Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-105/3/b/ii/solution
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 105 3 b ii Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-05
The assertion about all is false as printed in the original PDF. Take , , and . This is a nonzero smooth Dirichlet function, but and . In particularAlready at , is not defined on its stipulated domain, because . Extending to for this example would give zero, not an inner product. Additional boundary conditions are essential.
The corrected spaces are the Sobolev domains of powers of an elliptic Dirichlet operator. Let be the Dirichlet realization of an elliptic operator, and setIn particular , , and also requires . The formulas in the question define inner products on these spaces.
Here is the norm-equivalence proof on the corrected domains. The base cases are the inner product at and part (i) at . Standard Dirichlet elliptic regularity, together with invertibility from the Lax-Milgram theorem, gives for every integer The first estimate applies when ; invertibility absorbs the usual lower-order term. Moreover, is an isomorphism. To see surjectivity, solve with zero Dirichlet trace for ; regularity gives , and for each further required trace. The formulas satisfyInduction therefore givesInjectivity of each iterate of gives positive definiteness. These spaces are ; the spectral characterization of elliptic Dirichlet domains in the next part makes the half-powers precise. On the paper's uncorrected spaces the claim is valid at , but fails in general beyond them.
New to topics? Read the docs here!