A detail in Qing-Lin's Elliptic PDE book, section 3.4. I am reading Section 3.4 of Lin and Qing's textbook on elliptic PDEs.
Now I have a problem to show that the function
$$
\phi(r) := \int_{B_r(x_0)}|Du(y) - (Du)_{x_0,r}|^2 \,dy
$$
is nondecreasing in $r$, where $x_0$ is a fixed point in $B_1(0)$, $(Du)_{x_0,r}:= \frac{1}{m(B_r(x_0))} \int_{B_r(x_0)} Du(z)\,dz$ ($m$ means the Lebesgue measure) and $u$ is the $H^1(B_1)$ weak solution to
$$
-D_i(a^{ij} D_j u)+ cu = f,
$$
an equation with $1-n/q$-Holder continuous coefficient function $a^{ij}$, and the zero order coefficient and source term $c,f \in L^q$. (q > n.)
Such nondecreasing property is necessary to apply the iteration lemma 3.4 state there.
I also check the other books, for example, Giaquinta(1983,1993), Chen Ya-Zhe & Wu Lan-Cheng(1998) and Giquinta-Martinazzi (2012). But I did not find any explanation of the nondecreasing property, too.
I really appreciate any discussion or idea. Thanks.
Alternative proof to the Schauder theory claimed there is recorded in my personal errata sheet to Qing-Lin's book (page 62)
 A: I learn from the lecture by Spencer Frei that my question can be solved by its Lemma 12, page 24.
Theorem 1
If $f \in L^2$, then
$\int_{B(x_0,R)} |f- f_{x_0,R}|^2 = \inf_{a \in \mathbb{R}} \int_{B(x_0,R)} |f- a|^2$
Proof:
Let $F(a):= \int_{B(x_0,R)} |f- a|^2 $. Then $F'(a) = 2 \int_{B(x_0,R)} (a - f)$ with only one critical point $A = f_{x_0,R}$. By definition, $F$ is convex and hence the critical point is a global minimizer by the following theorem.
Theorem 2
Let $C \subset \mathbb{R}^n$ convex. Let $F: C \to \mathbb{R}$ has continuous first partial derivative. Then $F$ is convex if and only if $f(x) + \nabla f(x) \cdot (y-x) \leq f(y)$ for all $x,y \in C$.
Proof:
($\Rightarrow$) By convexity, 
$
\frac{f( x + \lambda (y-x) ) - f(x) }{\lambda} \leq f(y) - f(x).
$
Letting $\lambda \to 0$, we have the desired inequality.
($\Leftarrow$) This is proved by the following two inequalities:
$
f(x+ \lambda (y- x)) + \nabla f(x+ \lambda (y- x)) \cdot \lambda(y-x) \leq f(x)
$
$
f(x+ \lambda (y- x)) + \nabla f(x+ \lambda (y- x)) \cdot (1-\lambda)(x-y) \leq f(y)
$
