Inequalities That Imply the Norm of a Linear Space Is Induced by an Inner Product

Abstract

The aim of this paper is to investigate when a linear normed space is an inner product space. Several conditions in a linear normed space are formulated with the help of inequalities. Some of them are from the literature and others are new. We prove that these conditions are equivalent with the fact that the norm is induced by an inner product. One of the new results is the following: in an inner product space, the sum of opposite edges of a tetrahedron are the sides of an acute angled triangle. The converse of this result holds also. More precisely, this property characterizes inner product spaces. Another new result is the following: in a tetrahedron, the sum of squares of opposite edges are the lengths of a triangle. We prove also that this property characterizes inner product spaces. In addition, we give simpler proofs to some theorems already known from the publications of other authors.

1. Introduction

Every inner product $〈,〉$ gives rise to a norm defined by $∥x∥$=$\sqrt{〈x,x〉}$. But, a norm in a linear normed space does not necessarily arise from an inner product. There are linear normed spaces whose norm cannot be induced in the way above from any inner product. Consequently, the class of linear normed spaces is larger than the class of all (linear normed spaces with the norm induced by an inner product) inner product spaces.

In fact, there are several conditions of the norm in a normed space that guarantee that the norm is induced by an inner product. It is well known that in a linear normed space E the following Jordan–von Neumann parallelogram law

$${∥x+y∥}^2+{∥x-y∥}^2=2{∥x∥}^2+2{∥y∥}^2,x,y\in E$$

(1)

holds if and only if the norm $\parallel \parallel$ is induced by an inner product (cf. Jordan and von Neumann [1]). Condition (1) implies that a linear normed space is an inner product space if and only if every two dimensional subspace is an inner product space. Because of this, most computations in this paper can be performed in two-dimensional spaces.

In the literature, there are many other conditions that characterize inner product spaces among normed spaces. Early conditions can be found in Day [2], Lorch [3] and Ficken [4]. A large collection of such characterizations is contained in the famous book of Amir [5]. See also the books Istrăţescu [6], or Alsina and Sikorska [7].

Characterizations of inner product spaces with the help of strongly convex functions can be found in Nicodem and Pales [8]. Recent papers that study conditions which imply that norms are induced by inner products include Ahmad et al. [9], Cichon [10], Harandi et al. [11], Krnic and Minculete [12,13] and Liu et al. [14].

In the present paper, we investigate when a linear normed space is an inner product space. Several conditions in a linear normed space are formulated with the help of inequalities. Some of them are from the literature and others are new. We prove that these conditions are equivalent with the fact that the norm is induced by an inner product. One of the new results is the following: in an inner product space, the sum of opposite edges of a tetrahedron are the sides of an acute angled triangle. The converse of this result holds also. More precisely, this property characterizes inner product spaces.

In Ficken [4], a condition was introduced that characterizes inner product spaces. Ficken proved that a norm in a linear space that satisfies his condition is induced by an inner product. In [10] Cichon gave a new proof of Ficken’s result. Cichon’s paper was motivated by the fact that Ficken’s proof in [4] was nearly four pages long and the arguments lacked clarity. Since then, several correct proofs have appeared (see [10] for a discussion and references) but none are elementary and straightforward. While Cichon’s proof in [10] is clearer and simpler, in this paper we simplify its arguments and include more details. Conditions (P${}_4$), (P${}_9$), (P${}_{10}$), (P${}_{16}$) and (P${}_{17}$) formulated in Section 2 are new.

2. Main Results

We shall say that a linear normed space E has the inner product property (IPP) if its norm is induced by some inner product. In the following, we shall define 20 properties of a linear normed space E and we shall prove their equivalence.

(P${}_1$) E has (IPP)

(P${}_2$) If $x,y,z\in E$ then

$${∥x∥}^2+{∥y∥}^2+{∥z∥}^2+{∥x+y+z∥}^2={∥x+y∥}^2+{∥x+z∥}^2+{∥y+z∥}^2$$

(P${}_3$) If $x,y,z\in E$ then

$${∥x∥}^2+{∥y∥}^2+{∥z∥}^2+{∥x+y+z∥}^2\le {∥x+y∥}^2+{∥x+z∥}^2+{∥y+z∥}^2$$

(P${}_4$) If $x,y\in E$ and $∥x∥=∥y∥=1$ then

$${∥x+y∥}^2+{∥x-y∥}^2\ge 4$$

(P${}_5$) If $x,y,z\in E$ then

$$∥x∥∥y-z∥\le ∥y∥∥x-z∥+∥z∥∥x-y∥$$

(P${}_6$) If $x,y,z,t\in E$ then

$$∥x-t∥∥y-z∥\le ∥y-t∥∥x-z∥+∥z-t∥∥x-y∥$$

(P${}_7$) If $x,y\in E$ then

$$∥x+y∥∥x-y∥\le {∥x∥}^2+{∥y∥}^2$$

(P${}_8$) If $x,y,z\in E$ then

$$∥x+y∥∥y+z∥\le ∥y∥∥x+y+z∥+∥x∥∥z∥$$

(P${}_9$) If $x,y,z\in E$ then

$${\left(∥x∥+∥y-z∥\right)}^2\le {\left(∥y∥+∥x-z∥\right)}^2+{\left(∥z∥+∥x-y∥\right)}^2$$

(P${}_{10}$) If $x,y,z,t\in E$ then

$${\left(∥x-t∥+∥y-z∥\right)}^2\le {\left(∥x-t∥+∥y-z∥\right)}^2+{\left(∥z-t∥+∥x-y∥\right)}^2$$

