Lemma

Tag 08KR, permanent. Cite it as [Bourbaki, Tag 08KR].

Integration, chapter II, § 2. Linear forms on a Riesz space

— *Let E be a directed ordered vector space, $E^*$ its dual, ordered by taking as positive elements the positive linear forms. Let $(u_\alpha)$ be an increasing directed family of elements of $E^*$ . If, for every $x \geqslant 0$ in E, $\sup u_\alpha(x) < +\infty$ , then the family $(u_\alpha)$ has a supremum $u$ in $E^*$ and, for all $x \geqslant 0$ in E,

u(x) = \sup_{\alpha} u_\alpha(x).

In the set P of all $x \geqslant 0$ in E, define the mapping $u$ by the formula (2); it is immediate that $u(\lambda x) = \lambda u(x)$ for all $\lambda \geqslant 0$ and $x \in P$ ; to prove the lemma it therefore suffices, by Prop. 2 of No. 1, to show that

u(x + y) = u(x) + u(y)

for $x, y$ in P. But this is immediate on observing that $u(x) = \lim u_\alpha(x)$ with respect to the directed set of indices (monotone limit theorem).

From the formula (1), one deduces immediately that if $L$ and $M$ are two relatively bounded linear forms on E then, for every $x \geqslant 0$ ,

\begin{cases} \sup(L, M)(x) = \sup_{y \geqslant 0,\ z \geqslant 0,\ y+z=x} (L(y) + M(z)) \\ \inf(L, M)(x) = \inf_{y \geqslant 0,\ z \geqslant 0,\ y+z=x} (L(y) + M(z)). \end{cases}

In particular if, in the first of these formulas, $M$ is replaced by $-L$ , we get

|L|(x) = \sup_{y \geqslant 0,\ z \geqslant 0,\ y+z=x} L(y - z).

Now, if $x = y + z$ , $y \geqslant 0$ and $z \geqslant 0$ , then $-x \leqslant y - z \leqslant x$ ; conversely, the relation $|u| \leqslant x$ implies $L(u) \leqslant |L|(|u|) \leqslant |L|(x)$ . From this we deduce the formula

|L|(x) = \sup_{|y| \leqslant x} L(y) \quad \text{for } x \geqslant 0,

whence, in particular,

|L(x)| \leqslant |L|(|x|)

for all $x \in E$ .

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Prop. 2