I am struggling to understand how different definitions of lower semi-continuity compare to each other. I am mostly concerned with that concept regarding closed convex functions.
Therefore I am reading several books on convex analysis and get the following definitions:
1.) In the book "Convex Analysis" by Rockafeller, he defines lower semi-continuity as follows:
An extended-real-valued function $f$ given on $S\subseteq \mathbb{R}^n$ is said to be lower semi-continuous at a point $x$ of $S$ if:
$$f(x) \leq \lim_{n \to \infty} f(x_n)$$
for any sequence $(x_n)$ in $S$ that converges to $x$ and the limit of $f(x_1), f(x_2), … $ exists in $[-\infty, +\infty]$.
This condition may be expressed as:
$$f(x)= \liminf_{y \to x} f(y)=\lim_{\varepsilon \downarrow 0}(\inf{f(y)\mid |x-y|\leq \varepsilon}).$$
2.) Then in the book Fundamentals of Convex Analysis we have the definition:
A function $f$ is lower semi-continuous if, for each $x\in \mathbb{R}^n$,
$$ \liminf_{y \to x} f(y) \geq f(x)$$
This relation has to hold in $\mathbb{R} \cup {+\infty}$ (i.e., they exclude $-\infty$ here).
3.) In the book by Bertskas, the definition is:
Recall that $f$ (where $f: X\rightarrow [-\infty, +\infty]$) is called lower semi-continuous at a vector $x \in X$ if
$$f(x)\leq \liminf_{n\to \infty} f(x_n) $$
So my questions are as follows:
i) I can see that the Definitions 2) and 3) are the same. But how do they relate to 1)? Is 1) the same as 2) and 3) under certain conditions? I would say that 2) and 3) are not the same as 1) but in all three books they prove that their definition of lower semicontinuity is equivalent to the epigraph being closed and all sublevel sets being closed. So that would actually imply that their respective definitions of semi-continuity are also the same.
ii) Does the value $-\infty$ play a role in whether we need an inequality or an equality?
iii) How does Rockafeller in 1) get from his definition via an inequality to an equality?
iv) In fundamentals of convex analysis they define the closure of a function $f$ the function $cl(f): \mathbb{R}^n \rightarrow \mathbb{R}\cup \{\pm \infty\}$ as:
$$(cl(f))(x)= \liminf_{y \to x} f(y)$$
and say that this is equivalent to
$$epi(cl(f)) = cl(epi(f)).$$
How can we see that this is equivalent? (I think this only holds for proper convex functions though, as they only treat these in the book)
Best Answer
Rockafeller defines that $f : S \to [-\infty,+\infty]$ is lower semi-continuous at $x \in S$ if
$$f(x) \le \lim_{n \to \infty} f(x_n) \text{ for any sequence } (x_n) \text{ in } S \text{ converging to } x \\ \text{ such that the limit of } (f(x_n)) \text{ exists in } [-\infty, +\infty] . \tag{1}$$ The claim
is not true if we use the standard definition of $\liminf$ which is $$\liminf_{y \to x} f(y) = \lim_{\epsilon \to 0} I(f,x,\epsilon)$$ with $I(f,x,\epsilon) = \inf f( S \cap B(x,\epsilon) \setminus \{x\})$.
However, Rockafeller uses the follwing unusual definition which I denote by $\liminf^*$ for the sake of clarity:
$${\liminf}^*_{y \to x} f(y) = \lim_{\epsilon \to 0} I^*(f,x,\epsilon)$$ with $I^*(f,x,\epsilon) = \inf f(S \cap B(x,\epsilon))$.
Note that $f(S \cap B(x,\epsilon)) = f(S \cap B(x,\epsilon) \setminus \{x\}) \cup \{f(x)\}$, thus $$I^*(f,x,\epsilon) = \min(I(f,x,\epsilon), f(x) $$ and therefore $${\liminf}^*_{y \to x} f(y) = \min({\liminf}_{y \to x} f(y) , f(y)) .$$
Hence Rockafeller's $f(x) = {\liminf}^*_{y \to x} f(y)$ means nothing else than $$f(x) \le \liminf_{y \to x} f(y) . \tag{2}$$
Note that in general $\liminf_{y \to x} f(y) > {\liminf}^*_{y \to x} f(y)$. As an example consider the function $f : \mathbb R \to [-\infty,+\infty], f(x) = 1$ for $x \ne 0$ and $f(0) = 0$. We have $\liminf_{y \to 0} f(y) = 1 > 0 = {\liminf}^*_{y \to 0} f(y)$.
