Concrete semigroups and their infinitesimal generators

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If f∈H2f \in H^{2} (see Exercise 9) and f(z)=∑cnznf(z) = \sum c_{n} z^{n}, define

[Q(t)f](z)=∑n=0∞(n+1)−tcnzn(0≤t<∞) [Q(t) f](z) = \sum _{n = 0}^{\infty }(n+1)^{-t} c_{n} z^{n} \quad (0 \leq t < \infty )

Show that each Q(t)Q(t) is self-adjoint (and positive). Find the infinitesimal generator AA of the semigroup {Q(t)}\left\{ Q(t)\right\}. Is AA self-adjoint? Show that AA has pure point spectrum, at the points log⁡1,log⁡12,log⁡13,…\log 1, \log \frac{1}{2}, \log \frac{1}{3}, \ldots.

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For f∈L2(R),x∈R,0<y<∞f \in L^{2}(R), x \in R, 0 < y < \infty, define

[Q(y)f](x)=1π∫−∞∞y(x−ξ)2+y2f(ξ)dξ [Q(y) f](x) = \frac{1}{\pi } \int _{-\infty }^{\infty } \frac{y}{(x-\xi )^{2}+y^{2}} f(\xi ) d \xi

and put Q(0)f=fQ(0) f = f. Show that {Q(y):0≤y<∞}\left\{ Q(y): 0 \leq y < \infty \right\} satisfies the conditions imposed in Definition 13.34 and that ∥Q(y)∥=1\left\| Q(y)\right\| = 1 for all yy.

Find the domain of the infinitesimal generator AA of {Q(y)}\left\{ Q(y)\right\}, and prove that

Af=−Hf′ A f = -H f^{\prime }

where HH is the Hilbert transform (Chapter 7, Exercise 24). Prove that −A-A is positive and self-adjoint.

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Remark.
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The integral represents a harmonic function in the upper half-plane, with boundary values ff. The semigroup property of {Q(y)}\left\{ Q(y)\right\} can be deduced from this, as well as from a look at the Fourier transforms of the functions Q(y)fQ(y) f.

Let (Pt)t⩾0\left(P_{t}\right)_{t \geqslant 0} be the transition semigroup of a BMd\mathrm{BM}^{d} and denote by Lp,1⩽p<∞L^{p}, 1 \leqslant p < \infty, the space of pp th power integrable functions with respect to Lebesgue measure on Rd\mathbb {R}^{d}. Set un:=(−n)∨u∧nu_{n}: = ( - n) \vee u \wedge n for every u∈Lpu \in L^{p}. Verify that

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(a)

un∈Bb(Rd)∩Lp(Rd)u_{n} \in \mathcal{B}_{b}\left(\mathbb {R}^{d}\right) \cap L^{p}\left(\mathbb {R}^{d}\right) and Lp−lim⁡n→∞un=uL^{p} - \lim_{n \rightarrow \infty } u_{n} = u.

(b)

P~tu:=lim⁡n→∞Ptun\widetilde{P}_{t} u: = \lim_{n \rightarrow \infty } P_{t} u_{n} extends PtP_{t} and gives a contraction semigroup on LpL^{p}.

(c)

P~t\widetilde{P}_{t} is sub - Markovian, i.e. if u∈Lpu \in L^{p} and 0⩽u⩽10 \leqslant u \leqslant 1 a.e., then 0⩽P~tu⩽10 \leqslant \widetilde{P}_{t} u \leqslant 1 a. e.

(d)

P~t\widetilde{P}_{t} is strongly continuous, i.e. lim⁡t→0∥P~tu−u∥Lp=0\lim_{t \rightarrow 0}\left\| \widetilde{P}_{t} u - u\right\|_{L^{p}} = 0 for all u∈Lpu \in L^{p}.

Let A,B∈Rd×dA, B \in \mathbb {R}^{d \times d} and set Pt:=exp⁡(tA):=∑j=0∞(tA)j/jP_{t}: = \exp (t A): = \sum_{j = 0}^{\infty }(t A)^{j} / j !.

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(a)

Show that PtP_{t} is a strongly continuous semigroup. Is it contractive?

(b)

Show that ddtetA\frac{d}{d t} e^{t A} exists and that ddtetA=AetA=etAA\frac{d}{d t} e^{t A} = A e^{t A} = e^{t A} A.

(c)

Show that etAetB=etBetA∀t⟺AB=BAe^{t A} e^{t B} = e^{t B} e^{t A} \forall t \Longleftrightarrow A B = B A.

(d)

(Trotter product formula) Show that eA+B=lim⁡k→∞(eA/keB/k)ke^{A+B} = \lim_{k \rightarrow \infty }\left(e^{A / k} e^{B / k}\right)^{k}.

References

[Rud91] Rudin W. Functional Analysis. 2nd ed., McGraw-Hill, 1991.
[Sch21] Schilling R.L. Brownian Motion: A Guide to Random Processes and Stochastic Calculus. 3rd ed., De Gruyter, 2021.