Richard's inequality, Cauchy-Schwarz's inequality and approximate solutions of Sincov's equation

TL;DR

This paper explores the relationship between Richard's and Cauchy-Schwarz inequalities in inner product spaces and investigates the stability of Sincov's functional equation, showing that unbounded approximate solutions are exact.

Contribution

It establishes a connection between classical inequalities and proves super-stability of Sincov's equation for unbounded functions.

Findings

Unbounded approximate solutions to Sincov's equation are exact solutions.

A link between Richard's inequality and Cauchy-Schwarz inequality is demonstrated.

The super-stability of Sincov's equation is proven for unbounded mappings.

Abstract

We observe a connection between Cauchy-Schwarz' and Richard's inequalities in inner product spaces and a Ulam-type stability problem for multiplicative Sincov's functional equation. We prove that this equation is super-stable for unbounded mappings, i.e. every unbounded approximate solution is an exact solution.

Equations

$$\left|⟨a|x⟩\cdot ⟨x|b⟩-\frac{⟨a|b⟩\cdot \Vert x{\Vert }^2}{2}\right|\le \frac{\Vert a\Vert \cdot \Vert b\Vert \cdot \Vert x{\Vert }^2}{2}$$

$$\big{|}\left\langle a|x\right\rangle\cdot\left\langle x|b\right\rangle\big{|}\leq\frac{1}{2}\big{[}\|a\|\cdot\|b\|+|\left\langle a|b\right\rangle|\big{]}\cdot\|x\|^{2}.$$

$$F(u,v):=\frac{2⟨u|v⟩}{\Vert u\Vert \cdot \Vert v\Vert },u,v\in H\setminus \{0\}.$$

$$|F(a,x)\cdot F(x,b)-F(a,b)|\le 2,a,b,x\in H\setminus \{0\},$$

$$|F(u,v)|\le 2,u,v\in H\setminus \{0\}.$$

$$F(a,x)\cdot F(x,b)=F(a,b),a,b,x\in X.$$

$$|F(a,x)\cdot F(x,b)-F(a,b)|\le c,a,b,x\in X.$$

$$F(a,b):=\frac{a}{b+c},a,b\in X.$$

$$\sup \{F(a,b):a,b\in X\}=\frac{n^2}{n+c}$$

$$|F(a,x)\cdot F(x,b)-F(a,b)|=\frac{ac}{(b+c)(x+c)}\le \frac{n^2}{(n+c{)}^2}c<c,a,b,x\in X.$$

$$|F(a,x)F(x,a)-F(a,a)|$$

$$|F(x,a)F(a,x)-F(x,x)|$$

$$|F(x,x)|\le |F(x,a)\cdot F(a,x)|+c\le |F(a,a)|+2c.$$

$$|F(x_0,b)|\cdot |F(x_0,x_0)-1|\le c,b\in X.$$

$$|F(a_n,y)\cdot F(y,y_0)-F(a_n,y_0)|\le c.$$

$$F(u,v)=\frac{f(u)}{f(v)},u,v\in X.$$

$$\left|F(a,b)-f(a)g(b)\right|\le c,a,b\in X.$$

$$|F(a,x)f(x)g(b)-f(a)g(x)f(x)g(b)|\le c|f(x)|\cdot |g(b)|$$

$$|F(a,x)F(x,b)-F(a,x)f(x)g(b)|\le c|F(a,x)|\le c^2+c|f(a)|\cdot |g(x)|.$$

$$|F(a,x)F(x,b)-f(a)g(x)f(x)g(b)|\le c^2+c(|f(x)|\cdot |g(b)|+|f(a)|\cdot |g(x)|).$$

$$|f(a)g(x)f(x)g(b)-f(a)g(b)|$$

$$+|F(a,x)F(x,b)-F(a,b)|+|F(a,b)-f(a)g(b)|$$

$$\le c^2+2c+c(|f(x)|\cdot |g(b)|+|f(a)|\cdot |g(x)|).$$

$$|g(x)f(x)-1|\le \frac{c^2+2c}{|f(a)|\cdot |g(b)|}+c\frac{|f(x)|}{|f(a)|}+c\frac{|g(x)|}{|g(b)|}.$$

