A new look at Bernoulli's inequality

TL;DR

This paper introduces a novel generalization of Bernoulli's inequality using discrete fractional calculus, providing new insights into its mathematical properties.

Contribution

It presents a new approach to Bernoulli's inequality through discrete fractional calculus, which has not been explored before.

Findings

Established a generalized form of Bernoulli's inequality

Applied discrete fractional calculus to inequality analysis

Proposed a novel mathematical framework for inequalities

Abstract

In this work, a generalization of the well known Bernoulli inequality is obtained by using the theory of discrete fractional calculus. As far as we know our approach is novel.

Equations

$$(1+x{)}^n\ge 1+nx.$$

$$({\Delta }_{a+\nu -1}^{\nu }x)(t)$$

$$x(a+\nu -1)$$

$$x^{(y)}$$

$$({\Delta }_a^{0}f)(t)$$

$$({\Delta }_a^{-\nu }f)(t)$$

$$({\Delta }_a^{-1}f)(t)=\sum _{s=a}^{t-1}f(s).$$

$$({\Delta }_a^{\nu }f)(t)=(\Delta {\Delta }_a^{-(1-\nu )}f)(t),t\in {\mathbb{N}}_{a+\nu -1}.$$

$$x(t)=\frac{(t-a{)}^{(\nu -1)}}{\Gamma (\nu )}{x}_{a+\nu -1}+\frac{1}{\Gamma (\nu )}\sum _{s=a}^{t-\nu }(t-\sigma (s){)}^{(\nu -1)}f(s+\nu -1,x(s+\nu -1)),$$

$$({\Delta }_{a+\nu -1}^{\nu }x)(t)$$

$$x(a+\nu -1)$$

$$(T_y^{0}f)(t)$$

$$(T_y^{1}f)(t)$$

$$(T_y^{k+1}f)(t)$$

$$x(t)=\sum_{k=a}^{t-(\nu-1)}\left[\frac{x_{a+\nu-1}}{\Gamma{(\nu)}}\left(T^{k-a}_{y}(s-a)^{(\nu-1)}\right)(t)+\left(T^{k-a}_{y}\Delta^{-\nu}_{a}z(s+\nu-1)\right)(t)\right],\$$

$$x(t)=\frac{(t-a{)}^{(\nu -1)}}{\Gamma (\nu )}{x}_{a+\nu -1}+\left({\Delta }_a^{-\nu }[y(s+\nu -1)x(s+\nu -1)+z(s+\nu -1)]\right)(t),$$

$$\left({\Delta }_a^{-\nu }y(s+\nu -1)f(s+\nu -1)\right)(t),$$

$$E(t,a,\nu ,\beta ,c,\lambda )=\sum _{k=a}^{t-(\nu -1)}\frac{c^{k-a}}{\Gamma ((k-a)\nu +\beta )}(t-\lambda +(k-a)(\nu -1){)}^{((k-a)\nu +\beta -1)},$$

$$x(t)=x_{a+\nu -1}E(t,a,\nu ,\nu ,c,a)+\sum _{r=a}^{t-\nu }E(t,a,\nu ,\nu ,c,\sigma (r))z(r+\nu -1),t\in {\mathbb{N}}_{a+\nu -1}.$$

$$x(t)=x_{a+\nu -1}E(t,a,\nu ,\nu ,c,a)+KE(t,a,\nu ,\nu +1,c,a),t\in {\mathbb{N}}_{a+\nu -1}.$$

$${\Delta }_t(s-t{)}^{(r)}=-r(s-\sigma (t){)}^{(r-1)},s,r\in \mathbb{R}.$$

$$\sum _{k=a}^{b-1}\Delta f(k)=f(b)-f(a).$$

$$\sum _{r=a}^{t-\nu }E(t,a,\nu ,\nu ,c,\sigma (r))z(r+\nu -1)$$

$$=K\sum _{r=a}^{t-\nu }\sum _{k=a}^{t-(\nu -1)}\frac{c^{k-a}}{\Gamma ((k-a)\nu +\nu )}(t-\sigma (r)+(k-a)(\nu -1){)}^{((k-a)\nu +\nu -1)}$$

$$=K\sum _{k=a}^{t-(\nu -1)}\frac{c^{k-a}}{\Gamma ((k-a)\nu +\nu )}\sum _{r=a}^{t-(\nu -1)-1}(t-\sigma (r)+(k-a)(\nu -1){)}^{((k-a)\nu +\nu -1)}$$

$$=K\sum _{k=a}^{t-(\nu -1)}\frac{c^{k-a}}{\Gamma ((k-a)\nu +\nu )((k-a)\nu +\nu )}$$

