Let $U$ be an open set in $\mathbb{R}^n$,$f:U\to \mathbb{R},f\in W^{1,p}(U)$. Let $\tau_{h,i}f(x)=\frac{f(x+he_i)-f(x)}{h},h>0$
Given any compact $V\subset U$, show there exists $h_0>0$ such that $D(\tau_{h,i}f)=\tau_{h,i}(Df)$, for $h<h_0$
I try to prove it by definition.
Let $h_0=\frac{1}{2}d(\partial U,V)>0,h<h_0$, let $\phi\in C^1_c(V)\subset C^1_c(U)$. Write $Df=(\frac{\partial}{x_j}f)_{j=1}^n$ then by definition, $$\int_Vf(x)\frac{\partial \phi(x)}{\partial x_j}=-\int_V\frac{\partial f(x)}{\partial x_j} \phi(x)\tag{1}$$ Then $$\begin{align} \int_V\tau_{h,i}f(x)\frac{\partial \phi(x)}{\partial x_j}&=\int_V\frac{f(x+he_i)-f(x)}{h}\frac{\partial \phi(x)}{\partial x_j}\\ &=\int_V\frac{f(x+he_i)}{h}\frac{\partial \phi(x)}{\partial x_j}-\frac{f(x)}{h}\frac{\partial \phi(x)}{\partial x_j}\\ &=\frac{1}{h}\left[\color{red}{-\int_V\frac{\partial f(x+he_i)}{\partial x_j} \phi(x)}+\int_V\frac{\partial f(x)}{\partial x_j} \phi(x)\right]\\ &=-\int_V\tau_{h,i}\frac{ \partial f}{\partial x_j}(x) \phi(x) \end{align}$$
Then we are done. But the problem is I don't know whether the red colored text is correct. Since by definition, to use $(1)$, the functions should be evaluated at the same point, while in my proof, they are not.
Thanks for your help.