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Question Suppose $u$ and $v$ are in $L^1(0,T; X)$ where $X$ is Banach. Suppose v = u' in the distributional sense. I want to show that, for $w \in X^*$, that $$\frac{d}{dt}\langle w, u \rangle = \langle w, v \rangle$$ implies $$\langle w, \int_0^T \phi' u + \int_0^T \phi v\rangle = 0$$ for all $\phi \in C_c^\infty(0,T)$.

Attempt I can get no further than this (please check if I am right): By definition, we have $$\int_0^T \langle w, u \rangle \phi' = -\int_0^T \langle w, v \rangle \phi$$ hence $$\int_0^T \langle w, u \rangle \phi' + \langle w, v \rangle \phi \;dt= 0$$ which is $$\int_0^T w(u) \phi' + w(v)\phi \;dt= 0$$ and by linearity $$\int_0^T w(\phi' u) + w(\phi v) \;dt= 0$$ so $$\int_0^T w(\phi' u) + \int_0^T w(\phi v) = \int_0^T \langle w, \phi' u\rangle + \int_0^T \langle w, \phi v\rangle $$ Where to go from here? How can I put the integral inside the functional? Also I am not sure if I am using the linearity property of functionals correctly, since I am assuming that $\phi$ doesn't depend on $X$. Thanks

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1 Answer 1

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The only result needed now it that if $L\in X^*$ and $f\in L^1(0,T;X)$, then $$\int_{(0,t)}L(f(t))dt=L\left(\int_{(0,t)}f(t)dt\right).$$ To see that, we use $s_n(t)$ for $t\in (0,T)$, a simple function such that $s_n(t)\to f(t)$ for all $t$ and in $L^1$. So we just have to deal with the case $f$ simple, and in this case it follows from linearity of the integral.

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Thanks. If $f$ is simple, it means $f(t) = \sum_{i}a_i(t)\mathbb{1}_{A_i(t)}(x),$ right? If so, I can't get your claim to work. –  soup Oct 28 '12 at 12:58

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