The Cauchy-Schwarz Inequality in Complex Normed Spaces

TL;DR

This paper introduces a generalized product in complex normed spaces, proves the Cauchy-Schwarz inequality within this context, and offers a new proof that relies solely on the norm, extending understanding beyond inner product spaces.

Contribution

It defines a new product in complex normed spaces and proves the Cauchy-Schwarz inequality without using linearity of the inner product.

Findings

The generalized product satisfies the Cauchy-Schwarz inequality.

A new proof of the inequality that depends only on the norm.

Properties of the generalized product are explored.

Abstract

We introduce a product in all complex normed vector spaces, which generalizes the inner product of complex inner product spaces. Naturally the question occurs whether the Cauchy-Schwarz inequality is fulfilled. We provide a positive answer. This also yields a new proof of the Cauchy-Schwarz inequality in complex inner product spaces, which does not rely on the linearity of the inner product. The proof depends only on the norm in the vector space. Further, we present some properties of the generalized product.

Equations

$$<\vec{x}|\vec{y}>=\frac{1}{4}\cdot \left[\Vert \vec{x}+\vec{y}{\Vert }^2-\Vert \vec{x}-\vec{y}{\Vert }^2+\mathbf{i}\cdot \left(\Vert \vec{x}+\mathbf{i}\cdot \vec{y}{\Vert }^2-\Vert \vec{x}-\mathbf{i}\cdot \vec{y}{\Vert }^2\right)\right],$$

$$\Vert \vec{x}\Vert \cdot \Vert \vec{y}\Vert \cdot \frac{1}{4}\cdot \left[{\Vert \frac{\vec{x}}{\Vert \vec{x}\Vert }+\frac{\vec{y}}{\Vert \vec{y}\Vert }\Vert }^2-{\Vert \frac{\vec{x}}{\Vert \vec{x}\Vert }-\frac{\vec{y}}{\Vert \vec{y}\Vert }\Vert }^2+\mathbf{i}\cdot \left({\Vert \frac{\vec{x}}{\Vert \vec{x}\Vert }+\mathbf{i}\cdot \frac{\vec{y}}{\Vert \vec{y}\Vert }\Vert }^2-{\Vert \frac{\vec{x}}{\Vert \vec{x}\Vert }-\mathbf{i}\cdot \frac{\vec{y}}{\Vert \vec{y}\Vert }\Vert }^2\right)\right].$$

$$<e^{\mathbf{i}\cdot \phi }\cdot \vec{x}|\vec{x}>=e^{\mathbf{i}\cdot \phi }\cdot <\vec{x}|\vec{x}>,\text{and}<e^{\mathbf{i}\cdot \phi }\cdot \vec{x}|e^{\mathbf{i}\cdot \phi }\cdot \vec{y}>=<\vec{x}|\vec{y}>.$$

$$\left\|\left(\begin{array}[]{c}r+{\mathbf{i}}\cdot s\\ v+{\mathbf{i}}\cdot w\end{array}\right)\right\|_{\infty}\ =\ \max\left\{\sqrt{r^{2}+s^{2}},\sqrt{v^{2}+w^{2}}\right\}\ .$$

$$\vec{x}\ :=\ \frac{1}{4}\cdot\left(\begin{array}[]{cr}1+{\mathbf{i}}\cdot\sqrt{15}\\ 2+{\mathbf{i}}\cdot 2\end{array}\right)\ \quad\text{and}\ \ \ \vec{y}\ :=\ \frac{1}{4}\cdot\left(\begin{array}[]{c}2+{\mathbf{i}}\\ 3+{\mathbf{i}}\cdot\sqrt{7}\end{array}\right)\ .$$

$$<\vec{x}|\vec{y}>=\frac{1}{64}\cdot \left(19+4\cdot \sqrt{7}+2\cdot \sqrt{15}+\mathbf{i}\cdot \left[7-4\cdot \sqrt{7}+4\cdot \sqrt{15}\right]\right)\approx 0.583+\mathbf{i}\cdot 0.186.$$

$$\displaystyle e^{{\mathbf{i}}\cdot\varphi}\cdot\vec{x}\,=\,\frac{1}{8}\cdot\left(\begin{array}[]{c}1-\sqrt{45}+{\mathbf{i}}\cdot\left[\sqrt{3}+\sqrt{15}\right]\\ 2-2\cdot\sqrt{3}+{\mathbf{i}}\cdot\left[2+2\cdot\sqrt{3}\right]\end{array}\right),\ \text{and we compute the product}\ <e^{{\mathbf{i}}\cdot\varphi}\cdot\vec{x}\,|\,\vec{y}>,$$

$$<e^{\mathbf{i}\cdot \phi }\cdot \vec{x}|\vec{y}>=\left(p+\mathbf{i}\cdot q\right)/64\approx 0.113+\mathbf{i}\cdot 0.628,\text{where}p\text{and}q\text{abbreviate real numbers}$$

$$p=11+2\cdot \left(\sqrt{7}+\sqrt{21}-\sqrt{45}\right)-5\cdot \sqrt{3}+\sqrt{15},q=8+2\cdot \left(4\cdot \sqrt{3}-\sqrt{7}+\sqrt{15}+\sqrt{21}\right)+\sqrt{45}.$$

