Norm inequality of positive element of $C^*$-Algebra with norm less than 1

Question

We know that for two real positive numbers $a,b$, this property holds:

$$ \begin{equation} A+B<1\rightarrow\dfrac{A}{1-B}<1 \end{equation} $$

If $x, y$ are two positive elements of unital $C^*$-Algebra with $\|x+y\|<1$, is true that

$$ \begin{equation} \dfrac{\|x\|}{1-\|y\|}<1? \end{equation} $$

With inequality for positive element, it is obvious that $\|x\|,\|y\|<1$. I know it's sufficient to show that $\|x\|+\|y\|<1$, but it seems that I can't get there for some reason. Is there any clue to do this or maybe there has to be additional condition so that the inequality holds?

Thank you for your help.

Answer 1

This fails in any C$^*$-algebra other than $\mathbb C$. In $\mathbb C^2$, fix $c\in(0,1)$ and take $$ x=(c,0),\qquad\qquad y=(0,c). $$ Then $\|x+y\|=c<1$, while $\|x\|+\|y\|=c+c=2c$. So as long as $\frac12<c<1$, we get $\|x+y\|<1$ and $\frac{\|x\|}{1-\|y\|}>1$.

In an arbitrary C$^*$-algebra the same game can be played by taking a selfadjoint element with non-singleton spectrum and doing the above for two functions in $C(\sigma(a))$ such that their product is zero.

Answer 2

The claim does not hold. Consider $\mathbb{C}^2$ with the norm $$\|(x,y)\|=\max\{|x|,|y|\}$$ and pointwise multiplication and pointwise addition. Let $a=(r,0)$ and $b=(0,r)$ where ${1\over 2}<r<1.$ Then $\|a+b\|=r<$ and $\|a\|+\|b\|=2r>1.$