Consider a set $X$, and consider a sequence of functionals on $X$, that is maps $F_n: X \to \mathbb{R}$. We say that $F_n$ "$\Gamma$" converges to $F$, if the limit satisfies:
- $F(x) \leq \inf\{\liminf _{n \to \infty} F_n(x_n) : x_n \to x\}$
that is to say, for any sequence we $x_n \to x$ we choose, we have that $F(x) \leq \liminf_{n \to \infty} F_n(x_n) $. The second is the existence of a "$\Gamma$" realizing sequence, that is, there exists a sequence $x_n \to x$, that satisfies:
- $\limsup_{n\to \infty} F_n(x_n) \leq F(x)$
I have the following two questions about this limit.
First, is it accurate to replace condition $2$ with the condition that there exists a sequence $x_n$ such that $\lim_{n \to \infty} F_n(x_n) \to F(x)$ ? (I.e., replacing the limsup with a bona-fide limit).
Second, is the characterization of $F(x)$ as :$$ F(x) = \min\{\liminf_{n \to \infty} F(x_n) : x_n \to x\} $$ accurate? Is there an example of when this inf is not acheived? (i.e, we have no "realizing sequence"?
If we take the realizing sequence according to (2) and put it into (1), then $$ F(x) \le \liminf F_n(x_n) \le \limsup F_n(x_n)\le F(x), $$ which shows $F(x) = \lim F_n(x_n)$ for this particular sequence.
So, yes, given condition (1), we can replace 'lim sup' by 'lim' in (2).