What's the $L^1$ analogue of Stone's theorem saying that any strongly continuous 1-parameter unitary groups has a unique self-adjoint generator?

More precisely: let $X$ be a measure space ($\sigma$-finite, if you like). Say a linear operator

$$U : L^1(X) \to L^1(X)$$

is stochastic if

$$\int U \psi \; dx = \int \psi \; dx $$


$$\psi \ge 0 \quad \Rightarrow \quad U \psi \ge 0$$

for all $\psi \in L^1(X)$. (So, it sends probability distributions to probability distributions.)

Suppose we have 1-parameter family of stochastic operators

$$ U(t) : L^1(X) \to L^1(X) \qquad \mathrm{for} \;\; t \ge 0 $$


$$ U(0) = I $$

$$ U(t) U(s) = U(t+s) $$

and strong continuity:

$$ t_i \to t \quad \Rightarrow \quad U(t_i) \psi \to U(t)\psi $$

for any $\psi \in L^1(X)$. I would like to say it is of the form

$$ U(t) = \exp(t H)$$

for a unique infinitesimal stochastic operator $H$. And I would like a nice characterization of these operators! They should be some sort of densely-defined operators on $L^1(X)$.

If $X$ is a finite set with counting measure, I think the theorem is true: $H$ will be a square matrix, and I believe such a matrix counts as 'infinitesimal stochastic' if 1) the sum of the matrix entries in each column is zero and 2) the off-diagonal entries are nonnegative.

So, I want to see the generalization of this result to more exciting measure spaces $X$. I imagine somebody already knows this.

For more details, see my blog entry here:


  • 2
    $\begingroup$ I think that if one applies the general machinery of C_0-semigroups en.wikipedia.org/wiki/C0-semigroup then you get $U(t)=\exp(tH)$ where $H$ is densely defined, and we should get $\int H\psi = 0$ whenever $\psi\in {\mathcal D}(H)$. But I also suspect that people who know about Markov processes will be able to give a better answer $\endgroup$
    – Yemon Choi
    Commented Apr 11, 2011 at 8:36
  • 3
    $\begingroup$ Surely I am missing something but isn't this called Hille-Yosida theorem? See math.brown.edu/~schhita/Semigroups.pdf. Note that even in the finite dimensional case, not every square matrix such that your conditions (1) and (2) hold is a generator. See math.stackexchange.com/questions/31174/… $\endgroup$
    – Did
    Commented Apr 11, 2011 at 10:17
  • $\begingroup$ The Hille-Yosida theorem is surely relevant, but that's a general theorem about continuous one-parameter semigroups on Banach spaces; I'm trying to characterize a particular class of such semigroups. It sounds like over at stackexchange they're saying that even in the finite-dimensional case not every stochastic operator $U$ is of the form $\exp(tH)$. That's unsurprising. I don't see them exhibiting square matrices $H$ obeying conditions 1) and 2) such that $U = \exp(tH)$ fails to be stochastic. That would shock and interest me. $\endgroup$
    – John Baez
    Commented Apr 15, 2011 at 2:15
  • $\begingroup$ @John Baez: To be clear, some stochastic matrices $U$ cannot be written as $U=\exp(H)$, for any $H$ such that (1)+(2) holds, and some stochastic matrices $U$ can be written as $U=\exp(H)$ where $H$ is such that (1)+(2) does not hold. (For your interest, I discovered your comment by chance, so you might want to use this @ thing at the beginning of your comments to signal them to your interlocutor.) $\endgroup$
    – Did
    Commented Apr 20, 2011 at 15:22

3 Answers 3


Here are a few comments: the answer to the question as stated is indeed rather in the domain of the general one-parameter semigroup theory, and the characterisation of the generators you are asking about can be found for example in the article "Positive one-parameter semigroups on ordered Banach spaces" by Charles Batty and Derek Robinson (Acta Applicandae Mathematicae, Volume 2, 1984, Numbers 3-4, 221-296). The characterisation is neccessarily somewhat involved - there is no way round the analytic conditions of the Hille-Phillips type which tell you when a given unbounded operator generates a $C_0$-semigroup.