$$\cdot \sum _{r=a}^{t-(\nu -1)-1}-((k-a)\nu +\nu )(t-\sigma (r)+(k-a)(\nu -1){)}^{((k-a)\nu +\nu -1)}$$

$$=K\sum _{k=a}^{t-(\nu -1)}\frac{c^{k-a}}{\Gamma ((k-a)\nu +\nu +1)}[-(t-r+(k-a)(\nu -1){)}^{((k-a)\nu +\nu )}{]}_{r=a}^{r=t-(\nu -1)}$$

$$=K\sum _{k=a}^{t-(\nu -1)}\frac{c^{k-a}}{\Gamma ((k-a)\nu +\nu +1)}(t-a+(k-a)(\nu -1){)}^{((k-a)\nu +\nu )}$$

$$=KE(t,a,\nu ,\nu +1,c,a),$$

$$({\Delta }_{a+\nu -1}^{\nu }x)(t)$$

$$x(a+\nu -1)$$

$$({\Delta }_{a+\nu -1}^{\nu }x)(t)=c_1(t)x(t+\nu -1),t\in \{a,a+1,a+2,\dots \}$$

$$({\Delta }_{a+\nu -1}^{\nu }y)(t)=c_2(t)y(t+\nu -1),t\in \{a,a+1,a+2,\dots \},$$

$$x(t)\ge y(t),t\in {\mathbb{N}}_{a+\nu -1}.$$

$$({\Delta }_a^{\nu }f)(t)=\frac{1}{\Gamma (-\nu )}\sum _{k=a}^{t+\nu }(t-\sigma (k){)}^{(-\nu -1)}f(k),t\in {\mathbb{N}}_{a-\nu +1},$$

Full text

The author was supported by the “Fundação para a Ciência e a Tecnologia (FCT)” through the program “Investigador FCT” with reference IF/01345/2014.

1. Introduction

In classical analysis the following inequality is attributed to Bernoulli: for a real number $x>-1$ and a nonnegative integer $n$, it holds:

[TABLE]

One can find in the literature several (elementary) different proofs of inequality (1.1) (see e.g. [1, 10]). Moreover, various generalizations were also obtained throughout the years (cf. [9]) as well as different kinds of applications (see e.g. [7]).

In this work we obtain an inequality that generalizes (1.1) in a completely different direction than the ones mentioned before. The reasoning is that we use the theory of discrete fractional calculus [4], in particular, the (delta) Riemann–Liouville fractional operators which were introduced by Miller and Ross in 1988 [8] and for which real developments happened only in the past eight years. Therefore, we actually believe our main results to be new and obtained following a novel procedure.

In order to accomplish our desires we need to further develop the theory of linear fractional difference equations, which was initially started in the work [2] and generalized afterwards in [3]. More specifically, we solve explicitly the IVP111This result alone might obviously be used by researchers in other contexts.,

[TABLE]

and, after deducing some of its important consequences, we use a recent (comparison) result of [6] to deduce our Bernoulli-type inequality.

This paper is organized as follows: In Section 2 we provide the reader some background on the discrete fractional calculus theory. In Section 3 we present our achievements.

2. Preliminaries on Discrete Fractional Calculus

In this section we introduce the reader to basic concepts and results about discrete fractional calculus (the monograph [4], particularly Chapter 2, could be useful to that matter).

Throughout this work and as usual we assume that empty sums and products equal 0 and 1, respectively.

The power function is defined by

[TABLE]

For $a\in\mathbb{R}$ and $0<\nu\leq 1$ we define the set $\mathbb{N}_{a}=\{a,a+1,a+2,\ldots\}$. Also, we use the notation $\sigma(s)=s+1$ for the shift operator and $(\Delta f)(t)=f(t+1)-f(t)$ to the forward difference operator.

For a function $f:\mathbb{N}_{a}\rightarrow\mathbb{R}$, the discrete fractional sum of order $\nu\geq 0$ is defined as

[TABLE]

Remark 2.1*.*

Note that the operator $\Delta^{-\nu}_{a}$ with $\nu>0$ maps functions defined on $\mathbb{N}_{a}$ to functions defined on $\mathbb{N}_{a+\nu-1}$. Also observe that if $\nu=1$, then we get the summation operator:

[TABLE]

The discrete fractional derivative of order $\nu\in(0,1]$ is defined by

[TABLE]

Remark 2.2*.*

Note that if $\nu=1$, then the fractional derivative is just the forward difference operator.

3. Main Results

This section is devoted in great part to deduce our generalized Bernoulli’s inequality.

