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Let $L$ be the Ornstein-Uhlenbeck operator on $L^2(\gamma)$ where $\gamma$ is the Gaussian measure on $\mathbf R^d$. Hille-Yosida or Lumer-Phillips can be used to prove that $L$ generates a strongly continuous semigroup $e^{tA}$.

Now I have the following integral:

$$\int_0^\infty (t^2 L)^{N + 1} e^{\beta t^2 L} u \, \frac{\text{d}t}{t},$$

where $u \in L^2$, $\beta > 0$.

This doesn't look like a normal Bochner integral. I guess I need some kind of functional calculus. I would like to know how I could have this integral make sense.

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Well, I googled "Ornstein UhlENbeck functional calculus" and one of the first hits was this paper (among dozens of others). I have no clue if it's of any use to you. –  t.b. Jul 18 '11 at 21:00
    
@Theo: Scrolling through the paper diagonally I don't think it is of any use, but I'll study it a bit. Thanks. –  Jonas Teuwen Jul 18 '11 at 21:03
    
As I said, there are dozens of others, so maybe the big G will help –  t.b. Jul 18 '11 at 21:06
    
I have figured out how I can make sense of this thing, but I didn't use a functional calculus (first I did, but that was overkill). I'm not sure if it is of any use to someone so maybe I'd better delete the question? –  Jonas Teuwen Jul 19 '11 at 14:37
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1 Answer

up vote 2 down vote accepted

I have a "solution". This is a limit of Bochner integrals. I wanted to do a density argument with this operator so I had to prove that it is continuous.

First of all, note that we can scale out the squares and the $\beta$. Now we which to estimate the integral

$$\int_0^\infty (t L)^{N + 1} e^{t L} u \, \frac{\text{d}t}{t}$$ in $L^2(\gamma)$.

So, this is a Hilbert space (Luckily!) so we argue as follows

$$\begin{align} \left \|\int_0^\infty (t L)^{N + 1} e^{t L} u \, \frac{\text{d}t}{t}\right \|^2 &= \left<\int_0^\infty (t L)^{N + 1} e^{t L} u \, \frac{\text{d}t}{t}, \int_0^\infty (s L)^{N + 1} e^{s L} u \, \frac{\text{d}s}{s}\right>\\ &= \left < \int_{\mathbf R^2} (st L)^{N + 1} e^{(s + t)L}u \frac{\text{d}s}{s} \frac{\text{d}t}{t}, u \right>\\ &= (N!)^2 \left <u, \right >\\\ &= (N!)^2 \|u\|^2\end{align}$$

Which is what I wanted.

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