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?


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – Chee Han
    May 18, 2021 at 18:45
  • 2
    $\begingroup$ @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 $\epsilon x^2$ $\endgroup$
    – Iiro Ullin
    May 18, 2021 at 18:48
  • $\begingroup$ ...i.e., each $\epsilon x^2$ is coercive on its own, but the entire family is not equicoercive... $\endgroup$
    – Iiro Ullin
    May 18, 2021 at 18:57

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