Distance between invertible elements of a normed algebra

[Fredrik]

Is there a way to see that if $\|x-y\|$ is "small", then so is $\|x^{-1}-y^{-1}\|$? For example, if $\|x-y\|<r$, is there a function f such that $\|x^{-1}-y^{-1}\|<f(r)$

Edit: Nevermind. What I needed is just the operator version of (1/2-2/3)=(3-2)/6:

$$\|x^{-1}-y^{-1}\|=\|x^{-1}yy^{-1}-x^{-1}xy^{-1}\|=\|x^{-1}(y-x)y^{-1}\|\leq \|x^{-1}\|\|x-y\|\|y^{-1}\|$$

 

[disregardthat]

|x-y| can be made as small as you want while |x^-1+y^-1| simultaneously can be made as large as you want.

In R, take x=2/2^k, y=1/2^k, and let k grow without restriction.

 

[Fredrik]

Good point. I was however more interested in showing that you can make $\|x^{-1}-y^{-1}\|$ as small as you want by choosing y so that $\|x-y\|$ is small enough. The calculation I added in my edit is what I need for that.