Subdifferential boundary conditions: Testing pointwise or with $L^2$ functions

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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