The following question is motivated from the fact that I need to do some calculations in the weak sense, since I do not have enough regularity of the function $u$.
Let $ u \in L^2 (0,T; W^{1,2} (\Omega))$ and $ \partial_{t}u \in L^2 (0,T; {(W^{1,2} (\Omega)})^\prime)$. Also assume $ u \in L^\infty {((0,T) \times \Omega)}, \ u \geq 0$ a.e. in $(0,T) \times \Omega$ and $ \delta >0$ be a constant. Does the following hold? \begin{eqnarray*} < \partial_{t} \log (u+\delta), u+ \delta)>_{(W^{1,2} (\Omega))^\prime,W^{1,2} (\Omega)}=\left< \frac{1}{u+\delta}\partial_{t} u, u+\delta\right>_{(W^{1,2} (\Omega))^\prime,W^{1,2} (\Omega)} \\=<\partial_{t}u,1>_{(W^{1,2} (\Omega))^\prime,W^{1,2} (\Omega)} \end{eqnarray*} Clearly the above calculation is true if $u$ is smooth w.r.t. time and space. In that case one replaces the action of the functional by the $L^2$ inner product. Of course, the question is vague, because we explicitly did not define $\partial_{t} \log (u+\delta)$. But is there a natural way to define $\partial_{t} \log (u+\delta) \in (W^{1,2} (\Omega))^\prime $ such that the above calculation holds? For example, in distribution theory one can naturally define the multiplication of a distribution with a bump function. So can we make such definitions on $(W^{1,2} (\Omega))^\prime$ so that the above calculation holds?