Sharp Wirtinger's type inequalities for double integrals with Applications

TL;DR

This paper establishes sharp Wirtinger-type inequalities for double integrals and applies them to derive new sharp byev inequalities for absolutely continuous functions with second derivatives in L^2.

Contribution

It introduces novel sharp inequalities for double integrals and extends these results to byev inequalities for functions with specific smoothness.

Findings

Derived sharp Wirtinger inequalities for double integrals.

Proved new sharp byev inequalities for functions with second derivatives in L^2.

Extended classical inequalities using these new bounds.

Abstract

In this work, sharp Wirtinger type inequalities for double integrals are established. As applications, two sharp \v{C}eby\v{s}ev type inequalities for absolutely continuous functions whose second partial derivatives belong to $L^2$ space are proved.

Equations

$$2a_0^2+\sum _{n=1}^{\infty }\left(a_n^2+b_n^2\right)\le \frac{1}{\pi }{\int }_{-\pi }^{\pi }{f}^2\left(x\right)dx,$$

$${\int }_0^{2\pi }{f}^2\left(x\right)dx\le {\int }_0^{2\pi }{f}^{\prime 2}\left(x\right)dx,$$

$$\begin{gathered} {\int }_a^b{\left[f\left(x\right)-f\left(t_1\right)\right]}^{2}dx \\ \le \frac{4}{{\pi }^2}\max \left\{{\left(t_1-a\right)}^2,{\left(b-t_2\right)}^2,{\left(\frac{t_2-t_1}{2}\right)}^2\right\}{\int }_a^b{f}^{\prime 2}\left(x\right)dx, \end{gathered}$$

$${\int }_a^b{\left[f\left(x\right)-f\left(t\right)\right]}^{2}dx\le \frac{4}{{\pi }^2}{\left[\frac{b-a}{2}+\left|t-\frac{a+b}{2}\right|\right]}^2{\int }_a^b{f}^{\prime 2}\left(x\right)dx,$$

$$\mathcal{T}\left(f,g\right)=\frac{1}{b-a}{\int }_a^{b}f\left(t\right)g\left(t\right)dt-\frac{1}{b-a}{\int }_a^{b}f\left(t\right)dt\cdot \frac{1}{b-a}{\int }_a^{b}g\left(t\right)dt.$$

$$\left|\mathcal{T}\left(f,g\right)\right|\le C{\Vert f^{\prime }\Vert }_2{\Vert g^{\prime }\Vert }_2.$$

$$\left|\mathcal{T}\left(f,g\right)\right|\le \frac{b-a}{{\pi }^2}{\Vert f^{\prime }\Vert }_2{\Vert g^{\prime }\Vert }_2.$$

$${\int }_c^d{\int }_a^b{f}^2\left(x,y\right)dxdy\le C{\int }_c^d{\int }_a^b{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^{2}dxdy$$

$$\mathcal{T}\left(f,g\right)$$

$$-\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(t,s\right)dsdt\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}g\left(t,s\right)dtds,$$

$$\left|\mathcal{T}\left(f,g\right)\right|\le C^{\prime }{\Vert \frac{{\partial }^{2}f}{\partial x\partial y}\Vert }_2{\Vert \frac{{\partial }^{2}g}{\partial x\partial y}\Vert }_2$$

$$I^{-}:=I-\left\{b,d\right\}=\left[a,b\right)\times \left[c,d\right),\mathrm{and}{I}_{-}:=I-\left\{a,c\right\}=\left(a,b\right]\times \left(c,d\right]$$

$${\int }_c^d{\int }_a^b{f}^2\left(x,y\right)dxdy\le \frac{16}{{\pi }^4}{\left(b-a\right)}^2{\left(d-c\right)}^2{\int }_c^d{\int }_a^b{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^{2}dxdy$$

