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Let $\phi \colon \mathbb R^n \to \mathbb R$ be convex, proper and lower semi-continuous (lsc). Let $M$ be a measurable subset of $\mathbb R^n$.

We can define a functional $\Phi \colon L^2(M) \to \mathbb R$ through $$ \Phi(v) = \begin{cases} \int_M \Phi(v(x))\,dx & \text{if $\Phi(v) \in L^1(M)$} \\ \infty & \text{otherwise} \end{cases} $$

It is clear that this function is again convex and proper. That it is also lsc is shown in Inequality Problems in Mechanic and Applications: Convex and Nonconvex Energy Functions (Proposition 3.3.1).

Even more is proved there, however. Assume $f \in L^2(M)$. Proposition 3.3.2 says that the following two statements are equivalent:

  1. $\Phi(v) \ge \Phi(u) + \int_M f\cdot(v-u)$ for every $v \in L^2(M)$
  2. We have $\phi(v) \ge \phi(u(x)) + f(x)[v-u(x)]$ for every $v \in \mathbb R^n$ almost everywhere in $M$.

It is mentioned at the end of the corresponding section (namely p.113), that this result does not hold for the case of a functional $\tilde \Phi$ with

$$ \tilde \Phi(v) = \int_M \tilde \phi(v(x),x)\,dx $$

with a function $\tilde \phi$ that depends on $x$. Obviously, such functionals are still convex.

Q: If $\tilde \phi(\,\cdot\,,x)$ is non-negative and convex for every $x$, then $\tilde \Phi$ is lsc., see e.g. Modern Methods in the Calculus of Variations: Lp Spaces (Theorem 6.49). If we additionally assume e.g. $\tilde \phi(x,0) = 0$, then $\tilde \Phi$ is proper. Does equivalence of the above three statements (with an additional dependence on $x$) also hold?

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1 Answer 1

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According to Corollary 3E in

Rockafellar, R. Tyrrell. Integral functionals, normal integrands and measurable selections. Nonlinear operators and the calculus of variations (Summer School, Univ. Libre Bruxelles, Brussels, 1975), pp. 157--207. Lecture Notes in Math., Vol. 543, Springer, Berlin, 1976. DOI: 10.1007/bfb0079944

the conditions (1) and (2) are equivalent for any proper normal convex integrand.

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