What happens when you change space of test functions associated with weak derivatives?

Question

Recall that $u \in L^2(0,T;H^1)$ has weak derivative $u' \in L^2(0,T;H^{-1})$ iff $$\int_0^T uv' = -\int_0^T u'v$$ holds for all $v \in C_0^\infty(0,T).$

What happens if we only require that this holds for all $v \in C_0^1(0,T)$? What changes? What issues are there to consider?

Answer 1

The reason the definition of a distribution involves infinitely differentiable functions is to make it possible to define derivatives of all orders of any distribution. However, if you stick to distributions of orders at most $1$ (which is usual in the context of 2nd order elliptic problems), then the space of test functions can be taken to be $C^1$, as gerw explained. Unless you want to differentiate the elements of $H^{-1}$ once more, which is unlikely.

Answer 2

Nothing changes at all.

Define $W(0,T) = L^2(0,T; H^1) \cap H^1(0,T; H^{-1})$. Note that for $u \in W(0,T)$ the mapping $$ v \mapsto \int_0^T u \, v' + u' \, v \, \mathrm{d}t $$ is continuous from $H_0^1(0,T)$ to $H^{-1}$. Hence, if the mapping is zero on the dense subspace $C_0^\infty(0,T)$, it is zero on the whole space $H_0^1(0,T)$, in particular on the subspace $C^1_0(0,T)$.