$$f(x,y)=g_1(x)g_2(y)h(x,y)$$

$$g_1\left(x\right)=\sin {\omega }_1\left(x-a\right),\forall x\in [a,b)$$

$$g_2\left(y\right)=\sin {\omega }_2\left(y-c\right),\forall y\in [c,d)$$

$$\frac{\partial f}{\partial x}=g_1{g}_2\frac{\partial h}{\partial x}+g_{2}h{g}_1^{\prime },$$

$$\frac{{\partial }^{2}f}{\partial x\partial y}$$

$$=\frac{d}{dy}\left(g_1{g}_2\frac{\partial h}{\partial x}+g_1^{\prime }{g}_{2}h\right)$$

$$=g_2^{\prime }\left(g_1\frac{\partial h}{\partial x}+g_1^{\prime }h\right)+g_2\frac{d}{dy}\left(g_1\frac{\partial h}{\partial x}+g_1^{\prime }h\right).$$

$$\Phi :=\Phi \left(x,y\right)=g_1\frac{\partial h}{\partial x}+g_1^{\prime }h=\frac{f_x}{g_2}\Rightarrow g_2\Phi =f_x,$$

$$\frac{d}{dy}\left(g_1{g}_2\frac{\partial h}{\partial x}+g_1^{\prime }{g}_{2}h\right)=g_2^{\prime }\left(g_1\frac{\partial h}{\partial x}+g_1^{\prime }h\right)+g_2{\left(g_1\frac{\partial h}{\partial x}+g_1^{\prime }h\right)}^{\prime }=\Phi g_2^{\prime }+g_2{\Phi }_y^{\prime }.$$

$${\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^{2}dydx$$

$$={\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{\left(\Phi g_2^{\prime }\right)}^2{\left(1+\frac{g_2{\Phi }_y^{\prime }}{\Phi g_2^{\prime }}\right)}^{2}dydx$$

$$\ge {\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{\left(\Phi g_2^{\prime }\right)}^2\left(1+2\frac{g_2{\Phi }_y^{\prime }}{\Phi g_2^{\prime }}\right)dydx$$

$$={\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{\left(\Phi g_2^{\prime }\right)}^{2}dydx+2{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }\left(\Phi g_2^{\prime }\right)g_2{\Phi }_y^{\prime }dydx$$

$$={{g_2\left(g_2^{\prime }\right){\Phi }^2|}_{\gamma }^{\delta }|}_{\alpha }^{\beta }-{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }\left(g_2\left(g_2^{\prime \prime }\right)+(g_2^{\prime }{)}^2\right){\Phi }^{2}dydx$$

$$+{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{\left(\Phi g_2^{\prime }\right)}^{2}dydx$$

$$={{g_2\left(g_2^{\prime }\right){\Phi }^2|}_{\gamma }^{\delta }|}_{\alpha }^{\beta }-{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }\left(-{\omega }_2^2{g}_2^2+{\left(g_2^{\prime }\right)}^2\right){\Phi }^{2}dydx$$

$$+{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{\left(\Phi g_2^{\prime }\right)}^{2}dydx$$

$$={{g_2\left(g_2^{\prime }\right){\Phi }^2|}_{\gamma }^{\delta }|}_{\alpha }^{\beta }+{\omega }_2^2{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{g}_2^2{\Phi }^{2}dydx$$

$$-{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{\left(g_2^{\prime }\right)}^2{\Phi }^{2}dydx+{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{\left(\Phi g_2^{\prime }\right)}^{2}dydx$$

$$={{g_2\left(g_2^{\prime }\right){\Phi }^2|}_{\gamma }^{\delta }|}_{\alpha }^{\beta }+{\omega }_2^2{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{g}_2^2{\Phi }^{2}dydx$$

$$={{g_2\left(g_2^{\prime }\right){\Phi }^2|}_{\gamma }^{\delta }|}_{\alpha }^{\beta }+{\omega }_2^2{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{\left(\frac{\partial f}{\partial x}\right)}^{2}dydx$$

$${\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{\left(\frac{\partial f}{\partial x}\right)}^{2}dydx$$

