Factorization Theorem for C*-Algebra in terms of Bound on Modulus/Lemma

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra. 
Let $\le_A$ be the canonical preordering of $A$.
Let $\alpha, \beta \in \R$ be such that $\alpha + \beta > 1$. 
Let $a \in A$ be positive.
Let $x, y \in A$ be such that:
:$x^\ast x \le_A a^\alpha$
and:
:$y y^\ast \le_A a^\beta$
Let:
:$u_n = x \paren {n^{-1} {\mathbf 1}_A + a}^{-1/2} y$
where $\paren {n^{-1} {\mathbf 1}_A + a}^{-1/2}$ is given by the continuous functional calculus for $a$.
Then $\sequence {u_n}_{n \mathop \in \N}$ converges to $u \in A$ such that:
:$\norm u \le \norm a^{\frac 1 2 \paren {\alpha + \beta - 1} }$
== Proof ==
Define:
:$d_{n, m} = \paren {n^{-1} {\mathbf 1}_A + a}^{-1/2} - \paren {m^{-1} {\mathbf 1}_A + a}^{-1/2}$
From Product of Element of C*-Algebra with its Star is Positive, we have:
:${\mathbf 0}_A \le_A x^\ast x \le_A a^\alpha$
and:
:${\mathbf 0}_A \le_A y^\ast y \le_A a^\beta$ 
From Continuous Function applied to Normal Element of Unital C*-Algebra is Hermitian iff Function is Real-Valued, we have that $d_{n, m}$ is Hermitian for each $n, m \in \N$. 
We then have:
From Conjugation in C*-Algebra preserves Positivity: Corollary to $x^\ast x \le_A a^\alpha$, we have:
:${\mathbf 0}_A \le_A \paren {d_{n, m} y}^\ast x^\ast x \paren {d_{n, m} y} \le_A \paren {d_{n, m} y}^\ast a^\alpha \paren {d_{n, m} y}$
hence:
:${\mathbf 0}_A \le_A y^\ast d_{n, m} x^\ast x d_{n, m} y \le_A y^\ast d_{n, m} a^\alpha d_{n, m} y$
From Norm Preserves Ordering on Positive Elements of C*-Algebra, we have:
:$\norm {y^\ast d_{n, m} x^\ast x d_{n, m} y} \le \norm {y^\ast d_{n, m} a^\alpha d_{n, m} y}$
From Power of Power of Positive Element of Unital C*-Algebra, we have:
:$\paren {a^{\alpha/2} }^2 = a^\alpha$
We then have:
Applying Conjugation in C*-Algebra preserves Positivity: Corollary again to $y y^\ast \le_A a^\beta$ we have:
:${\mathbf 0}_A \le_A \paren {a^{\alpha/2} d_{n, m} }^\ast y y^\ast \paren {a^{\alpha/2} d_{n, m} } \le_A \paren {a^{\alpha/2} d_{n, m} }^\ast a^\beta \paren {a^{\alpha/2} d_{n, m} }$
so that:
:${\mathbf 0}_A \le_A a^{\alpha/2} d_{n, m} y y^\ast d_{n, m} a^{\alpha/2} \le_A a^{\alpha/2} d_{n, m} a^\beta d_{n, m} a^{\alpha/2}$
Hence from Norm Preserves Ordering on Positive Elements of C*-Algebra, we have:
:$\norm {a^{\alpha/2} d_{n, m} y y^\ast d_{n, m} a^{\alpha/2} } \le \norm {a^{\alpha/2} d_{n, m} a^\beta d_{n, m} a^{\alpha/2} }$
By the definition of the continuous functional calculus, $a^{\alpha/2}$, $d_{n, m}$ and $a^\beta$ all commute.
Hence we have:
:$a^{\alpha/2} d_{n, m} a^\beta d_{n, m} a^{\alpha/2} = d_{n, m}^2 a^{\alpha + \beta}$
from Product of Powers of Positive Element of Unital C*-Algebra.
