Suppose that we have a sequence of functions $u_j$ that are in $L^{\infty}(0,1)$. Then the sequence of maps $N_j(s) := \|u_j(s)\|^2$ are also in $L^{\infty}(0,1)$. Hence they give rise to distributions and therefore has a distributional derivative. What is the explicit formula for $DN_j$? Is it related to the classical formula $2\langle u_j , Du_j\rangle$?
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1$\begingroup$ What is the norm $\|\cdot\|$ here? $\endgroup$– AndrewCommented Jun 30, 2012 at 17:12
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$\begingroup$ Euclidean norm. $\endgroup$– dcs24Commented Jul 1, 2012 at 9:16
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First, I do not understand why do you need a sequence of functions when the question involves an individual function. Suppose that $u$ is real valued. Then the product of the distributions $u$ and $u'$ may not even be defined. (This is the case when $u$ is the Heaviside function.) However, if the distributional derivative of $u$ is Lebesgue integrable, then
$$ \frac{d}{dt}(\; u^2\;) = 2u u'. $$
answered Jul 1, 2012 at 9:06