$$={\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }\left\{{\left(g_1^{\prime }{g}_{2}h\right)}^2{\left(1+\frac{g_1{g}_2\frac{\partial h}{\partial x}}{g_1^{\prime }{g}_{2}h}\right)}^2\right\}dydx$$

$$\ge {\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }\left\{{\left(g_1^{\prime }{g}_{2}h\right)}^2\left(1+2\frac{g_1{g}_2\frac{\partial h}{\partial x}}{g_1^{\prime }{g}_{2}h}\right)\right\}dydx$$

$$={\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }{\left(g_1^{\prime }{g}_{2}h\right)}^{2}dydx+2{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }\left(g_1^{\prime }{g}_{2}h\right)\left(g_1{g}_2\frac{\partial h}{\partial x}\right)dydx$$

$$={{g_2^2{g}_1\left(g_1^{\prime }\right)h^2|}_{\gamma }^{\delta }|}_{\alpha }^{\beta }-{\int }_{\alpha }^{\beta }{\int }_{\gamma }^{\delta }\left(g_1{g}_1^{\prime \prime }+(g_1^{\prime }{)}^2\right)g_2^2{h}^{2}dydx$$

Full text

1. Introduction

The theory of Fourier series has a significant role in almost all branches of mathematical and numerical analysis. A very interesting connection between inequalities and Fourier series has been made along more than a hundred year ago. The celebrated Bessel’s integral inequality

[TABLE]

was named after Bessel death and considered from that time as the first adobe in this connection and started point for other related works after the end of 18-th century.

In 1916, Wirtinger [8] credibly proved his inequality regarding square integrable periodic functions, which reads:

Theorem 1**.**

Let $f$ be a real valued function with period $2\pi$ and $\int_{0}^{2\pi}f\left({x}\right)dx=0$. If $f^{\prime}\in L^{2}[0,2\pi]$, then

[TABLE]

with equality if and only if $f(x)=A\cos x+B\sin x$, $A,B\in\mathbb{R}$.

Many authors have considered a main attention for Wirtinger’s inequality and therefore, several generalizations, counterparts and refinements was collected in a chapter of the book [20].

In 1967, Diaz and Metcalf [10] have extended and generalized Wirtinger inequality and they proved the following result:

Theorem 2**.**

Let $f$ be continuously differentiable on $(a,b)$. Suppose $f(t_{1})=f(t_{2})$ for $a\leq t_{1}\leq t_{2}\leq b$, then the inequality

[TABLE]

holds. In particular, if $t_{1}=t_{2}=t$, then

[TABLE]

For other related results see [6], [7] and [19].

One of the most direct applicable usage of (1.3) were considered in several works regarding the famous Čebyšev functional

[TABLE]

which compare or measure the difference between the integral product of two functions with their corresponding integrals product.

In 1970, Ostrowski [16] proved that if $f^{\prime},g^{\prime}\in L^{2}[a,b]$, then there exists a constant $C$, $0\leq C\leq\frac{b-a}{8}$, such that

[TABLE]

After that in 1973, A. Lupaş [15] has improved the result of Ostrowski’s (1.6) and proved that

[TABLE]

where, the constant $\frac{1}{\pi^{2}}$ is the best possible.

In this work we deal with the problem: what is the best possible constant $C$ would the inequality

[TABLE]

holds, whenever $f,g\in\mathfrak{L}^{2}(I)$. This question is a natural extension of Diaz-Metcalf inequality (1.3), as well as the complementary works of Beesack and Milovanović in one variable, see [6], [7] and [19].

Accordingly, for the Čebyšev functional

[TABLE]

what is the best possible constant $C^{\prime}$ would the inequality

[TABLE]

holds, and this is an extension of the Lupaş inequality (1.7).

