Let $A,B$ are two $p\times p$ positive definite matrices such that $0<\delta\_0\leq \min\{\lambda\_{\min}(A), \lambda\_{\min}(B)\}\leq \max\{\lambda\_{\max}(A), \lambda\_{\max}(B)\}\leq \delta\_1$. Also assume that $\Vert A-B\Vert\_{op}\leq \varepsilon$. Can we upper bound $\Vert A^{-1} - B^{-1} \Vert\_{op}$ in terms of $(\delta\_0, \delta\_1, \varepsilon)$. I tried several things with the definition
$$
\Vert A^{-1} - B^{-1} \Vert\_{op} = \max\_{u,v\in \mathbb{S}^{p-1}} u^\top (A^{-1}-B^{-1})v.
$$
One easy observation is that $\Vert A^{-1}-B^{-1}\Vert\_{op}\leq \vert 1/\lambda\_{\min}(A) - 1/\lambda\_{\max}(B)\vert$. But I think this does not help me in any way. Any help will be appreciated.


**Edit:**
It could be possible that one can not get such bound. For a related question see [here](https://math.stackexchange.com/questions/2800167/matrix-inverse-is-a-uniformly-continuous-function-for-uniformly-positive-definit).



If $\Vert A-B\Vert_{op}\leq \varepsilon$ then $A^{-1}$ and $B^{-1}$ are uniformly close
$A^{-1}-B^{-1}=B^{-1}(B-A)A^{-1}$ so
$\|A^{-1}-B^{-1}\| \le \delta\_0^{-2} \epsilon $. This is the best possible bound in terms of the given parameters, as you can see by considering 2 by 2 diagonal matrices: Consider
$A=$diag$(\delta\_0,\delta\_1)$ and $B=$diag$(\delta\_0+\epsilon,\delta\_1+\epsilon)$ where $\delta\_1$ is large and $\epsilon \to 0$.


