1
$\begingroup$

I am struggling with the proof of a property of epi-convergence.

We need the following definitions:

For a sequence of sets $(C^\nu)_\nu$ in $\mathbb R^n$, the outer limit is the set $\limsup_\nu C^\nu = \{x:~ \exists N \in \mathcal N_\infty^\#, \exists x^\nu\in C^\nu:~ x^\nu\stackrel{N}\to x\} $. The inner limit is the set $\liminf_\nu C^\nu = \{x:~ \exists N \in \mathcal N_\infty, \exists x^\nu\in C^\nu:~ x^\nu\stackrel{N}\to x\} $

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 $\{M,M+1,M+2,\ldots\}$.

Also, 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):

$$ epi( e-\liminf_\nu f^\nu) := \limsup_\nu (epi (f^\nu))\qquad \text{ ( = outer limit of epigraphs)}$$ and

$$ epi( e-\limsup_\nu f^\nu) := \liminf_\nu (epi (f^\nu))\qquad \text{ ( = inner limit of epigraphs)}$$

Then we can define $e-\liminf_\nu f^\nu$ and $e-\limsup_\nu f^\nu$ by extracting the graph from the epigraph. If those two functions coincide, we call this the epilimit $e-\lim_\nu f^\nu$.

My problem is now the proof of the following proposition (from Rockafellar, Wets, "Variational Analysis", Chapter 7, page 241):

enter image description here

Now I don't understand at all everything from "the first formula is thereby obvious".

Even if we have $N\in \mathcal N_\infty^\#$ and $x^\nu \stackrel{N}{\to} x$, $\alpha^\nu \stackrel{N}{\to} \alpha$, how do we construct a sequence $x^\nu \to x$?

And even if the first two equations in the proposition are proven, how do we obtain equation 7(3) from there?

$\endgroup$
0

2 Answers 2

3
$\begingroup$

Partial Answer:

Even if we have $N\in \mathcal N_\infty^\#$ and $x^\nu \stackrel{N}{\to} x$, $\alpha^\nu \stackrel{N}{\to} \alpha$, how do we construct a sequence $x^\nu \to x$?

You can construct any sequence you want as long as it contains that subsequence and still converges to $x$. Because of the $\liminf$, this will suffice: If we have $\alpha^\nu\geq f^\nu(x^\nu)$ for a subsequence $N$, then it follows that $$ \alpha = \liminf_{\nu\in N} \alpha^\nu \geq \liminf_{\nu\in N} f^\nu(x^\nu) \geq \liminf_{\nu\in \mathbb{N}} f^\nu(x^\nu) $$ Here, the last inequality is true because we only add more elements to the sequence. This inequality implies that $\alpha$ is greater or equal to the right-hand side of the formula.

For the other direction, assume that $\alpha$ is greater or equal to the right-hand side of the formula. Thus, there is a sequence $x^\nu\to x$ with $\liminf_\nu f^\nu(x^\nu)\leq \alpha$. Now we choose the subsequence $N$ such that $f^\nu(x^\nu)\stackrel{N}{\to} \liminf_\nu f^\nu(x^\nu)$ and $\alpha^\nu := \max(f^\nu(x^\nu),\alpha)$. Thus, by the first "if and only if" statement in the proof, it follows that $\alpha\geq (e-\liminf_\nu f^\nu)(x)$.

$\endgroup$
3
  • $\begingroup$ Thanks, that was revealing. So that proves $(e-\liminf_\nu f^\nu)(x) = \min\{\alpha \in \mathbb R|~ \exists x^\nu \to x \text { with } \liminf_\nu f^\nu(x^\nu) \leq \alpha\}$. How do we get the equality in the bracket? $\endgroup$ Oct 30, 2018 at 10:47
  • $\begingroup$ Because we take the smallest $\alpha$ (using $\min$), it does not matter if we use $=$ or $\leq$ in brackets. $\endgroup$
    – supinf
    Oct 30, 2018 at 10:50
  • $\begingroup$ I see, of course. But how does that line of reasoning work for the limsup? We have $n\in \mathcal N_\infty$ and then $\alpha = \limsup_{\nu\in N} \alpha^\nu \geq \limsup_{\nu\in N} f^\nu(x^\nu)$ and as $N$ is actually just a tail of $\mathbb N$, we can equate the last term to $\limsup_{\nu \in \mathbb N}f^\nu(x^\nu)$? $\endgroup$ Oct 30, 2018 at 10:54
0
$\begingroup$

Ok, so with help from supinf I managed to write a more complete version of the proof (which I can now understand). See https://pwacker.com/proof_epiconvergence.html for a hierarchical version of it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.