2. Wirtinger’s type inequalities

Let $I$ be a two dimensional interval and denotes $I^{\circ}$ its interior, for $a,b,c,d\in\mathbb{R}$, we consider the subset $\mathbb{D}:=\left\{{\left({x,y}\right):a\leq x\leq b,c\leq y\leq d}\right\}\subseteq\mathbb{R}^{2}$ such that $\mathbb{D}\subset I^{\circ}$. Also, define the subsets $I_{-}$ and $I^{-}$ of $I$ as follows:

[TABLE]

Throughout of this section we assume that $f:I\to\mathbb{R}$ satisfies the boundary conditions: $f(a,\cdot)=f(\cdot,c)=0$, $f_{x}(a,\cdot)=f_{x}(\cdot,c)=0$, $f_{y}(a,\cdot)=f_{y}(\cdot,c)=0$ on $I^{-}$. Also, we assume $f(b,\cdot)=f(\cdot,d)=0$, $f_{x}(b,\cdot)=f_{x}(\cdot,d)=0$, $f_{y}(b,\cdot)=f_{y}(\cdot,d)=0$ on $I_{-}$, and both conditions on $I^{\circ}$.

Let $\mathfrak{L}^{2}(I)$ be the space of all functions $f$ which are absolutely continuous on $I$, with $\int_{c}^{d}\int_{a}^{b}{\left|{\frac{{\partial^{2}f}}{{\partial x\partial y}}}\right|^{2}dxdy}<\infty$.

Theorem 3**.**

Let $f\in\mathfrak{L}^{2}(I^{-})$. Then the inequality

[TABLE]

is valid. The constant $\frac{16}{\pi^{2}}$ is the best possible, in the sense that it cannot be replaced by a smaller one.

Proof.

Let $a\leq x<b$ and $c\leq y<d$, since $f$ is absolutely continuous then we can write $f\left({x,y}\right)=\int_{c}^{y}\int_{a}^{x}{f_{ts}\left({t,s}\right)dtds}$. If $a$ and $c$ are real numbers this is equivalent to saying that $f(a,c)=0$ and $f$ is absolutely continuous on $[a,b)\times[c,d)$. Setting

[TABLE]

where,

[TABLE]

with $\omega_{1}=\lambda_{1}^{1/2}$ and $\lambda_{1}=\frac{{\pi^{2}}}{{4\left({b-a}\right)^{2}}}$, and

[TABLE]

with $\omega_{2}=\lambda_{2}^{1/2}$ and $\lambda_{2}=\frac{{\pi^{2}}}{{4\left({d-c}\right)^{2}}}$.

Firstly, let us observe that since $g^{\prime}_{1}\left(x\right)=\omega_{1}\cos\omega_{1}\left({x-a}\right),$ and so that

$g^{\prime\prime}_{1}\left(x\right)=-\omega_{1}^{2}g_{1}\left(x\right).$ Similarly, we have $g^{\prime\prime}_{2}\left(y\right)=-\omega_{2}^{2}g_{2}\left(y\right)$.

For simplicity, since

[TABLE]

then

[TABLE]

Setting

[TABLE]

therefore

[TABLE]

Now, if $a<\alpha<\beta<b$, and  $c<\gamma<\delta<d$, we have

[TABLE]

where, in (2.4) we integrate by parts, assuming that $u=g_{2}\left({g^{\prime}_{2}}\right)$ and $dv=2\Phi_{y}^{\prime}\Phi$. Now, we also have

[TABLE]

Substitute (2.4) in (2.3), we get

[TABLE]

Hence,

[TABLE]

Now, since

[TABLE]

then

[TABLE]

as $\alpha\longrightarrow a^{+}$ and $\gamma\longrightarrow c^{+}$, i.e., $f^{2}(\alpha,\gamma)=0$ and so that

[TABLE]

Similarly,

[TABLE]

then

[TABLE]

as $\gamma\longrightarrow c^{+}$, i.e., $f^{2}(\alpha,\gamma)=0$ and so that

[TABLE]

Then, from (2.5) it follows

[TABLE]

where $a<\beta<b$ and $c<\delta<d$. Now let $\alpha\longrightarrow a^{+}$, $\beta\longrightarrow b^{-}$ and $\gamma\longrightarrow c^{+}$, $\delta\longrightarrow d^{-}$ to obtain the inequality (2.1).