$$|<\vec{x}|\vec{y}>|\le \Vert \vec{x}\Vert \cdot \Vert \vec{y}\Vert$$

$$|<\vec{a}|\vec{b}>|>\Vert \vec{a}\Vert \cdot \Vert \vec{b}\Vert .$$

$$|<\vec{x}|\vec{y}>|>1.$$

$$\left<\left(\begin{array}[]{c}1\\ 0\end{array}\right)\ |\ \left(\begin{array}[]{c}0\\ 1\end{array}\right)\right>\ =\ \ \frac{1}{4}\ \cdot\ \left[\;\left\|\left(\begin{array}[]{c}1\\ 1\end{array}\right)\right\|^{2}\ -\ \left\|\left(\begin{array}[]{c}1\\ -1\end{array}\right)\right\|^{2}\ +\ {\mathbf{i}}\ \cdot\left(\left\|\left(\begin{array}[]{c}1\\ {\mathbf{i}}\end{array}\right)\right\|^{2}\ -\ \left\|\ \left(\begin{array}[]{c}1\\ -{\mathbf{i}}\end{array}\right)\right\|^{2}\right)\;\right]\ .$$

$$\left\|\left(\begin{array}[]{c}1\\ 1\end{array}\right)\right\|\ =:\ \frac{1}{s}\ ,\ \ \left\|\left(\begin{array}[]{c}1\\ -1\end{array}\right)\right\|\ =:\ \frac{1}{t}\ ,\ \ \left\|\left(\begin{array}[]{c}1\\ {\mathbf{i}}\end{array}\right)\right\|\ =:\ \frac{1}{v}\ ,\ \ \left\|\left(\begin{array}[]{c}1\\ -{\mathbf{i}}\end{array}\right)\right\|\ =:\ \frac{1}{w}\ ,$$

$$\left<\left(\begin{array}[]{c}1\\ 0\end{array}\right)\ |\ \left(\begin{array}[]{c}0\\ 1\end{array}\right)\right>\ =\ \ \frac{1}{4}\ \cdot\ \left[\;\left(\frac{1}{s}\right)^{2}\ -\ \left(\frac{1}{t}\right)^{2}\ +\ {\mathbf{i}}\cdot\left(\left(\frac{1}{v}\right)^{2}\ -\ \left(\frac{1}{w}\right)^{2}\right)\;\right]\ .$$

$$\left|4\cdot\left<\left(\begin{array}[]{c}1\\ 0\end{array}\right)\ |\ \left(\begin{array}[]{c}0\\ 1\end{array}\right)\right>\right|^{2}\ =\ \left[\;\left(\frac{1}{s}\right)^{2}\ -\ \left(\frac{1}{t}\right)^{2}\;\right]^{2}\ +\ \left[\left(\frac{1}{v}\right)^{2}\ -\ \left(\frac{1}{w}\right)^{2}\;\right]^{2}\ \ \leq\ \ 16\ .$$

$$\frac{1}{s}\le 2\cdot \sqrt{1+2\cdot b^2-2\cdot b}+\sqrt{2}\cdot |b|\cdot \frac{1}{w},\frac{1}{v}\le 2\cdot \sqrt{1+2\cdot b^2-2\cdot b}+\sqrt{2}\cdot |b|\cdot \frac{1}{t}.$$

$$(\mathsf{A}):{\left(\frac{1}{s}\right)}^2\le 2+\frac{\sqrt{4\cdot w^2-1}}{w^2}\text{and}(\mathsf{B}):{\left(\frac{1}{v}\right)}^2\le 2+\frac{\sqrt{4\cdot t^2-1}}{t^2}.$$

$$\mathcal{R}(b):=2\cdot \sqrt{1+2\cdot b^2-2\cdot b}+\sqrt{2}\cdot |b|\cdot \frac{1}{w}.$$

$${\mathcal{R}}^{\prime }(b)=\frac{4\cdot b-2}{\sqrt{1+2\cdot b^2-2\cdot b}}+\frac{\sqrt{2}}{w},\text{for all}b\ge 0.$$

$$b_E=\frac{1}{2}\cdot \left[1-\frac{1}{\sqrt{4\cdot w^2-1}}\right](\text{We have}w>1/2\text{by the lemmas}\text{ref}allererstesLemma\text{and}\text{ref}zweitesLemma).$$

$$\frac{2\cdot w^2}{4\cdot w^2-1}.$$

$$\left(b_E,\mathcal{R}(b_E)\right)=\left(\frac{1}{2}\cdot \left[1-\frac{1}{\sqrt{4\cdot w^2-1}}\right],\frac{\sqrt{2}}{2\cdot w}\cdot \left[1+\sqrt{4\cdot w^2-1}\right]\right).$$

$${\left(\mathcal{R}(b_E)\right)}^2=2+\frac{\sqrt{4\cdot w^2-1}}{w^2},$$

$$\displaystyle\left|4\cdot\left<\left(\begin{array}[]{c}1\\ 0\end{array}\right)\ |\ \left(\begin{array}[]{c}0\\ 1\end{array}\right)\right>\right|^{2}\leq\$$

$$\displaystyle=\$$

$$(\mathsf{C}):$$

$$(\mathsf{D}):$$

$$\displaystyle\left(\frac{p+1}{2}-1\right)\cdot\sqrt{z}\ +\ \left(\frac{z+1}{2}-1\right)\cdot\sqrt{p}\ \ \leq\$$

$$\displaystyle\Longleftrightarrow\ \ \ p\cdot\sqrt{z}-\sqrt{z}\ +\ z\cdot\sqrt{p}-\sqrt{p}\ \ \leq\$$

$$\displaystyle\ \ \ h^{2}\cdot k-k+k^{2}\cdot h-h\ \ \leq\ \ \frac{1}{2}\cdot\left(h^{2}\cdot k^{2}+h^{2}+k^{2}+1\right)\$$

$$\displaystyle\Longleftrightarrow\$$

$$h^2\cdot \left(-2\cdot k+k^2+1\right)-2\cdot h\cdot (k^2-1)+\left(2\cdot k+k^2+1\right)$$

Full text

Keywords and phrases: complex normed space, complex inner product space, Cauchy-Schwarz inequality

AMS subject classification: 46B99

1 Introduction

We deal with vector spaces $X$ over the complex field ${\mathbb{C}}$, provided with a norm $\|\cdot\|$. As a motivation we begin with the special case of an inner product space $\left(X,<\cdot\,|\,\cdot>\right)$. The inner product $<\cdot\,|\,\cdot>$ generates a norm by $\|\vec{x}\|=\sqrt{<\vec{x}\,|\,\vec{x}>}$, for all $\vec{x}\in X$. By the same token it is well known that the inner product can be expressed by this norm, namely for $\vec{x},\vec{y}\in X$ we can write

[TABLE]

where the symbol ‘${\mathbf{i}}$’ means the imaginary unit.