(P${}_{11}$) If $x,y\in E$ then

$${∥x+y∥}^2+{∥x-y∥}^2\ge 4∥x∥∥y∥$$

(P${}_{12}$) If $x,y\in E$, $∥x∥=∥y∥$ and $a,b\in \mathbb{R}$ then

$$∥ax+by∥=∥bx+ay∥$$

(P${}_{13}$) If $x,y\in E$, $∥x∥=∥y∥$ and $a\in \left(0,\infty \right)$ then

$$∥ax+y∥=∥x+ay∥$$

(P${}_{14}$) If $x,y\in E$, $∥x∥=∥y∥=1$ and $a\in \left(0,\infty \right)$ then

$$∥ax+y∥=∥x+ay∥$$

(P${}_{15}$) If $x,y\in E$, $∥x∥=∥y∥=1$ and $a\in \left(0,\infty \right)$ then

$$∥ax+a^{-1}y∥\ge ∥x+y∥$$

(P${}_{16}$) Let $ϵ\in \left(0,1\right).$ If $x,y\in E$, $∥x∥=∥y∥=1$ and $a\in \left(1-ϵ,1+ϵ\right)$ then

$$∥ax+a^{-1}y∥\ge ∥x+y∥$$

(P${}_{17}$) Let $ϵ\in \left(0,1\right)$. If $x,y\in E$, and for every $a\in \left(1-ϵ,1+ϵ\right)$ we have

$$∥ax+a^{-1}y∥\ge ∥x+y∥$$

then $∥x∥=∥y∥$.

(P${}_{18}$) If $x,y\in E$ and for every $a\in \left(0,\infty \right)$ we have

$$∥ax+a^{-1}y∥\ge ∥x+y∥$$

then $∥x∥=∥y∥$.

(P${}_{19}$) Let $a\ge 0$ and define ${\psi }_a\left(x,y,z\right)={∥x∥}^2+a∥x∥∥y-z∥+{∥y-z∥}^2$, $x,y,z\in E.$ If $x,y,z\in E$, then

$${\psi }_a\left(x,y,z\right)\le {\psi }_a\left(y,z,x\right)+{\psi }_a\left(z,x,y\right)$$

(P${}_{20}$) Let us define $\theta \left(x_1,x_2,x_3,x_4\right)={∥x_1-x_2∥}^2+{∥x_3-x_4∥}^2,x_1,x_2,x_3,x_4\in E$. If $x_1,x_2,x_3,x_4\in E$, then

$$\theta \left(x_1,x_2,x_3,x_4\right)\le \theta \left(x_1,x_3,x_2,x_4\right)+\theta \left(x_1,x_4,x_2,x_3\right).$$

Recall Ptolemy’s inequality. If $abcd$ is a convex quadrilateral, then we have

$$∥ac∥∥bd∥\le ∥ab∥∥cd∥+∥bc∥∥ad∥,$$

with equality if and only if the quadrilateral is inscribed in a circle. Here, by $∥ac∥$,$∥bd∥$,…we denoted the lengths of the segments $ac$, $bd$, …

If a, b, c, d are vectors in the plane, then the above inequality can be written as

$$∥a-c∥∥b-d∥\le ∥a-b∥∥c-d∥+∥b-c∥∥a-d∥$$

In [15] Schoenberg has shown that Ptolemy’s inequality holds if $a,b,c,d$ are any four points in a real inner product space. In [16] Schoenberg also proved that every real semi-normed Ptolemaic space (i.e., where Ptolemy’s inequality holds) must arise from a real inner product space. An interesting connection between Ptolemy’s inequality and the chordal metric can be found in Apostol [17] and Klamkin and Meir [18]. We recall that a linear normed space E that satisfies property (P${}_5$) is called Ptolemaic. Note that (P${}_6$) is equivalent with the following property of the tetrahedron “The products of opposite edges of a tetrahedron are the sides of a triangle”. We show in Theorem 1 that (P${}_5$)⇔(P${}_6$). Property (P${}_8$) occurs in Zbaganu [19]. In the above-mentioned paper, it is proved that (P${}_8$)⇔(P${}_1$). The equivalence (P${}_1$)⇔(P${}_2$) was proved in Fréchet [20]. Implication (P${}_{13}$)⇒(P${}_1$) was proved in Falkner [21]. Properties (P${}_{10}$), (P${}_{16}$), (P${}_{17}$), (P${}_{19}$) and (P${}_{20}$) are new and occur for the first time. Note that (P${}_{10}$) is equivalent with the following property of the tetrahedron “The sum of opposite edges of a tetrahedron are the sides of an acute angled triangle”. Also, observe that (P${}_{20}$) is equivalent with the following property: “In a tetrahedron the sum of squares of opposite edges are the lengths of a triangle”.

Lemma 1. 

Let E be an inner product space and define the function $\varphi :E\setminus \left\{0\right\}\to \mathbb{R}$, $\varphi \left(x\right)=\frac{x}{{∥x∥}^2},x\in E\setminus \left\{0\right\}$. Then, the following identity holds:

$$∥\varphi \left(x\right)-\varphi \left(y\right)∥=\frac{∥x-y∥}{∥x∥·∥y∥},x,y\in E\setminus \left\{0\right\}$$

Proof. 

Note that

$$\begin{array}{cc}\hfill {∥\varphi \left(x\right)-\varphi \left(y\right)∥}^2& ={∥\varphi \left(x\right)∥}^2+{∥\varphi \left(y\right)∥}^2-2\frac{〈x,y〉}{{∥x∥}^2·{∥y∥}^2}=\frac{1}{{∥x∥}^2}+\frac{1}{{∥y∥}^2}-2\frac{〈x,y〉}{{∥x∥}^2·{∥y∥}^2}\hfill \\ \hfill & =\frac{{∥x∥}^2+{∥y∥}^2-2〈x,y〉}{{∥x∥}^2·{∥y∥}^2}={\left(\frac{∥x-y∥}{∥x∥·∥y∥}\right)}^2\hfill \end{array}$$

This ends the proof. □

Lemma 2. 

Let E be a strictly convex linear normed space, $x,y\in E$, $x\ne 0$ and $\psi \left(t\right)=∥tx+y∥$, $t\in \mathbb{R}$. Then, the following assertions hold:

1${}^0$. The function $\psi \left(t\right)=∥tx+y∥$, $t\in \mathbb{R}$ is convex

2${}^0$. If ψ attains its minimum at $t_0\in \mathbb{R}$ then from $\psi \left(t_0\right)=\psi \left(t_1\right)$ then $t_0=t_1$.

Proof. 

