Opial's inequality in $q$-Calculus revisited

TL;DR

This paper revisits Opial's inequality within q-calculus, providing corrected proofs using elementary inequalities and the Gauchman q-restricted integral, offering a more straightforward approach.

Contribution

It offers a fundamentally corrected and simplified proof of Opial's inequality in q-calculus, improving upon previous methods.

Findings

Corrected proofs of Opial's inequality in q-calculus

Introduction of a simple proof method based on elementary inequalities

Application of Gauchman q-restricted integral

Abstract

We have fundamentally corrected the proofs of the theorems from our paper [9] by giving an entirely different approach, using quite a simple method based on applications of some elementary inequalities, well-known H\"older's inequality, and the Gauchman $q$-restricted integral.

Equations

$${\int }_0^h|f(x)f^{\prime }(x)|dx\le \frac{h}{4}{\int }_0^h(f^{\prime }(x){)}^{2}dx$$

$$[n{]}_q=\frac{1-q^n}{1-q}=q^{n-1}+\cdots +q+1,\dots ,n\in \mathbb{N}.$$

$$(D_{q}f)(x)=\frac{f(x)-f(qx)}{x-qx},x\ne 0;D_{q}f(0)=\underset{x\to 0}{\lim }{D}_{q}f(x).$$

$${\int }_0^{a}f(x)d_{q}x=a(1-q)\sum _{j=0}^{\infty }{q}^{j}f(a{q}^j).$$

$${\int }_a^{b}f(x)d_{q}x={\int }_0^{b}f(x)d_{q}x-{\int }_0^{a}f(x)d_{q}x,$$

$${\int }_a^{b}f(x)d_{q}x={\int }_{b{q}^n}^{b}f(x)d_{q}x=b(1-q)\sum _{j=0}^{n-1}{q}^{j}f(q^{j}b).$$

$${\int }_a^b|D_{q}f(x)||f(x){|}^p{d}_{q}x\le (b-a{)}^p{\int }_a^b|D_{q}f(x){|}^{p+1}{d}_{q}x.$$

$${\int }_a^b|D_{q}f(x)||f(x){|}^p{d}_{q}x$$

$$\sum _{j=0}^{n-1}|f(b{q}^j)-f(b{q}^{j+1})||f(b{q}^j){|}^p\le |f(b{q}^n){|}^p\sum _{j=0}^{n-1}|f(b{q}^j)-f(b{q}^{j+1})|.$$

$$|f(bq^{n})|^{p}=\Big{|}\sum_{j=0}^{n-1}f(bq^{j+1})-f(bq^{j})\Big{|}^{p}\leq\Big{(}\sum_{j=0}^{n-1}\big{|}f(bq^{j})-f(bq^{j+1})\big{|}\Big{)}^{p},$$

$$|f(bq^{n})|^{p}\sum_{j=0}^{n-1}\big{|}f(bq^{j})-f(bq^{j+1})\big{|}\leq\Big{(}\sum_{j=0}^{n-1}\big{|}f(bq^{j})-f(bq^{j+1})\big{|}\Big{)}^{p+1}.$$

$$\int_{a}^{b}|D_{q}f(x)||f(x)|^{p}d_{q}x\leq\Big{(}\sum_{j=0}^{n-1}\big{|}f(bq^{j})-f(bq^{j+1})\big{|}\Big{)}^{p+1}.$$

$$\displaystyle\Big{(}\sum_{j=0}^{n-1}\big{|}f(bq^{j})-f(bq^{j+1})\big{|}\Big{)}^{p+1}$$

$$\sum_{j=0}^{n-1}(q^{j})^{\frac{p}{p+1}}(q^{j})^{\frac{1}{p+1}}\Big{|}\frac{f(bq^{j})-f(bq^{j+1})}{bq^{j}-bq^{j+1}}\Big{|}\leq\Big{(}\sum_{j=0}^{n-1}\big{(}(q^{j})^{\frac{p}{p+1}}\big{)}^{\frac{p+1}{p}}\Big{)}^{\frac{p}{p+1}}\times\\ \Big{(}\sum_{j=0}^{n-1}q^{j}\Big{|}\frac{f(bq^{j})-f(bq^{j+1})}{bq^{j}-bq^{j+1}}\Big{|}^{p+1}\Big{)}^{\frac{1}{p+1}}.$$

$$\Big{(}\sum_{j=0}^{n-1}q^{j}\Big{|}\frac{f(bq^{j})-f(bq^{j+1})}{bq^{j}-bq^{j+1}}\Big{|}\Big{)}^{p+1}\leq\Big{(}\sum_{j=0}^{n-1}q^{j}\Big{)}^{p}\sum_{j=0}^{n-1}q^{j}\Big{|}\frac{f(bq^{j})-f(bq^{j+1})}{bq^{j}-bq^{j+1}}\Big{|}^{p+1}.$$

