The first result of this section is the trapezoid type inequality related to (4).
Theorem 2.1
Consider a set
\(I\subset \mathbb{R}^{2}\)
with
\(D(C,R)\subset I^{ \circ }\). Suppose that the mapping
\(f:D(C,R)\to \mathbb{R}\)
has continuous partial derivatives in the disk
\(D(C,R)\)
with respect to the variables
r
and
θ
in polar coordinates. If, for any constant
\(\theta \in [0,2\pi ]\), the function
\(|\frac{\partial f}{\partial r} |\)
is convex with respect to the variable
r
on
\([0,R]\), then
$$ \biggl\vert \frac{1}{2\pi R} \int _{\partial (C,R)}f(\gamma )\,dl(\gamma )-\frac{1}{ \pi R^{2}} \iint _{D(C,R)}f (x,y)\,dx\,dy \biggr\vert \leq \frac{1}{6\pi } \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert ( \gamma )\,dl( \gamma ). $$
(5)
Proof
For a constant \(\theta \in [0,2\pi ]\), if we consider
$$ x(r)=a+r\cos \theta $$
and
$$ y(r)=b+r\sin \theta , $$
then we have \(([\dot{x}(r)]^{2}+[\dot{y}(r)]^{2} )^{ \frac{1}{2}}= (\sin ^{2}(\theta )+\cos ^{2}(\theta ) )^{ \frac{1}{2}}=1\), where ẋ, ẏ are the derivatives of x, y, respectively, with respect to the variable r on \([0,R]\). So, by the use of integration by parts, we have the following equalities:
$$\begin{aligned} & \int _{0}^{R} \frac{\partial f}{\partial r} (a+r\cos\theta ,b+r \sin \theta )r^{2} \,dr=r^{2} f (a+r\cos\theta ,b+r\sin\theta ) \bigg|_{0}^{R} \\ &\quad {}-2 \int _{0}^{R} f (a+r\cos\theta ,b+r\sin\theta )r\, dr=R^{2}f (a+R\cos\theta ,b+R\sin\theta ) \\ &\quad {}-2 \int _{0}^{R} f (a+r\cos\theta ,b+r\sin\theta )r\, dr. \end{aligned}$$
(6)
The integration of (6) with respect to θ on \([0,2\pi ]\) implies that
$$\begin{aligned} &R^{2} \int _{0}^{2\pi }f (a+R\cos\theta ,b+R\sin\theta )\, d \theta -2 \int _{0}^{2\pi } \int _{0}^{R}f (a+r\cos\theta ,b+r\sin\theta )r\, dr \, d \theta \\ &\quad = \int _{0}^{2\pi } \int _{0}^{R}\frac{\partial f}{\partial r} (a+r\cos \theta ,b+r \sin\theta )r^{2} \, dr\, d\theta . \end{aligned}$$
Since \(|\frac{\partial f}{\partial r}|\) is convex with respect to the variable r on \([0,R]\) for any \(\theta \in [0,2\pi ]\), then
$$\begin{aligned} & \biggl\vert R^{2} \int _{0}^{2\pi }f (a+R\cos\theta ,b+R\sin\theta )\, d \theta -2 \int _{0}^{2\pi } \int _{0}^{R}f (a+r\cos\theta ,b+r\sin\theta )r\, dr \, d\theta \biggr\vert \\ &\quad \leq \int _{0}^{2\pi } \int _{0}^{R} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert (a+r\cos\theta ,b+r\sin\theta )r^{2} \, dr\, d\theta \\ &\quad = \int _{0}^{2\pi } \int _{0}^{R} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert \biggl(\frac{r}{R} (a+R\cos\theta ,b+R\sin\theta )+ \biggl(1- \frac{r}{R} \biggr) (a,b ) \biggr)r^{2} \, dr\, d\theta \\ &\quad \leq \int _{0}^{2\pi } \int _{0}^{R}\frac{r^{3}}{R} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert (a+R\cos\theta ,b+R\sin\theta )\, dr\, d\theta \\ &\qquad {}+ \int _{0}^{2\pi } \int _{0}^{R}r^{2} \biggl(1- \frac{r}{R} \biggr) \biggl\vert \frac{ \partial f}{\partial r} \biggr\vert (C )\,dr \,d\theta \\ &\quad =\frac{R^{3}}{4} \int _{0}^{2\pi } \biggl\vert \frac{\partial f}{\partial r} \biggr\vert (a+R\cos\theta ,b+R\sin\theta )\, d\theta+\frac{\pi R^{3}}{6} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert (C ). \end{aligned}$$
(7)
Now, consider the curve \(\gamma : [0, 2\pi ] \to \mathbb{R}^{2}\) given by