We use an idea in [3] to generate a continuous product in all complex normed vector spaces $(X,\|\cdot\|)$, which is just the inner product in the special case of a complex inner product space.

Definition 1.1**.**

*Let $\vec{x},\vec{y}$ be two arbitrary elements of $X$. In the case of  $\vec{x}=\vec{0}$ or $\vec{y}=\vec{0}$ we set  $<\vec{x}\>|\>\vec{y}>\ :=\,0$, and if both  $\vec{x},\vec{y}\neq\vec{0}\ (\text{i.e.}\ \|\vec{x}\|\cdot\|\vec{y}\|>0\;)$  we define the complex number

$<\vec{x}\>|\>\vec{y}>\ :=$

[TABLE]

It is easy to show that the product fulfills the conjugate symmetry ($<\vec{x}\>|\>\vec{y}>$ $\ =\ \overline{<\vec{y}\>|\>\vec{x}>}$), where $\overline{<\vec{y}\>|\>\vec{x}>}$ means the complex conjugate of $<\vec{y}\>|\>\vec{x}>$, the positive definiteness ($<\vec{x}\>|\>\vec{x}>\ \geq\ 0,\ \text{and}\ <\vec{x}\>|\>\vec{x}>\ =0\ \text{only for}\ \vec{x}=\vec{0}$), and the homogeneity for real numbers ($<r\cdot\vec{x}\,|\,\vec{y}>\ =r\ \cdot<\vec{x}\,|\,\vec{y}>\ =\ <\vec{x}\,|\,r\cdot\vec{y}>$), and the homogeneity for pure imaginary numbers ($<r\cdot{\mathbf{i}}\cdot\vec{x}\,|\,\vec{y}>\ =\ r\ \cdot{\mathbf{i}}\,\cdot<\vec{x}\,|\,\vec{y}>\ =\ <\vec{x}\,|\,-r\cdot{\mathbf{i}}\cdot\vec{y}>$),  for $\vec{x},\vec{y}\in X,\ r\in{\mathbb{R}}$.  Further, for $\vec{x}\in X$ it holds  $\|\vec{x}\|=\sqrt{<\vec{x}\>|\>\vec{x}>}$.

The product from Definition 1.1 opens the possibility to define a generalized ‘angle’ both in real normed spaces, see [4], and in complex normed spaces, see [5]. In this paper we turn our focus on the product. We prove the famous Cauchy-Schwarz-Bunjakowsky inequality, or briefly the Cauchy-Schwarz inequality. Further we notice some properties of the product.

Let $(X,\|\cdot\|)$ be an arbitrary complex normed vector space. In Definition 1.1 we defined a continuous product  $<\cdot\>|\>\cdot>$  on $X$. This is an inner product in the case that the norm $\|\cdot\|$ generates this product by the equation of line (1.1).

Generally, for spaces $X\neq\{\vec{0}\}$, the codomain of the product from Definition 1.1 is the entire complex plane ${\mathbb{C}}$, i.e. we have a surjective map  $<\cdot\,|\,\cdot>:X^{2}\rightarrow{\mathbb{C}}$. If we restrict the domain of the product $<\cdot\,|\,\cdot>$ on unit vectors of $X$, it is easy to see that the codomain changes into the ‘complex square’  $\{r+{\mathbf{i}}\cdot s\in{\mathbb{C}}\ |\ -1\leq r,s\leq+1\}$. We can improve this statement: Actually the codomain is the complex unit circle $\{r+{\mathbf{i}}\cdot s\in{\mathbb{C}}\ |r^{2}+s^{2}\leq 1\}$. This is a consequence of the Cauchy-Schwarz-Bunjakowsky inequality or ‘${\mathsf{CSB}}$ inequality’ .

First we show that for a proof of this inequality we can restrict our research on the two dimensional complex vector space ${\mathbb{C}}^{2}$, provided with all possible norms.

2 General Definitions and Properties

Let  $(X,\|\cdot\|)$ be an arbitrary complex vector space provided with a * norm* $\|\cdot\|$, this means that there is a continuous map $\|\cdot\|$:  $X\longrightarrow$ ${\mathbb{R}}^{+}\cup\{0\}$ which fulfills the following axioms  $\|z\cdot\vec{x}\|\ =|z|\cdot\|\vec{x}\|$  (‘absolute homogeneity’),  $\|\vec{x}+\vec{y}\|\leq\|\vec{x}\|+\|\vec{y}\|$ (‘triangle  inequality’), and  $\|\vec{x}\|=0$ only for $\vec{x}=\vec{0}$ (‘positive definiteness’),  for   $\vec{x},\vec{y}\in X$ and $z\in{\mathbb{C}}$.

Let $<\cdot\>|\>\cdot>\ :\ X^{2}\longrightarrow{\mathbb{C}}$ be a map from the product space $X\times X$ into the field ${\mathbb{C}}$.  Such a map is called a

• product*.