$$\left|F(u,v)-\frac{f(u)}{f(v)}\right|\le c,u,v\in X.$$

$$\left|F(u,v)-\frac{F(a_n,v)}{F(a_n,u)}\right|\le \frac{c}{|F(a_n,u)|}.$$

$$\underset{n\to \infty }{\lim }\frac{F(a_n,v)}{F(a_n,u)}=F(u,v).$$

$$\underset{n\to \infty }{\lim }\frac{F(a_n,v)-\frac{f(a_n)}{f(v)}}{F(a_n,u)}=\underset{n\to \infty }{\lim }\frac{F(a_n,u)-\frac{f(a_n)}{f(u)}}{F(a_n,v)}=0.$$

$$F(u,v)=\underset{n\to \infty }{\lim }\frac{F(a_n,v)}{F(a_n,u)}=\underset{n\to \infty }{\lim }\frac{\frac{f(a_n)}{f(v)}}{\frac{f(a_n)}{f(u)}}=\frac{f(u)}{f(v)}.$$

$$F(u,v)=\left[\begin{array}[]{cc}\frac{u}{v}&0\\ 0&c_{0}\end{array}\right],\quad u,v\in X.$$

$$\Vert F(a,x)\cdot F(x,b)-F(a,b)\Vert =|c_0^2-c_0|=ca,b,x\in X.$$

Full text

1. Inequalities

Let $H$ be an inner product space over the field $\mathbb{K}\in\{\mathbb{R},\mathbb{C}\}$. In the present note we are concerned with Richard’s inequality (see [11]):

[TABLE]

for all $a,b,x\in H$. Earlier, Ch. Blatter [3, Lemma 2] obtained a one-sided estimate of (1) in a real inner product space, but his result remained unknown for several years. Inequality (1) improves Buzano inequality (see [4]):

[TABLE]

for all $a,b,x\in H$. Observe that for $a=b$ inequality (2) reduces to Cauchy-Schwarz’s inequality. Recently, a number of papers regarding inequalities (1) and (2) appeared. We refer the reader to a recent article by S.S. Dragomir [5] and references therein.

Let us define a map $F\colon(H\setminus\{0\})\times(H\setminus\{0\})\to\mathbb{K}$ by the formula

[TABLE]

With the aid of this notation inequality (1) can be rewritten as

[TABLE]

whereas Cauchy-Schwarz’s inequality is equivalent to

[TABLE]

The main purpose of the paper is to study a connection between (3) and (4) in case $F$ is an arbitrary mapping defined on a non-empty set. Moreover, we will replace the number $2$ appearing on the right-hand sides of both estimates by an arbitrary non-negative constant.

2. Stability of Sincov’s functional equation

2.1. The problem

Let $X$ be a non-empty set, $F\colon X\times X\to\mathbb{C}$ be an arbitrary mapping and assume that $c\geq 0$ is a fixed constant.

Multiplicative Sincov’s functional equation is of the form

[TABLE]

Equation (5) has an old history and is of significant importance. For details we refer the reader to papers of D. Gronau [7, 8].

In this section we will study approximate solutions of (5). We are concerned with the following Ulam-type stability problem:

[TABLE]

Observe that (6) is a direct abstract version of (3) and we assume that mapping $F\colon X\times X\to\mathbb{C}$ is arbitrary.

Dealing with a Ulam-type stability problem one wants to establish a relation between approximate solutions of a given equation and exact solutions. Several types of stability behaviour are known (main types are: stability, super-stability, hyper-stability). For more informations on the stability of functional equations the reader is referred to a survey article of Z. Moszner [10]. On page 64 of [10] some variants of Sincov’s equation together with their stabilities appear.

2.2. Bounded solutions

In this subsection we exploit the case of bounded mappings satisfying (6). First, we give some examples. Next, some conditions related to the boundedness of solution of (6) are provided.

Our first easy remark shows that every bounded mapping is a solution of (6).

