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?
> 
> 
> 



Fundamental Theorem of Gamma-Convergence
**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.


