Timeline for Existence of unique convolution semigroups of probability measures on more general spaces then $\mathbb R^d$
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| Nov 19, 2020 at 19:39 | comment | added | Mateusz Kwaśnicki | That question is another example ot the XY problem: it should ask for a reference or a proof of the formula for the generator of the Banach space valued Lévy process, but instead it asks for an alternative proof for real-valued ones. | |
| Nov 19, 2020 at 17:59 | comment | added | 0xbadf00d | I already asked a separate question: mathoverflow.net/q/376585/91890. At least the "standard proof" given by Schilling doesn't carry over to the Banach space case. | |
| Nov 19, 2020 at 9:01 | comment | added | Mateusz Kwaśnicki | In this case I suggest that you ask a new question with a clear explanation of what you need, rather than pose a dozen of closely related, but different questions. I do not see the result you need in Linde's book. I bet the standard proof carries over with minor modifications (all necessary information on Lévy measures etc. seems to be given in the book), but I have no time now to see if everything really works as expected. | |
| Nov 17, 2020 at 11:35 | comment | added | 0xbadf00d | I want to prove that if $(X_t)_{t\ge0}$ is a Lévy process taking values in a Banach space $E$ and $f:E\to\mathbb R$ is such that $f,f',f''$ exist and are bounded and uniformly continuous, then $f$ is in the domain of the generator $A$ of the transition semigroup of $(X_t)_{t\ge0}$ (which is a strongly continuous semigroup on the space of bounded uniformly continuous functions on $E\to\mathbb R$) and derive an expression for $Af$. This is done, for example, in the reference of Schilling. And he (as many other references I've seen so far) is using the Fourier inversion theorem on $\mathbb R^d$. | |
| Nov 17, 2020 at 11:27 | comment | added | Mateusz Kwaśnicki | @0xbadf00d: I fail to understand what are you trying to prove. An expression for the generator in what context? Banach-space-valued Lévy processes? What class of functions? What for? This looks like a typical example of the XY problem. | |
| Nov 17, 2020 at 11:23 | comment | added | 0xbadf00d | On the other hand, $\operatorname E_x[f(X_t)]=(\kappa_tf)(x)$. But now the proof in the reference of Schilling uses the Fourier inversion formula on $\mathbb R^d$ in order to rewrite $f(x)$. I'm not sure if we can generalize this. Maybe Bochner's theorem is of use? | |
| Nov 17, 2020 at 11:23 | comment | added | 0xbadf00d | I'm not sure how the Fourier transform approach generalizes to spaces other than $\mathbb R^d$. The basic idea is clear to me. If $\mu_t=\mathcal L(X_t)$ and $\varphi_{\mu_1}=e^\psi$, we easily see that the law of $X_t$ under the probability measure $\operatorname P_x$ for which $\operatorname P_x\circ X_0=\delta_x$, is given by $\operatorname P_x\circ X_t^{-1}=\tau_x(\mu_t)$, where $\tau_x(y):=y+x$ for $x,y\in E$. With this result we see that $\varphi_{\mu_t}(x')=e^{{\rm i}\langle x',\:x\rangle}\varphi_{\operatorname P_x\circ X_t^{-1}}=\varphi_{\mu_t}(x')$. | |
| Nov 17, 2020 at 11:23 | comment | added | 0xbadf00d | It's not that I insist on an "alternative proof" without a good reason. The reference you gave in your answer, doesn't provide a result for the generator of a Lévy process. All references I've found either use the decomposition which I've described in this answer (see, for example, Theorem 5.4 in Peszat/Zabczyk) or they are considering the Fourier transform (see, for example, math.tu-dresden.de/sto/schilling/sources/papers/…). | |
| Nov 17, 2020 at 9:09 | comment | added | Mateusz Kwaśnicki | I do not know what Corollary 15.20 is all about, but I guess one has weak convergence as an assumption. And of course if $\mu_t^{\star 1/t} = \mu_1$, then also $\mu_t^{\star 1/t} \to \mu_1$ weakly. | |
| Nov 17, 2020 at 4:58 | comment | added | 0xbadf00d | Could you say something about the issue in Kallenberg? | |
| Nov 16, 2020 at 20:35 | comment | added | Mateusz Kwaśnicki | @0xbadf00d: I always had troubles understanding people that insist on "alternative proofs", so I guess I am not the right person to ask for advice. | |
| Nov 16, 2020 at 11:58 | comment | added | 0xbadf00d | I've asked for that here: mathoverflow.net/q/376585/91890. | |
| Nov 16, 2020 at 11:58 | comment | added | 0xbadf00d | I'm asking all these questions, cause I'm searching for an alternative proof for the characterization of the generator. They either use Fourier transforms on $\mathbb R^d$ or consider the decomposition of the process, which I've described in my answer to the other question. I really would like to know whether there is a more semigroup-theoretic proof or (maybe) a proof in terms of martingale problems. Maybe one can use that a process $M:=X-\int_0^{\;\cdot}Y_s\:{\rm d}s$ is a martingale iff $N_t:=e^{-\lambda t}X_t+\int_0^te^{-\lambda s}(\lambda X_s-Y_s)\:{\rm d}s$ is a martingale. | |
| Nov 16, 2020 at 5:40 | comment | added | 0xbadf00d | Thank you. This looks exactly like the reference, I was looking for. Meanwhile, I've found a claim in the book of Kallenberg, which confused me: i.sstatic.net/6Ykkz.png. If $μ_t$ is the distribution of $X_t$ and $X$ is a Lévy process, then $(μ_t)_{t\ge0}$ is a continuous convolution semigroup. The characteristic function $φ_{μ_t}$ of $μ_t$ is equal to $e^{tψ}$, where $ψ$ is the characteristic exponent of $μ_1$. So shouldn't we have $μ_t^{*1/t}=μ_1$ (since $φ_{μ_t^{\ast1/t}}=e^f=φ_{μ_1}$)? So, it's weird that he's claiming that $μ_t^{*1/t}$ weakly converges to $μ_1$ as $t\to0$. | |
| Nov 16, 2020 at 5:33 | vote | accept | 0xbadf00d | ||
| Nov 15, 2020 at 21:27 | history | answered | Mateusz Kwaśnicki | CC BY-SA 4.0 |
