Passing to weak-strong limit in pointwise inclusions

Question

Let $F:\mathbb R^m\rightrightarrows\mathbb R^n$ be a set-valued map (or multi-function, correspondence) with $F(x)\ne\emptyset$ for all $x\in \mathbb R^m$.

Let $I\subset\mathbb R$ be an interval. Let be sequences of functions $(y_n)$ and $(x_n)$ in $L^2(I,\mathbb R^n)$ and $L^2(I,\mathbb R^m)$ be given such that

1. $x_n(t) \to x(t)$ for almost all $t\in I$,
2. $y_n \rightharpoonup y$ in $L^2(I,\mathbb R^n)$
3. $y_n(t) \in F(x_n(t))$ for almost all $t\in I$.

If $F$ satisfies certain properties then this implies $$ y(t) \in \overline{conv} ( F(x(t))) $$ for almost all $t\in I$.

This is proven in the book of Aubin & Frankowska under the assumption that $F$ is outer semicontinuous (i.e., the graph of $F$ is closed) and local boundedness of $F$ (each point $x$ has a small neighborhood $U$ such that $F(U)$ is bounded). In the book by Aubin & Cellina, $F$ is assumed to be upper hemicontinuous ($x\mapsto \sup_{y\in F(x)}y^Tp$ is upper semicontinuous for all $p$).

The following mapping $F$ does not satisfy these assumptions: $$ F(x) = \begin{cases} \{0\} & x\le 0\\ \{0,\frac 1x\} & x>0\end{cases}. $$ My question is: is the statement of the theorem true or false for this kind of map? Can one find a counterexample? Is there a complete characterization of properties of $F$ to reach the conclusion?

Answer 1

I think for your specific example there is no counterexample. However, I am not sure which properties of $F$ are important for this result, or how to characterize those $F$ such that the statement of the theorem holds.

sketch of proof: We can consider the case where $I$ is a measurable set with finite measure, not just an interval.

By Egorovs theorem we know that $x_n\to x$ uniformly on a set $B_\varepsilon$ which is equal to $I$ up to a set of measure $\varepsilon$.

Thus, wlog we can assume that $x_n\to x$ uniformly on $I$.

We only consider the "interesting" case where $x=0$, $x_n\geq 0$, $y>0$. (Otherwise we can solve the problem on measurable subsets of $I$, e.g. $\{t : y(t)=0 \},\{t:y(t)>0\}$, $\{t:x(t)<0\}$, etc.).

We have $0\leq x_n\leq s_n$ for some $s_n\in \mathbb R$ with $s_n\to0$. Due to the definition of $F$, we can see that $$ y_n(t) = \frac1{x_n(t)} \chi_{A_n} $$ for a suitable measurable set $A_n\subset I$.

Then we have the inquality $$ \| y_n \|_{L^2} \geq \int_I y_n(t) \frac1{s_n} \mathrm dt. $$ Since $y_n$ converges weakly we have $\int_I y_n(t) \mathrm dt\to \int_I y \mathrm dt>0$. Thus $y_n$ is unbounded due to the previous inequality. This is a contradiction to to the weak convergence of $y_n$.

Answer 2

Here is a supposed proof with weaker assumptions on $F$:

Let $I$ be $\sigma$-finite. Assume $F$ is outer semicontinuous. Suppose $y_n\rightharpoonup y$ in $L^p(I)$, $1<p<\infty$. Assume $0\in F(x(t))$ for almost all $t$. Then the conclusion of the theorem in OP is valid.

Assume $I$ has finite measure. Let $m>0$. Define $$ I_{m,n} = \{t: \ |y_n(t)|>m\}, \quad \chi_{m,n}:= \chi_{I_{m,n}}. $$ Then $\|\chi_{m,n}\|_{L^1(I)} \le m^{-p} \|y_n\|_{L^p(I)}^p \le c m^{-p}$ for all $m,n$ with $c$ independent of $n$. In addition, we have the inclusion $$ (1-\chi_{m,n})y_n \in \left( F(x_n(t))\cap \overline{B_M(0)}\right) \cup\{0\}. $$ Since $((1-\chi_{m,n})y_n)_n$ is bounded in $L^p(I)$, there is a weakly converging subsequence with limit $y^m$. Passing to the limit in the inclusion we get by the result of Aubin and Frankowska $$ y^m(t) \in \overline{conv}\left( \left( F(x_n(t))\cap \overline{B_M(0)}\right) \cup\{0\} \right) \subset\overline{conv}{ F(x(t))}. $$ It remains to show that $y_m$ converges to $y$. To this end, consider $$ (1-\chi_{m,n})y_n - y_n = \chi_{m,n} y_n. $$ By the estimates of $\chi_{m,n}$, we have $$ \|\chi_{m,n} y_n\|_{L^1(I)} \le \|y_n\|_{L^p(I)} \|\chi_{m,n}\|_{L^{p'}(I)} \le c \ m^{-p/p'}, $$ where $c$ is independent of $n$. This shows $\|y^m - y\|_{L^1(I)} \le c \ m^{-p/p'}$, which converges to zero for $m\to \infty$. Hence, a subsequence of $(y^m)$ converges pointwise a.e. to $y$, which proves the claim.