$$\Big{(}b(1-q)\sum_{j=0}^{n-1}q^{j}\Big{|}\frac{f(bq^{j})-f(bq^{j+1})}{bq^{j}-bq^{j+1}}\Big{|}\Big{)}^{p+1}\leq b^{p}(1-q^{n})^{p}\times\\ b(1-q)\sum_{j=0}^{n-1}q^{j}\Big{|}\frac{f(bq^{j})-f(bq^{j+1})}{bq^{j}-bq^{j+1}}\Big{|}^{p+1}.$$

$$\int_{a}^{b}|D_{q}f(x)||f(x)|^{p}d_{q}x\leq(b-a)^{p}\Big{(}b(1-q)\sum_{j=0}^{n-1}q^{j}\Big{|}\frac{f(bq^{j})-f(bq^{j+1})}{bq^{j}-bq^{j+1}}\Big{|}^{p+1}\Big{)}.$$

$$b(1-q)\sum_{j=0}^{n-1}q^{j}\Big{|}\frac{f(bq^{j})-f(bq^{j+1})}{bq^{j}-bq^{j+1}}\Big{|}^{p+1}=\int_{a}^{b}|D_{q}f(x)|^{p+1}d_{q}x,$$

$${\int }_a^b|D_{q}f(x)||f(x)|d_{q}x\le (b-a){\int }_a^b|D_{q}f(x){|}^2{d}_{q}x.$$

$$\int_{a}^{b}\big{(}f(x)D_{q}g(x)+g(qx)D_{q}f(x)\big{)}d_{q}x\\ \leq\frac{b-a}{2}\int_{a}^{b}\big{(}(D_{q}f(x))^{2}+(D_{q}g(x))^{2}\big{)}d_{q}x.$$

$$\int_{a}^{b}\big{(}f(x)D_{q}g(x)+g(qx)D_{q}f(x)\big{)}d_{q}x,$$

$${\int }_{b{q}^n}^b\left(f(x)\frac{g(x)-g(qx)}{x-qx}+g(qx)\frac{f(x)-f(qx)}{x-qx}\right)d_{q}x,$$

$$b(1-q)$$

$$f(bq^{n})=\sum_{j=0}^{n-1}\big{(}f(bq^{j+1})-f(bq^{j})\big{)},\quad g(bq^{n})=\sum_{j=0}^{n-1}\big{(}g(bq^{j+1})-g(bq^{j})\big{)},$$

$$-f(bq^{n})g(bq^{n})\leq\frac{1}{2}\Bigg{(}\Big{(}\sum_{j=0}^{n-1}\big{(}f(bq^{j+1})-f(bq^{j})\big{)}\Big{)}^{2}+\Big{(}\sum_{j=0}^{n-1}\big{(}g(bq^{j+1})-g(bq^{j})\big{)}\Big{)}^{2}\Bigg{)}$$

$$\Big{(}b(1-q)\sum_{j=0}^{n-1}q^{j}\frac{f(bq^{j+1})-f(bq^{j})}{bq^{j}-bq^{j+1}}\Big{)}^{2}\leq b(1-q^{n})b(1-q)\sum_{j=0}^{n-1}q^{j}\Big{(}\frac{f(bq^{j})-f(bq^{j+1})}{bq^{j}-bq^{j+1}}\Big{)}^{2}$$

$$\Big{(}b(1-q)\sum_{j=0}^{n-1}q^{j}\frac{g(bq^{j+1})-g(bq^{j})}{bq^{j}-bq^{j+1}}\Big{)}^{2}\leq b(1-q^{n})b(1-q)\sum_{j=0}^{n-1}q^{j}\Big{(}\frac{g(bq^{j})-f(bq^{j+1})}{bq^{j}-bq^{j+1}}\Big{)}^{2}.$$

$$\int_{a}^{b}\big{(}f(x)D_{q}g(x)+g(qx)D_{q}f(x)\big{)}d_{q}x\leq\frac{b-a}{2}\int_{a}^{b}\big{(}(D_{q}f(x))^{2}+(D_{q}g(x))^{2}\big{)}d_{q}x,$$

$$\int_{a}^{b}|f(x)|^{s}|g(x)|^{t}d_{q}x\\ \leq(b-a)^{s+t}\int_{a}^{b}\Big{(}\frac{s}{s+t}|D_{q}f(x)|^{s+t}d_{q}x+\frac{t}{s+t}|D_{q}g(x)|^{s+t}d_{q}x\Big{)}.$$

$${\int }_a^b|f(x){|}^s|g(x){|}^t{d}_{q}x=b(1-q)\sum _{i=0}^{n-1}{q}^i|f(b{q}^i){|}^s|g(b{q}^i){|}^t.$$

Full text

1 Introduction and preliminaries

In the recent paper [9] a generalization of the Opial integral inequality

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in $q$-calculus was given. Here we eliminate some inaccuracies by simplifying and modifying the proofs of the theorems.