Assume that the complex vector space  $X$  is provided with a norm $\|\cdot\|$, and further there is a product $<\cdot\>|\>\cdot>:X\times X\rightarrow{\mathbb{C}}$. We say that the triple $(X,\|\cdot\|,<\cdot\>|\>\cdot>)$ satisfies the * Cauchy-Schwarz-Bunjakowsky Inequality* or ‘${\mathsf{CSB}}$ inequality’, or briefly the * Cauchy-Schwarz Inequality*, if and only if for all $\vec{x},\vec{y}\in X$ there is the inequality

$|<\vec{x}\>|\>\vec{y}>|\ \leq\ \|\vec{x}\|\cdot\|\vec{y}\|\;$.

It is well known that a complex normed space $(X,\|\cdot\|)$, where the product of Definition 1.1 is actually an inner product, fulfills the ${\mathsf{CSB}}$ inequality.

Let $(X,\|\cdot\|)$ be an arbitrary complex normed vector space. $<\cdot\>|\>\cdot>$  on $X$. In the introduction we already mentioned that the product of Definition 1.1 is an inner product in the case that the norm $\|\cdot\|$ generates this product by the equation in line (1.1).

Proposition 2.1**.**

*For all vectors  $\vec{x},\vec{y}\in(X,\|\cdot\|)$ and for real numbers  $r$  the product  $<\cdot\>|\>\cdot>$  of Definition 1.1 has the following properties.

${\mathrm{(a)}}\ <\vec{x}\>|\>\vec{y}>\ =\ \overline{<\vec{y}\>|\>\vec{x}>}$ (conjugate symmetry),

${\mathrm{(b)}}\ <\vec{x}\>|\>\vec{x}>\ \ \geq\ 0,\ \text{and}\ <\vec{x}\>|\>\vec{x}>\ =0\ \text{only for}\ \vec{x}=\vec{0}$ (positive definiteness),

{\mathrm{(c)}}\ <r\cdot\vec{x}\,|\,\vec{y}>\ =\ r\,\cdot\,<\vec{x}\,|\,\vec{y}>\ =\ <\vec{x}\,|\,r\cdot\vec{y}>\ (homogeneity for real numbers),

${\mathrm{(d)}}\ <r\cdot{\mathbf{i}}\cdot\vec{x}\,|\,\vec{y}>\ =\ r\cdot{\mathbf{i}}\,\cdot<\vec{x}\,|\,\vec{y}>\ =\ <\vec{x}\,|\,-r\cdot{\mathbf{i}}\cdot\vec{y}>$ (homogeneity for pure imaginary   

numbers), 

${\mathrm{(e)}}\ \|\vec{x}\|=\sqrt{<\vec{x}\>|\>\vec{x}>}$ (the norm can be expressed by the product).*

Proof.

We use Definition 1.1, and the proofs for ${\mathrm{(a)}}$ and ${\mathrm{(b)}}$ are easy. For positive $r\in{\mathbb{R}}$ the point ${\mathrm{(c)}}$ is trivial. We can prove $<-\vec{x}|\vec{y}>=-<\vec{x}|\vec{y}>=<\vec{x}|-\vec{y}>$, and ${\mathrm{(c)}}$ follows immediately. The point ${\mathrm{(d)}}$ is similar to ${\mathrm{(c)}}$, and ${\mathrm{(e)}}$ is clear. ∎

Lemma 2.2**.**

For a pair $\vec{x},\vec{y}\in X$ of unit vectors, i.e. $\|\vec{x}\|=1=\|\vec{y}\|$, it holds that both the real part and the imaginary part of $<\vec{x}\>|\>\vec{y}>$  are in the interval $[-1,1]$.

Proof.

The lemma can be proven easily with the triangle inequality. ∎

Corollary 2.3**.**

Lemma 2.2 means, that  $\left\{<\vec{x}\>|\>\vec{y}>\ |\ \vec{x},\vec{y}\in X,\ \|\vec{x}\|=1=\|\vec{y}\|\right\}$ is a subset of the ‘complex square’   $\{r+{\mathbf{i}}\cdot s\in{\mathbb{C}}\ |\ -1\leq r,s\leq+1\}$.  Immediately, it follows for unit vectors $\vec{x},\vec{y}$ the estimate  $\left|<\vec{x}\>|\>\vec{y}>\right|\ \leq\ \sqrt{2}$.

Now we notice a few facts about the general product $<\cdot\,|\,\cdot>$ from Definition 1.1.

Lemma 2.4**.**

In a complex normed space $(X,\|\cdot\|)$ for $\vec{x},\vec{y}\in X$ and real $\varphi$ there are identities

[TABLE]

Proof.

To prove the first equation take an unit vector $\vec{x}$, and write  $e^{{\mathbf{i}}\,\cdot\varphi}=\cos(\varphi)+{\mathbf{i}}\,\cdot\sin(\varphi)$, and use Definition 1.1. The second identity comes directly from Defintion 1.1. ∎

Corollary 2.5**.**

For an unit vector $\vec{x}\in X$ we have that the set $\{<e^{{\mathbf{i}}\cdot\varphi}\cdot\vec{x}\,|\,\vec{x}>\ |\ \varphi\in[0,2\,\pi]\}$ is the complex unit circle, since  $<e^{{\mathbf{i}}\cdot\varphi}\cdot\vec{x}\,|\,\vec{x}>\ =\ e^{{\mathbf{i}}\ \cdot\varphi}\cdot<\vec{x}\,|\,\vec{x}>\ =\ e^{{\mathbf{i}}\cdot\varphi}$.

The next example shows that in a complex normed space $(X,\|\cdot\|)$ generally we have the inequality  $<e^{{\mathbf{i}}\cdot\varphi}\cdot\vec{x}\,|\,\vec{y}>\ \neq\ e^{{\mathbf{i}}\,\cdot\varphi}\cdot<\vec{x}\,|\,\vec{y}>$. This statement seems to be ‘probable’, but we need an example, which we yield in the proof of the following lemma.

