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Say we have a function $F(\lambda) = \|f(\lambda)\|_{H^{-1}}$ where $\lambda \in \mathbb{R}$ and $H^{-1}(\mathbb{R}^n)$ is the usual dual of the Sobolev space $H^{1}_{0}(\mathbb{R}^n)$. Suppose we are interested in minimizing the function $F$, then is the following justified: $$\partial_{\lambda} F = \|\partial_{\lambda} f\|_{H^{-1}}$$ assuming $f$ smoothly depends on the parameter $\lambda?$

Edit [2]: Based on the comments, it does not make sense to take the norm of the derivative. How else might one proceed to minimize the $H^{-1}$ norm? As a toy example, let
$$f(\lambda) = \Delta u + \lambda u^p$$ for some $p>0, \lambda \in \mathbb{R}$ and $u\in H^{1}_0(\mathbb{R}^n).$ Then we can write, $$F(\lambda) = \|\Delta u + \lambda u^p\|_{H^{-1}}.$$ For what value of $\lambda$ does $F(\lambda)$ achieve it's minimum?

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    $\begingroup$ You are asking if the derivative of the norm is equal to the norm of the derivative. This is false, even in finite dimensions. $\endgroup$
    – Michael Renardy
    Commented Mar 17, 2021 at 11:26
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    $\begingroup$ To make Michael Renardy's comment extremely explicit. The function $f: \mathbb{R}\ni \theta \mapsto (\cos\theta,\sin\theta) \in \mathbb{R}^2$ has the property that $\|f(\lambda)\|$ is constant, but $\| \partial_\lambda f(\lambda)\|$ is $1$. You can cook up similar examples in any normed space of dimension $\geq 2$. It is not that taking the derivative of norm is not justified; it is that what you hope to prove is plainly false. $\endgroup$
    – Willie Wong
    Commented Mar 17, 2021 at 13:32
  • $\begingroup$ In your specific case [Edit 2]: I am assuming $u$ is a fixed function? assume both $\Delta u$ and $u^p$ are in $H^{-1}$. (The former is, the latter probably only is true for some range of $p$.) You are asking to minimize the norm along a line inside a HIlbert space. This happens when with respect to the Hilbert space inner product $\Delta u + \lambda u^p \perp u^p$ (velocity is orthogonal to position). $\endgroup$
    – Willie Wong
    Commented Mar 17, 2021 at 15:23
  • $\begingroup$ So this can be solved explicitly as $\lambda = - \langle \Delta u, u^p \rangle_{H^{-1}} / \langle u^p, u^p\rangle_{H^{-1}}$. $\endgroup$
    – Willie Wong
    Commented Mar 17, 2021 at 15:24
  • $\begingroup$ (BTW, this is true in general in Hilbert spaces. Critical points of $\| f(\lambda)\|_H$ in some real HIlbert space satisfies $\langle f(\lambda), \partial_\lambda f(\lambda)\rangle_H = 0$.) $\endgroup$
    – Willie Wong
    Commented Mar 17, 2021 at 15:29

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