5. Real-valued functions

1. Real-valued functions

By an abuse of language analogous to that mentioned in § 4, no. 2, mappings of $X$ into $\overline{\mathbf{R}}$ will also be called real-valued functions defined on $X$ in this and the following section. Mappings of $X$ into $\mathbf{R}$ will be called finite real-valued functions.

If $f$ and $g$ are two real-valued functions defined on $X$, the relation $f \leq g$ is by definition equivalent to "$f(x) \leq g(x)$ for all $x \in X$;” this relation is an ordering on the set $\overline{\mathbf{R}}^X$ of all real-valued functions on $X$. The set $\overline{\mathbf{R}},^X$ ordered by this relation, is a lattice; for if $f$ and $g$ are any two real-valued functions, the function $h$ defined by $h(x) = \sup (f(x), g(x))$ for all $x \in X$ is the smallest of the real-valued functions on $X$ which are both $\geq f$ and $\geq g$; in accordance with general notation, we denote this function (which is the least upper bound of $f$ and $g$ in $\overline{\mathbf{R}}^X$) by $\sup (f, g)$. Similarly, the function whose value at each $x \in X$ is $\inf (f(x), g(x))$ is denoted by $\inf (f, g)$.

Note that $\sup (f, g)$ is the composition of the mapping

A real-valued function $f$ defined on a set $X$ is said to be bounded above (resp. bounded below) in $X$, if $f(X)$ is a subset of $A'' = [-\infty, +\infty]$ and is bounded above (resp. if $f(X)$ is a subset of $A' = ]-\infty, +\infty]$ and is bounded below). $f$ is said to be bounded in $X$ if it is bounded both above and below, that is if $f(X)$ is a bounded subset of $\mathbf{R}$.

Every bounded function is therefore finite. The converse is false, as is shown by the function $1/x$ on $\mathbf{R}_+^* = ]0, +\infty[$.

2. Real-valued functions defined on a filtered set

3. Limits on the right and on the left of a function of a real variable

Let $A$ be a non-empty subset of $\overline{\mathbf{R}}$ and let $a \neq -\infty$ be a point of $\overline{\mathbf{R}}$ lying in the closure of the set $B = A \cap [-\infty, a[$. The set $B$ is directed with respect to the relation $\leq$, and its section filter $\mathfrak{F}$ is the same as the trace on $B$ of the neighbourhood filter of $a$ in $\overline{\mathbf{R}}$.

(*) This statement assumes implicitly that the ordering in $X$ is written $x \leq y$. If this relation is written $x(\sigma) y$, where $(\sigma)$ is a certain sign or group of signs characteristic of the relation envisaged, then the word "directed" in the statement must be replaced by "directed with respect to $(\sigma)$".

Likewise, if $a$ is in the closure of the set $A \cap ]a, + \infty]$, we define a limit on the right (if it exists) of $f$ at the point $a$, and denote it by

The following proposition is an immediate consequence of Theorem 2:

4. Bounds of a real-valued function

In particular, if A is a non-empty subset of $\overline{\mathbf{R}}$, then

The number $a = \sup_{x \in A} f(x)$ is characterized by the following two properties: (i) For all $x \in A, f(x) \leq a$. (ii) For every $b < a$, there exists $x \in A$ such that $b < f(x) \leq a$.

The numbers $\sup_{x \in A} f(x)$ and $\inf_{x \in A} f(x)$ belong to the closure of $f(A)$ in $\overline{\mathbf{R}}$. We have $\inf_{x \in A} f(x) \leq \sup_{x \in A} f(x)$; and these two numbers are equal if and only if $f$ is constant on A.

A real-valued function, defined on a set X, is bounded above (resp. bounded below) in a non-empty subset A of X if and only if $\sup_{x \in A} f(x) < + \infty$ [resp. $\inf_{x \in A} f(x) > -\infty$]. $f$ is bounded in A if and only if $|f|$ is bounded above in A, hence if and only if $\sup_{x \in A} |f(x)| < + \infty$.

