Proof of an elliptic equation. I'd like to see a proof of the theorem

Theorem:Let $u \in H^1(B_1)$ a weak solution of 
  \begin{equation}
- \operatorname{div}(a_{ij}(x)\nabla u(x)) = 0 \quad \text{in} \quad B_1
\end{equation}
  where.
  Then, given $0< \alpha <1$ there exists $\varepsilon = \varepsilon(n,\lambda,\Lambda,\alpha)$ such that if $\| a_{ij} -A\|_{L^\infty(B_1)}< \varepsilon $ and $A$ is constant matrix satisfying $\lambda \le A \le \Lambda$. Then $u \in C^\alpha(B_{1/2}).$

I will be very grateful. You can proof or show me a link. The notation used is the usual. But you need ask me about. Thank you.
 A: What you are looking for is known as a Cordes-Nirenberg type estimate, originally in papers by Cordes and Nirenberg in the 1950s. A good reference for all things relating to 2nd order elliptic PDES is Gilbarg and Trudinger. 
However, in this case, an easy proof follows if you read through Chapter 3 of Lin and Han's book "Elliptic Partial Differential Equations," in particular Theorem 3.8, replacing their modulus of continuity estimate $\tau(r)$ with your hypothesis. I've sketched out the argument below:
In any ball, let $u$ be a weak solution of your equation, and $w$ be the function which matches $u$ along the boundary of the ball and satisfies the constant coefficient equation in the interior. Let $v = u-w$. Then we have in all interior balls $B_r$
$$\int_{B_r} (A\nabla w)\cdot \nabla \phi = 0$$
and 
$$\int_{B_r} (a_{ij} \nabla u) \cdot \nabla \phi = 0$$
And hence 
$$\int_{B_r} (A \nabla v) \cdot \nabla \phi = \int_{B_r} (A - a_{ij}) \nabla u \cdot \nabla \phi$$
Taking $v$ to be the test function, we obtain
$$ \int_{B_r} |\nabla v|^2 \leq C \varepsilon^2 \int_{B_r} |\nabla u|^2 $$
where $C$ depends on the ellipticity of the matrix $A$. 
Next, use the following general property of $H^1$ functions on the ball, namely that
$$ \int_{B_\rho} |\nabla u|^2 \leq C \left( \left(\frac{\rho}{r}\right)^n \int_{B_r} |\nabla u|^2 + \int_{B_r} |\nabla v|^2 \right) $$
Substituting, we find
$$ \int_{B_\rho} |\nabla u|^2 \leq C \left( \left(\frac{\rho}{r}\right)^n + C \epsilon^2 \right) \int_{B_r} |\nabla u|^2$$
Hence as a function of $\rho$, the integral $D(\rho) = \int_{B_\rho} |\nabla u|^2$ satisfies 
$$D(\rho) \leq C \left( (\frac{\rho}{r})^n + C\epsilon^2 \right) D(r) $$
We apply an elementary inequality (see Lemma 3.4 in Lin and Han, for example) to see that
$$D(\rho) \leq C D(r) \left(\frac{\rho}{r}\right)^{n-2+2\alpha}$$
if $\epsilon$ is sufficiently small, where your final $C$ depends on the original constant, $n$, and $\alpha$. That control on the square norm of the gradient places $u$ in appropriate Morrey-Campanato space, which embeds into the Holder space $C^\alpha$. 
