Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 12G / d / Solution

Put $r=\Vert X\Vert <1$. By submultiplicativity of the operator norm, $\Vert X^k\Vert \le r^k$, so the Neumann series converges absolutely in the complete finite-dimensional matrix space. Its partial sums satisfyTaking limits gives the inverse and a quantitative remainder bound:To establish two actual Frechet derivatives, use the resolvent identityFor fixed $X\in U$ and sufficiently small $H$, factor $I-X-H=(I-X)(I-f(X)H)$ and apply the Neumann series to the second factor. This proves that $X+H\in U$, that $f(X+H)$ is locally bounded, and that $f(X+H)-f(X)=O(\Vert H\Vert )$. Subtracting $f(X)Hf(X)$ in the identity leaves $O(\Vert H{\Vert }^2)$. Thus $Df(X)[H]=f(X)Hf(X)$ is the Frechet derivative.

Near zero, $f(X)=I+X+O(\Vert X{\Vert }^2)$, uniformly in the operator norm. Hence, as a linear operator in $H$,This proves differentiability of $Df$ at zero, givingThe order of multiplication matters: the second Frechet derivative is the symmetric bilinear map $HK+KH$, not $2HK$.