Since $d_{n, m}^2 a^{\alpha + \beta}$ is Hermitian, we have:
:$\norm {d_{n, m}^2 a^{\alpha + \beta} } = \norm {d_{n, m} a^{\paren {\alpha + \beta}/2} }^2$
by $(\text C^\ast 5)$ in the definition of a $\text C^\ast$-algebra.
To recall, so far we have shown that:
:$\norm {u_n - u_m} \le \norm {d_{n, m} a^{\paren {\alpha + \beta}/2} }$
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\map \CC {\map {\sigma_A} a}$. 
Now define $f_n : \map {\sigma_A} a \to \R$ by:
:$\ds \map {f_n} t = \frac 1 {\sqrt {n^{-1} + t} } t^{\paren {\alpha + \beta}/2}$
for each $t \in \map {\sigma_A} a$.
Also define $f : \map {\sigma_A} a \to \R$ by:
:$\map f t = t^{\paren {\alpha + \beta - 1}/2}$
for each $t \in \map {\sigma_A} a$. 
For $n \ge m$, we have $n^{-1} \le m^{-1}$ and hence:
:$\ds \frac 1 {\sqrt {n^{-1} + t} } t^{\paren {\alpha + \beta}/2} \ge \frac 1 {\sqrt {m^{-1} + t} } t^{\paren {\alpha + \beta}/2}$ for each $t \in \map {\sigma_A} a$. 
Hence for each $t \in \map {\sigma_A} a$, $\sequence {f_n}_{n \mathop \in \N}$ is increasing.
Further for $t \in \map {\sigma_A} a \setminus \set 0$, we have:
:$\ds \frac 1 {\sqrt {n^{-1} + t} } \to t^{-1/2}$
and hence:
:$\ds \frac 1 {\sqrt {n^{-1} + t} } t^{\paren {\alpha + \beta}/2} \to t^{\frac {\alpha + \beta - 1} 2}$ 
Further, $\map {f_n} 0 = 0$ for each $n \in \N$ and $\map f 0 = 0$, so convergence also holds for $t = 0$. 
So $\sequence {f_n}_{n \mathop \in \N}$ converges pointwise to $f$.
Since $\sequence {f_n}_{n \mathop \in \N}$ is increasing, we have that:
:$\norm {f_n - f}_\infty \to 0$
by Dini's Theorem.
Since the continuous functional calculus is an isometry, we have:
:$\norm {d_{n, m} a^{\paren {\alpha + \beta}/2} } = \norm {f_n - f_m}_\infty$
Let $\epsilon > 0$. 
Since $\sequence {f_n}_{n \mathop \in \N}$ converges, there exists $N \in \N$ such that:
:$\norm {f_n - f_m}_\infty < \epsilon$ for $n, m \ge N$.
Hence $\norm {u_n - u_m} < \epsilon$ for $n, m \ge N$.
Hence $\sequence {u_n}_{n \mathop \in \N}$ is a Cauchy sequence.
Since $A$ is a Banach algebra, $\sequence {u_n}_{n \mathop \in \N}$ converges to $u \in A$.
We just need to prove the bound on $\norm u$.
For brevity, define $g_n : \map {\sigma_A} a \to \R$ such that:
:$\map {g_n} t = \paren {n^{-1} + t}^{-1/2}$
for each $t \in \map {\sigma_A} a$.
Then:
:$\map {f_n} t = \map {g_n} t t^{\paren {\alpha + \beta - 1}/2}$
for each $t \in \map {\sigma_A} a$.
Almost identically (hence we suppress details) to before we have:
Hence:
:$\ds \norm {u_n} \le \norm a^{\paren {\alpha + \beta + 1}/2}$ for each $n \in \N$.
From Modulus of Limit: Normed Vector Space, we have:
:$\ds \norm u \le \norm a^{\paren {\alpha + \beta + 1}/2}$
by taking $n \to \infty$.
== Sources ==
* : Lemma $4.5$
Category:Factorization Theorem for C*-Algebra in terms of Bound on Modulus