$\Gamma$-convergence of minimizers

Question

Let $X$ be a metric space and $f_{n}:X\to (-\infty,\infty)$. Assume for any $n\in \mathbb{N}$ there exists a minimizer $x_n\in X$, that is $\min_{x\in X} f_n(x)=f_n(x_n)$. If $f_n$ $\Gamma$-converges to $f$ and the sequence of minimizers $x_n\to x$, it follows that $\min_{x\in X} f(x)$ exists and $f(x)=\min_{z\in X}f(z)=\lim_{n\to \infty}\min_{z\in X} f_n(z)$. I would be thankful for some hints how to prove this.

Answer 1

Let $f_n \stackrel{\Gamma}{\to }f$. Let $x_n\to x$, where $x_n$ are minimizes of $f_n$. Take $z\in X$. Then there is $z_n\to z$ such that $\limsup f_n(z_n) \le f(z)$. Then $$ f(x) \le \liminf f_n(x_n) \le \liminf f_n(z_n) \le \limsup f_n(z_n) \le f(z)$. $$