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
$$\begin{aligned} \biggl\vert \frac{1}{2\pi R} \int _{\partial (C,R)}f(\gamma )\,dl(\gamma )-\frac{1}{ \pi R^{2}} \iint _{D(C,R)}f (x,y)\,dx\,dy \biggr\vert \leq \frac{1}{6\pi } \int _{\partial (C,R)} \biggl\vert \frac{\partial f}{\partial r} \biggr\vert ( \gamma )\,dl( \gamma ). \end{aligned}$$
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