If $\alpha ,\beta \in \left[0,1\right]$, $\alpha +\beta =1$ and $s,t\in \mathbb{R}$ then

$$\begin{array}{cc}\hfill \psi \left(\alpha t+\beta s\right)& =∥\left(\alpha t+\beta s\right)x+\left(\alpha +\beta \right)y∥=∥\alpha \left(tx+y\right)+\beta \left(sx+y\right)∥\hfill \\ \hfill & \le \alpha ∥tx+y∥+\beta ∥sx+y∥=\alpha \psi \left(t\right)+\beta \psi \left(s\right),\hfill \end{array}$$

hence, $\psi$ is convex. In order to prove 2${}^0$. suppose that $\psi$ attains its minimum at $t_0\in \mathbb{R}.$ Let $t_1\in \mathbb{R}$ be such that $\psi \left(t_0\right)=\psi \left(t_1\right)$. Then, by 1${}^0$ we obtain:

$$\psi \left(\frac{t_0+t_1}{2}\right)\le \frac{\psi \left(t_0\right)+\psi \left(t_1\right)}{2}=\psi \left(t_0\right)\le \psi \left(\frac{t_0+t_1}{2}\right),$$

hence,

$$\psi \left(t_0\right)=\psi \left(\frac{t_0+t_1}{2}\right)$$

If $t_0\ne t_1$ then, by strict convexity of E we have

$$\begin{array}{cc}\hfill \psi \left(\frac{t_0+t_1}{2}\right)& =∥\frac{t_0+t_1}{2}x+y∥=∥\frac{\left(t_{0}x+y\right)+\left(t_{1}x+y\right)}{2}∥\hfill \\ \hfill & <\frac{∥t_{0}x+y∥+∥t_{1}x+y∥}{2}=\frac{\psi \left(t_0\right)+\psi \left(t_1\right)}{2}=\psi \left(t_0\right).\hfill \end{array}$$

We obtained a contradiction. Hence, $t_0=t_1$. □

Lemma 3. 

Let E be a linear normed space with the property that

$$∥x∥=∥y∥\mathit{implies}\mathit{that}∥ax+by∥=∥bx+ay∥\mathit{for}\mathit{every}a,b\in \mathbb{R}$$

(2)

Then the following assertions hold:

1${}^0$. $∥x∥=∥y∥=∥\frac{x+y}{2}∥$ implies $x=y$

2${}^0$. E is strictly convex.

3${}^0$. If $∥x-y∥=∥x+y∥$ then, the function $\psi \left(t\right)=∥tx+y∥$, $t\in \mathbb{R}$ is convex, even and $\psi \left(t\right)\ge \psi \left(0\right)$ for every $t\in \mathbb{R}.$

4${}^0$. If $x\ne 0$ and $\psi \left(t_0\right)=\psi \left(0\right)$ then, $t_0=0$.

Proof. 

In order to prove assertion 1${}^0$ suppose that the equalities $∥x∥=∥y∥=∥\frac{x+y}{2}∥$ hold. We shall prove by induction that for every $n\ge 0$

$$∥\left(2n-1\right)\left(x-y\right)-2y∥=2∥x∥$$

(3)

One can easily see that in the case $n=0$ the above equality becomes $∥x+y∥=2∥x∥$. which holds by the hypothesis. Suppose now that (3) holds for some n. Denote by $u=\left(2n-1\right)\left(x-y\right)-2y$, $v=2y$. By (3) it follows that $∥u∥=∥v∥$. By (2) it follows that

$$∥2u+\left(2n+1\right)v∥=∥\left(2n+1\right)u+2v∥$$

hence

$$∥\left(2n+1\right)\left(x-y\right)-2y∥=2∥x∥$$

By induction it follows that (3) holds for every $n\ge 0$. From (3), by the triangle inequality, we deduce that for every $n\ge 1$ the following inequality holds:

$$∥\left(2n-1\right)\left(x-y\right)∥-2∥y∥\le 2∥x∥$$

hence,

$$∥x-y∥\le \frac{2\left(∥x∥+∥y∥\right)}{2n-1}\mathrm{for}\mathrm{every}n\ge 1.$$

By letting $n\to \infty$ in the above inequality we obtain $∥x-y∥=0$, hence $x=y$ and assertion 1${}^0$ is proved. The fact that assertion 2${}^0$ follows from 1${}^0$ was proved by Lin [22].

In order to prove assertion 3${}^0$ suppose that $∥x-y∥=∥x+y∥$. If $t\in \mathbb{R}$ then by (2) it follows that

$$∥\frac{t+1}{2}\left(x+y\right)+\frac{t-1}{2}\left(x-y\right)∥=∥\frac{t-1}{2}\left(x+y\right)+\frac{t+1}{2}\left(x-y\right)∥$$

hence,

$$\psi \left(t\right)=∥tx+y∥=∥tx-y∥=\psi \left(-t\right)$$

We proved that $\psi$ is even. The convexity proof of $\psi$ follows from the preceding lemma. By the convexity of $\psi$ it follows that for every $t\in \mathbb{R}$ the following inequality holds:

$$\psi \left(t\right)=\frac{\psi \left(t\right)+\psi \left(-t\right)}{2}\ge \psi \left(\frac{t-t}{2}\right)=\psi \left(0\right)$$

Assertion 4${}^0$ follows from the preceding lemma. □

Lemma 4. 

Let E be a linear normed space. For any fixed $x,y\in E$, the function f given by $f\left(t\right)=∥tx+t^{-1}y∥$, $t\in \left(0,\infty \right)$ is quasiconvex.

Proof. 

