A characterization of numerical lower semi-continuity

One can define numerical lower semi-continuity using a sequential criterion: for $\newcommand{\RR}{\mathbb{R}} \renewcommand{\bar}{\overline} f : X \to \bar{\RR},$ we say that $f$ is lower semi-continuous at $x \in X$ if for any sequence $x_n \to x$ we have

$$f(x) \leq \liminf_{n \to \infty} f(x_n).$$

The idea is that $f(x_n)$ is eventually bounded below by $f(x)$; that is, for every $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that whenever $n \geq N$ we have

$$f(x_n) - f(x) < \varepsilon.$$

Unsurprisingly, if $f$ is lower semi-continuous it follows that for any convergent sequence $x_n \to x$ we have

$$f(x) \leq \limsup_{n \to \infty} f(x_n),$$

since

$$\begin{align} f(x) & \leq \liminf_{n \to \infty} f(x_n) \\ &\leq \limsup_{n \to \infty} f(x_n). \end{align}$$

What’s more surprising is that these conditions are equivalent!

Proposition. If for any convergent sequence $x_n \to x$ we have

$$f(x) \leq \limsup_{n \to \infty} f(x_n)$$

then $f$ is lower semi-continuous at $x$.

I find this surprising because the property above relates to the limiting upper bounds of a sequence. Why should that have any influence on the limiting behavior of the lower bounds? There is sort of a mathematical feint at work here, because the upper limit is not where the strength of the hypothesis lies. Instead, it’s hidden in the fact that this upper limit bound holds for every convergent sequence. Let’s find out why.

Proof. By the contrapositive. Suppose $f$ is not lower semi-continuous, so that there exists a sequence $x_n \to x$ with

$$\liminf_{n \to \infty} f(x_n) = s < f(x).$$

Now, $s$ is a limit point of the sequence $( f(x_n) )$, which means we can fix a subsequence $x_{n_k}$ for which $\lim _ {k \to \infty} f(x_{n_k}) = s$. But since $( x _ {n_k} )$ is a subsequence of $(x _ n)$, it remains true that $x _ {n_k} \to x$. This sequence converges to $x$ but also has the property that

$$\begin{align} \limsup _ {k \to \infty} f(x _ {n_k}) &= \lim _ {k \to \infty} f(n_k) \\ &= s \\ &< f(x) \end{align}$$

which implies that the hypothesis of the proposition is false. By taking the contrapositive, we obtain the desired result. $\square$

I suspect that this is actually a highly practical result, because it provides the flexibility of looking at either the upper or lower limits of arbitrary sequences. That can (and soon will) come in handy for some of my work.

Published on June 28th, 2018 by David Kraemer