$$ \gamma : \textstyle\begin{cases} x(\theta )=a+R\cos \theta, \\ y(\theta )=b+R\sin \theta , \end{cases}\displaystyle \quad \theta \in [0,2\pi ]. $$
Then \(\gamma ([0, 2\pi ] ) =\partial (C,R)\), and we write (integrating with respect to arc length)
$$\begin{aligned} \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert ( \gamma )\,dl(\gamma )&= \int _{0}^{2\pi } \biggl\vert \frac{\partial f}{\partial r} \biggr\vert \bigl(x(\theta ),y(\theta ) \bigr) \bigl(\bigl[\dot{x}(\theta ) \bigr]^{2}+\bigl[\dot{y}( \theta )\bigr]^{2} \bigr)^{\frac{1}{2}}\, d\theta \\ &=R \int _{0}^{2\pi } \biggl\vert \frac{\partial f}{\partial r} \biggr\vert (a+R\cos \theta ,b+R\sin\theta )\, d\theta. \end{aligned}$$
(8)
From (7) and (8) we obtain
$$\begin{aligned} & \biggl\vert R^{2} \int _{0}^{2\pi }f (a+R\cos\theta ,b+R\sin\theta )\,d \theta -2 \int _{0}^{2\pi } \int _{0}^{R}f (a+r\cos\theta ,b+r\sin\theta )r\,dr \,d\theta \biggr\vert \\ &\quad \leq \frac{R^{2}}{4} \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{ \partial r} \biggr\vert ( \gamma )\,dl(\gamma )+\frac{\pi R^{3}}{6} \biggl\vert \frac{ \partial f}{\partial r} \biggr\vert (C ). \end{aligned}$$
(9)
Also using the convexity of \(|\frac{\partial f}{\partial r}|\) in (4) we have
$$\begin{aligned} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert (C )&\leq \frac{1}{ \pi R^{2}} \int _{0}^{2\pi } \int _{0}^{R} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert (a+r\cos\theta ,b+r\sin\theta )\,dr\,d\theta \\ &\leq \frac{1}{2\pi R} \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{ \partial r} \biggr\vert ( \gamma )\,dl(\gamma ). \end{aligned}$$
(10)
So by replacing (10) in (9) we obtain
$$\begin{aligned} & \biggl\vert R \int _{\partial (C,R)}f(\gamma )\,dl(\gamma )-2 \int _{0}^{2\pi } \int _{0}^{R}f (a+r\cos\theta ,b+r\sin\theta )r\,dr \,d\theta \biggr\vert \\ &\quad \leq \frac{R^{2}}{3} \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{ \partial r} \biggr\vert ( \gamma )\,dl(\gamma ). \end{aligned}$$
(11)
Finally dividing (11) with \(2\pi R^{2}\) we get
$$\begin{aligned} \biggl\vert \frac{1}{2\pi R} \int _{\partial (C,R)}f(\gamma )\,dl(\gamma )-\frac{1}{ \pi R^{2}} \iint _{D(C,R)}f (x,y)\,dx\,dy \biggr\vert \leq \frac{1}{6\pi } \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert ( \gamma )\,dl( \gamma ). \end{aligned}$$
□
Example 2.2
Consider the bifunction \(f(x,y)=R-\sqrt{(x-a)^{2}+(y-b)^{2}}\) defined on the disk \(D(C,R)\). In polar coordinates we have that
$$ f(a+r\cos\theta ,b+r\sin \theta )=R-r $$
for \(0\leq r\leq R\), \(\theta \in [0,2\pi ]\) and specially \(f(a+R\cos \theta ,b+R\sin \theta )=0\) for all \(\theta \in [0,2\pi ]\). So
$$\begin{aligned} & \biggl\vert \frac{1}{2\pi R} \int _{\partial (C,R)}f(\gamma )\,dl(\gamma )-\frac{1}{ \pi R^{2}} \iint _{D(C,R)}f (x,y)\,dx\,dy \biggr\vert \\ &\quad =\frac{1}{\pi R^{2}} \iint _{D(C,R)}f (x,y)\,dx\,dy=\frac{1}{ \pi R^{2}} \int _{0}^{2\pi } \int _{0}^{R} (R-r)r\,dr\,d\theta = \frac{R}{3}. \end{aligned}$$
(12)
On the other hand, it is not hard to see that \(|\frac{\partial f}{ \partial r}|(a+R\cos\theta ,b+R\sin\theta )=1\) for all \(\theta \in [0,2 \pi ]\), and so
$$ \frac{1}{6\pi } \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert ( \gamma )\,dl(\gamma )=\frac{R}{3}. $$
(13)
Then identities (12) and (13) show that inequality (5) is sharp.