Consider the functions $g\left(t\right)=∥tx+y∥$ and $h\left(t\right)=\frac{{∥tx+y∥}^2}{t}$, $t\in \left(0,\infty \right)$. By Lemma 2, g is convex. We prove that h is convex. Let $\alpha ,\beta \in [0,1]$ with $\alpha +\beta =1$ and $t_1,t_2\in (0,\infty ),$ and define $u(\alpha ,\beta ,t_1,t_2)=h\left(\alpha t_1+\beta t_2\right)-\alpha h\left(t_1\right)-\beta h\left(t_2\right).$ Note that

$$u(\alpha ,\beta ,t_1,t_2)=\frac{g^2\left(\alpha t_1+\beta t_2\right)}{\alpha t_1+\beta t_2}-\alpha \frac{g^2\left(t_1\right)}{t_1}-\beta \frac{g^2\left(t_2\right)}{t_2}=\frac{A}{t_1{t}_2\left(\alpha t_1+\beta t_2\right)},$$

where

$$A=t_1{t}_2{g}^2\left(\alpha t_1+\beta t_2\right)-\alpha \left(\alpha t_1+\beta t_2\right)t_2{g}^2\left(t_1\right)-\beta \left(\alpha t_1+\beta t_2\right)t_1{g}^2\left(t_2\right).$$

By the convexity of g it follows that

$$\begin{array}{cc}\hfill A& \le t_1{t}_2{\left[\alpha g\left(t_1\right)+\beta g\left(t_2\right)\right]}^2-\alpha \left(\alpha t_1+\beta t_2\right)t_2{g}^2\left(t_1\right)-\beta \left(\alpha t_1+\beta t_2\right)t_1{g}^2\left(t_2\right)\hfill \\ \hfill & =-\alpha \beta {\left(t_{2}g\left(t_1\right)-t_{1}g\left(t_2\right)\right)}^2\le 0,\hfill \end{array}$$

hence $u(\alpha ,\beta ,t_1,t_2)\le 0$. Consequently, h is convex. Clearly, h is also quasiconvex, since

$$h\left(\alpha x+\beta y\right)\le \alpha h\left(x\right)+\beta h\left(y\right)\le max\{h\left(x\right),h\left(y\right)\}.$$

Note that

$$f\left(t\right)=\frac{∥t^{2}x+y∥}{t}=\sqrt{h\left(t^2\right)},t\in \left(0,\infty \right).$$

Let $\varphi :\left(0,\infty \right)\to \left(0,\infty \right)$,$\varphi \left(t\right)=t^2$, $t\in \left(0,\infty \right)$. Since $\varphi$ is monotone, ${\varphi }^{-1}$ is increasing and $f={\varphi }^{-1}\circ h\circ \varphi$ it follows that f is quasiconvex. Indeed, let $\alpha ,\beta \in [0,1]$ be such that $\alpha +\beta =1$ and $t_1,t_2\in (0,\infty ).$ Then

$$min\left(\varphi \left(t_1\right),\varphi \left(t_2\right)\right)\le \varphi \left(\alpha t_1+\beta t_2\right)\le max\left(\varphi \left(t_1\right),\varphi \left(t_2\right)\right).$$

Since h is quasiconvex it follows that

$$\begin{array}{cc}\hfill h\left(\varphi \left(\alpha t_1+\beta t_2\right)\right)& \le max\left(h\left(min\left(\varphi \left(t_1\right),\varphi \left(t_2\right)\right)\right),h\left(max\left(\varphi \left(t_1\right),\varphi \left(t_2\right)\right)\right)\right)\hfill \\ \hfill & =max\left(h\left(\varphi \left(t_1\right)\right),h\left(\varphi \left(t_2\right)\right)\right).\hfill \end{array}$$

Since ${\varphi }^{-1}$ is increasing it follows that

$$\begin{array}{cc}\hfill f\left(\alpha t_1+\beta t_2\right)& ={\varphi }^{-1}\left(h\left(\varphi \left(\alpha t_1+\beta t_2\right)\right)\right)\hfill \\ \hfill & \le {\varphi }^{-1}\left(max\left(h\left(\varphi \left(t_1\right)\right),h\left(\varphi \left(t_2\right)\right)\right)\right)=max\left({\varphi }^{-1}\left(h\left(\varphi \left(t_1\right)\right)\right),{\varphi }^{-1}\left(h\left(\varphi \left(t_2\right)\right)\right)\right)\hfill \\ \hfill & =max\left(\left(f\left(t_1\right)\right),\left(f\left(t_2\right)\right)\right).\hfill \end{array}$$

□

Lemma 5. 

Let E be a linear normed space, $x,y\in E\setminus \left\{0\right\}$, $f\left(t\right)=∥tx+t^{-1}y∥$, $t\in \left(0,\infty \right)$. Then f is not constant on every interval.

Proof. 

Suppose by absurd that f is constant on the subinterval $\left(\alpha ,\beta \right)$ of $\left(0,\infty \right)$. Then, there exists a constant $c\ge 0$ such that

$$\frac{∥t^{2}x+y∥}{t}=c\mathrm{for}\mathrm{every}t\in \left(\alpha ,\beta \right)$$

This implies that function $\varphi \left(t\right)=∥tx+y∥-c\sqrt{t}$, $t\in \left({\alpha }^2,{\beta }^2\right)$ is vanishing on $\left({\alpha }^2,{\beta }^2\right)$. If $c>0$ then $\varphi$ is strictly convex hence $\varphi$ cannot vanish over an interval. If $c=0$ then $\varphi \left(t\right)=∥tx+y∥=0$ for every $t\in \left({\alpha }^2,{\beta }^2\right)$. This is impossible as $x\ne 0$ and $y\ne 0$. □

Lemma 6. 

Let E be a linear normed space, $\epsilon \in \left(0,1\right)$ and $x,y\in E\setminus \left\{0\right\}$. If

$$∥tx+t^{-1}y∥\ge ∥x+y∥\mathit{for}\mathit{every}t\in \left(1-\epsilon ,1+\epsilon \right)$$

then

$$∥tx+t^{-1}y∥\ge ∥x+y∥\mathit{for}\mathit{every}t\in \left(0,\infty \right).$$

Proof. 

Let $f\left(t\right)=∥tx+t^{-1}y∥$, $t\in \left(0,\infty \right)$. The first inequality from the statement of the Lemma shows that $t=1$ is a local minimum for f. The second inequality from the statement shows that $t=1$ is a global minimum for f.

