# almost unitaries are close to a unitary element

I need help to prove the following exercise:

Let $\epsilon >0$. Show that there exists $\delta >0$ with the property: If $A$ is a unital $C^*$-algebra and $x\in A$ such that $\|x^*x-1\|<\delta,\;\|xx^*-1\|<\delta$, then there is a unitary element $u\in A$ with $\|x-u\|<\epsilon$.

To prove this, I use continuous functional calculus for the element $x^*x$ and $xx^*$ to define $u$. The assumptions $\|x^*x-1\|<\delta,\;\|xx^*-1\|<\delta$ imply $\sigma(x^*x), \sigma(xx^*)\subseteq [1-\delta , 1+\delta]$, where $\sigma(x^*x)$ denotes the spectrum of $x^*x$.

If $x$ were invertible I could try something like $u=x|x|^{-1}$, where $|x|=(x^*x)^{\frac{1}{2}}$(polar decomposition), but $x$ could be non-invertible I think. But $x^*x$ is invertible if $\delta <1$, but I also have to choose a number $\delta$ which depends on $\epsilon$ .

Here I'm stuck. Which continuous function is suitable for u or how to choose $u$?

• Invertibility of $x$ indeed follows, if you take $\delta \leq 1$. – user42761 Jul 7 '16 at 12:36

If $\delta\leq1$, then $x^*x$ and $xx^*$ are invertible. In particular, we can do the polar decomposition $x=u(x^*x)^{1/2}$ and we will have $u\in A$. Also, $u$ is a unitary because $$u^*u=(x(x^*x)^{-1/2})^*(x(x^*x)^{1/2} =(x^*x)^{-1/2}x^*x(x^*x)^{-1/2}=1,$$ $$uu^*=x(x^*x)^{-1/2}(x^*x)^{-1/2}x^*=x(x^*x)^{-1}x^*=1.$$ This last equality is not obvious, but it is not hard: we have $$x(x^*x)x^*=(xx^*)^2=(xx^*)^{1/2}xx^*(x^*x)^{1/2};$$ it follows that $$xp(x^*x)x^*=(x^*x)^{1/2}p(xx^*)(x^*x)^{1/2}$$ for any polynomial $p$, and then $$xf(x^*x)x^*=(x^*x)^{1/2}f(xx^*)(x^*x)^{1/2}$$ for any continuous function $f$. Then $$x(x^*x)^{-1}x^*=(xx^*)^{1/2}(xx^*)^{-1}(xx^*)^{1/2}=1.$$

Now $$\|x-u\|=\|u(x^*x)^{1/2}-u\|=\|u(x^*x)^{1/2}-1)\|=\|(x^*x)^{1/2}-1\|.$$ Since $\sigma(x^*x)\subset(1-\delta,1+\delta)$, we have that $$\sigma((x^*x)^{1/2})\subset((1-\delta)^{1/2},(1+\delta)^{1/2}\subset(1-\delta,1+\delta).$$ Then $\|(x^*x)^{1/2}-1\|<\delta$. So we can take $$\delta=\min\{\varepsilon,1\}.$$

• Is $\lVert (x^*x)^{1/2} - 1\rVert \leq \lVert x^*x - 1 \rVert$ ? Since only the latter one is bounded by $\delta$. – user42761 Jul 6 '16 at 15:30
• Good point. Someone mentioned something about that in a talk I heard today... I'll delete while I think about it. – Martin Argerami Jul 6 '16 at 15:49
• I still don't remember what I heard this morning. And the natural inequality is the other way than needed, with coefficient $1+\|x\|$. But I added an argument that makes it work. – Martin Argerami Jul 6 '16 at 16:01
• Nice solution. But I think there are a few typos. The exponents should be $1/2$ everywhere I guess, but they are mixed up. You want to conclude that $xf(x^*x)x^* = (xx^*)^{1/2} f(xx^*) (xx^*)^{1/2}$. – user42761 Jul 7 '16 at 10:52
• nice! I wasn't sure if I can do polar decomposition and if I can choose u in this way, but now I see why there are no problems. Thank you – Sabrina G. Jul 7 '16 at 11:43

Another similar approach (inspired by Martin Argerami) is the following:

If $x \in A$ and $x^*x, xx^*$ are invertible, then $x$ is invertible. To see this, check that $(x^*x)^{-1}x^* = x^*(xx^*)^{-1}$. This is then the inverse of $x$. Now $x = u \sqrt{x^*x}$ for $u$ unitary since $x$ is invertible. Then the last argument of Martin applies, i.e. that $\lVert \sqrt{x^*x} - 1\rVert < \delta$.

However, this assumes that you know that in this case $u = x \sqrt{x^*x}^{-1}$ is unitary.

• thank you for the helpful answer. For x invertible I can apply polar decomposition on x directly and then I get such an unitary $u$ as in Martins answer (I combine both arguments now). – Sabrina G. Jul 7 '16 at 13:16
• Yes and you know that $x$ is invertible since you may choose $\delta \leq 1$. I am not sure if you realized that $x$ is invertible in any case if you take such a $\delta$. – user42761 Jul 7 '16 at 14:00