Epiconvergence

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

We start with the necessary definitions:

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.

Step 1: Proving an equivalent form of the epigraphs of the epi-limits.

\[\begin{aligned} \operatorname{epi}\{\operatorname{e-liminf}_\nu f^\nu\} &= \left\{\begin{pmatrix} x\\ \alpha \end{pmatrix}\Big|~ \exists x^\nu \to x: \liminf_\nu f^\nu(x^\nu) \leq \alpha \right\}\\ \operatorname{epi}\{\operatorname{e-limsup}_\nu f^\nu\} &= \left\{\begin{pmatrix} x\\ \alpha \end{pmatrix}\Big|~ \exists x^\nu \to x: \limsup_\nu f^\nu(x^\nu) \leq \alpha \right\} \end{aligned} \]

Proof: By checking inclusions elementwise.

Step 1.1: "\(\subset\)" of the first equation.

It holds that \(\operatorname{epi}\{\operatorname{e-liminf}_\nu f^\nu\} \subset \left\{\begin{pmatrix} x\\ \alpha \end{pmatrix}\Big|~ \exists x^\nu \to x: \liminf_\nu f^\nu(x^\nu) \leq \alpha \right\}\)

Proof: We choose an arbitrary pair \((x, \alpha) \in \operatorname{epi}\{\operatorname{e-liminf}_\nu f^\nu\}.\)
By definition of \(\operatorname{epi}\{\operatorname{e-liminf}_\nu f^\nu\} = \operatorname{limsup}_\nu \operatorname{epi} f^\nu\), this means that there is a sequence of integers \(N \in \mathcal N_\infty^\#\), and sequences \(x^\nu \stackrel{N}\to x, \alpha^\nu \stackrel{N}\to \alpha\) with \(f^\nu(x^\nu) \leq \alpha^\nu.\)
We extend the sequences \((x^\nu)_\nu, (\alpha^\nu)_\nu\) to all of \(\mathbb N\) by
Then \[\alpha = \liminf_{\nu\in N} \alpha^\nu \geq \liminf_{\nu \in N} f^\nu(x^\nu) = \liminf_{\nu\in \mathbb N} f^\nu(x^\nu).\] This proves the relation "\(\subset\)".

Step 1.2: "\(\supset\)" of the first equation.

It holds that \(\operatorname{epi}\{\operatorname{e-liminf}_\nu f^\nu\} \supset \left\{\begin{pmatrix} x\\ \alpha \end{pmatrix}\Big|~ \exists x^\nu \to x: \liminf_\nu f^\nu(x^\nu) \leq \alpha \right\}\)

Proof: Let \((x,\alpha)\in \text{[r.h.s.]}\), i.e. there exists a sequence \(x^\nu \to x\) such that \(\liminf_\nu f^\nu(x^\nu) = \alpha .\)
Considering the definition of \(\operatorname{epi}\{\operatorname{e-liminf}_\nu f^\nu\} = \operatorname{limsup}_\nu \operatorname{epi} f^\nu\), we need to construct a integer sequence \(N \in \mathcal N_\infty^\#\) with \(x^\nu \stackrel{N}\to x\) and a sequence \(\alpha^\nu \stackrel{N}\to \alpha\) with \(f^\nu(x^\nu) \leq \alpha^\nu.\)
We obtain \(N\) by extracting a subsequence of \(\mathbb N\) such that \(f^\nu (x^\nu) \stackrel{N}\to \liminf_\nu f^\nu(x^\nu) \leq \alpha.\)
The sequence \((\alpha^\nu)_\nu\) is constructed by setting \(\alpha^\nu := \max\{f^\nu(x^\nu), \alpha\}\). Then indeed \(\alpha^\nu \stackrel{N}\to \alpha\) and \(\alpha^\nu \geq f^\nu(x^\nu).\) This proves the relation "\(\supset\)".

Step 1.3: "\(\subset\)" of the second equation.

It holds that \(\operatorname{epi}\{\operatorname{e-limsup}_\nu f^\nu\} \subset \left\{\begin{pmatrix} x\\ \alpha \end{pmatrix}\Big|~ \exists x^\nu \to x: \limsup_\nu f^\nu(x^\nu) \leq \alpha \right\}\)

Proof: Similar to 1.1.

Step 1.4: "\(\supset\)" of the second equation.

It holds that \(\operatorname{epi}\{\operatorname{e-limsup}_\nu f^\nu\} \supset \left\{\begin{pmatrix} x\\ \alpha \end{pmatrix}\Big|~ \exists x^\nu \to x: \limsup_\nu f^\nu(x^\nu) \leq \alpha \right\}\)

Proof: Similar to 1.2.

Step 2: Proving an equivalent form of the epi-limits.

