On page 250 Evans' PDE book proves that smooth functions are dense in the space of weak derivatives.
One step that confuses me is
$$D^\alpha u^\epsilon(x) = \int_U D^\alpha_x \eta_\epsilon(x-y)u(y) dy = (-1)^{|\alpha|} \int_U D^\alpha_y \eta_\epsilon(x-y)u(y) dy $$
Here $\eta_\epsilon$ is the standard mollifier and $u^\epsilon$ is the convolution of $\eta_\epsilon$ and $u.$
I do not understand how he switched the derivative from $x$ to $y$. If it is $x \to -y$, why is the integral still with respect to $dy$? this is a multi-index derivative
Consider the one-dimensional case with $\alpha=1$. Then we have $$ D_x \eta_\varepsilon(x-y). = \eta_\varepsilon'(x-y) = -D_y \eta_\varepsilon(x-y) $$ The multi-dimensional case is similar, but the factor is not $-1$ but $(-1)^{|\alpha|}$ because you have to do the same trick $|\alpha|$ times.
A 2-dimensional example: Next, consider the two-dimensional case with $\alpha=(1,1)$.
We have to be careful to not confuse the derivatives with respect to a variable (like $\tfrac{d}{dx_1}$) with partial derivatives of functions. For the function $\eta_\varepsilon:\Bbb R^2\to\Bbb R$ we will denote its partial derivatives by $\partial_1$ and $\partial_2$.
We have $$ D_x^\alpha\eta_\varepsilon(x-y) = \tfrac{d}{dx_1}\tfrac{d}{dx_2} \eta_\varepsilon(x-y) = \partial_1\partial_2\eta_\varepsilon(x-y) \\\qquad\qquad= \partial_1(-\tfrac{d}{dy_2}\eta_\varepsilon(x-y)) = -\tfrac{d}{dy_1}(-\tfrac{d}{dy_2}\eta_\varepsilon(x-y)) = (-1)^2 D_y^\alpha\eta_\varepsilon(x-y). $$