First of all, we present necessary definitions and facts from the $q$-calculus, where $q$ is a real number satisfying $0<q<1$, and $q$-natural number is defined by

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Definition 1.1**.**

Let $f$ be a function defined on an interval $(a,b)\subset\mathbb{R}$, so that $qx\in(a,b)$ for all $x\in(a,b)$. For $0<q<1$, we define the $q$-derivative as

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In the paper [8], Jackson defined $q$-integral, which in the $q$-calculus bears his name.

Definition 1.2**.**

The $q$-integral on $[0,a]$ is

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On this basis, in the same paper, Jackson defined an integral on $[a,b]$

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For a positive integer $n$ and $a=bq^{n}$, using the left-hand side integral of (3), in the paper [7], Gauchman introduced the $q$-restricted integral

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Definition 1.3**.**

The real function $f$ defined on $[a,b]$ is called $q$-increasing ($q$-decreasing)* on $[a,b]$ if$f(qx)\leq f(x)$ ($f(qx)\geq f(x)$) for $x,\,qx\in[a,b]$.*

It is easy to see that if the function $f$ is increasing (decreasing), then it is $q$-increasing ($q$-decreasing) too.

2 Results and discussions

Our main results are contained in three theorems.

Theorem 2.1**.**

Let $f(x)$ be $q$-decreasing function on $[a,b]$ with $f(bq^{0})=0$. Then, for any $p\geq 0$, there holds

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Proof.

Using Definition 1.1 and (4), we have

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whence, taking into account that $f(x)$ is $q$-decreasing, we have

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In view of $f(bq^{n})=\sum\limits_{j=0}^{n-1}{f(bq^{j+1})-f(bq^{j})}$, we obtain

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so that

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Thus

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The right-hand side of this inequality we can write in the form of

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After rewriting $q^{j}=(q^{j})^{\frac{p}{p+1}}(q^{j})^{\frac{1}{p+1}}$, and applying Hölder’s inequality to the last sum, we have

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After raising both sides to the power $p+1$, we find

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Multiplying this inequality by $b^{p+1}(1-q)^{p+1}$, and relying on the formula for the sum of the first $n$ terms of the geometric series, we arrive at the inequality

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Considering that $b^{p}(1-q^{n})^{p}=(b-bq^{n})^{p}=(b-a)^{p}$, taking into account (6), we have proved the inequality

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Referring to (4), there holds

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whereby we prove the theorem. ∎

Remark 2.2**.**

In particular, by taking $p=1$, the inequality (5) in Theorem 2.1 reduces to the following Opial’s inequality in $q$-Calculus.

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The following theorems are concerned with $q$-monotonic functions.

Theorem 2.3**.**

If $f(x)$ and $g(x)$ are $q$-decreasing functions on $[a,b]$ satisfying $f(bq^{0})=0$ and $g(bq^{0})=0$, then there holds the inequality

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Proof.

Replacing (2) in the integral

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we obtain

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whence, using Gauchman $q$-restricted integral, we have

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Using the elementary inequality $-xy\leq\frac{1}{2}(x^{2}+y^{2}),\,x,y\in\mathbb{R}$, and considering that

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we find

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Applying (7) for $p=1$, knowing that $f(x)$ and $g(x)$ are $q$-decreasing, we obtain

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as well as

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Since $b(1-q^{n})=b-a$, making use of (4), we have

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whereby (8) is proved. ∎

Theorem 2.4**.**

If $f(x)$ and $g(x)$ are $q$-decreasing functions on $[a,b]$ satisfying $f(bq^{0})=g(bq^{0})=0$, then there holds the inequality

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Proof.

First, we apply (4) to the left-hand side of (9), and have

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For real numbers $z,w\geq 0$ and $s,t>0$, we rely on the elementary inequality

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After setting $z=(q^{i})^{\frac{1}{s+t}}|f(bq^{i})|,\,w=(q^{i})^{\frac{1}{s+t}}|g(bq^{i})|$, we find

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Considering that $f$ and $g$ are $q$-decreasing functions, so $|f(bq^{i})|^{s+t}\leq|f(bq^{n})|^{s+t}$ and $|g(bq^{i})|^{s+t}\leq|g(bq^{n})|^{s+t}$, the last inequality becomes

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However, there holds

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so that we have

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Here we follow the same procedure as in the proof of Theorem 2.1. So, after rewriting $q^{i}=(q^{i})^{\frac{s+t-1}{s+t}}(q^{j})^{\frac{1}{s+t}}$, and applying Hölder’s inequality to both sums on the right side of the last inequality, for the first sum we have

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and for the second as well

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We multiply both inequalities by $b(1-q)$, then raise them to the power $s+t$. Thus, we obtain

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and similarly

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so that, because $b(1-q^{n})=b-a$, from (10) there follows (9), whereby we complete the proof. ∎

Remark 2.5**.**

In the special case when $s=t=r$ and $f(x)=g(x)=h(x)$, the inequality established in (9) reduces to the $q$-Wirtinger-type inequality

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