We shall prove that $(1)$ is equivalent to $(2)$.
Observe that $I(f,x,\epsilon') \ge I(f,x,\epsilon)$ for $\epsilon' \le \epsilon$. Thus
$\liminf_{y \to x} f(y) \ge I(f,x,\epsilon)$ for all $\epsilon > 0$.
$\liminf_{y \to x} f(y) = -\infty$ if and only if $I(f,x,\epsilon) = -\infty$ for all $\epsilon > 0$.
If $(\epsilon_n)$ is any sequence of positive numbers converging to $0$, then $\lim_{n \to \infty} I(f,x,\epsilon_n) = \liminf_{y \to x} f(y)$.
$(1) \implies (2)$:
Let us first consider the case $\liminf_{y \to x} f(y) = -\infty$. Then $I(f,x,1/n) = -\infty$ for all $n$. We can therefore pick $x_n \in S \cap B(x,1/n) \setminus \{x\}$ such that $f(x_n) < -n$. Then $\lim_{n \to \infty} x_n = x$ and $\lim_{n \to \infty} f(x_n) = -\infty$. From $(1)$ we conlude that $f(x) = -\infty$, thus $(2)$ is satisfied.
If $\liminf_{y \to x} f(y) > -\infty$, we know that $I(f,x,\epsilon_0) > -\infty$ for some $\epsilon_0 > 0$. Hence $I(f,x,\epsilon) \ge I(f,x,\epsilon_0) > -\infty$ for all $\epsilon \le \epsilon_0$. Pick $x_n \in S \cap B(x,\epsilon_0/n) \setminus \{x\}$ such that $f(x_n) < I(f,x,\epsilon_0/n) + 1/n$. Then $\lim_{n \to \infty} x_n = x$ and $\lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} I(f,x,\epsilon_0/n) = \liminf_{y \to x} f(y)$. Thus $(1)$ shows $f(x) \le \liminf_{y \to x} f(y) $.
$(2) \implies (1)$:
If $f(x) = -\infty$, then $(1)$ is trivially satisfied. So let $f(x) > -\infty$. Then $(2)$ excludes $\liminf_{y \to x} f(y) =-\infty$. Consider any sequence $(x_n)$ in $S$ converging to $x$ such that $(f(x_n))$ converges to some $\eta \in [-\infty,\infty]$. If $\eta = \infty$, we are done. If infinitely many $x_n = x$, we must have $\eta = f(x)$ and we are done again. So let us consider the case that only finitely many $x_n = x$. Assume that $f(x) > \eta$, i.e. $r = f(x) - \eta > 0$. Then $f(x) - r/2 > f(x_n)$ for $n \ge n_0$. Clearly each $S \cap B(x,\epsilon) \setminus \{x\}$ contains infinitely many $x_n$, thus $I(f,x, \epsilon) < f(x) - r/2 $ and thus $\liminf_{y \to x} f(y) \le f(x) - r/2$ which contradicts $(2)$.
Update:
Bertskas defines that $f : S \to [-\infty,+\infty]$ is lower semi-continuous at $x \in S$ if
$$f(x) \le \liminf_{n \to \infty} f(x_n) \text{ for any sequence } (x_n) \text{ in } S \text{ converging to } x . \tag{3}$$
$(3) \implies (1)$:
Consider a sequence $(x_n)$ converging to $x$ such that $\lim_{n \to \infty} f(x_n)$ exists and has the value $\eta \in[-\infty,+\infty]$. Then $\liminf_{n \to \infty} f(x_n) = \eta = \lim_{n \to \infty} f(x_n)$ because $\eta$ is the only accumulation point of $(f(x_n))$. Thus from $(3)$ we get $f(x) \le \lim_{n \to \infty} f(x_n)$.
$(1) \implies (3)$:
Let $(x_n)$ be any sequence converging to $x$ and $\eta = \liminf_{n \to \infty} f(x_n)$. There exists a subsequence $(x_{n_k})$ of $(x_n)$ such that $\lim_{k \to \infty} f(x_{n_k}) = \eta$. Since clearly $(x_{n_k})$ converges to $x$, $(1)$ applies to give $f(x) \le \lim_{k \to \infty} f(x_{n_k}) = \eta$. Thus $f(x) \le \liminf_{n \to \infty} f(x_n)$.