Remark 1*.*

Assume that $X$ is a non-empty set, $F\colon X\times X\to\mathbb{C}$ is a bounded mapping and define $c_{0}:=\sup\{|F(x,y)|:x,y\in X\}.$ Then $F$ solves (6) with $c=c_{0}^{2}+c_{0}$. In general the value $c=c_{0}^{2}+c_{0}$ cannot be replaced by a smaller constant. To see this take arbitrary $c_{0}>0$ and define $F\colon X\times X\to\mathbb{C}$ as a constant mapping equal to $-c_{0}$. Then, the left-hand side of (6) is equal to $c_{0}^{2}+c_{0}$.

Next, we provide an example which shows that for a given $c>0$ there exist bounded but arbitrary large solutions of (6) which do not satisfy equation (5).

Example 1*.*

Let $c>0$ and $n\in\mathbb{N}$ be fixed constants. Take $X=\{n,n+1,\ldots,n^{2}\}$ and define $F\colon X\times X\to\mathbb{C}$ by

[TABLE]

Then

[TABLE]

and

[TABLE]

Another example showing a similar effect in a different settings is due to Z. Moszner [10, p. 64]. His claim was that multiplicative Sincov’s equation is not stable in a sense of Ulam in the class of mappings which attain positive values only. In order to justify this he observed that if $n>2$ is a positive integer such that $(1-1/n)/n\leq c$, then a constant function $F_{c}$ which is equal to $1/n$ satisfies (6). Further, it is easy to observe that every solution of (5) with non-zero values is equal to $1$ on the diagonal and, as a consequence, can’t be uniformly close to $F_{c}$. Note however, that if we do not restrict ourselves to the class of positive mappings, then the function constantly equal to zero is a solution of (5) which uniformly approximates $F_{c}$.

Next, we will establish some basic properties of solutions of (6) related to boundedness.

Proposition 1**.**

Assume that $c\geq 0$ is fixed, $F\colon X\times X\to\mathbb{C}$ solves (6) and $a,x\in X$ are arbitrary. Then:

• (i)

$|F(a,a)-F(x,x)|\leq 2c$,

• (ii)

the value $|F(x,x)|$ is bounded by a constant depending upon $F$,

• (iii)

the product $|F(a,x)\cdot F(x,a)|$ is bounded by a constant depending upon $F$.

Proof.

Directly from (6) we obtain

[TABLE]

which proves (i).

To justify (ii) it is enough to observe that by (7) and (8) we have

[TABLE]

Now, (iii) follows from the above computation and from (ii). ∎

Proposition 2**.**

Assume that $c\geq 0$ is fixed and $F\colon X\times X\to\mathbb{C}$ solves (6). If for a certain $x_{0}\in X$ we have $F(x_{0},x_{0})\neq 1$, then both functions $F(x_{0},\cdot)$ and $F(\cdot,x_{0})$ are bounded.

Proof.

Suppose that we are given $x_{0}\in X$ such that function $F(x_{0},\cdot)$ is unbounded. Then apply (6) for $a=x=x_{0}$ to get

[TABLE]

Clearly, since the value $|F(x_{0},b)|$ can be as large as we want, we get $F(x_{0},x_{0})=1$.

Proof in case $F(\cdot,x_{0})$ is unbounded is similar. ∎

2.3. Unbounded solutions

In this subsection we prove that multiplicative Sincov’s functional equation is super-stable. This means that unbounded mappings which satisfy inequality (6) solve equation (5) and, as a consequence are of some prescribed form.

Lemma 1**.**

Assume that $c\geq 0$ is fixed, $F\colon X\times X\to\mathbb{C}$ is a solution of (6) and there exists a $y_{0}\in X$ such that the map $F(\cdot,y_{0})$ is unbounded. Then $F$ never vanishes and for every $y\in X$ the map $F(\cdot,y)$ is unbounded.

Proof.

Fix a sequence $(a_{n})_{n\in\mathbb{N}}\subset X$ such that $|F(a_{n},y_{0})|\to\infty$ and an arbitrary point $y\in X$. From (6) we have

[TABLE]

Thus $|F(a_{n},y)\cdot F(y,y_{0})|\to\infty$, which implies that $|F(a_{n},y)|\to\infty$ and $F(y,y_{0})\neq 0$. To get that $F$ never vanishes it is enough to repeat the foregoing reasoning and apply the already proven part with $y_{0}$ and $y$ replaced by two arbitrary points. ∎

The next lemma can be proven analogously.

