Let $u(x) \in H^1_0$ is a complex function in sobolev space extension by zero.
How to find $J'(u)$ for
$$
J(u)= \int\limits_0^l |u(x)|^2\operatorname{d\!}x\;??
$$
In $L_2$ it's easy:
$$
J'(u) = \left(\int\limits_0^l|u(x)|^2 \operatorname{d\!}x\right)'=\big(\|u(x)\|^2\big)'= 2 u(x),
$$
but it does not work with $H^1_0$, where
$$
\|u(x)\|^2_{H^1_0} = \int\limits_0^l |u'(x)|^2 \operatorname{d\!}x
$$
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$\begingroup$ Your assertion that $J'(u)=2u(x)$ is wrong and actually makes no sense, for several reasons, including the following: (i) the $x$ there is undefined; (ii) $J'(u)$ is a linear functional, while $u(x)$ is a number (if $x$ is a point between $0$ and $l$); (iii) we can identify $J'(u)$, in a certain manner, with $2u$ only if $u$ is real valued. See the answer below for details. $\endgroup$– Iosif PinelisCommented Mar 12, 2023 at 21:30
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1 Answer
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We shall assume that $l\in(0,\infty)$, so that for any $h\in H_0^1$ we have $$\int_0^l|h|^2=\int_0^l dx\,\Big|\int_0^x h'\Big|^2 \le\int_0^l dx\,\Big(\int_0^l|h'|\Big)^2 \\ \le\int_0^l dx\,l\,\int_0^l|h'|^2=l^2\|h\|^2_{H_0^1}. \tag{1}\label{1}$$
Next, for any $u$ and $h$ in $H_0^1$, in view of \eqref{1}, $$J(u+h)-J(u)-2\int_0^l\Re(\bar uh)=\int_0^l|h|^2=o(\|h\|_{H_0^1})$$ as $\|h\|_{H_0^1}\to0$. So, the (Fréchet) derivative $J'(u)$ of $J$ at $u\in H_0^1$ is the linear functional on $H_0^1$ given by the formula $$J'(u)(h)=2\int_0^l\Re(\bar uh)$$ for $h\in H_0^1$.
answered Mar 12, 2023 at 21:22
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$\begingroup$ thank you very much! I may not have fully understood, but why is there a dependence on h in the Frechet derivative in the last equation? $\endgroup$– anon.forCommented Mar 13, 2023 at 20:20
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1$\begingroup$ @anon.for : The Fréchet derivative $J'(u)$ of $J$ at $u$ is a linear functional on $H_0^1$, and this linear functional depends only on $u$. However, the value $J'(u)(h)$ of the linear functional $J'(u)$ at $h\in H_0^1$ depends, of course, on $h$. $\endgroup$ Commented Mar 13, 2023 at 23:45