Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-7/1/solution

Consider the operators on all of or , extending the displayed positive-function formula linearly. Here is the space of continuous functions vanishing at infinity. The multiplier has modulus at most one, so , and multiplication of exponentials gives and . For strong continuity on , choose a compact outside which is small. On the continuous is bounded, making uniform; outside , . For , use pointwise convergence and the dominated convergence theorem with bound . Thus both are contraction semigroups, even when is unbounded.
The generator of a multiplication semigroup is
Indeed pointwise difference quotients converge to . A norm limit therefore forces the displayed domain condition, using an almost-everywhere convergent subsequence for the case. Conversely , so the dominated convergence theorem proves the limit when . In write the error as times , with the value at supplied by continuity. That second factor tends uniformly to zero where is bounded and has modulus at most one; the compact-set and small-tail argument applied to proves uniform convergence. To prove directly that is a closed operator, take and . Pointwise limits in , or a common almost-everywhere convergent subsequence in , give . Thus is in the appropriate generator domain and . In the domain is dense because ; in compactly supported continuous functions are a dense subspace contained in the domain.
A real multiplication operator on its maximal domain is an unbounded self-adjoint operator. One direct verification is to test the adjoint relation against functions supported on . If is in the adjoint operator domain with representing vector , those tests force on each such set, hence . The converse follows by integration. On the sigma-finite measure spaces used below, the spectrum of a real multiplication operator is its essential range; spectral values are detected by unit vectors supported where the multiplier is arbitrarily close to that value, while outside the essential range its reciprocal is a bounded resolvent multiplier.
For the heat semigroup on , take the normalization . The unitary Fourier transform converts it to multiplication by , and converts the semigroup to multiplication by . Consequently
The Sobolev space domain is exactly . The continuous spectrum comes from the full essential range of , not from square-integrable Fourier eigenvectors.
For the Ornstein-Uhlenbeck semigroup on standard Gaussian measure, the normalized Probabilists' Hermite polynomials form an orthonormal basis. The construction below gives and , so the Hermite coefficient map turns it into a real multiplication operator on . Therefore
Finally, the heat semigroup on the whole real line has no positive spectral gap, whereas the standard Ornstein-Uhlenbeck semigroup has gap one. For the first claim, take a nonzero smooth compactly supported and set . Then , but . This disproves a uniform whole-line Poincare inequality ; one can also choose . Lebesgue measure here is infinite, so it has no normalized probability mean to subtract. For Gaussian measure, the Hermite expansion instead gives the sharp Gaussian Poincaré inequality
Equality holds for affine functions. The identity on the right is interpreted on the energy-form domain, which is larger than the full operator domain.

New to topics? Read the docs here!