Let $\Omega \subset \mathbb{R}^n$ be a compact smooth hypersurface. Suppose $\varphi \in C\_c^\infty(0,T; H^1(\Omega))$ is a $H^1(\Omega)$-valued test function (so $\varphi(t) \in H^1(\Omega)$ for each $t$ and $\varphi(0) = \varphi(T)= 0$), and $f \in C^1([0,T] \times \Omega)$. Let $w \in L^2(0,T;H^1(\Omega))$ with weak time derivative $w' \in L^2(0,T;H^{-1}(\Omega))$ (i.e. it satisfies $\int\_0^T w(t)\varphi'(t) = -\int\_0^T w'(t)\varphi(t)$ for all $\varphi \in C\_c^\infty(0,T;H^1(\Omega))$.)


How do I show that
$$\int\_0^T \langle w', f\varphi \rangle\_{H^{-1}, H^1} = \int\_0^T \langle fw', \varphi \rangle\_{H^{-1}, H^1}$$


where $\langle f,u \rangle\_{H^{-1}, H^1} := f(u)$ for $f \in H^{-1}$ and $u \in H^1.$ 


I tried writing the pairing as a functional and used RRT but to no avail. Appreciate any help..


(My question in MSE received no attention, so I posted it here. <https://math.stackexchange.com/questions/260897/weak-derivative-and-continuous-functions-functionals-distributions>)


**Edit** Maybe I am asking the wrong question. Because what does $fw'$ as a functional mean? Perhaps it's *defined* to satisfy $fw'(\varphi) = w'(f\varphi)$? If so, is this an appropriate thing to do?



weak derivative and continuous function
Multiplication of a distribution by a smooth function is defined in the way you indicate. So there is nothing to prove.