This inequality means, that the set of products  $\left\{<e^{{\mathbf{i}}\cdot\varphi}\cdot\vec{x}\,|\,\vec{y}>\ |\ \varphi\in[0,2\pi]\right\}$  commonly does not generate a proper Euclidean circle (with radius $\ |<\vec{x}\,|\,\vec{y}>|$) in ${\mathbb{C}}$.

If we take $\varphi\in\{\pi,\pi/2,$ $\pi\cdot 3/2\}$, however, we get with Proposition 2.1 three identities

$<-\vec{x}\,|\,\vec{y}>\ =\ -<\vec{x}\,|\,\vec{y}>\ ,\ <{\mathbf{i}}\cdot\vec{x}\,|\,\vec{y}>\ =\ {\mathbf{i}}\,\cdot<\vec{x}\,|\,\vec{y}>\ ,\ \text{and}\ <-{\mathbf{i}}\cdot\vec{x}\,|\,\vec{y}>\ =\ -{\mathbf{i}}\ \cdot<\vec{x}\,|\,\vec{y}>$.

Lemma 2.6**.**

*In a complex normed space $(X,\|\cdot\|)$ generally it holds the inequality

$<e^{{\mathbf{i}}\cdot\varphi}\cdot\vec{x}\,|\,\vec{y}>\ \neq\ e^{{\mathbf{i}}\,\cdot\varphi}\cdot<\vec{x}\,|\,\vec{y}>$,  even their moduli are different.

Proof.

We use the most simple non-trivial example of a complex normed space, let  $(X,\|\cdot\|):=$ $\left({\mathbb{C}}^{2},\|\cdot\|_{\infty}\right)$, where for two complex numbers  $r+{\mathbf{i}}\cdot s\,,\ v+{\mathbf{i}}\cdot w\in{\mathbb{C}}$  we have its norm $\|\cdot\|_{\infty}$ by

[TABLE]

The following calculations are easy, but tiring. We define two unit vectors  $\vec{x},\vec{y}$  of  $\left({\mathbb{C}}^{2},\|\cdot\|_{\infty}\right)$,

[TABLE]

Some calculations yield the complex number

[TABLE]

We choose  $e^{{\mathbf{i}}\cdot\varphi}:=1/2\cdot\left(1+{\mathbf{i}}\cdot\sqrt{3}\right)$  from the complex unit circle, and we get approximately  $e^{{\mathbf{i}}\,\cdot\,\varphi}\,\cdot<\vec{x}\,|\,\vec{y}>\ \approx\ 0.130+{\mathbf{i}}\cdot 0.598.$   After that we take the unit vector

[TABLE]

[TABLE]

This proves the inequality  $<e^{{\mathbf{i}}\cdot\varphi}\cdot\vec{x}\,|\,\vec{y}>\ \neq\ e^{{\mathbf{i}}\,\cdot\varphi}\cdot<\vec{x}\,|\,\vec{y}>$,  and the lemma is confirmed. ∎

The above lemma suggests the following conjecture. One direction is trivial.

Conjecture 2.7**.**

In a complex normed space $(X,\|\cdot\|)$ for all  $\vec{x},\vec{y}\in X,\,\varphi\in{\mathbb{R}}$, it holds

• $<e^{{\mathbf{i}}\cdot\varphi}\cdot\vec{x}\,|\,\vec{y}>\ =\ e^{{\mathbf{i}}\,\cdot\varphi}\,\cdot\,<\vec{x}\,|\,\vec{y}>$*

if and only if its product $<\cdot\;|\;\cdot>$ from Definition 1.1 is actually an inner product, i.e. $(X,<\cdot\;|\;\cdot>)$ is an inner product space.

3 The Cauchy-Schwarz-Bunjakowsky Inequality

In this section we deal with the famous Cauchy-Schwarz-Bunjakowsky inequality or ‘${\mathsf{CSB}}$ inequality’ , or briefly the Cauchy-Schwarz inequality. Another name is the ‘Polarization Inequality’. Let $X$ be a complex normed space, let $\|\cdot\|$ be the norm on $X$ and let $<\cdot\>|\>\cdot>$ be the product from Definition 1.1. We ask whether in the triple $(X,\|\cdot\|,<\cdot\>|\>\cdot>)$ the inequality

[TABLE]

is fulfilled for all $\vec{x},\vec{y}\in X$. The answer is positive.

Theorem 3.1**.**

The Cauchy-Schwarz-Bunjakowsky inequality in line (3.1) holds in all complex vector spaces $X$, provided with a norm $\|\cdot\|$ and the product $<\cdot\>|\>\cdot>$ from Definition 1.1.

Remark 3.2**.**

This theorem is the main contribution of the paper. The proof of the Cauchy-Schwarz inequality in inner product spaces is well documented in many books about functional analysis by using the linearity of the inner product, see for instance [7], p.204. This new proof of the Cauchy-Schwarz inequality depends only on the norm in the vector space.

Proof.

First we need a lemma, which shows that for a complete answer it suffices to investigate the complex vector space ${\mathbb{C}}^{2}$, provided with all possible norms.

Lemma 3.3**.**

*The following two statements ${\mathrm{(1)}}$ and ${\mathrm{(2)}}$ are equivalent.

${\mathrm{(1)}}$ There exists a complex normed vector space $(X,\|\cdot\|)$ and two vectors $\vec{a},\vec{b}\in X$  with*

[TABLE]

${\mathrm{(2)}}$* There is a norm $\|\cdot\|$ on ${\mathbb{C}}^{2}$ and two unit vectors $\vec{x},\vec{y}\in{\mathbb{C}}^{2}$  with*

[TABLE]

Proof.

(1)   $\Leftarrow$    (2)     Trivial.

