The locally integrable function has distributional derivative equal to the principal-value reciprocal distribution. Symmetric integration by parts makes the boundary term tend to zero.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 327 2 i Solution Created 2026-10-03 Updated 2026-10-05
The space of test functions is . A sequence converges to when all functions eventually have compact support in one compact set and every derivative converges uniformly:A distribution is a continuous linear functional on this space of test functions. Equivalently, for each compact set , there are and a finite integer such thatThe integer is permitted to depend on . The usual weak convergence of distributions isThus the convergence convention on the distribution space is its weak dual topology.
For the principal-value reciprocal distribution, the symmetric truncations can be writtenThe numerator is near zero by the mean value theorem, and the integrand vanishes for large because has compact support. For ,Consequently the Cauchy principal value defines a distribution of order of a distribution at most one.
The function has local integrability, since , and therefore defines a distribution. For its distributional derivative, remove and use integration by parts on both remaining intervals:The boundary term is and tends to zero. The omitted integral of tends to zero by local integrability. Hence the distributional derivative of the logarithmic modulus satisfiesSymmetric truncation is essential to this normalization of the principal-value reciprocal distribution.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 327 2 b Solution Created 2026-10-03 Updated 2026-10-05
The principal-value reciprocal distribution isThe limit exists because the constant part of the numerator cancels symmetrically at zero. Multiplying by gives .
If is any other solution, obeys . To identify this kernel of multiplication by a coordinate, choose a cutoff function equal to one near zero. Every test function has the form with . Thus , proving . ThereforeFor real distributions the constants are real.