In order to prove that the first inequality from the statement implies the validity of the second inequality we shall suppose by absurd that $t=1$ is not a global minimum for f. Hence, there exists $t_0\in$$\left(0,1\right)\cup \left(1,\infty \right)$ such that $f\left(t_0\right)<f\left(1\right)$. In fact $t_0\in$$\left(0,1-ϵ\right)\cup \left(1+ϵ,\infty \right)$. We shall study two cases: the case $t_0\in \left(0,1-ϵ\right)$ and the case $t_0\in \left(1+ϵ,\infty \right)$. In the case $t_0\in \left(0,1-ϵ\right)$ we shall prove that f is constant on $\left(1-\epsilon ,1\right)$. Suppose by absurd that f is not constant on $\left(1-\epsilon ,1\right)$. Then there exists $t_1\in \left(1-\epsilon ,1\right)$ such that $f\left(t_1\right)>$ $f\left(1\right)$. This contradicts with quasiconvexity of f. Since $t_0<t_1<1$ and $f\left(1\right)<f\left(t_1\right)\le max\left(f\left(t_0\right),f\left(1\right)\right)=f\left(1\right)<f\left(t_1\right)$. It results that $f\left(t\right)=f\left(1\right)$ for every $t\in \left(1-\epsilon ,1\right)$. This contradicts with Lemma 5. Hence $t=1$ is a global minimum for f. The treatment of the case $t_0\in \left(1+ϵ,\infty \right)$ is analogous. □

Theorem 1. 

Properties (P${}_1$)–(P${}_{20}$) are equivalent.

Proof. 

Implication (P${}_1$)⇒(P${}_2$) can be easily checked by a computation. Implication (P${}_2$)⇒(P${}_3$) is obvious. In order to prove (P${}_3$)⇒(P${}_4$) we substitute $z=-y$ in (P${}_3$). In the case $∥x∥=∥y∥=1$ we find (P${}_4$). Implication (P${}_4$)⇒(P${}_1$) was proved by Schoenberg ([16], Theorem 2). In order to prove implication (P${}_1$)⇒(P${}_5$) suppose that (P${}_1$) holds. If $x,y,z\in E\setminus \left\{0\right\}$ then by the triangle inequality we have:

$$∥\frac{y}{{∥y∥}^2}-\frac{z}{{∥z∥}^2}∥\le ∥\frac{y}{{∥y∥}^2}-\frac{x}{{∥x∥}^2}∥+∥\frac{x}{{∥x∥}^2}-\frac{z}{{∥z∥}^2}∥$$

From Lemma 1 we obtain

$$\frac{∥y-z∥}{∥y∥·∥z∥}\le \frac{∥x-y∥}{∥x∥·∥y∥}+\frac{∥x-z∥}{∥x∥·∥z∥},$$

hence (P${}_5$) holds.

We shall prove now the equivalence (P${}_5$)⇔(P${}_6$). One can easily see that if we take $t=0$ in (P${}_6$) we obtain (P${}_5$). If in (P${}_5$) we substitute $x\to x-t$, $y\to y-t$, $z\to z-t$ we obtain (P${}_6$). In order to prove (P${}_5$)⇒(P${}_8$) we shall make the substitution $z\to -z$ and $x\to x+y$ in (P${}_5$).

Implication (P${}_8$)⇒(P${}_7$) follows if we make in (P${}_8$) the substitution $z\to -x$. Implication (P${}_7$)⇒(P${}_{11}$) follows if we make in (P${}_7$) the substitution $x\to u+v$, $y\to u-v$.

In order to prove (P${}_9$)⇒(P${}_7$) we shall make the substitution $x\to y+z$ in (P${}_9$). We shall obtain

$${\left(∥y-z∥+∥y+z∥\right)}^2\le 4{∥y∥}^2+4{∥z∥}^2$$

By the preceding inequality and by the following inequality

$$4∥y-z∥∥y+z∥\le {\left(∥y-z∥+∥y+z∥\right)}^2$$

one obtains the validity of (P${}_7$).

We shall prove now the implication (P${}_1$)⇒(P${}_9$). If E is an inner product space and $x,y,z\in E$ then

$$\begin{array}{cc}\hfill & {\left(∥y∥+∥x-z∥\right)}^2+{\left(∥z∥+∥x-y∥\right)}^2-{\left(∥x∥+∥y-z∥\right)}^2\hfill \\ \hfill & ={∥y∥}^2+{∥x-z∥}^2+2∥y∥∥x-z∥+{∥z∥}^2+{∥x-y∥}^2+2∥z∥∥x-y∥\hfill \\ \hfill & -{∥x∥}^2-{∥y-z∥}^2-2∥x∥∥y-z∥={∥y∥}^2+{∥x∥}^2+{∥z∥}^2-2〈x,z〉\hfill \\ \hfill & +2∥y∥∥x-z∥+{∥z∥}^2+{∥x∥}^2+{∥y∥}^2-2〈x,y〉+2∥z∥∥x-y∥\hfill \\ \hfill & -{∥x∥}^2-{∥y∥}^2-{∥z∥}^2+2〈y,z〉-2∥x∥∥y-z∥\hfill \\ \hfill & ={∥x∥}^2+{∥y∥}^2+{∥z∥}^2-2〈x,z〉-2〈x,y〉+2〈y,z〉\hfill \\ \hfill & +2\left(∥y∥∥x-z∥+∥z∥∥x-y∥-∥x∥∥y-z∥\right)\hfill \\ \hfill & ={∥x-y-z∥}^2+2\left(∥y∥∥x-z∥+∥z∥∥x-y∥-∥x∥∥y-z∥\right)\ge 0.\hfill \end{array}$$

The last inequality follows from the fact that an inner product space is a Ptolemaic space cf implication (P${}_1$)⇒(P${}_5$) that was already proved.

We shall prove now the equivalence (P${}_9$)⇔(P${}_{10}$). One can easily see that if we take $t=0$ in (P${}_{10}$) we obtain (P${}_9$). If in (P${}_9$) we substitute $x\to x-t$, $y\to y-t$, $z\to z-t$ we obtain (P${}_{10}$). The equivalence (P${}_{13}$)⇔(P${}_{14}$) is trivial.

In order to prove implication (P${}_{15}$)⇒(P${}_{18}$) we suppose that (P${}_{15}$) holds.

Consider the following property:

(P’${}_{15}$) If $x,y\in E$ and $∥x∥=∥y∥$ then $∥ax+a^{-1}y∥\ge ∥x+y∥$, $a\in (0,\infty )$.

