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Let

  • $\lambda$ denote the Lebesgue measure on $\mathbb R$
  • $T>0$
  • $E$ be a $\mathbb R$-Banach space
  • $u,v\in L^1_{\text{loc}}((0,T),E)$

In Navier-Stokes Equations: Theory and Numerical Analysis by Roger Temam I've read in Lemma 1.1 of Chapter III that the following statements are equivalent:

  1. $u\in W^1((0,T),E)$ with $u'=v$
  2. Let $\eta\in E'$ $\Rightarrow$ $$\frac{\rm d}{{\rm d}t}\langle\eta(u)\rangle_{\mathcal D'((0,T))}=\langle\eta(v)\rangle_{\mathcal D'((0,T))}\tag 1$$

where $W^1((0,T),E)$ is the set of $L^1_{\text{loc}}((0,T),E)$-functions which admit a weak derivative in $L^1_{\text{loc}}((0,T),E)$ and $$\langle f\rangle_{\mathcal D'((0,T))}(\phi):=\int_{(0,\:T)}\phi f\;{\rm d}\lambda$$ denotes the distribution generated by $f\in L^1_{\text{loc}}((0,T),E)$.


I wonder why he is taking the derivative in 2. in the distributional sense. If I'm not terribly wrong, we should have $\eta(u)\in W^1((0,T))$ for all $\eta\in E'$, if $u\in W^1((0,T),E)$.

Assume that 1. holds and let $\eta\in E'$. In fact, it's easy to see that $$\eta(u)\in L^1_{\text{loc}}((0,T))\;.\tag 2$$ Now, we should have

\begin{equation} \begin{split} -\int_{(0,\:T)}\phi\eta(v)\;{\rm d}\lambda&=-\int_{(0,\:T)}\eta(\phi v)\;{\rm d}\lambda\\ &=\eta\left(-\int_{(0,\:T)}\phi v\;{\rm d}\lambda\right)\\ &=\eta\left(\int_{(0,\:T)}\phi'u\;{\rm d}\lambda\right)\\ &=\int_{(0,\:T)}\eta(\phi'u)\;{\rm d}\lambda\\ &=\int_{(0,\:T)}\phi'\eta(u)\;{\rm d}\lambda \end{split}\tag 3 \end{equation}

So, 1. and 2. should be equivalent to

  1. Let $\eta\in E'$ $\Rightarrow$ $\eta(u)\in W^1((0,T))$ with $$\frac{\rm d}{{\rm d}t}\eta(u)=\eta(v)\tag 4$$

too. Am I wrong?

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