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I have a question about the following definition:

A probability measure $\mu$, such that the Markov semigroup $e^{Lt} \in \mathcal{L}(L^2)$ exists and is symmetric, satisfies the Sobolev inequality iff for some $p \in (2, \infty)$ and two finite constants $(a,b) \in [0,\infty)^2$, we have

$$\|f\|_p^2 \le a \|\Gamma_1(f,f)\|_1 + b \|f\|_2^2$$ for any measurable function $f$ such that the right-hand side exists ($\Gamma_1$ is the carré du champ) and, in case that $b=0$ so that $\int f d\mu=0$. The carré du champ is basically defined as $||\Gamma_1(f,f)||_1 = - \langle f,Lf \rangle$ where $L$ is the negative generator of the semigroup, as written above.

I took this definition from the following paper [1, Definition 3.1] and I was wondering whether $a,b$ can depend on $p$ (maybe you have a guess, my intuition says yes.) and where this strange condition $b=0$ then $\int f d\mu=0$ comes from?

The reason why I am asking is that I could nowhere else find (exactly) this equation and I thought that this equation might be too localized for MSE.

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    $\begingroup$ 1) The link is broken so I could not access the paper. 2) Looks to me that you are talking about Meyer's inequality in Malliavin calculus. The book of Malliavin or the book of Bogachev on Gaussian measures. $\endgroup$ Commented Oct 16, 2015 at 22:07
  • $\begingroup$ @LiviuNicolaescu link is fixed now, I will have a look at the book although I could not find this equation under the name you gave me here using google $\endgroup$ Commented Oct 16, 2015 at 22:14
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    $\begingroup$ See the references at this site math.bu.edu/people/bourguin/malliavin_calculus_F2015/… $\endgroup$ Commented Oct 17, 2015 at 0:05
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    $\begingroup$ If $b=0$ then this strange condition $\int f d\mu =0 $ is sufficient to avoid contradictions. Otherwise test the inequality on the functions $f=const$ then the right hand side is zero and the left hand side is not. For your first question: Given semigroup and given $p$ if you find some $a$ and $b$ such that inequality holds then (I believe) it is possible that for other $p$ you can find again some other $a$ and $b$ so that inequality holds but this new choice of $a$ and $b$ will now depend on the choice of $p$. $\endgroup$ Commented Oct 17, 2015 at 0:15

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You can find many details and references on this inequality (and related ones) in Chapter 6 of the recent book Analysis and Geometry of Markov diffusion operators by Bakry, Gentil and Ledoux. The constants $a$ and $b$ do indeed depend on $p$. In this book the inequalities are stated without the recentering, so that if $\mu$ is a probability measure, then $b$ is always larger than $1$. When $b=1$ the inequality is said to be tight. In that case it implies a Poincaré inequality with an explicit constant $C_P(a,p)$ (see proposition 6.2.2 in the book), so for centered functions one can write $$ \|f\|_p^2 \leq (a+C_P) \| \Gamma(f,f)\|_1 $$ and recover the form you mentioned ($b=0$ and centered functions).

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