity of the form ${\sum }_{B\subseteq \overline{n}}{c_B\Vert x_B\Vert }^2=0$ holds in $\mathcal{H}$ with the $c_B\ne 0$ for some $B\ne$ θ then an identity of the form $\sum _{B\subseteq A}{a}_B{x}_B^2=0$ with $a_A\ne 0$ and $A\ne$ θholds in $\mathcal{H}$ .

2. If an identity of the form ${\sum }_{B\subseteq A}{a}_B{\Vert x_B\Vert }^2=0$ with $a_A\ne 0$ and $A\ne$ θ holds in $\mathcal{H}$ . Then an identity of the form $\sum _{B⊊\mathrm{A}}{\left(-1\right)}^{\left|A\right|-\left|B\right|}{x}_B^2=0$ also holds in $\mathcal{H}$ for some $A\ne$ θ.

Since $A$ is maximal $D\cap A=A$ ,i.e. $D\supseteq A$ , and $c_D\ne 0$ implies that $D=A$ so the coefficient of $x_A^2$ is $c_A\ne 0$ and by our choice of $A$ , $A\ne \mathrm{\varnothing }$ . Thus, the identity we obtained above has desired type.

1. Let $N$ be the minimum element of the nonempty set