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I'm trying to solve a functional minimization problem of the following form:

$$\arg\min_{f:\mathbb{R}\rightarrow [0,1]} h(f)$$ where $h$ is some expression in terms of several integrals over $f$.

I have a heuristic calculation that seems indeed to work (to find at least a local minimum): I simply treat $f:\mathbb{R}\rightarrow [0,1]$ as a continuum of variables $f(x)$ for each $x \in \mathbb{R}$, "take the derivative" of $h$ with respect to each variable $f(x)$, set it to zero, and solve. Together, this gives me a functional form for $f$. In my setting, this indeed seems to work, but obviously this technique is not rigorous as stated.

My question is, are there any conditions under which functional minimization can be done in this way? When can this technique be made rigorous?

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2 Answers 2

This is pretty easy : find out an appropriate Banach space for f (the one that makes your integrals well-defined; usually, some kind of Sobolev space), prove that h is C^1 (with respect to differentiation in Banach spaces), and then standard arguments apply. Keywords you might want to look up are: differentiation in Banach spaces, weak solutions, Sobolev spaces.

For instance, if $h(f) = \int f^2 + f'^2$, then $h$ is $C^1$ as a functional in $H^1$, and the solution is a (weak) solution to $f'' + f = 0$

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A useful and fairly complete reference on this and related questions is Morrey Multiple Integrals in the Calculus of Variations. http://books.google.com/books?id=-QNKm1PBohsC

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