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?

  • $\begingroup$ I think your question will be answered once you clarify the definitions of various objects in the expressions. $\endgroup$
    – timur
    Commented Dec 18, 2012 at 23:55
  • $\begingroup$ @timur I added some more details, I hope it helps. $\endgroup$
    – user28178
    Commented Dec 19, 2012 at 20:09

1 Answer 1


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

  • $\begingroup$ I see. Thanks. Can you recommend any literature where this is discussed? I can't find much about vector-valued distributions. $\endgroup$
    – user28178
    Commented Dec 20, 2012 at 11:00

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