I have an open set $A\subset \mathbb R^2$ and $f\in C^1(A,\mathbb R^2)$ and I want to show that :
$$(\frac{\partial f_1}{\partial x_1} + \frac{\partial f_2}{\partial x_2}) (x) = \lim_{\epsilon \to 0^+}\frac{1}{4\epsilon ^2}\int_{x_1-\epsilon}^{x_1+\epsilon} \int_{x_2-\epsilon}^{x_2+\epsilon} \frac{\partial f_1}{\partial x_1} + \frac{\partial f_2}{\partial x_2} d x_2 dx_1 $$
I am now sure how I can show this equality however... I mean I can split the integral into two parts and then start by integrating w.r.t the variable that $f_i$ was originally derived to. but then I am still left with an integral over the other variable, and since I do now know $f$ I do not know how I could possibly simplify that...
Any help and hints are greatly appreciated!
I assume that by $\frac{\partial f}{x_1}, \frac{\partial f}{x_2}$, you mean $\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}.$ If this is the case, this is part of a larger result.
More generally, for any continuous function $g : \mathbb R^2 \to \mathbb R$, we have $$g(x_1,x_2) = \lim_{\epsilon \to 0^+} \frac{1}{4\epsilon^2}\int^{x_1 + \epsilon}_{x_1 - \epsilon} \int^{x_2+\epsilon}_{x_2 - \epsilon} g(s,t) dt ds.$$ Indeed, for any $\epsilon > 0$, we see that $$g(x_1,x_2) = \frac{1}{4\epsilon^2}\int^{x_1 + \epsilon}_{x_1 - \epsilon} \int^{x_2+\epsilon}_{x_2 - \epsilon} g(x_1,x_2) dt ds$$ since the integrand is constant in $s,t$ and the area of integration is $4\epsilon^2$. Thus for $\epsilon > 0$, we have \begin{align*} \left \lvert g(x_1,x_2) -\frac{1}{4\epsilon^2}\int^{x_1 + \epsilon}_{x_1 - \epsilon} \int^{x_2+\epsilon}_{x_2 - \epsilon} g(s,t) dt ds \right \rvert &= \frac{1}{4\epsilon^2} \left\lvert \int^{x_1 + \epsilon}_{x_1 - \epsilon} \int^{x_2+\epsilon}_{x_2 - \epsilon} \big(g(x_1,x_2) - g(s,t)\big) dt ds\right \rvert \\ &\le \frac{1}{4\epsilon^2} \int^{x_1 + \epsilon}_{x_1 - \epsilon} \int^{x_2+\epsilon}_{x_2 - \epsilon} \lvert g(x_1,x_2) - g(s,t) \rvert dt ds \end{align*} Since $g$ is continuous, for any $\delta > 0$, there is $\epsilon > 0$ small enough that $$(s,t) \in (x_1 - \epsilon, x_1+ \epsilon) \times (x_2 - \epsilon, x_2 + \epsilon) \,\,\,\, \implies \,\,\,\, \lvert g(x_1,x_2) - g(s,t) \rvert < \delta.$$ Thus for such $\epsilon$, we have $$\left \lvert g(x_1,x_2) -\frac{1}{4\epsilon^2}\int^{x_1 + \epsilon}_{x_1 - \epsilon} \int^{x_2+\epsilon}_{x_2 - \epsilon} g(s,t) dt ds \right \rvert < \frac{1}{4\epsilon^2}\int^{x_1 + \epsilon}_{x_1 - \epsilon} \int^{x_2+\epsilon}_{x_2 - \epsilon} \delta\,\, dt ds = \delta.$$ That is, as $\epsilon > 0$ gets arbitrarily small, we also see that $$\left \lvert g(x_1,x_2) -\frac{1}{4\epsilon^2}\int^{x_1 + \epsilon}_{x_1 - \epsilon} \int^{x_2+\epsilon}_{x_2 - \epsilon} g(s,t) dt ds \right \rvert$$ is made arbitrarily small. By definition, this means $$g(x_1,x_2) = \lim_{\epsilon \to 0^+} \frac{1}{4\epsilon^2}\int^{x_1 + \epsilon}_{x_1 - \epsilon} \int^{x_2+\epsilon}_{x_2 - \epsilon} g(s,t) dt ds.$$ Now just apply this result to $\text{div} f$ for your $f$.