liminf and limsup requirements for Gamma-convergence

Question

A sequence of functionals $f_n$ $\Gamma$-converges to a functional $f$ if:

1. for every sequence $x_n \rightarrow x$, $$ \liminf_n f_n(x_n) \geq f(x).$$

2. there exists some sequence $x_n \rightarrow x$ such that $$ \limsup_n f_n(x_n) \leq f(x).$$

The point of Gamma-convergence is to ensure that minimizers of $f_n$ converge to minimizers of $f$. So why do these two requirements make this happen? (I know there has to be an additional compactness requirement, so let's just assume that's true).

I can see that for the first condition, we don't want any $f_n(x_n)$ to be lower than $f(x)$ because, then we wouldn't have the convergence of the minimum values of $f_n$ to the minimum value of $f$. Is that right? And what about the second condition?

Answer 1

Let $(x_n^*)$ be a sequence of minimizers of $f_n$ such that $x_n^* \to x^*$. Let $x$ be given. Then according to (2) there is a sequence $x_n \to x$ such that $\limsup f_n(x_n) \le f(x)$.

Now lets prove that $x^*$ is a minimizer. From the minimizing property of the $x_n^*$ we get $$ f_n(x_n^*) \le f_n(x_n). $$ Now (1) and (2) imply $$ f(x^*) \le \liminf f_n(x_n^*) \le \limsup f_n(x_n) \le f(x). $$ As $x$ was arbitrary, this shows that $x^*$ is a global minimum of $f$.

So (1) is needed to control the limit of the minimum value $f_n(x_n^*)$, while (2) is needed to get from feasible points to $f$ suitable feasible points to $f_n$.