Γ-convergence

Remarks

The two conditions together give $F(x) = \lim_n F_n(x_n)$ along the recovery sequence. A Γ-limit is unique when it exists, and it is always lower semicontinuous. For a family $(F_\varepsilon)_{\varepsilon > 0}$, one says that $F_\varepsilon$ Γ-converges to $F$ as $\varepsilon \to 0$ if $F_{\varepsilon_k}$ Γ-converges to $F$ for every sequence $\varepsilon_k \to 0$. The same definition with neighbourhoods instead of sequences works in any topological space, and it agrees with this one in metric spaces.

Example: $F_n(x) = \sin(nx)$ on $\R$ Γ-converges to the constant $F = -1$. The lower bound holds because $\sin(n x_n) \ge -1$ for every $n$. For a recovery sequence, pick $x_n$ with $\abs{x_n - x} \le 2\pi/n$ and $\sin(n x_n) = -1$; then $x_n \to x$ and $F_n(x_n) = -1$.

Non-example: let $F(0) = 1$ and $F(x) = 0$ for $x \ne 0$, and let $F_n = F$ for all $n$. The constant sequence does not Γ-converge to $F$: for $x_n = 1/n \to 0$ one has $\liminf_n F_n(x_n) = 0 < 1 = F(0)$, so the lower bound fails at $0$. Its Γ-limit is the function $0$, the lower semicontinuous envelope of $F$.