The following representation is true: \[ \begin{aligned} \operatorname{e-liminf}_\nu f^\nu(x) &= \min\left\{\alpha\in \overline \mathbb R : ~\exists x^\nu \to x \text{ with } \liminf_\nu f^\nu(x^\nu) = \alpha \right\}\\ \operatorname{e-limsup}_\nu f^\nu(x) &= \min\left\{\alpha\in \overline \mathbb R : ~\exists x^\nu \to x \text{ with } \limsup_\nu f^\nu(x^\nu) = \alpha \right\} \end{aligned} \]

Proof: This is valid by extracting the graph from the epigraph representation in step 1. The inequality can be seen to be an equality as we take the minimum over all \(\alpha\).

Step 3: "\(\Rightarrow\)"

If \(f^\nu \stackrel {\operatorname{epi}}\to f\), then \[ \begin{aligned} \text{For every sequence } x^\nu \to x, \quad &\liminf_\nu f^\nu(x^\nu) \geq f(x) \\ \text{There is a sequence } x^\nu \to x, \quad &\limsup_\nu f^\nu(x^\nu) \leq f(x). \end{aligned} \]

By using Step 2, this statement is already fulfilled.

Step 3.1: The first statement.

Recall that we want to prove that for all sequences \(x^\nu \to x,\) \[\liminf_\nu f^\nu(x^\nu) \geq f(x). \] From Step 2 we know that \[\operatorname{e-liminf}_\nu f^\nu(x) = \min\left\{\alpha\in \overline \mathbb R : ~\exists x^\nu \to x \text{ with } \liminf_\nu f^\nu(x^\nu) = \alpha \right\}.\] But \(f^\nu \stackrel {\operatorname{epi}}\to f\) and thus \(\operatorname{e-liminf}_\nu f^\nu(x) = f(x).\) Made explictly, this means \[ f(x) = \min \{\liminf_\nu f^\nu(x^\nu):~ \exists x^\nu \to x \},\] which is exactly what we want to prove.

Step 3.1: The second statement.

Recall that we want to prove that there is a sequence \(x^\nu \to x,\) \[\limsup_\nu f^\nu(x^\nu) \leq f(x). \] From Step 2 we know that \[\operatorname{e-limsup}_\nu f^\nu(x) = \min\left\{\alpha\in \overline \mathbb R : ~\exists x^\nu \to x \text{ with } \limsup_\nu f^\nu(x^\nu) = \alpha \right\}.\] But \(f^\nu \stackrel {\operatorname{epi}}\to f\) and thus \(\operatorname{e-limsup}_\nu f^\nu(x) = f(x).\) We can extract the sequence \(x^\nu\) which yields the minimal value of \(\alpha\) and obtain equality (and hence also \(\leq\).)

Step 4: "\(\Leftarrow\)"

If \[ \begin{aligned} \text{For every sequence } x^\nu \to x, \quad &\liminf_\nu f^\nu(x^\nu) \geq f(x) \\ \text{There is a sequence } x^\nu \to x, \quad &\limsup_\nu f^\nu(x^\nu) \leq f(x), \end{aligned} \] then \(f^\nu \stackrel {\operatorname{epi}}\to f.\)

Proof: By using the representation in step 1.

Step 4.1

It suffices to prove that \[\operatorname{epi}(\operatorname{e-liminf}_\nu f^\nu)\subset \operatorname{epi}(\operatorname{e-limsup}_\nu f^\nu)\]

Proof: \[\operatorname{epi}(\operatorname{e-liminf}_\nu f^\nu)\supset \operatorname{epi}(\operatorname{e-limsup}_\nu f^\nu)\] is always fulfilled and \[\operatorname{epi}(\operatorname{e-liminf}_\nu f^\nu) = \operatorname{epi}(\operatorname{e-limsup}_\nu f^\nu)\] implies epiconvergence.

Step 4.2

QED.

Proof: Choose \((x,\alpha)\in \operatorname{epi}(\operatorname{e-liminf}_\nu f^\nu),\) then by step 1, there is a sequence \(x^\nu \to x\) such that \(\operatorname{liminf}_\nu f^\nu(x^\nu) \leq \alpha.\) By the first assumption of step 4, we have a lower bound for \(\operatorname{liminf}_\nu f^\nu(x^\nu)\) and obtain \[f(x) \leq \alpha. \] By the second assumption of step 4, there exists a sequence \(y^\nu\to y\) with \(\operatorname{limsup}_\nu f^\nu(y^\nu) \leq f(x)\) and hence \[ \operatorname{limsup}_\nu f^\nu(y^\nu) \leq \alpha.\] This is exactly the condition for \((x,\alpha)\) to be an element of \(\operatorname{epi}(\operatorname{e-limsup}_\nu f^\nu).\)
This shows that \[\operatorname{epi}(\operatorname{e-liminf}_\nu f^\nu) \subset \operatorname{epi}(\operatorname{e-slimsup}_\nu f^\nu)\] and together with 4.1 this concludes the proof.

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