We have

This relation reduces all properties of the greatest lower bound to those of the least upper bound; hence in general we shall speak only of the latter.

5. Envelopes of a family of real-valued functions

The upper envelope of the family $(f_i)$ is thus the least upper bound of this family in the lattice $\overline{\mathbf{R}}$ of real-valued functions defined on $X$, and this justifies the notation $\sup_{i \in I} f_i$.

Furthermore, if we endow $\overline{\mathbf{R}}^X$ with the topology which is the product of the topologies of its factors (all identical with $\overline{\mathbf{R}}$), we have the following proposition:

Thus $(f_i)$ is uniformly bounded above in $X$ if and only if the upper envelope of this family is bounded above in $X$. $(f_i)$ is uniformly bounded in $X$ if and only if the upper envelope of the family $(|f_i|)$ is bounded above in $X$ [i.e. if and only if there is a finite real number $a \geq 0$ such that $|f_i(x)| \leq a$ for all $x \in X$ and all $i \in I$].

6. Upper limit and lower limit of a real-valued function with respect to a filter

Let $f$ be a real-valued function defined on a set $X$ filtered by a filter $\mathfrak{G}$. $\mathfrak{G}$ is a directed set with respect to the relation $\supseteq$ (Chapter I, § 6). For each $M \in \mathfrak{G}$ consider the real number $\sup_{x \in M} f(x)$: we have a function $M \to \sup_{x \in M} f(x)$ of $\mathfrak{G}$ into $\overline{\mathbf{R}}$, which is a decreasing function on $\mathfrak{G}$, by Proposition 7 of no. 4. Hence, by Theorem 2 of no. 2, it has a limit with respect to the directed set $\mathfrak{G}$.

The lower limit of $f$ with respect to the filter $\mathfrak{G}$ is defined similarly, and is denoted by $\lim \inf_{\mathfrak{G}} f$ or $\lim \inf_{x, \mathfrak{G}} f(x)$. Thus we have

(10)

Often the filter $\mathfrak{G}$ is suppressed from the notation, and we write simply $\lim \sup f$ or $\lim \sup_{x} f(x)$, or $\lim \sup f(x)$ when there is no risk of confusion.

From formulae (10) and Theorem 1 we have

(11)

By Theorem 2 of no. 2 we may also write

(12)

Also we may replace the filter $\mathfrak{G}$, on the right-hand sides of formulae (10) and (12), by any base $\mathfrak{B}$ of $\mathfrak{G}$.

From (2) and (10),

and therefore we need consider only the upper limit.

7. Algebraic operations on real-valued functions

Let $f$ and $g$ be two real-valued functions defined on a set $X$; if the sum $f(x) + g(x)$ [resp. the product $f(x)g(x)$] is defined for all $x \in X$, then we denote by $f + g$ (resp. $fg$) the real-valued function

This last function is therefore defined provided $f$ does not take the value 0; when $f$ takes its values in the interval $[0, +\infty]$ (resp. in $[-\infty, 0]$) $1/f(x)$ as everywhere defined by putting $1/0 = +\infty$ (resp. $1/0 = -\infty$); in this case the function $1/f$ is defined.

Suppose that $X$ is filtered by a filter $\mathfrak{F}$, and that $\lim_{\mathfrak{F}} f$ and $\lim_{\mathfrak{F}} g$ exist. If on the one hand the function $f + g$ (resp. $fg$, $1/g$) is defined, and if on the other hand the expression $\lim_{\mathfrak{F}} f + \lim_{\mathfrak{F}} g$ (resp. $\lim_{\mathfrak{F}} f \cdot \lim_{\mathfrak{F}} g$, $1/\lim_{\mathfrak{F}} f$) has a sense, then $\lim_{\mathfrak{F}} (f + g)$ [resp. $\lim_{\mathfrak{F}} fg$, $\lim_{\mathfrak{F}} (1/f)$] exists and is equal to this expression by reason of the continuity of the function $x + y$ (resp. $xy$, $1/x$) at points where it is defined.