# A hierarchical proof of an epiconvergence lemma.

Posted on November 19, 2018

Beware: This is more like a personal note for myself, less an accessible introduction to epiconvergence.

For a sequence of sets $$(C^\nu)_\nu$$ in $$\mathbb R^n$$, the outer and inner limit are defined as the sets \begin{aligned} \limsup_\nu C^\nu &= \{x:~ \exists N \in \mathcal N_\infty^\#, \exists x^\nu\in C^\nu:~ x^\nu\stackrel{N}\to x\} \\ \liminf_\nu C^\nu &= \{x:~ \exists N \in \mathcal N_\infty, \exists x^\nu\in C^\nu:~ x^\nu\stackrel{N}\to x\}. \end{aligned} Here, $$\mathcal N_\infty^\#$$ is the set of subsequences of $$\mathbb N$$ and $$\mathcal N_\infty$$ is the set of "tails" of $$\mathbb N$$, i.e. sets of the form $$\{M,M+1,M+2,\ldots\}$$.
The lower and upper epi-limit of a sequence of functions $$f^\nu: \mathbb R^n\to \mathbb R$$ is defined as (first by their epigraph): \begin{aligned} \operatorname{epi}\{\operatorname{e-liminf}_\nu f^\nu\} &:= \limsup_\nu (epi (f^\nu))\qquad \text{ ( = outer limit of epigraphs)} \\ \operatorname{epi}\{\operatorname{e-limsup}_\nu f^\nu\} &:= \liminf_\nu (epi (f^\nu))\qquad \text{ ( = inner limit of epigraphs)}\end{aligned} Then we can define $$\operatorname{e-liminf}_\nu f^\nu$$ and $$\operatorname{e-limsup}_\nu f^\nu$$ by extracting the graph from the epigraph. If those two functions coincide, we call this the epilimit $$e-\lim_\nu f^\nu$$.

We show the following:

$$f \stackrel{\operatorname{epi}} \to f$$ if and only if \begin{aligned} \text{For every sequence } x^\nu \to x, \quad &\liminf_\nu f^\nu(x^\nu) \geq f(x) \\ \text{There is a sequence } x^\nu \to x, \quad &\limsup_\nu f^\nu(x^\nu) \leq f(x) \end{aligned}

This is a hierarchical proof (as proposed by Leslie Lamport in "How to write a 21st century proof"). You can interact with elements by clicking on and
Proof
Sketch of proof: We use the geometric definition of the epilimit of $$f^\nu$$ directly to construct suitable sequences.

Thanks to mathoverflow user supinf who filled some of the gaps in my understanding.