In [3] it was shown that, for $t\in\mathbb{N}_{a+\nu-1}$,

[TABLE]

is the solution of the following nonlinear fractional difference initial value problem:

[TABLE]

For our purposes we need the solution of (3.1) in the particular case when $f(t,x)=y(t)x+z(t)$. The next result is obtained following the same procedure as for the case $f(t,x)=y(t)x$ done in [3] and, therefore, we leave the details of its proof to the reader.

Theorem 3.1**.**

Let $a\in\mathbb{R}$ and $\nu\in(0,1]$. Suppose that $y:\mathbb{N}_{a+\nu-1}\rightarrow\mathbb{R}$ is a function. Define an operator $T$ by

[TABLE]

for $t\in\mathbb{N}_{a+\nu-1}$. Then, the function

[TABLE]

is the solution of the summation equation

[TABLE]

for all $t\in\mathbb{N}_{a+\nu-1}$.

Remark 3.2*.*

In Theorem 3.1, the notation

[TABLE]

stands for $\left(\Delta^{-\nu}_{a}g\right)(t)$ where $g(s)=y(s+\nu-1)f(s+\nu-1)$.

Let us now introduce some notation. We define a function $E:\mathbb{N}_{a+\nu-1}\times\mathbb{R}^{5}\rightarrow\mathbb{R}$ (it can be thought of as a discrete Mittag–Leffler function) by:

[TABLE]

whenever the right hand side makes sense.

Corollary 3.3**.**

If $y(t)=c$ for some $c\in\mathbb{R}$ and all $t\in\mathbb{N}_{a+\nu-1}$ in Theorem 3.1, then the solution given by (3.2) is

[TABLE]

It is pertinent to formulate the following consequence of Corollary 3.3.

Corollary 3.4**.**

Suppose that the function $z$ is a constant equal to $K$ in Corollary 3.3. Then,

[TABLE]

Proof.

Let us first note that (cf. [4, Theorem 1.8]):

[TABLE]

Moreover,

[TABLE]

Therefore,

[TABLE]

and the proof is done. ∎

We now need to address the question of the sign of the function $E(t,a,\nu,\nu,c,a)$. First we note that it is the solution of the fractional IVP:

[TABLE]

Let us now recall a recent result proved by Jia et.al:

Theorem 3.5**.**

[6, Theorem 4.2.]** Assume $c_{1}(t)\geq c_{2}(t)\geq-\nu$, $0<\nu<1$ and $x(t)$, $y(t)$ are solutions of the equations

[TABLE]

and

[TABLE]

respectively, satisfying $x(a+\nu-1)\geq y(a+\nu-1)>0$. Then,

[TABLE]

A closer look to the proof of the previous theorem permits us to conclude immediately that $y(a+\nu-1)$ might be equal to zero and the result still remains. Moreover, the representation for the Riemann–Liouville fractional difference,

[TABLE]

is used to prove the result and, hence, the restriction $\nu\neq 1$. Nevertheless, M. Holm showed in [5, Theorem 2.2.] the continuity of the fractional difference operator (3.3) with respect to $\nu$ and, therefore, one may also consider $\nu=1$. In conclusion, for a real number $a$, $0<\nu\leq 1$ and $c\geq-\nu$:

[TABLE]

We are now ready to prove our main result:

Theorem 3.6**.**

(Generalized Bernoulli inequality) Let $\nu\in(0,1]$, $c\in[-\nu,\infty)$ and $a\in\mathbb{R}$. Then, the following inequality holds:

[TABLE]

Proof.

Let $c\geq-\nu$ and $x:\mathbb{N}_{a+\nu-1}\rightarrow\mathbb{R}$ be the function defined by:

[TABLE]

Then $x(a+\nu-1)=0$ and $(\Delta^{\nu}_{a+\nu-1}x)(t)=(\Delta^{\nu}_{a+\nu}x)(t)=c$ by [4, Theorem 2.40]. Therefore,

[TABLE]

Define the function $m$ by:

[TABLE]

which is nonnegative. By Corollary 3.3 we get (note that $x_{a}=0$)

[TABLE]

Hence,

[TABLE]

Finally, by (3.4) we conclude that

[TABLE]

which is equivalent to

[TABLE]

The proof is done. ∎

We finish this work showing that inequality (3.5) truly generalizes the Bernoulli inequality, i.e. when we let $\nu=1$ (and $a=0$) in (3.5), then we get (1.1):

[TABLE]

where the last equivalency follows from [3, Remark 3.7.].

Acknowledgements

The author would like to thank the referees for their careful reading of the manuscript and their suggestions that undoubtedly contributed to its final draft.