In most cases of interest to probabilists it is however natural to assume in addition that each operator $U(t)$ is $L^2$-selfadjoint, i.e. satisfies the condition

$ \int f$ $ U(t) g = \int U(t) f $ $ g $,

for square integrable functions $f,g$, or at least its approximate version (to be explained below). Then the problem can be transferred to the Hilbert space $L^2(X)$, and one obtains two new powerful tools:

  • theory of quadratic forms (or more specifically Dirichlet forms) making it possible to study unbounded generators in an easier framework;

  • the interpolation, which means that in a sense one deals at the same time with the semigroup on all $L^p$-spaces.

A very good modern treatment of this can be found in the first two chapters of "Analysis of Heat Equations on Domains" by El Maati Ouhabaz. Ouhabaz describes in fact the more general construction, when the operators $U(t)$ are not selfadjoint, but the `non-symmetric' part of the whole semigroup is controlled by the symmetric one. On the level of the $L^2$-stochastic generators this translates into the generator being a relatively bounded perturbation of an (unbounded) selfadjoint operator.

  • $\begingroup$ That is an interesting point what you write. Why is it usual to assume $L^2$-selfadjointness? I am not familiar whith this, but in the matrix case doesn't this mean that you have symmetric matrices? $\endgroup$ Commented Apr 11, 2011 at 10:44
  • $\begingroup$ Indeed, it does - if the measure on the finite set you consider is the uniform one. Symmetry assumption is sometimes viewed as a version of symmetry under the time reversal; this however is not a universal point of view. In the infinite-dimensional framework the symmetry (or at least some aspect of it, as described in the main text) makes it possible to work with semigroups on all $L^p$-spaces, which seems to be a very natural requirement. $\endgroup$ Commented Apr 11, 2011 at 10:59
  • $\begingroup$ Thanks very much, Adam. I'd really like to avoid the symmetry assumption you mention, because the stochastic processes I'm interested in don't usually satisfy it. Over on my blog, Martin Gisser recommended Introduction to the Theory of (Non-Symmetric) Dirichlet Forms by Z. M. Ma and M. Röckner. I haven't looked at it yet, but the title hints that the usual theory of Dirichlet forms is part of a bigger theory that drops the symmetry assumption. $\endgroup$
    – John Baez
    Commented Apr 15, 2011 at 13:06
  • $\begingroup$ I have only now noticed John's response - hence the delay. The book of Ma and Rockner deals precisely with the generalization of the symmetric case I described in the last paragraph of my answer above - the standing assumption throughout the book is that the non-symmetric part (of the semigroup/form/generator) is dominated by the symmetric one. $\endgroup$ Commented Jun 18, 2011 at 14:18

Expanding the remark of Yemon and Didier, what you need is the theory of stochastic $C_0$-semigroups, which is a rich theory. You may find an exhastive discussion of these topics in Nagel (ed.): One-parameter semigroups of positive operators, Springer, 1986, or a nice introduction in

Bobrowski: Functional analysis for Probobality and Stochastic Processes, Cambridge UP, 2005, Chapter 8.

You shoud be able to find all what you need in this last book.


The most important thing is ψ≥0 ⇒ U(t)ψ≥0. (The stochastician in me would be content with a substochastic semigroup (Markov processes might die), and the Hilbert spaced part of me would add: well, that's simply the condition of the generator being negative definite.)

It seems the definite treatment of ψ≥0 ⇒ U(t)ψ≥0 is theorem 1.6 in

Wolfgang Arendt, Kato's Inequality: A Characterisation of Generators of Positive Semigroups, Proc. R. Ir. Acad. Vol. 84A No. 2 (1984), 155-174

   ~ . ~ . ~

( Arendt also has the semigroup domination theorem 4.3 plus important ramifications. Until yesterday I called it the Kato-Simon-Shigekawa criterion. But no, according to Arendt he learned it from Kato himself. I've studied a bit of this stuff ca. 1995 (told Shigekawa about Simon's proof, and still (2011) have the most elegant verification of Kato's inequalities for general manifolds...) - but never did I come across Arendt's paper! (Well, these studies were laying so dormant in my ol brains that John's question of April 2011 couldn't shake them fully awake. It took a second hit from somewhere else.) Mestupid did quite some digging this week (plus, 3 failed attempts of proof of Arendt's theorem (in $L^1$) plus a total recall of all my higher analysis from gone times) to finally hit the paper... )

   ~ . ~ . ~

This comment dedicated in memoriam Johann Schneidermeier


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