Lemma 2**.**

Assume that $c\geq 0$ is fixed, $F\colon X\times X\to\mathbb{C}$ is a solution of (6) and there exists a $x_{0}\in X$ such that the map $F(x_{0},\cdot)$ is unbounded. Then $F$ never vanishes and for every $x\in X$ the map $F(x,\cdot)$ is unbounded.

Our next example shows that assumptions of both previous lemmas are independent.

Example 2*.*

Let $X=[1,+\infty)$ and define $F\colon X\times X\to\mathbb{C}$ as $F(x,y)=\frac{x}{y}$. Then $F$ solves (6) with every $c\geq 0$, for every $x_{0}\in X$ the map $F(x_{0},\cdot)$ is bounded and for every $y_{0}\in X$ the map $F(\cdot,y_{0})$ is unbounded.

Now, we are ready to state and prove our main theorem.

Theorem 1**.**

Assume that $c\geq 0$ is fixed, $F\colon X\times X\to\mathbb{C}$ is a solution of (6) and there exist $x_{0},y_{0}\in X$ such that the maps $F(x_{0},\cdot),F(\cdot,y_{0})$ are unbounded. Then, there exists a mapping $f\colon X\to\mathbb{C}\setminus\{0\}$ such that

[TABLE]

Proof.

Define $f,g\colon X\to\mathbb{C}$ as $f:=F(\cdot,x_{0})$ and $g:=F(x_{0},\cdot)$. In view of Lemma 1 and Lemma 2 both mappings are unbounded and never vanish. Apply (6) with $x=x_{0}$ to get

[TABLE]

Our first claim is that $f(x)g(x)=1$ for every $x\in X$. In order to show this, fix arbitrary $a,b,x\in X$. Apply (10) to obtain

[TABLE]

and

[TABLE]

Adding these inequalities together we get

[TABLE]

Using the triangle inequality and then (11), (6) and (10) we derive

[TABLE]

From this we have

[TABLE]

Since both mappings $f$ and $g$ are unbounded, then keeping $x$ fixed one can choose $a,b\in X$ such that the right-hand side of (12) is arbitrary small. This proves the desired equality $f(x)g(x)=1$. This, together with (10) gives us the estimate.

[TABLE]

To finish the proof fix $u,v\in X$ arbitrarily. By Lemma 1 there exists a sequence $(a_{n})_{n\in\mathbb{N}}\subset X$ such that $|F(a_{n},u)|\to\infty$. From (6) we get

[TABLE]

From this we see that

[TABLE]

On the other hand, since $|F(a_{n},u)|\to\infty$ and $|F(a_{n},v)|\to\infty$, then by (13) we have

[TABLE]

Consequently, we obtain

[TABLE]

∎

Remark 2*.*

It is easy to see that function $f$ appearing in formula (9) is determined uniquely up to a multiplicative constant.

3. Conclusions and final remarks

Remark 3*.*

Our Theorem 1 is in line with well-known results on the superstability of the exponential Cauchy equation and related equations of J.A. Baker, J. Lawrence, F. Zorzitto [1], J.A. Baker [2], R. Ger, P. Šemrl [6], T. Kochanek, M. Lewicki [9], among others. However, taking into account the case of bounded solutions of (6) no full analogy can be observed, see Example 1. Therefore, we can conclude that the multiplicative Sincov’s equation is super-stable, however, there is no control over bounded solutions.

The following example is motivated by an example due to J.A. Baker from [2] and shows that in higher dimension the super-stability effect for Sincov’s equation does not hold.

Example 3*.*

Denote by $M_{2\times 2}(\mathbb{R})$ the algebra of all real $2\times 2$ matrices equipped with the usual matrix norm. Take $X=(0,\infty)$ and fix a constant $c>0$. Choose $c_{0}>0$ such that $|c_{0}^{2}-c_{0}|=c$ and let $F\colon X\times X\to M_{2\times 2}(\mathbb{R})$ be defined as

[TABLE]

It is easy to check that

[TABLE]

Therefore $F$ does not solve equation (5), is unbounded and it satisfies the stability inequality (6).

4. Acknowledgements

The work performed in this study was supported by the National Science Centre, Poland (under Grant No. 2015/19/B/ST6/03259).