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It's well known that a functional of the form $u \mapsto \int f(u) dx$ is continuous with respect to weak convergence (say weak* convergence in $L^\infty$) if and only if the function $f$ is affine. An even more interesting thing to study is functionals which involve derivatives $u \mapsto \int f(du) dx$. The function $f$ needs to be quasiconvex to be lower semicontinuous.

I'm interested in functionals which also depend on the $x$ variable, like $\int f(x, u, du) dx$. Can anyone tell me a good place to read about continuity and semicontinuity with respect to weak convergence for functionals of this form. For example, must the $f(x,u)$ in $\int f(x,u) dx$ be convex / affine in the $u$ for almost every point $x$?

EDIT: I should clarify the last question. Dorian has given me a good reference regarding weak continuity/semicontinuity of functionals which involve derivatives and depend on the spatial variables. The question that remains is simpler: when you have a functional $\int f(x,u) dx$ that is, say, continuous with respect to weak-* convergence in $L^\infty$, is it necessarily the case that $f(x,\cdot)$ is affine in $u$ at almost every point $x$? This question is purely measure theoretic and has nothing to do with derivatives (so without loss of generality I am asking about the measure space $I = [0,1]$ since I am interested in the case without atoms). By using a sequence of the form $\chi_{E_k} u + (1 - \chi_{E_k}) v$ for suitable characteristic functions $E_k$, we know that $f(x, \theta u + (1- \theta) v) = \theta f(x, u) + (1-\theta) f(x,v)$ for all measurable functions $\theta : I \to [0,1]$ and all bounded, measurable $u, v$ on $I$. Probably, this is enough to imply $f(x, u)$ is affine in $u$ for almost every $x$. Similarly, lower semicontinuity should imply convexity at almost every point.

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I just remembered when I read the book of A. Braides 'Gamma-convergence for beginners', I noticed several remarks like your question, maybe you can check page 55 for some information and the reference there. –  Shaoming Guo Aug 9 '11 at 9:05
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1 Answer

I am not going to try to find the most general conditions under which lower semi-continuity holds but for that I suggest the standard reference for all of this is "Direct methods in the Calculus of Variations" by Bernard Dacorogna which covers all of this in full detail. I will give a brief outline of the answer however.

In order for $\int f(x,u,du)dx$ to be lower semi-continuous with respect to weak convergence one does not in general need any sort of convexity in the $u$ variable and what is more important generally is

1) Coercivity (see below) of the functional $f$ in the $du$ variable.

2) The right embedding along with continuity of $f$ in the $u$ variable with respect to the topology of strong convergence in this space.

Example: A good example is $E(u) = \int_0^1 |\nabla u |^2 + (1-u^2)^2$ with $u = 0$ on $\partial \Omega$ in $\mathbb{R}^d$. Observe that when $d \leq 4$ one has $H^1(\Omega) \subset \subset L^4(\Omega)$ and consequently the non-convex term depending on $u$ is lower order. Therefore one can pass to the limit in any minimizing sequence even though $(1-u^2)^2$ is very non convex (but it is continuous with respect to strong convergence). Notice however that for the energy $\bar E(u) = \int_0^1 |\nabla u|^2 + u^4$ in $\mathbb{R}^5$ with $u$ prescribed on $\partial \Omega$, the second term is not lower order but here we may use convexity to conclude lower semi-continuity of the second term.

The function $p \mapsto f(x,z,p)$ is coercive if there exists some constant $C > 0$ so that $f(x,z,p) \geq C|p|^q$ for some range of $q$. It then depends on what spaces one is working in but the goal is to use an embedding theorem such as $L^q \subset\subset W^{1,p}$ for $1 \leq q < p^*$ and to conclude that for a minimizing sequence $u_n$, there is in fact a strongly convergent subsequence in $L^q$ for some $q$. Then one will generally expect some sort of continuity of $f$ in the $u$ variable.

There are many more interesting examples but in almost all cases in practice the goal is to show that one has strong convergence of the $u_n$s in your minimizing sequence in some $L^p$ space. The book by Braides focuses mostly on asymptotics of functionals which depend on some large (or small) parameter and I'm not sure how much he talks about the assumptions needed in the direct method.

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Thanks for the reference and all of the tips! I think you meant to say that $H^1 \subseteq \subseteq L^4$. I looked at Dacorogna for a little while. One thing I wanted to understand, which oddly enough does not seem to be in the book but maybe I'll be able to figure it out from the methods, is the lower semicontinuity of functionals $\int f(x, u(x)) dx$ which do not involve the derivatives.... –  Phil Isett Sep 12 '11 at 2:37
I can show that continuity with respect to, say, $L^\infty$ weak-* convergence implies $f(x, \theta u + (1-\theta) v ) = \theta f(x, u) + (1-\theta) f(x, v)$ for all bounded measurable functions $u, v$ and measurable functions $\theta \in [0,1]$; so now I'm wondering: does it follow $f(x,u)$ is convex in $u$ for almost every $x$? –  Phil Isett Sep 12 '11 at 2:39
Hi Phil. In fact the reason lower semi-continuity is hard for functionals like $\int f(x,u(x))dx$ is that assuming you have a minimizing sequence, you don't have any coercivity. You don't conclude any bounds in say $H^1$ on your functions $u_n$ from assuming $\limsup_{n} \int f(x,u_n(x)) dx < +\infty$ and so you don't have sufficient compactness. Does that make sense? And do you mean continuity or lower semi-continuity? In the scalar valued case you are considering, lower semi-continuity always implies convexity in that variable. For the vector valued case, convexity always implies –  Dorian Sep 13 '11 at 1:40
semi-continuity but not vice versa (the correct notion there is quasiconvexity). –  Dorian Sep 13 '11 at 1:40
Hi Dorian, I think that the theorems you are quoting about lower-semicontinuity etc. regard functionals which include derivatives. Functionals without derivatives make sense at a purely measure-theoretic level, where the notion of quasiconvexity does not make sense. I should clarify the question. –  Phil Isett Oct 3 '11 at 14:25
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