I am learning that Young measures can be useful for variational problems. When a set of functions isn't compact in function space (so they don't have a convergent subsequence), we can associate those functions with Young measures, and then use a sort of compactness on the set of Young measures.
Here is a nice explanation I found introducing Young measures and a version of Prohorov's theorem for Young measures.
It says that given a bounded sequence $u_n \in L^\infty(\Omega)$, the sequence of corresponding Young measures $\nu^n$ (which can be "sliced" into $\nu^n_x = \delta_{u_n(x)}$ for all $x\in \mathbb{R}$) has a convergent subsequence. Furthermore, the limit $\nu$ is also a Young measure.
My question is
- Is $\nu$ also of the form $\nu_x = \delta_{u(x)}$ for some function $u$? I want to say no (or maybe $u$ isn't $L^\infty$) because $L^\infty$ isn't compact.
- If not, then can we describe it other other than the most general definition of Young measure (stated in the link)?
- And if not, why is this useful for variational problems?
If $\Omega=(0,1)$ and $u_n(x) = \sin(n\pi x)$. Then the Young measure limit is $$ \nu_x = \frac12 \delta_{-1} + \frac12 \delta_{+1} $$ in the sense that $$ \int_\Omega f( u_n(x))dx \to \frac 12 (f(-1) + f(1)) $$ for all continuous $f$. This example is also discussed on the website you linked.
Why is this usefull for variational problems? Problems posed wrt Young measures have solutions under much weaker conditions. In addition, the Young measure minimum coincides with the infimum of the original problem.