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Let $V \subset H \subset V^*$ be a Gelfand triple (eg. $H^1 \subset L^2 \subset H^{-1}$).

Let $u \in L^2(0,T;V)$ have a distributional derivative $u' \in L^2(0,T;V^*)$. So $\int_0^T u(t)\varphi'(t) = -\int_0^T u'(t)\varphi(t)$ holds for all $\varphi \in C_c^\infty(0,T)$ as an equality in $V^*$.

Let $f:\mathbb{R} \to \mathbb{R}$ be a differentiable function with $f'$ bounded.

Suppose $u' \in L^2(0,T;L^2)$; then the chain rule formula for the weak derivative $(f(u))' = f'(u)u' \in L^2(0,T;L^2)$ makes sense.

But if $u' \in L^2(0,T;H^{-1})$ only, how to make sense of $(f(u))'$? Is it right to define the derivative as $$\langle (f(u))', v \rangle := \langle u', f'(u)v \rangle$$ if for all $v \in L^2(0,T;H^1)$, $f'(u)v \in L^2(0,T;H^1)$?

Where may I find more details about this sort of thing? I'd like to avoid BV spaces and measures because this is simpler. Thank you.

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    $\begingroup$ I don't think the question being asked makes sense. You are treating $u$ as an $L^2$ valued function, then you cannot plug $u$ into $f$, which has domain $\mathbb{R}$. Perhaps you meant to think about $u$ as a space-time function, and $f$ as really acting on the final output. But then it is not compatible with the form of the chain rule you are suggesting. (In fact, if the spatial domain is non-compact, and $f(0) \neq 0$, you don't even have $f(u)$ is an element of $L^2(0,T;L^2)$.) So I think the first thing is to for you to think about and specify what exactly is this object $f(u)$. $\endgroup$
    – Willie Wong
    Commented Mar 26 at 4:57

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For the $H^1 \subset L^2 \subset H^{-1}$ case.

Let $K_\epsilon$ be a mollifier then for $u\in L^2(0,T;L^2)$ we have that $u_\epsilon = K_\epsilon\ast u$ is actually smooth. In fact, $\lim_{\epsilon \rightarrow 0} u_\epsilon$ converges to $u$ in $L^p_{loc}$ for all $p\in [1,\infty)$. Since $f$ is differentiable, then it is continuous. So, we have the $\lim_{\epsilon \rightarrow 0} f(u_\epsilon) = f(u)$ in the same spaces as before.

Consider the following, \begin{align*} \langle v',f(u_\epsilon)\rangle &= -\langle v, f'(u_\epsilon) u_\epsilon' \rangle \\ &= -\langle v,f'(u) u_\epsilon' \rangle + \langle v,\left[f'(u) - f'(u_\epsilon)\right] u_\epsilon'\rangle \end{align*} A useful fact to use is that since $f$ has bounded derivative then $f$ is Lipshitz. Using this fact and taking $\epsilon$ to zero gives you what you want.

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    $\begingroup$ Do you mean the brackets to be in space or space-time? I'm trying to understand the meaning of the displayed equation ... $\endgroup$
    – Mark Peletier
    Commented Jun 13, 2014 at 13:58
  • $\begingroup$ It should only be space, no? I meant it as mathias did in his question and I thought he had taken it to only be a space integral. The idea is to use a smoothing argument and take limits. $\endgroup$
    – k3thomps
    Commented Jun 13, 2014 at 14:17
  • $\begingroup$ If the brackets are in space, then it seems the full time derivative is missing .. $\endgroup$
    – Mark Peletier
    Commented Jun 13, 2014 at 14:24
  • $\begingroup$ Shouldn't the second bracket read $\langle v,f'(u_\epsilon)u'_\epsilon \rangle$? $\endgroup$
    – Mark Peletier
    Commented Jun 13, 2014 at 14:30
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    $\begingroup$ @No-one: are you sure the proof of the linked MO question applies here? At the level the OP is asking about I don't even believe $f(u)$ makes sense as an $L^2$ function.... $\endgroup$
    – Willie Wong
    Commented Mar 26 at 4:40

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