    (1)   $\Rightarrow$    (2)    Easy.  Let us consider the two-dimensional subspace $\mathsf{{U}}$ of $X$ which is spaned by the linear independent vectors $\vec{a},\vec{b}$. This space $\mathsf{{U}}$ is isomorphic to ${\mathbb{C}}^{2}$. We take the norm from $X$ on $\mathsf{{U}}$. We normalize $\vec{a},\vec{b}$, i.e. we define unit vectors $\vec{x}:=\vec{a}/\|\vec{a}\|$, and $\vec{y}:=\vec{b}/\|\vec{b}\|$. Hence the inequality (3.2) turns into (3.3). ∎

The lemma means, that we can restrict our investigations on the complex vector space ${\mathbb{C}}^{2}$. By a transformation of coordinates we state that instead of the unit vectors $\vec{x},\vec{y}$ of inequality (3.3) we set $(1,0):=\vec{x}$, and $(0,1):=\vec{y}$. With Definition 1.1 the product   $<(1,0)|(0,1)>$   has the presentation

[TABLE]

We take four suitable real numbers  $s,t,v,w$, and we define four positive values

[TABLE]

or, equivalently, we have four unit vectors    $(s,s),(t,-t),(v,{\mathbf{i}}\cdot v),(w,-{\mathbf{i}}\cdot w),\ \text{ i. e.}$

$1=\|(s,s)\|=\|(t,-t)\|=\|(v,{\mathbf{i}}\cdot v)\|=\|(w,-{\mathbf{i}}\cdot w)\|$.

Hence the product   $<(1,0)|(0,1)>$   changes into

[TABLE]

Further, instead of the ${\mathsf{CSB}}$ inequality $|<(1,0)|(0,1)>|\leq 1$, for an easier handling we can deal with the equivalent inequality

[TABLE]

Lemma 3.4**.**

All four numbers  $s,t,v,w$  are greater or equal $1/2$.

Proof.

For instance to show $1/2\leq s$, use the equation $(s,s)=s\cdot(1,0)+s\cdot(0,1)$. Apply the triangle inequality, and note  $\|(s,s)\|=1$,  and also  $\|(0,1)\|=1=\|(1,0)\|$. ∎

The next lemma means, that we can assume that both the real part and the imaginary part of   $<(1,0)|(0,1)>$   are positive.

Lemma 3.5**.**

Without restriction of generality we assume $s<t$ and $v<w$.

Proof.

In the case of $s=t$, the first sumand of the middle term in line (3.7) is zero. From Lemma 3.4 follows $1/v\leq 2$. Hence $|4\cdot<(1,0)|(0,1)>|^{2}\leq(1/v)^{4}\leq(2)^{4}=16$, it holds (3.7).

In the case of $s>t,\ \text{i.e.}\ 1/s<1/t$, i.e. we have a negative real part of $<(1,0)|(0,1)>$, we consider instead $<(-1,0)|(0,1)>$. By Proposition 2.1(c), we get a positive real part. With a transformation of coordinates we rename $-(1,0)$ into $(1,0)$, to get a representation $<(1,0)|(0,1)>$ with positive real part. In the case that the imaginary part of $<(1,0)|(0,1)>$ is still negative, we take the product $<(0,1)|(1,0)>$. Now, by Proposition 2.1(a), also the imaginary part is positive. We make a second transformation of coordinates, and in new coordinates we call this $<(1,0)|(0,1)>$. ∎

The following propositions Proposition 3.6 and Proposition 3.7 collect general properties of the product $<(1,0)|(0,1)>$ from line (3.4). The proofs always rely on the triangle inequality of a normed space, which is equivalent to the fact that its unit ball is convex.

The next proposition looks weird, but it will give the deciding hint for the proof.

Proposition 3.6**.**

We get for each $b\in{\mathbb{R}}$ the following two inequalities.

[TABLE]

Proof.

Please see the following Proposition 3.10. From the line (3.32) we get

$1/s\ \leq\ \sqrt{(1-a)^{2}+b^{2}}\ +\ \sqrt{(1-b)^{2}+a^{2}}\ +\ \sqrt{a^{2}+b^{2}}\,/w$ ,

which is true for arbitrary real numbers $a,b$. If we choose $b=a$, it follows the first inequality of Proposition 3.6. The second inequality uses the corresponding equation of line (3.33) . ∎

The above Proposition 3.6 has an important consequence.

Proposition 3.7**.**

If  $\frac{1}{2}\cdot\sqrt{2}\leq w$  it holds inequality ${\mathsf{(A)}}$, and in the case of  $\frac{1}{2}\cdot\sqrt{2}\leq t$  it holds inequality ${\mathsf{(B)}}$, where

[TABLE]

Proof.

To prove inequality ${\mathsf{(A)}}$   we consider again Proposition 3.6, and we investigate the right hand side of the first inequality in line (3.8). For all constants $w\geq 1/2$, we define a function ${\cal R}(b),\ b\in{\mathbb{R}}$,

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Obviously we get the limits  $\lim_{b\rightarrow+\infty}{\cal R}(b)=\lim_{b\rightarrow-\infty}{\cal R}(b)=+\infty$ and since the parabola $1+2\cdot b^{2}-2\cdot b$ has only positive values, we state that ${\cal R}$ has a codomain of positive numbers,

${\cal R}:{\mathbb{R}}\rightarrow{\mathbb{R}}^{+}\ .$

By Proposition 3.6 it holds $1/s\leq{\cal R}(b)$ for all $b$, hence we are interested in minimums of ${\cal R}$, to get an estimate for $1/s$ as small as possible. Since ${\cal R}(-b)>{\cal R}(b)$ for all positive $b$, the minimum must occur at a non negative $b$. Therefore, we consider the function ${\cal R}$ for non negative $b$. The search for a minimum is the standard method, we have

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In the case of  $\frac{1}{2}\cdot\sqrt{2}<w$   the equation  ${\cal R}^{\prime}(b_{E})=0$  has one positive solution  $b_{E}$,

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Recall that we are looking for positive $b_{E}\,^{\prime}s$, hence we are investigating the positive part in the definition (3.9) of ${\cal R}(b)$, i.e. $b\geq 0$. Note that the condition $0\leq b_{E}$ holds if $\frac{1}{2}\cdot\sqrt{2}\leq w$. We add this as an assumption, i.e. in this Proposition we assume  $\frac{1}{2}\cdot\sqrt{2}\leq w,t$.

