I'm working with Evans PDE book and I can't understand this:
Let $U\subset \mathbb{R}^n$ open and $u\in L^{2}([0,T], H_0^1(U))$ with $u' \in L^2([0,T] , H^{-1}(U))$ and now we consider the mollifications of $u$ and $u'$.
Why this is correct?
For $\epsilon, \delta >0$, $$\frac{d}{dt} \|u^{\epsilon}(t)- u^{\delta}(t) \|^2_{L^2(U)} = 2\langle u^{\epsilon'} - u^{\delta'},u^{\epsilon} - u^{\delta}\rangle $$
Once you convolve $u, u'$ with say a gaussian, they become smooth $\forall t > 0$ (with compact support). Then this equality comes down to "moving the derivative inside the integral": \begin{eqnarray*} \frac{d}{dt} \|u^{\epsilon}(t)- u^{\delta}(t) \|^2_{L^2(U)} &=& \frac{d}{dt} \int_U |u^{\epsilon}(t)(x) - u^{\delta}(t)(x) |^2 dx \\ &=& \int_U \frac{d}{dt} |u^{\epsilon}(t)(x) - u^{\delta}(t)(x) |^2 dx \\ &=& \int_U 2(u^{\epsilon'}(t)(x)- u^{\delta'}(t)(x))(u^{\epsilon}(t)(x)- u^{\delta}(t)(x)) dx \\ &=& 2\langle u^{\epsilon'}(t) - u^{\delta'}(t),u^{\epsilon}(t) - u^{\delta}(t)\rangle \\ \end{eqnarray*} You can move the derivative inside due to Leibniz' Integral Rule.