We shall prove that (P${}_{15}$)⇔(P’${}_{15}$). The implication (P’${}_{15}$)⇒(P${}_{15}$) is trivial. In the following we show that (P${}_{15}$)⇒(P’${}_{15}$). Let $x,y\in E$ be such that $∥x∥=∥y∥\ne 0$. If $x_1=\frac{x}{∥x∥}$ and $y_1=\frac{y}{∥y∥}$ then $∥x_1∥=∥y_1∥=1$. By (P${}_{15}$) we have

$$∥a{x}_1+a^{-1}{y}_1∥\ge ∥x_1+y_1∥\mathrm{for}\mathrm{every}a\in \left(0,\infty \right).$$

Note that

$$∥ax+a^{-1}y∥=∥x∥·∥a{x}_1+a^{-1}{y}_1∥\ge ∥x∥·∥x_1+y_1∥=∥x+y∥$$

Consequently, the implication (P${}_{15}$)⇒(P’${}_{15}$) was proved.

Let $x,y\in E$ and suppose that

$$∥ax+a^{-1}y∥\ge ∥x+y∥\mathrm{for}\mathrm{every}a\in \left(0,\infty \right)$$

If $y=0$ then $∥ax∥\ge ∥x∥$ for every $a\in \left(0,\infty \right)$. Hence $x=y=0$. Suppose now that $x\ne 0$ and $y\ne 0$. Let $a=\frac{∥y∥}{∥x∥}$, $u=x$, $v=∥x∥\frac{y}{∥y∥}$. Note that $∥u∥=∥v∥$ and by (P’${}_{15}$) the following sequence of equalities and inequalities hold:

$$\begin{array}{cc}\hfill ∥x+y∥& =∥x+a∥x∥\frac{y}{∥y∥}∥=∥u+av∥=a^{1/2}∥a^{-1/2}u+a^{1/2}v∥\hfill \\ \hfill & \ge a^{1/2}∥u+v∥=a^{1/2}∥x+a^{-1}y∥=∥a^{1/2}x+a^{-1/2}y∥.\hfill \end{array}$$

Let $\alpha =\frac{1}{2}\left(1+a^{1/2}\right)$, $\beta =\frac{1}{2}\left(1+a^{-1/2}\right).$ Suppose by absurd that $a\ne 1$. We have the following:

$$\begin{array}{cc}\hfill ∥x+y∥& \ge \frac{1}{2}∥x+y∥+\frac{1}{2}∥a^{1/2}x+a^{-1/2}y∥\ge ∥\alpha x+\beta y∥\hfill \\ \hfill & =\sqrt{\alpha \beta }∥\sqrt{\frac{\alpha }{\beta }}x+\sqrt{\frac{\beta }{\alpha }}y∥\ge \sqrt{\alpha \beta }∥x+y∥.\hfill \end{array}$$

Since $x+y\ne 0$, it follows that $\alpha \beta \le 1$. Note that

$$\alpha \beta -1=\frac{1}{4}\left(1+a^{1/2}\right)\left(1+a^{-1/2}\right)-1=\frac{1}{4}{\left(a^{1/2}-a^{-1/2}\right)}^2>0$$

The contradiction we arrived implies that $a=1$, hence $∥x∥=∥y∥$.

In order to prove implication (P${}_{18}$)⇒(P${}_{13}$) suppose that $x,y\in E$, $∥x∥=∥y∥$ and (P${}_{18}$) holds. If $x=0$ then $y=0$, hence (P${}_{13}$) holds. Let $a,b\in \left(0,\infty \right)$, $\alpha =ab+a^{-1}{b}^{-1}$, $\beta =a{b}^{-1}+a^{-1}b$, $u=ax+a^{-1}y$, $v=a^{-1}x+ay$. By (P${}_{18}$) we have:

$$∥bu+b^{-1}v∥=∥b\left(ax+a^{-1}y\right)+b^{-1}\left(a^{-1}x+ay\right)∥=∥\alpha x+\beta y∥=$$

$$=\sqrt{\alpha \beta }∥\sqrt{\frac{\alpha }{\beta }}x+\sqrt{\frac{\beta }{\alpha }}y∥\ge \sqrt{\alpha \beta }∥x+y∥=$$

$$=\sqrt{\left(ab+a^{-1}{b}^{-1}\right)\left(a{b}^{-1}+a^{-1}b\right)}∥x+y∥=$$

$$=\sqrt{a^{-2}+a^2+b^2+b^{-2}}∥x+y∥\ge \sqrt{{\left(a+a^{-1}\right)}^2}∥x+y∥=$$

$$=\left(a+a^{-1}\right)∥x+y∥=∥\left(ax+a^{-1}y\right)+\left(a^{-1}x+ay\right)∥=∥u+v∥.$$

Here, the inequality

$$∥\sqrt{\frac{\alpha }{\beta }}x+\sqrt{\frac{\beta }{\alpha }}y∥\ge ∥x+y∥$$

follows at once from the implication (P${}_{18}$)⇒(P’${}_{15}$).

By (P${}_{18}$) we have $∥u∥=∥v∥$, hence $∥ax+a^{-1}y∥=∥a^{-1}x+ay∥$, thus $∥a^{2}x+y∥=∥x+a^{2}y∥$ for every $a\in \left(0,\infty \right)$. Hence, we proved that

$$∥ax+y∥=∥x+ay∥\mathrm{for}\mathrm{every}a\in \left(0,\infty \right),$$

that is (P${}_{13}$) holds.

In order to prove implication (P${}_{18}$)⇒(P${}_{15}$) suppose that $x,y\in E$ and (P${}_{18}$) holds. If $y=0$ in (P${}_{18}$) and $x\ne 0$ we obtain a contradiction. Thus, if $y=0$ then $x=0$. Suppose now that $x,y\in E\setminus \left\{0\right\}$ and $∥x∥=∥y∥$. Since

$$\underset{a\to 0}{lim}∥ax+a^{-1}y∥=\underset{a\to \infty }{lim}∥ax+a^{-1}y∥=+\infty ,$$

it follows that there exists $a_0\in \left(0,\infty \right)$ such that

$$∥ax+a^{-1}y∥\ge ∥a_{0}x+a_0^{-1}y∥\mathrm{for}\mathrm{every}a\in \left(0,\infty \right).$$

Note that

$$∥{\left(a{a}_0\right)}^{-1}{a}_{0}x+\left(a{a}_0\right)a_0^{-1}y∥=∥a^{-1}x+ay∥\ge ∥a_{0}x+a_0^{-1}y∥.$$

From (P${}_{18}$) we obtain that $∥a_{0}x∥=∥a_0^{-1}y∥$, hence $a_0=a_0^{-1}$ thus $a_0=1$.

