Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 12G (Analysis II)

A norm on a real or complex vector space $V$ is a function $\Vert \cdot \Vert :V\to [0,\infty )$ with $\Vert v\Vert =0$ exactly when $v=0$, $\Vert \lambda v\Vert =|\lambda |\Vert v\Vert$, and the triangle inequality $\Vert v+w\Vert \le \Vert v\Vert +\Vert w\Vert$. These are positivity, absolute homogeneity and the triangle inequality.

For a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$, define the operator norm byIt is finite. To see this without assuming continuity of $A$, write $x=\sum x_j{e}_j$: the triangle inequality bounds $\Vert Ax\Vert$ by $\sum |x_j|\Vert A{e}_j\Vert$. Every norm in finite dimensions is equivalent to the Euclidean norm. One proof first uses this same coordinate bound to establish continuity of the norm on the Euclidean unit sphere; compactness then gives a strictly positive minimum there. Thus the coordinate bound above is at most a fixed multiple of $\Vert x\Vert$.

Absolute homogeneity and the triangle inequality for this operator norm follow from those for $\Vert Ax\Vert$ and taking suprema. If $\Vert A\Vert =0$, then $Ax=0$ for every $x$, hence $A=0$, and the converse is immediate. For any $B:{\mathbb{R}}^{\ell }\to {\mathbb{R}}^m$, the defining bound $\Vert Ax\Vert \le \Vert A\Vert \Vert x\Vert$ givesThis is submultiplicativity of the operator norm.

The function $X\mapsto \det (I-X)$ is a polynomial in the matrix entries, hence continuous. Consequently $U$ is the inverse image of the open set $\mathbb{R}\setminus \{0\}$ and is an open set.

If $\Vert X\Vert <1$ and $(I-X)v=0$, then $v=Xv$ implies $\Vert v\Vert \le \Vert X\Vert \Vert v\Vert$. This forces $v=0$. The square matrix $I-X$ is therefore injective, hence an invertible matrix, soThe argument works with any of the induced operator norms in part (b).

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 $2HK$.