# Alternative characterization of epi-convergence

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): 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?

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 $$\alpha\geq (e-\liminf_\nu f^\nu)(x)$$.
• 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? – Mercury Bench Oct 30 '18 at 10:47
• Because we take the smallest $\alpha$ (using $\min$), it does not matter if we use $=$ or $\leq$ in brackets. – supinf Oct 30 '18 at 10:50
• 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 $\limsup_{\nu \in \mathbb N}f^\nu(x^\nu)$? – Mercury Bench Oct 30 '18 at 10:54