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Let $\Omega$ be an open set from $\mathbb{R}^N$. How can we prove that if $u,v\in H^1(0,T,L^2(\Omega))$ (in Bochner sense) then $(u\cdot v)'\in L^2(0,T,L^1(\Omega))$ with $(uv)'=u'v+v'u$ and the integration by parts formula holds:

$$\int_0^T u'v\ dt=u(T)v(T)-u(0)v(0)-\int_{0}^T uv'\ dt $$

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I see this formula applied in this context, but I found no proof for it.

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    $\begingroup$ You can't. Did you perhaps mean that $u\cdot v \in H^1(0,T,L^1(\Omega))$? The product of two $L^2$ functions is generally not in $L^2$. $\endgroup$
    – Willie Wong
    Commented Jan 11 at 7:20
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    $\begingroup$ This is an exception. See, the book of Brezis, page 215. $\endgroup$
    – Bogdan
    Commented Jan 11 at 7:41
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    $\begingroup$ It is not, Brezis p. 215 is about Sobolev functions on an interval which are automatically bounded and this is exactly what you need. In the question, $(uv)(t)$ is generically only an $L^1(\Omega)$ function and there is no real way out without further assumptions. $\endgroup$
    – Hannes
    Commented Jan 11 at 8:52
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    $\begingroup$ The statement is true and it is a generalization of Brezis, p. 215. $\endgroup$
    – Bogdan
    Commented Jan 11 at 9:11
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    $\begingroup$ No, it is wrong for $\Omega = (0, 1)$ and $u(t,x) = v(t,x) = x^{-1/4}$. $\endgroup$
    – Keba
    Commented Jan 11 at 10:20

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