# Cube root in $C^{*}$-algebra.

Let $A$ be a $C^*\text{-algbera}$ and $x\in A$. I'm trying to show that

a)for $0<\alpha<\frac{1}{2}$, there exists $u\in A$ with $x=u(x^*x)^{\alpha}$ and $u^*u=(x^*x)^{1-2\alpha}$.

b) there exists $y\in A$ such that $x=yy^*y$ ($\text{"cube root"}$ of x) and such $y$ is unique.

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Is this a homework? –  Martin Brandenburg Mar 25 '13 at 0:13
@MartinBrandenburg, Yes, its form a homework. –  i.a.m Mar 25 '13 at 0:19
I would start with the polar decomposition. But this is really just a guess. Try to understand the case $A=\mathbb{C}$ and try to generalize it. –  Martin Brandenburg Mar 25 '13 at 0:32

For the proof of first part see $C^*$-algebras and their automorphisms group by G. K. Pedersen pages 11-12, lemmas 1.4.4-1.4.5.
For the proof of the second part use previous result with $\alpha=1/3$. Then you get $u\in A$ such that $x=u(x^*x)^{1/3}$ and $u^*u=(x^*x)^{1/3}$. Hence $x=uu^*u$. Proof of uniqueness I leave to you since this is a homework.
@i.a.m since $a$ is self adjoint then (by continuous functional calculus) so $a^{1/2-\alpha}$ and $(n^{-1}+a)^{-1/2}$. Hence \begin{align}u_n^*&=a^{1/2-\alpha}(n^{-1}+\alpha)x^*\\u_n^*u_n&=a^{1/2-\alpha}(‌​n^{-1}+\alpha)^{-1/2}x^*x(n^{-1}+a)^{-1/2}a^{1/2-\alpha}\\&=a^{1/2-\alpha}(n^{-1}+‌​\alpha)^{-1/2}a(n^{-1}+a)^{-1/2}a^{1/2-\alpha}\end{align} Since $\lim\limits_{n\to\infty}u_n=u$, then $u^*u=\lim\limits_{n\to\infty}u_n^*u_n$. –  Norbert Mar 30 '13 at 6:53
By Dini's theorem $f_n(t)=t^{1/2-\alpha}(n^{-1}+t)^{-1/2}t(n^{-1}+t)^{-1/2}t^{1/2-\alpha}$ uniformly converges to $t^{1-2\alpha}$, then $$u^*u=\lim\limits_{n\to\infty}u_n^*u_n=a^{1-2\alpha}=(x^*x)^{1-2\alpha}$$ –  Norbert Mar 30 '13 at 7:59