# Weak derivative and homeomorphisms commute

Suppose $\phi:H^1(S) \to H^2(R)$ is a linear homeomorphism, where $S$ and $R$ are compact surfaces in $\mathbb{R}^n$. Let $\varphi \in D(0,T;H^1(S))$ be a $H^1(S)$-valued $C_c^\infty(0,T)$ function. So $\varphi(t) \in H^1(S)$ for each $t$.

Recall that a weak derivative of a function $u$ is a function $u'$ that satisfies $$\int_0^T uv' = -\int_0^T u'v$$ for all $v \in C_c^\infty(0,T)$.

I want to show that the weak derivative of the homeomorphism of $\varphi$ is the homeomorphism of the weak derivative of $\varphi$, i.e. that $$\phi(\varphi'(t)) = [\phi(\varphi(t))]'$$

How do I show this rigorously? I can sort of see that since $\phi$ just acts in space, the prime on the RHS goes straight through to the argument of $\phi$, but that is not a proper argument.

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You can show (using approximation by riemann sums and continuity and linearity of $\phi$) that for continuous $\alpha\colon [0,T] \to H^1(R)$ you have $$\phi\left(\int_0^T \alpha(t)\, dt\right) = \int_0^T (\phi \circ \alpha)(t)\, dt.$$ Your result follows. –  martini Nov 20 '12 at 16:40
@martini thanks! –  soup Nov 20 '12 at 22:11