# Fundamental Theorem of Gamma-Convergence

Let $$(X, d)$$ be a metric space and $$F_\varepsilon, F\colon X\to [-\infty, \infty]$$. Suppose $$F_\varepsilon$$ is an equicoercive sequence of functions on $$X$$, i.e. for all $$t\in\mathbb{R}$$ there exists a compact set $$K_t$$ such that $$\{ x\colon F_\varepsilon(x)\le t \} \subset K_t$$ for every $$\varepsilon > 0$$. Suppose $$F_\varepsilon\xrightarrow{\Gamma} F$$, i.e. for all $$u\in X$$ we have that

1. for every sequence $$\{u_\varepsilon\}$$ converging to $$u$$ it holds that $$F(u)\le\displaystyle\liminf_{\varepsilon\to 0}F_\varepsilon(u_\varepsilon)$$.
2. there exists a sequence $$\{u_\varepsilon\}$$ converging to $$u$$ such that $$F(u) = \displaystyle\lim_{\varepsilon\to 0}F_\varepsilon(u_\varepsilon)$$.

Question: Is $$F$$ coercive on X?

Yes. Take any sequence $$\{x_i\}$$ for which $$F(x_i)$$ is bounded. For each $$x_i$$ consider the sequence satisfying (2), $$\{x^\epsilon_i\}$$ s.t. $$x^\epsilon_i\to x_i$$. Consider $$x_i^{\epsilon_i}$$ such that $$F_{\epsilon_i}(x_i^{\epsilon_i})$$ is bounded. Due to equicoercivity, $$x_i^{\epsilon_i}$$ has to be in a compact set, so it has a converging subsequence, thus $$x_i$$ must also have a converging subsequence.
Comment: equicoercivity is essential, e.g., take $$F_\epsilon=\epsilon x^2$$. $$F_\epsilon\stackrel{\Gamma}{\to}0$$, which is not coercive.
• Why is $F_\varepsilon$ not equicoercive? I might have misunderstood the definition of equicoercive. Isn't equicoercive equivalent to showing that any uniformly bounded sublevel sets has a convergent subsequence? May 18, 2021 at 18:45
• @Chee Han According to your definition, sublevels sets for $F_\epsilon$ should be contained in the same compact set $K$ for all $\epsilon$, which is not the case for $\epsilon x^2$ May 18, 2021 at 18:48
• ...i.e., each $\epsilon x^2$ is coercive on its own, but the entire family is not equicoercive... May 18, 2021 at 18:57