If $A$ is a C*-algebra, $a\in A$ positive and $\varepsilon>0$, then $\|a(a+\varepsilon)^{-1}\|\leq1$.

Question

Suppose that $a$ is a (not necessarily unital) C*-algebra and let $a\in A$ be a positive element. Let $\varepsilon>0$ be any real number. Then $0\notin\sigma(a+\varepsilon)$ since $\sigma(a)\subset[0,\infty)$. So $a+\varepsilon$ is invertible in the unitization $\widetilde{A}$. Thus $b:=a(a+\varepsilon)^{-1}\in A$ because $A$ is an ideal in $\widetilde{A}$. Is it true that $\|b\|\leq1$? For continuous function spaces, i.e. the case where $A$ is of the form $C(\Omega)$ for some compact Hausdorff space $\Omega$, this feels intuitive, but I can't prove it in the general case. All I can say is that $\|b\|\leq\|a\|\|(a+\varepsilon)^{-1}\|$. Any suggestions are greatly appreciated!

Answer 1

Here is a different argument, using that the norm can be determined via states. We have $$ \|b\|=\sup\{|f(b)|: f \text{ is a state}\}. $$ Because $a+\varepsilon$ is invertible, the family of maps $x\longmapsto f((a+\varepsilon)x)/f(a+\varepsilon)$ is again the family of all states. Then \begin{align} \|b\|&=\sup\left\{\frac{f((a+\varepsilon)b)}{f(a+\varepsilon)}:\ f\ \text{ is a state}\right\}\\[0.3cm] &=\sup\left\{\frac{f(a)}{f(a+\varepsilon)}:\ f\ \text{ is a state}\right\}\\[0.3cm] \end{align} Because $a\geq0$ and $f$ is a state, $$ f(a)\leq f(a)+\varepsilon=f(a+\varepsilon), $$ so $f(a)/f(a+\varepsilon)\leq1$ for all states $f$ and thus $\|b\|\leq1$.

If you have enough theory to know that $A$ represents on a Hilbert space (or if you already have that $A$ is a concrete C$^*$-algebra inside $B(H)$), then the above argument can be done spatially. That is, $$ \begin{align} \|b\|^2 &=\sup\{\|bx\|^2:\ \|x\|=1\} =\sup\left\{\frac{\langle bx,bx\rangle}{\|x\|^2}:\ x\ne0 \right\}\\[0.3cm] &=\sup\left\{\frac{\langle b(a+\varepsilon)x,b(a+\varepsilon)x\rangle}{\|(a+\varepsilon)x\|^2}:\ x\ne0 \right\}\\[0.3cm] &=\sup\left\{\frac{\langle ax,ax\rangle}{\|(a+\varepsilon)x\|^2}:\ x\ne0 \right\}\\[0.3cm] &=\sup\left\{\frac{\langle a^2x,x\rangle}{\langle(a+\varepsilon)^2x,x\rangle}:\ x\ne0 \right\}.\\[0.3cm] \end{align} $$ And now we can use $$ \langle (a+\varepsilon)^2x,x\rangle=\langle a^2x,x\rangle+ 2\varepsilon \langle ax,x\rangle+\varepsilon^2\,\langle x,x\rangle\geq\langle a^2x,x\rangle $$ to conclude that $\|b\|^2\leq1$.

Answer 2

Perhaps the easiest way to do this is to use functional calculus, and the properties of it.

Consider the function $f:[0,\infty)\to[0,\infty)$ given by $f(x)=\frac{x}{x+\varepsilon}$. For all $x\in[0,\infty)$, we have $|f(x)|\leq 1$. Since $f(0)=0$, we have $f(a)\in A$, regardless of whether or not $A$ is unital. Now we have $$\|a(a+\varepsilon)^{-1}\|=\|f(a)\|=\sup_{\lambda\in\sigma(a)}|f(\lambda)|\leq 1$$