Implication (P${}_1$)⇒(P${}_{16}$) is trivial. In order to prove the implication (P${}_1$)⇒(P${}_{17}$) let $\epsilon \in \left(0,1\right)$, $x,y\in E$ be such that $∥ax+a^{-1}y∥\ge ∥x+y∥$ for every $a\in \left(1-\epsilon ,1+\epsilon \right)$. Suppose that (P${}_1$) holds. Then, for every $a\in \left(1-\epsilon ,1+\epsilon \right)$ we have:

$$\begin{array}{cc}\hfill 0& \le {∥ax+a^{-1}y∥}^2-{∥x+y∥}^2=\left(a^2-1\right){∥x∥}^2+\left(a^{-2}-1\right){∥y∥}^2\hfill \\ \hfill & =a^{-2}\left(a^2-1\right)\left(a^2{∥x∥}^2-{∥y∥}^2\right)=a^{-2}\left(a+1\right)\left(a∥x∥+∥y∥\right)\left(a-1\right)\left(a∥x∥-∥y∥\right),\hfill \end{array}$$

hence,

$$\left(a-1\right)\left(a∥x∥-∥y∥\right)\ge 0.$$

Let $a_n=1-\frac{\epsilon }{n},b_n=1+\frac{\epsilon }{n},n\ge 1.$ Note that for every $n\ge 1$ we have $a_n,b_n\in \left(1-\epsilon ,1+\epsilon \right)$ and

$$\left(a_n-1\right)\left(a_n∥x∥-∥y∥\right)\ge 0,\left(b_n-1\right)\left(b_n∥x∥-∥y∥\right)\ge 0.$$

From the preceding inequalities we obtain:

$$a_n∥x∥\le ∥y∥\le b_n∥x∥,n\ge 1.$$

By letting $n\to \infty$ in the above inequality we obtain $∥x∥=∥y∥.$ We proved that (P${}_{17}$) holds. The implications (P${}_{18}$)⇒(P${}_{17}$) and (P${}_{16}$)⇒(P${}_{15}$) follow from Lemma 6.

We shall prove now the implication (P${}_{17}$) ⇒(P${}_{18}$). Suppose that (P${}_{17}$) holds. Let $\epsilon >0$, $x,y\in E$ be such that

$$∥ax+a^{-1}y∥\ge ∥x+y∥\mathrm{for}\mathrm{every}a\in \left(0,\infty \right).$$

Then, the above inequality holds for also for every $a\in \left(1-\epsilon ,1+\epsilon \right)$. From (P${}_{17}$) it follows that $∥x∥=∥y∥$. We proved (P${}_{18}$).

In the following, we shall prove (P${}_{12}$)⇒(P${}_1$). One can easily see that it suffices to consider only the case dim$\left(E\right)=2$. Let $e_1\in E$ be such that $∥e_1∥=1$. Denote by $S\left(0,1\right)$ the unit sphere from E. That is, $S\left(0,1\right)=\left\{x\in E:∥x∥=1\right\}$. Consider the function $f:S\left(0,1\right)\to \mathbb{R}$, $f\left(x\right)=∥e_1+x∥-∥e_1-x∥$, $x\in S\left(0,1\right)$. Since f is continuous and $f\left(-x\right)=-f\left(x\right)$ for every $x\in S\left(0,1\right)$ it follows that there exists $e_2\in S\left(0,1\right)$ such that $f\left(e_2\right)=0$. Since $∥e_1∥=∥e_2∥=1$ and $∥e_1+e_2∥=∥e_1-e_2∥$ it follows that $e_1$ and $e_2$ are linearly independent. Let $u\in E\setminus \left\{e_1,e_2,-e_1,-e_2\right\}$. Then. there exist $a,b\in \mathbb{R}\setminus \left\{0\right\}$ such that $u=a{e}_1+b{e}_2$ and $∥u∥=1$. By (P${}_{12}$) we have

$$∥e_1∥=∥-\frac{b}{a}{e}_2+\frac{1}{a}u∥=∥\frac{1}{a}{e}_2-\frac{b}{a}u∥=∥-\frac{1}{b}{e}_1+\frac{1-b^2}{ab}u∥=$$

$$=∥\frac{1-b^2}{ab}{e}_1-\frac{1}{b}u∥=∥\frac{1-a^2-b^2}{ab}{e}_1-e_2∥=∥e_2∥$$

By statement $4^0$ of Lemma 3, we obtain $\frac{1-a^2-b^2}{ab}=0$, hence $a^2+b^2=1$. We proved that $∥a{e}_1+b{e}_2∥=1$ implies $a^2+b^2=1$. Let $\alpha$,$\beta \in \mathbb{R}$ such that ${\alpha }^2+{\beta }^2\ne 0$. One can easily see that $∥\alpha e_1+\beta e_2∥=t\ne 0$. We obtain

$$∥\frac{\alpha }{t}{e}_1+\frac{\beta }{t}{e}_2∥=1,$$

hence,

$${\left(\frac{\alpha }{t}\right)}^2+{\left(\frac{\beta }{t}\right)}^2=1,$$

thus,

$$∥\alpha e_1+\beta e_2∥=t=\sqrt{{\alpha }^2+{\beta }^2}.$$

If $x=x_1{e}_1+x_2{e}_2$ and $y=y_1{e}_1+y_2{e}_2$ are two elements of E then we shall define the inner product as follows:

$$〈x,y〉=x_1{y}_1+x_2{y}_2.$$

This completes the proof of the implication (P${}_{12}$)⇒(P${}_1$).