The following result is the mid-point type inequality related to (4).
Theorem 2.3
Consider a set
\(I\subset \mathbb{R}^{2}\)
with
\(D(C,R)\subset I^{ \circ }\). Suppose that the mapping
\(f:D(C,R)\to \mathbb{R}\)
has continuous partial derivatives in the disk
\(D(C,R)\)
with respect to the variables
r
and
θ
in polar coordinates. If, for any constant
\(\theta \in [0,2\pi ]\), the function
\(|\frac{\partial f}{\partial r} |\)
is convex with respect to the variable
r
on
\([0,R]\), then
$$ \biggl\vert \frac{1}{\pi R^{2}} \iint _{D(C,R)}f (x,y)\,dx\,dy-f(C) \biggr\vert \leq \frac{2}{3\pi } \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{ \partial r} \biggr\vert ( \gamma )\,dl(\gamma ). $$
(14)
Proof
As we have seen in the proof of Theorem 2.1, for a constant \(\theta \in [0,2\pi ]\), if we consider \(x(r)=a+r\cos \theta \) and \(y(r)=b+r\sin \theta \), then we have \(([\dot{x}(r)]^{2}+[\dot{y}(r)]^{2} )^{\frac{1}{2}}=1\). So from fundamental theorem of calculus we have
$$ \int _{0}^{R} \frac{\partial f}{\partial r} (a+r\cos\theta ,b+r \sin \theta ) \,dr=f (a+R\cos\theta ,b+R\sin\theta )-f(C). $$
Hence
$$\begin{aligned} & \int _{0}^{2\pi } \int _{0}^{R}\frac{\partial f}{\partial r} (a+r\cos \theta ,b+r \sin\theta )\,dr\,d\theta \\ &\quad = \int _{0}^{2\pi }f (a+R\cos\theta ,b+R\sin\theta )\,d \theta -2 \pi f(C), \end{aligned}$$
which implies that
$$ \int _{0}^{2\pi } \int _{0}^{R}\frac{\partial f}{\partial r} (a+r\cos \theta ,b+r \sin\theta )\,dr\,d\theta =\frac{1}{R} \int _{\partial (C,R)}f( \gamma )\,dl(\gamma )-2\pi f(C). $$
(15)
Now from (15) we obtain
$$\begin{aligned} & \biggl\vert \frac{1}{2\pi R} \int _{\partial (C,R)}f(\gamma )\,dl(\gamma )-f(C) \biggr\vert \\ &\quad \leq \frac{1}{2\pi } \int _{0}^{2\pi } \int _{0}^{R} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert (a+r\cos\theta ,b+r\sin\theta )\,dr\,d\theta. \end{aligned}$$
Since \(|\frac{\partial f}{\partial r}|\) is convex, then it follows that
$$\begin{aligned} & \biggl\vert \frac{1}{2\pi R} \int _{\partial (C,R)}f(\gamma )\,dl(\gamma )-f(a,b) \biggr\vert \\ &\quad \leq \frac{1}{2\pi } \biggl[ \int _{0}^{2\pi } \int _{0}^{R} \biggl\vert \frac{ \partial f}{\partial r} \biggr\vert \biggl(\frac{r}{R} (a+R\cos\theta ,b+R\sin \theta )+ \biggl(1-\frac{r}{R} \biggr) (a,b ) \biggr)\,dr\,d\theta \biggr] \\ &\quad \leq \frac{1}{2\pi } \biggl[ \int _{0}^{2\pi } \int _{0}^{R}\frac{r}{R} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert (a+R\cos\theta ,b+R\sin \theta )\,dr\,d\theta \\ &\qquad {}+ \int _{0}^{2\pi } \int _{0}^{R} \biggl(1-\frac{r}{R} \biggr) \biggl\vert \frac{ \partial f}{\partial r} \biggr\vert (C )\,dr\,d\theta \biggr] \\ &\quad = \frac{1}{4 \pi } \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert ( \gamma )\,dl(\gamma )+\frac{R}{2} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert (C). \end{aligned}$$