As an intermediate step we mention that for the term  $1+2\cdot b^{2}-2\cdot b$   for   $b=b_{E}$   we get the value

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Finally we get an expression of ${\cal R}\ \text{at}\ b_{E}$, we have

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By Proposition 3.6 we get an estimate for $1/s$, but actually we are more interested in an estimate for $(1/s)^{2}$. We calculate

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which finishes the proof of Proposition 3.7. ∎

The above propositions may be a useful tool for further computations, but we do not know whether the list is complete. For our purpose it will be sufficient. With Proposition 3.7 we are able to do the final stroke. We are still proving Theorem 3.1, i.e. we try to confirm the Cauchy-Schwarz inequality (3.7). To prove the theorem, we need to distinguish between three cases $\mathsf{(Case\,a),(Case\,b),(Case\,c)}$; only the third will be difficult.

$\mathsf{(Case\,a)}$: Let both $t,w$ be in the closed interval  $\left[1/2,\,\sqrt{2}/2\right]$. Hence   $|4\cdot<(1,0)|(0,1)>|^{2}=$  $\left[(1/s)^{2}-(1/t)^{2}\right]^{2}+\left[(1/v)^{2}-(1/w)^{2}\right]^{2}\,\leq\,2\cdot\left[4-(2/\sqrt{2})^{2}\right]^{2}\,=\,2\cdot\left[2\right]^{2}=8\ .$

$\mathsf{(Case\,b)}$: Let  $1/2\,<\,t\,\leq\,\sqrt{2}/2\,<\,w$  or the contrary   $1/2\,<\,w\,\leq\,\sqrt{2}/2\,<\,t$.

We use the estimate ${\mathsf{(A)}}$ or ${\mathsf{(B)}}$ from Proposition 3.7, and we compute

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and it is trivial that the last sum is less than 16, hence the Cauchy Schwarz inequality (3.7) is confirmed for $\mathsf{(Case\,b)}$.  Before we deal with the last case $\mathsf{(Case\,c)}$, one more lemma is necessary.

Lemma 3.8**.**

*Let  $|t|,|w|\geq\frac{1}{2}$.

  The following two inequalities $\mathsf{(C)}$ and $\mathsf{(D)}$ are equivalent, and both are true.*

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Proof.

Starting with $\mathsf{(C)}$, the proof of the equivalence is straightforward.

The last step is to confirm the second inequality $\mathsf{(D)}$ for all $|t|,|w|\geq\frac{1}{2}$. This needs two tricky substitutions. The first is  $p:=4\cdot t^{2}-1$  and  $z:=4\cdot w^{2}-1$. The inequality $\mathsf{(D)}$ leads to

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The second substitution is  $h:=\sqrt{p}$  and  $k:=\sqrt{z}$.  It follows the equivalent inequalities

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We multiply it by ($-2$), and we get

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Obviously, the last inequality is true. Hence, both inequalities ${\mathsf{(C)}}$ and ${\mathsf{(D)}}$ in the lemma are also correct, for real  $t,w$   with $|t|,|w|\geq\frac{1}{2}$. ∎

Remark 3.9**.**

In the above Lemma 3.8 in the second inequality $\mathsf{(D)}$ it occurs equality if and only if   $h\cdot(k-1)=k+1$, or equivalently if the two variables $t$ and $w$ fulfill the relation

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Note that in this formula the variables   $t$ and $w$   can be exchanged.

Further we remark that in inequality $\mathsf{(D)}$, for   $|t|=\frac{1}{2}\cdot\sqrt{2}$   or   $|w|=\frac{1}{2}\cdot\sqrt{2}$   both sides of $\mathsf{(D)}$ have a constant difference of 1.

Now we regard the last case.

$\mathsf{(Case\,c)}$: Let  $\sqrt{2}/2\,<\,t,w$.

From Proposition 3.7 we have the estimates ${\mathsf{(A)}}$ and ${\mathsf{(B)}}$, hence we get the inequality

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Together with Lemma 3.8 this yields the last step to prove the Cauchy-Schwarz-Bunjakowsky inequality, since from inequality $\mathsf{(C)}$ in Lemma 3.8 follows the ${\mathsf{CSB}}$ inequality, i.e. Theorem 3.1 finally is confirmed. ∎

We add a few propositions which we do not use (except the lines (3.31), (3.32) and (3.33)) in the proof of Theorem 3.1. We define the map $G,\ G:{\mathbb{R}}\times{\mathbb{R}}\longrightarrow{\mathbb{R}}^{+}$,

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We look for the infimum $\underline{{\cal M}}$ of $G$. This infimum $\underline{{\cal M}}$ must be a minimum, since it is easy to see that it occurs in the closed square

$\{(a,b)\in{\mathbb{R}}\times{\mathbb{R}}\ |\ 0\leq a,b\leq 1\}$.

The values of $G$ can be interpreted as the sum of three hypotenuses of three rectangle triangles. We are not able to find $\underline{{\cal M}}$, but if we restrict our search on the diagonal $a=b$, we find there with elementary analysis at  $a=b=(3-\sqrt{3})\,/\,6\ \approx\ 0.211$  the minimum ${{\cal M}}$, where

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This also might be the true global minimum $\underline{{\cal M}}$ of the map $G$. Please note that we just considered here the special case $w=1$ from line (3.9) in Proposition 3.7.

