## Appendix A: completeness ofIn this appendix we will verify explicitly that the set of
functions defined in Eq. (15) is complete, in the sense that Eqs. (3)
and (4) can be recovered. For the purpose, we will apply the Fourier
theorem (Brezis 1987). In the following, we will assume that
is a Let us consider a solution of the form (6). Then, because of the orthonormality condition (13), we have , so that Now we observe that the previous equation holds for any : then it holds also for any linear combination of . Thus In the last step we have integrated by parts ( is the unit vector orthogonal to the boundary of ). We now use this equation to show that the chosen set of functions,
described by Eq. (15), is For the purpose, we observe that if cosines are used as set of
functions, we can "build" any function In a similar manner, now that we have "disposed of" the first term,
we observe that using cosines we can build a function One might worry that, on the boundary, the chosen set of functions
of Eq. (15) has zero derivative in the direction of
, i.e.
. This might suggest that the
boundary condition (4) cannot be reproduced. In reality, although this
is true Suppose that we measure a constant field on a square field of side . The obvious solution for in this case is . Suppose now that we try to use a set of functions made of sines. Then the corresponding coefficients would be proportional to the integrals . Hence, every coefficient vanishes. This proves that a set based on sines is not complete (condition (12) is not satisfied or, equivalently, a curl-free vector field leads to a vanishing mass map). On the other hand, if we use the set (15), the coefficients do not vanish and the corresponding mass distribution is given by This function can be shown to reduce to . © European Southern Observatory (ESO) 1999 Online publication: July 16, 1999 |