Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 10 / 2 / a / Solution

Write $X=I+E$. Use a submultiplicative Sobolev algebra norm for which the Fourier projections are contractive, and its compatible matrix/column norms. The numeric hypothesis $\Vert E\Vert <1$ is understood in this algebra norm; the unscaled Fourier Sobolev norm need not have multiplication constant one.

Let $A_{>0}$ and $A_{<0}$ denote the strictly positive and strictly negative Fourier subspaces. The equation $y_{+}=-P_{>0}(E+E{y}_{+})$ is a contraction on $M_n(A_{>0})$, so it has a unique solution. Put $Y_{+}=I+y_{+}$ and $Z_{-}=X{Y}_{+}$. The equation says $Z_{-}\in M_n(A_{-})$, and $Y_{+}(0)=I$. Similarly the contraction $v_{-}=-P_{<0}(E+v_{-}E)$ gives $V_{-}=I+v_{-}$ with $Z_{+}=V_{-}X\in M_n(A_{+})$. No invertibility of these solutions has yet been assumed.

Now $V_{-}{Z}_{-}=Z_{+}{Y}_{+}$ belongs to both Fourier subalgebras, hence is a constant matrix $C$. Taking constant coefficients on either side shows $C=Z_{-}(0)=Z_{+}(0)$. We claim $C$ is invertible. If $Cv=0$, then the Hardy compression $T_X=P_{\ge 0}X$ annihilates $Y_{+}v$, because $P_{\ge 0}{Z}_{-}v=Cv=0$. But $T_X=I+P_{\ge 0}E$ on the Hardy Sobolev space has inverse given by a Neumann series, since the perturbation has norm less than one. Thus $Y_{+}v=0$, and its constant coefficient is $v$, forcing $v=0$. Finite-dimensionality proves the claim.

Consequently $C^{-1}{Z}_{+}$ is a left inverse of $Y_{+}$, and $C^{-1}{V}_{-}$ is a left inverse of $Z_{-}$. Matrix entries are functions in a commutative algebra, so pointwise finite-dimensional inversion makes these left inverses two-sided inverses in their respective Fourier subalgebras. SetThen $X=X_{-}{X}_{+}$, both factors are invertible in the stated algebras, and $X_{+}(0)=I$. If two normalized factorizations exist, ${\tilde{X}}_{-}^{-1}{X}_{-}={\tilde{X}}_{+}{X}_{+}^{-1}$ lies in $A_{-}\cap A_{+}$ and is constant; evaluating the analytic side at zero makes it $I$. This proves uniqueness. The argument is near-identity Sobolev loop factorization and, crucially, proves invertibility rather than treating the contraction solution itself as automatically invertible.