Given $X,Y$ two Banach spaces, I know the set of bounded operators $L(X,Y)$ is Banach with the operator norm $\|A\|=\sup_{\|x\|\leq1}\|Ax\|$. I know the set of bounded operators with bounded inverse $\mathrm{Inv}(X,Y)$ is open in $L(X,Y)$ with the norm topology. I have proved that, if $J(A):=A^{-1}$ is the inversion map, then it is Fréchet differentiable and $\mathrm{d}J(A)(B)=-A^{-1}BA^{-1}$. But the exercise I was given, in the midst of trying my luck at which I proved the above differentiability, was to prove $J$ was smooth. So I tried the second derivative, and I am stuck. I am supposed to find a bilinear form for which:
$$\lim_{\|T\|\to0}\frac{\|-(A+T)^{-1}B(A+T)^{-1}+A^{-1}BA^{-1}-D(T,B)\|}{\|T\|}=0,$$
and then $D$ would be the second derivative of $J$ at $A$. But how can I? I mean, If I try adding stuff to break the sum into two I get a piece which I can get back to the limit for the first derivative in the identity, but the other piece I get is similar but with $(A+T)^{-1}$ on the right (or left depending on the terms I add), meaning I cannot just stick it into the differential since it is not linear in $T$. Any hints? Is there any sensible way of proving smoothness? Perhaps inductively?
Using what you have, $$ (A+B)^{-1}-A^{-1}= -A^{-1}BA^{-1}+o(B). $$ So the derivative of the derivative examines \begin{align} &-(A+T)^{-1}B(A+T)^{-1}\\ &=-\{ -A^{-1}TA^{-1}+A^{-1}+o(T)\}B\{-A^{-1}TA^{-1}+A^{-1}+o(T)\} \\ &=+A^{-1}TA^{-1}BA^{-1}+A^{-1}BA^{-1}TA^{-1}+A^{-1}BA^{-1}+o(T) \end{align} And that looks about right. The required symmetry between $T$ and $B$ is there, for example. So the second derivative would be $$ D^2J(T,B)=A^{-1}TA^{-1}BA^{-1}+A^{-1}BA^{-1}TA^{-1} $$