Proof of Lemma 4.2 in [G-T] pg 55 Let $\Omega$ be an open bounded subset of $\mathbb{R}^n (n\geq 3)$ and let $\Omega_0$ be any domain containing $\Omega$ for which the divergence theorem is true. Let $f$ be bounded and locally Holder continuous (with exponent $\alpha\leq 1$) in $\Omega$ and $f$ is extended to vanish outside $\Omega$. 
We also fix a function $\eta\in C^1(\mathbb{R})$ satisfying:


*

*$0\leq\eta\leq 1$,

*$0\leq \eta'\leq 2$ and

*$\eta (t)=0$ for $t\leq 1$ and $\eta=1$ for $t\geq 2$.


Define for $\varepsilon>0$:
\begin{equation}
v_{\varepsilon}(x)\equiv \int_{\Omega} D_i\Phi(x-y)\eta_{\varepsilon}f(y) \text{ d}y\quad x\in\Omega,
\end{equation}
where $\Phi$ is the fundamental solution to Laplace's equation, $\eta_{\varepsilon}=\eta(\vert x-y\vert /\varepsilon)$ and the operator $D_i=\frac{\partial}{\partial x_i}$.
In one of the steps of the proof of lemma 4.2 in Gilbarg-Trudinger's book, Elliptic PDE of Second Order pg 55, they differentiate $v_{\varepsilon}$ to obtain:
\begin{align}
D_j v_{\varepsilon}(x)&=\int_{\Omega}D_j(D_i\Phi\eta_{\varepsilon})f(y)\text{ d} y\\
&=\int_{\Omega}D_j(D_i\Phi\eta_{\varepsilon})(f(y)-f(x))\text{ d} y+f(x)\int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})\text{ d} y\\
&=\int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})(f(y)-f(x))\text{ d} y-f(x)\int_{\partial\Omega_0}D_i\Phi\nu^j(y)\text{ d} S_y
\end{align} 
I am trying to justify these steps, particularly the last two equations and I am having a bit of difficulty.
Below is my work.
Since $f=0$ on $\Omega_0\setminus\Omega$ we can write
\begin{equation}
D_j v_{\varepsilon}(x)=\int_{\Omega}D_j(D_i\Phi\eta_{\varepsilon})f(y)\text{ d} y=\int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})(f(y)-f(x))\text{ d} y+f(x)\int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})\text{ d} y.
\end{equation}
Now I am trying to justify why
\begin{equation}
\int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})\text{ d} y=-\int_{\partial\Omega_0}D_i\Phi\nu^j(y)\text{ d} S_y.
\end{equation}
We note the following:
\begin{align}
\Phi (x-y)&=\frac{\vert x-y\vert^{2-n}}{n\alpha(n)(2-n)}\\
\Phi_{x_i}&=-\Phi_{y_i}\\
\Phi_{y_i x_j}&=-\Phi_{y_iy_j}\\
\frac{\partial\eta_{\varepsilon}}{\partial x_j}&=-\frac{\partial\eta_{\varepsilon}}{\partial y_j}
\end{align}
Thus we have (using the above equations and integration by parts formula as appropriate):
\begin{align}
\int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})\text{ d} y&=\int_{\Omega_0}\frac{\partial}{\partial x_j}\left[\frac{\partial\Phi}{\partial x_i} \eta_{\varepsilon}\right]\text{ d}y\\
&=-\int_{\Omega_0}\frac{\partial}{\partial x_j}\left[\frac{\partial\Phi}{\partial y_i} \eta_{\varepsilon}\right]\text{ d}y\\
&=\int_{\Omega_0}\frac{\partial^2\Phi}{\partial y_j\partial y_i} \eta_{\varepsilon}\text{ d}y-\int_{\Omega_0}\frac{\partial\Phi}{\partial y_i} \frac{\partial\eta_{\varepsilon}}{\partial x_j}\text{ d}y\\
&=\int_{\partial\Omega_0}\frac{\partial\Phi}{\partial y_i} \eta_{\varepsilon}\nu^{j}(y)\text{ d}S_y+\int_{\Omega_0}\frac{\partial\Phi}{\partial y_i} \frac{\partial\eta_{\varepsilon}}{\partial y_j}\text{ d}y\\
&=-\int_{\partial\Omega_0}\frac{\partial\Phi}{\partial x_i} \nu^{j}(y)\text{ d}S_y+\int_{\Omega_0}\frac{\partial\Phi}{\partial y_i} \frac{\partial\eta_{\varepsilon}}{\partial y_j}\text{ d}y.\\
\end{align}
I can't seem to justify why the last integral disappears. In the above $\varepsilon$ is such that $\text{dist}(x, \partial\Omega)>2\varepsilon$ so that $B(x, 2\varepsilon)\subset\Omega$.
 A: The last integral will not be zero in general. In fact it will not even tend to zero as $\epsilon \rightarrow 0$ in general. The partial derivative of $\eta_{\epsilon}$ is of order $\epsilon^{-1}$ while the partial derivative of $\Phi$ is of order $|x-y|^{-n+1}$ which, since the integral vanishes outside $B_{\epsilon}$, is of order $\epsilon^{-n+1}$ the term in the last integral is therefore of order $\epsilon^{-n}$ and is being integrated over (up to some constant multiple of the radius) $B_{\epsilon}$ which has volume $\epsilon^n$. Therefore the integral will have constant order and will not tend to zero as $\epsilon \rightarrow 0$.
Actually your mistake is a simple one, you have just applied the integration by parts formula incorrectly. The integral
$\displaystyle\int_{\Omega_0} \dfrac{\partial^2 \Phi}{\partial y_j \partial y_i} \eta_{\epsilon}$
does not equal
$\displaystyle\int_{\partial \Omega_0} \dfrac{\partial \Phi}{\partial y_i} \eta_{\epsilon} \nu^j dS$.
There is another term from the integration by parts formula which exactly cancels the last integral term in your last equation.
