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 $\{x\colon F_\varepsilon(x)\le t\}\subset K_t$. 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?

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?