To obtain the sharpness, assume that (2.1) holds with another constant $K>0$,

[TABLE]

Define the function $f:\left[{a,b}\right)\times\left[{c,d}\right)\to\mathbb{R}$, given by

[TABLE]

therefore, we have

[TABLE]

$\int_{a}^{b}{\int_{c}^{d}{f^{2}\left({x,y}\right)dydx}}=\frac{{\left({b-a}\right)\left({d-c}\right)}}{4}$, and $\int_{a}^{b}{\int_{c}^{d}{\left({\frac{{\partial^{2}f}}{{\partial x\partial y}}}\right)^{2}dydx}}=\frac{{\pi^{4}}}{{64\left({b-a}\right)\left({d-c}\right)}}$, substitute in (2.7)

[TABLE]

which means that $K\geq\frac{16}{\pi^{4}}$, thus the constant $\frac{16}{\pi^{4}}$ is the best possible and the inequality is sharp (2.1). ∎

Corollary 1**.**

If $f\in\mathfrak{L}^{2}(I_{-})$, then the inequality (2.1) still holds, and the inequality is sharp.

Proof.

The proof goes likewise the proof of Theorem 3, with few changes in the auxiliary function ‘$\sin$’ in both variable $x$ and $y$ defined on the bidimensional interval $I_{-}$. To obtain the sharpness, define the function $f:\left({a,b}\right]\times\left({c,d}\right]\to\mathbb{R}$, given by

[TABLE]

where $C$ is constant. ∎

Corollary 2**.**

Let $f\in\mathfrak{L}^{2}(I)$. Under the assumptions of Theorem 3 and Corollary 1 together, the inequality

[TABLE]

is valid for all $(\xi,\eta)\in\mathbb{D^{\circ}}$. The constant $\frac{1}{\pi^{4}}$ is the best possible.

Proof.

Applying Theorem 3 and Corollary 2 on the right hand side of the equation

[TABLE]

and the make the substitution $h(x,y)=\left|{f\left({x,y}\right)-f\left({\xi,\eta}\right)}\right|^{2}$. To obtain the sharpness define $f:\left[{a,b}\right]\times\left[{c,d}\right]\to\mathbb{R}$, given by

[TABLE]

where $K_{0}$, $K_{1}$, $K_{2}$, $K_{3}$ and $K_{2}$ are arbitrary constants, $\tau=2\xi-a-b$, $\psi=2\eta-c-d$ and $u_{s}\left(t\right)$ is the unit step function. ∎

3. Sharp bounds for the Čebyšev functional

The Čebyšev functional

[TABLE]

has interesting applications in the approximation of the integral of a product as pointed out in the references below.

In order to represent the remainder of the Taylor formula in an integral form which will allow a better estimation using the Grüss type inequalities, Hanna et al. [14], generalized the Korkine identity for double integrals and therefore Grüss type inequalities were proved.

In 2002, Pachpatte [17] has established two inequalities of Grüss type involving continuous functions of two independent variables whose first and second partial derivatives are exist, continuous and belong to $L_{\infty}(\mathbb{D})$; for details see [17]. For more results about multivariate and multidimensional Grüss type inequalities the reader may refer to [2]–[5], [11]–[14] and [18].

Recently, the author of this paper [1] established various inequalities of Grüss type for functions of two variables under various assumptions of the functions involved.

In viewing of Corollary 2, we may state the following result.

Theorem 4**.**

If $f,g\in\mathfrak{L}^{2}(\mathbb{D})$, then

[TABLE]

$\frac{1}{\pi^{4}}$* is the best possible.*

Proof.

By the triangle inequality and then using the Cauchy-Schwartz inequality, we get

[TABLE]

Now, since

[TABLE]

where $\Delta:=(b-a)(d-c)$.