Proposition 3.10**.**

It holds

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Proof.

We show the first of the two inequalities. Please see the equation

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which is true for arbitrary $a,b\in{\mathbb{R}}$. By the triangle inequality, we get

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Either $\|(1,-{\mathbf{i}})\|\leq 1$ or $\|(1,-{\mathbf{i}})\|>1$. It follows the first inequality in this proposition. The other inequality uses the corresponding equation of the next line (3.33).

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∎

Proposition 3.11**.**

Let  $s>1/2$  and  $v>1/2$,  respectively. It holds

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Proof.

We prove $s\leq\frac{s}{2\cdot s-1}$. This is equivalent to $s\,\leq\,1$.  Due to the proof of the following proposition this is always true. ∎

Proposition 3.12**.**

It holds both

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Proof.

Please see the equation  $(s,0)=1/2\cdot(s,s)+1/2\cdot(s,-s)$. Note the norms   $\|(s,0)\|=s,\ \|(s,s)\|=1=\|(t,-t)\|$. Since $s\leq t$ we have $\|(s,-s)\|\leq 1$. Now apply the triangle inequality to the equation. ∎

Proposition 3.13**.**

*Let  $s,t>1/2$.    The following two tripels

$(t,-t),(1,0),\left(\frac{t}{2\cdot t-1},\frac{t}{2\cdot t-1}\right)$  and  $(s,s),(1,0),\left(\frac{s}{2\cdot s-1},\frac{-s}{2\cdot s-1}\right)$ are collinear. (On two different lines, of course, except the special cases  $2\cdot s\cdot t=s+t$).*

Proof.

The first statement is proven by  $(t,-t)-(1,0)\ =\ (1-2\cdot t)\cdot\left[\frac{t}{2\cdot t-1}\cdot(1,1)-(1,0)\right]$.  The second statement uses  $(s,s)-(1,0)\ =\ (1-2\cdot s)\cdot\left[\frac{s}{2\cdot s-1}\cdot(1,-1)-(1,0)\right]$. ∎

Proposition 3.14**.**

$\min\{t,w\}\ \leq\ \sqrt{2}$.

Proof.

We use $e^{{\mathbf{i}}\cdot\pi/4}$ from the complex unit circle, where

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First assume $w\leq t$. We write

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By $w\leq t$, we have the norms $\|(-{\mathbf{i}}\cdot w,{\mathbf{i}}\cdot w)\|=\|(w,-w)\|\leq\|(t,-t)\|=1$, and  $\|(w,-{\mathbf{i}}\cdot w)\|=1$.  By the triangle inequality it follows  $\frac{1}{2}\cdot\sqrt{2}\cdot w\leq\frac{1}{2}+\frac{1}{2}$ .  The case $t\leq w$ is treated with the corresponding equation

$\frac{1}{2}\cdot\sqrt{2}\cdot t\cdot\left(1,0\right)\ \ =\ \ \left(\ \frac{1}{2}\cdot\sqrt{2}\ +\ {\mathbf{i}}\cdot\frac{1}{2}\cdot\sqrt{2}\right)\ \cdot\ \left[\frac{1}{2}\cdot\left(-{\mathbf{i}}\cdot t,{\mathbf{i}}\cdot t\right)\ +\ \frac{1}{2}\cdot\left(t,-{\mathbf{i}}\cdot t\right)\right]$  .

∎

Proposition 3.15**.**

There are two inequalities

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Proof.

For instance,   $1/v\leq 1/s+1/t+1/w$   is proven by the following equation,

$(1,{\mathbf{i}})=(1,1)+(1,-1)-(1,-{\mathbf{i}})$.

∎

Proposition 3.16**.**

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Proof.

For the first inequality use two times the equation   $(1,1)=(2,0)-(1,-1)$. ∎

Proposition 3.17**.**

Let  $\alpha\in\{s,t\}$,  and  $\gamma\in\{v,w\}$.

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Proof.

We need to show  $\frac{1}{v}-\frac{1}{t}\leq\sqrt{2}$  and  $\frac{1}{s}-\frac{1}{w}\leq\sqrt{2}$  and  $\sqrt{2}\leq\ \frac{1}{t}+\frac{1}{w}$.

Note $|1+{\mathbf{i}}|=|1-{\mathbf{i}}|=\sqrt{2}$,  and consider

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Hence it follows  $\frac{1}{v}\ \leq\ \frac{1}{t}+\sqrt{2}$.  Please see also   $(1,1)=(1,-{\mathbf{i}})+(1+{\mathbf{i}})\cdot(0,1)$  and   $(1-{\mathbf{i}})\cdot(0,1)=(1,-{\mathbf{i}})-(1,-1)$ . ∎

Proposition 3.18**.**

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Proof.

We show  $s\geq\sqrt{2}\,\cdot\left[\frac{v\cdot w}{v+w}\right]$.  We use

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The other inequality needs  \left(\begin{array}[]{c}1\\ {\mathbf{i}}\end{array}\right)\ =\ \frac{1+{\mathbf{i}}}{2\cdot s}\ \cdot\ \left(\begin{array}[]{c}s\\ s\end{array}\right)\ +\ \ \frac{1-{\mathbf{i}}}{2\cdot t}\cdot\left(\begin{array}[]{c}t\\ -t\end{array}\right)  . ∎

**Acknowledgements: ** We wish to thank Prof. Dr. Eberhard Oeljeklaus for a careful reading of the paper and some helpful calculations, and Dr. Malte von Arnim, who found the substitutions in the proof of inequality $\mathsf{(D)}$ in Lemma 3.8.