Proposition 18

Let A be an involutive algebra. The ordered set $A^{<1}_+$ is directed to the right (E, III, p. 12, def. 7), and the filter of its sections (TG, I, p. 38, example 2) is an increasing approximate unit of A.

Let $\widetilde{A}$ be the involutive algebra deduced from A by adjunction of a unit element denoted by 1 (def. 4 of I, p. 106).

Let $g$ be the function from $[0,1[$ into $\mathbf{R}_+$ defined by $g(t) =t(1-t)^{-1}$. It is a continuous increasing bijection, its inverse bijection being given by $t\mapsto 1-(1 +t)^{-1}$.

Let us prove that the ordered set $A^{<1}_+$ is directed to the right. Let $x$ and $y$ be elements of $A^{<1}_+$. Since $g(0) = 0$ and since Sp$'_A(x)$ and Sp$'_A(y)$ are contained in $[0,1[$, the elements $g(x)$ and $g(y)$ are defined; they are positive, so that $g(x) +g(y)\geqslant 0$. We can therefore form the element $z=g^{-1}(g(x) +g(y))$ of A. We have Sp$'_A(z)\subset [0,1[$, and hence $z\in A^{<1}_+$.

We have $0\leqslant g(x)\leqslant g(z)$, whence $1\leqslant 1+g(x)\leqslant 1+g(z)$. Assertion b) of lemma 14 of I, p. 119 implies that $1 +g(x)$ and $1 +g(z)$ are invertible and that $(1 +g(z))^{-1}\leqslant (1 +g(x))^{-1}$. Consequently, $z= 1-(1 +g(z))^{-1}\geqslant$ $1-(1 +g(x))^{-1}=x$. Analogously, we have $z\geqslant y$. In consequence, $z$ majorizes $x$ and $y$ in $A^{<1}_+$. The ordered set $A^{<1}_+$ is therefore directed to the right. Let us denote by $\mathfrak{F}$ the filter of its sections

Let $x$ be a positive element of A. For every integer $n\geqslant 1$, let $e_n=$ $g^{-1}(nx)$; we have $e_n\in A^{<1}_+$. Let $h_n$ be the continuous function on $\mathbf{R}_+$ defined for every $t\in \mathbf{R}_+$ by $h_n(t) =t^2(1-g^{-1}(nt)) =t^2/(1 +nt)$. We have $|h_n(t)|\leqslant t/n$ for every $t\geqslant 0$, and hence $\|x(1-e_n)x\|=\|h_n(x)\|\leqslant$ $\|x\|/n$. In particular, $x(1-e_n)x$ tends to 0 when $n$ tends to infinity.

Let $\varepsilon >0$ be a real number. Let $n$ be an integer such that $\|x(1-e_n)x\|< \varepsilon$. For every $f\in A^{<1}_+$ such that $f\geqslant e_n$, one has then

Moreover, since $0\leqslant f\leqslant 1$, one has $(1-f)-(1-f)^2= (1-f)f\geqslant 0$ (Lemma 12, d) of I, p. 116), hence $(1-f)^2\leqslant 1-f$. Since $1-f\leqslant 1-e_n$, it follows that $0\leqslant (1-f)^2\leqslant 1-e_n$. By Lemma 14, a) and c) of I, p. 119, it follows that

Thus lim$_{f,\mathfrak{F}}f x=x$ for every $x\in A_+$. The filter $\mathfrak{F}$ is therefore an approximate unit of A by Proposition 17.