(16)
From the triangle inequality and (16) we get
$$\begin{aligned} & \biggl\vert \frac{1}{\pi R^{2}} \int _{0}^{2\pi } \int _{0}^{R}f (a+r\cos \theta ,b+r\sin\theta )\,dr \,d\theta -f(C) \biggr\vert \\ &\quad \leq \frac{1}{4\pi } \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{ \partial r} \biggr\vert ( \gamma )\,dl(\gamma )+\frac{R}{2} \biggl\vert \frac{\partial f}{ \partial r} \biggr\vert (C) \\ &\qquad {}+ \biggl\vert \frac{1}{\pi R^{2}} \int _{0}^{2\pi } \int _{0}^{R}f (a+r\cos \theta ,b+r\sin\theta )r\,dr \,d\theta -\frac{1}{2\pi R} \int _{\partial (C,R)}f(\gamma )\,dl(\gamma ) \biggr\vert . \end{aligned}$$
(17)
Since \(|\frac{\partial f}{\partial r}|\) satisfies the Hermite–Hadamard inequality (4), then
$$ \biggl\vert \frac{\partial f}{\partial r} \biggr\vert (C)\leq \frac{1}{2\pi R} \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert ( \gamma )\,dl( \gamma ). $$
So, by replacing (5) and the inequality in (17) above, we obtain
$$ \biggl\vert \frac{1}{\pi R^{2}} \iint _{D(C,R)}f (x,y)\,dx\,dy-f(C) \biggr\vert \leq \frac{2}{3\pi } \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{ \partial r} \biggr\vert ( \gamma )\,dl(\gamma ). $$
□
Remark 2.4
If the functions f and \(|\frac{\partial f}{\partial r} |\) are convex on \(D(C,R)\), then by the use of inequalities (5), (14), and (4) we have
$$ 0\leq \frac{1}{\pi R^{2}} \iint _{D(C,R)}f (x,y)\,dx\,dy-f(C)\leq \frac{2}{3 \pi } \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert ( \gamma )\,dl(\gamma ) $$
and
$$ 0\leq \frac{1}{2\pi R} \int _{\partial (C,R)}f(\gamma )\,dl(\gamma )-\frac{1}{ \pi R^{2}} \iint _{D(C,R)}f (x,y)\,dx\,dy\leq \frac{1}{6\pi } \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert ( \gamma )\,dl( \gamma ). $$
Example 2.5
There exists a function satisfying all the conditions of Remark 2.4 as well. Consider the function \(f(x,y)=x^{2}+y^{2}\) with \((x,y)\in \mathbb{R}^{2}\) defined on a disk \(D((0,0),R)\). It is clear that \(f(r,\theta )=r^{2}\) and \(|\frac{\partial f}{\partial r} |=2r\), which is equivalent to \(f(x,y)=2\sqrt{x^{2}+y^{2}}\) with \((x,y)\in \mathbb{R}^{2}\) defined on a disk \(D ((0,0),R )\). As we can see in Fig. 3, the functions f and \(|\frac{\partial f}{ \partial r} |\) are convex.