Therefore, from (3.3)

[TABLE]

Therefore,

[TABLE]

Applying (2.7), we get

[TABLE]

Thus,

[TABLE]

In a similar argument we can observe that

[TABLE]

Finally, since

[TABLE]

which proves (3.2). To obtain the sharpness, assume that (3.2) holds with another constant $K>0$,

[TABLE]

Define the functions $f,g:\mathbb{D}\to\mathbb{R}$, given by

[TABLE]

therefore, we have

[TABLE]

[TABLE]

and

[TABLE]

Substituting in (3.6)

[TABLE]

which means that $K\geq\frac{1}{\pi^{4}}$, thus the constant $\frac{1}{\pi^{4}}$ is the best possible and the inequality is sharp (3.2). ∎

Theorem 5**.**

Let $f\in\mathfrak{L}^{2}(\mathbb{D})$ and $g:\mathbb{D}\to\mathbb{R}$ satisfies that there exists the real numbers $\gamma,\Gamma$ such that $\gamma\leq g(x,y)\leq\Gamma$ for all $(x,y)\in\mathbb{D}$, then

[TABLE]

The constant $\frac{4}{\pi^{2}}$ is the best possible.

Proof.

Since

[TABLE]

Taking the absolute value for both sides and making of use the triangle inequality, we get

[TABLE]

As in Theorem 1 in [1], we have observed that since there exists $\gamma,\Gamma\geq 0$ such that $\gamma\leq g(x,y)\leq\Gamma$ for all $(x,y)\in\mathbb{D}$, then

[TABLE]

On the other hand, using the elementary inequality

[TABLE]

for all $A,B,C,D\geq 0$, we also have

[TABLE]

Applying (2.7) for each integral above and simplify we get

[TABLE]

Combining the inequalities (3.9) and (3.10) with (3.8) we get the desired result (3.7).

To prove the sharpness of (3.7) holds with constant $C>0$, i.e.,

[TABLE]

and consider the functions $f,g:\mathbb{D}\to\mathbb{R}$ be defined as

[TABLE]

As in the proof of Theorem 4, $f\in\mathcal{L}^{2}(\mathbb{D})$, and $\Gamma-\gamma=2$, $\int_{a}^{b}{\int_{c}^{d}{g\left({t,s}\right)dsdt}}=0$,

[TABLE]

and

[TABLE]

Making use of (3.11) we get $\frac{4}{\pi^{2}}\leq C$, which proves that $\frac{4}{\pi^{2}}$ is the best possible and thus the proof is completely finished. ∎

3.1. An inequality of Ostrowski’s type

The mean value theorem for double integrals reads that: If $f$ is continuous on $[a,b]\times[c,d]$, then there exists $(\eta,\xi)\in[a,b]\times[c,d]$ such that

[TABLE]

What about if one needs to measure the difference between the image of an arbitrary point $(x,y)\in[a,b]\times[c,d]$ and the average value $\frac{1}{{\left({b-a}\right)\left({d-c}\right)}}\int_{c}^{d}{\int_{a}^{b}{f\left({t,s}\right)dtds}}$?

In this way Ostrowski introduced his famous inequality regarding differentiable functions and its average values. In [9] and [11]–[14] and other related works many authors have studied the Ostrowski’s type inequalities for various type of functions of several variables.

In the following, we present a bound belongs to $L_{2}$ norm for the Ostrowski inequality.

Theorem 6**.**

Let $f\in\mathfrak{L}^{2}(\mathbb{D})$ , then

[TABLE]

for all $(x,y)\in[a,b]\times[c,d]$. In special case, choose $(x,y)=\left({\frac{a+b}{2},\frac{c+d}{2}}\right)$

[TABLE]

Proof.

Since

[TABLE]

Taking the modulus, applying the triangle inequality and then use the Cauchy-Schwarz’s inequality, we get

[TABLE]

which follows by (2.7), and this proves (3.13). ∎

Conflict of Interests. The author declares that there is no conflict of interests regarding the publication of this paper.