The proof of (P${}_{11}$)⇒(P${}_4$) is obvious. We shall prove in the following the implication (P${}_{13}$)⇒(P${}_{12}$). Suppose that (P${}_{13}$) holds and let $x,y\in E$ such that $∥x∥=∥y∥$ and $a,b\in \mathbb{R}$. If $ab=0$ then (P${}_{12}$) clearly holds. If $ab\ne 0$ and $\epsilon \in \left\{-1,1\right\}$ is the sign of $a/b$ then

$$\begin{array}{cc}\hfill ∥ax+by∥& =\left|b\right|·∥\frac{a}{b}x+y∥=\left|b\right|·∥\left|\frac{a}{b}\right|\left(\epsilon x\right)+y∥=\left|b\right|·∥\epsilon x+\left|\frac{a}{b}\right|y∥\hfill \\ & =\left|b\right|·∥x+\left|\frac{a}{b}\right|\left(\epsilon y\right)∥=\left|b\right|·∥x+\frac{a}{b}y∥=∥bx+ay∥.\hfill \end{array}$$

We used (P${}_{13}$) and the fact that $∥x∥=∥y∥=∥\epsilon x∥=∥\epsilon y∥$.

Implication (P${}_1$)⇒(P${}_{19}$) can be proved by direct calculus. Now, let us prove implication (P${}_{19}$)⇒(P${}_1$). Suppose that (P${}_{19}$) holds and take $x=y+z$. We obtain the following:

$${∥y+z∥}^2+a∥y-z∥∥y+z∥+{∥y-z∥}^2\le \left(a+2\right)\left({∥y∥}^2+{∥z∥}^2\right)$$

Substitute in the above inequality $y=\frac{u+v}{2},z=\frac{u-v}{2}$, we get $u=y+z$, $v=y-z$ so

$${∥u∥}^2+a∥u∥∥v∥+{∥v∥}^2\le \frac{1}{4}\left(a+2\right)\left({∥u+v∥}^2+{∥u-v∥}^2\right)$$

In the case $∥u∥=∥v∥=1$ the above inequality becomes:

$$a+2\le \frac{1}{4}\left(a+2\right)\left({∥u+v∥}^2+{∥u-v∥}^2\right)$$

hence,

$$4\le {∥u+v∥}^2+{∥u-v∥}^2$$

The above property is property (P${}_4$). Since (P${}_4$)⇒(P${}_1$) it follows that the implication (P${}_{19}$)⇒(P${}_1$) is proved.

Implication (P${}_1$)⇒(P${}_{20}$) can be proved by direct calculus. Now let us prove implication (P${}_{20}$)⇒(P${}_1$). Suppose that (P${}_{20}$) holds.

Then, for every $x_1,x_2,x_3\in E$ the following inequality holds:

$$\theta \left(x_1,x_2,x_3,0\right)\le \theta \left(x_1,x_3,x_2,0\right)+\theta \left(x_1,0,x_2,x_3\right)$$

hence, ${∥x_1-x_2∥}^2+{∥x_3∥}^2\le {∥x_1-x_3∥}^2+{∥x_2∥}^2+{∥x_1∥}^2+{∥x_2-x_3∥}^2.$

If in the above inequality we substitute $x_3=x_1+x_2$ we obtain the following:

$${∥x_1-x_2∥}^2+{∥x_1+x_2∥}^2\le 2\left({∥x_1∥}^2+{∥x_2∥}^2\right).$$

We proved that (P${}_{20}$) implies the validity of the above inequality for every $x_1,x_2\in E$. If in the above inequality we make the substitutions: $x_1+x_2=u$, $x_1-x_2=v$ we obtain

$$2\left({∥u∥}^2+{∥v∥}^2\right)\le {∥u-v∥}^2+{∥u+v∥}^2.$$

Hence, the above inequality holds for every $u,v\in E$. The last two inequalities are equivalent with the parallelogram property, so (P${}_1$) holds. □

3. Conclusions

The aim of the present paper is to investigate the equivalence of several conditions in a linear normed space that imply that the norm is induced by an inner product.

Regarding the equivalences between the conditions studied in the paper, one can see that some metric relations (identities or inequalities) in elementary geometry, which can be formulated in linear normed spaces, hold only in inner product spaces. This is the case of

(i)

Parallelogram law.

(ii)

Parallelogram law with inequality of type 1. (The sum of the squares of the lengths of the four sides of a parallelogram is greater than the sum of the squares of the lengths of the two diagonals).

(iii)

Parallelogram law with inequality of type 2. (The inequality in (ii) is reversed).

(iv)

Property (P${}_6$) which is equivalent with the following property of the tetrahedron: “The products of opposite edges of a tetrahedron are the sides of a triangle”.

(v)

Property (P${}_{20}$) which is equivalent with the following property: “In a tetrahedron, the sum of squares of opposite edges are the lengths of a triangle”.

(vi)

Properties (P${}_9$) and (P${}_{10}$) which are equivalent with the following property “The sum of opposite edges of a tetrahedron are the sides of an acute triangle”.

The main contributions of the present paper are the following:

(i)

Several new conditions that imply that a norm of a linear space is induced by an inner product were introduced. See conditions (${\mathrm{P}}_{10}$), (${\mathrm{P}}_{16}$), (${\mathrm{P}}_{17}$), (${\mathrm{P}}_{19}$) and (${\mathrm{P}}_{20}$) that appear in this paper for the first time.

(ii)

Several proofs appeared in the literature were simplified and made clearer. See, for example, the implication (${\mathrm{P}}_{12}$) ⇒(${\mathrm{P}}_1$), which is a proof that outperforms the proof given in Ficken [4] and Cichon [10]. Other proofs of implications that appear for the first time in this paper are (${\mathrm{P}}_{18}$)⇒(${\mathrm{P}}_{13}$), (${\mathrm{P}}_{18}$)⇒(${\mathrm{P}}_{15}$), (${\mathrm{P}}_{17}$)⇒(${\mathrm{P}}_{18}$), (${\mathrm{P}}_{12}$)⇒(${\mathrm{P}}_1$), (${\mathrm{P}}_{19}$)⇒(${\mathrm{P}}_1$) and (${\mathrm{P}}_{20}$)⇒(${\mathrm{P}}_1$).