Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 10 2 a Solution Created 2026-10-03 Updated 2026-10-14
Write . Use a submultiplicative Sobolev algebra norm for which the Fourier projections are contractive, and its compatible matrix/column norms. The numeric hypothesis is understood in this algebra norm; the unscaled Fourier Sobolev norm need not have multiplication constant one.
Let and denote the strictly positive and strictly negative Fourier subspaces. The equation is a contraction on , so it has a unique solution. Put and . The equation says , and . Similarly the contraction gives with . No invertibility of these solutions has yet been assumed.
Now belongs to both Fourier subalgebras, hence is a constant matrix . Taking constant coefficients on either side shows . We claim is invertible. If , then the Hardy compression annihilates , because . But on the Hardy Sobolev space has inverse given by a Neumann series, since the perturbation has norm less than one. Thus , and its constant coefficient is , forcing . Finite-dimensionality proves the claim.
Consequently is a left inverse of , and is a left inverse of . 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 , both factors are invertible in the stated algebras, and . If two normalized factorizations exist, lies in and is constant; evaluating the analytic side at zero makes it . 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.