Infinitesimal Generators and Test-Function Domains

From Transitions to an Infinitesimal Generator

A transition semigroup records how a Markov process moves over a positive time interval. Its infinitesimal generator asks a different question: how quickly does the expected value of a test function change when that interval is very short? The limiting operation depends on the function being tested and on the meaning of convergence. A smooth test-function class is often convenient for calculations, but it is not automatically the entire domain of the generator.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Let $X_t$ be a time-homogeneous Markov process with state space $\mathbb{R}$, and write $P_t(x,A)$ for its transition probability from state $x$ into a Borel set $A$ after time $t$. For a measurable function $f$, define $P_t f(x)$ by integrating $f$ against this transition probability whenever the integral of $|f|$ is finite. Equivalently, $P_t f(x)$ is the expectation of $f(X_t)$ when the process starts at $x$.

Definition: The pointwise generator domain $D_{\mathrm{pt}}(A)$ consists of measurable functions $f$ for which $P_t|f|(x)$ is finite for every $x$ and all sufficiently small $t>0$, and for which the finite limit $$ Af(x)=\lim_{t\downarrow0}\frac{P_t f(x)-f(x)}{t} $$ exists for every $x$. The function $Af$ is the pointwise infinitesimal generator of $f$.

The requirement is pointwise: for each fixed state $x$, the quotient must have a finite limit. It does not require that convergence be uniform over all states. The domain depends on the transition semigroup and on the class of functions for which expectations are defined.

There is also a norm-based notion used for semigroups on a function space. For example, on $C_0(\mathbb{R})$, the space of continuous functions vanishing at infinity with norm $\|f\|_\infty=\sup_x|f(x)|$, the strong generator domain consists of those $f\in C_0(\mathbb{R})$ for which $(P_t f-f)/t$ converges in this norm to an element of $C_0(\mathbb{R})$. Norm convergence implies pointwise convergence, but pointwise convergence alone does not imply norm convergence. Nor can a function outside $C_0(\mathbb{R})$ belong to this particular strong domain. These are different domain questions, not conflicting definitions.

A Smooth Test-Function Class for Brownian Motion

For Brownian motion started at $x$, the transition is $x+B_t$, where $B_t$ is independent of the starting state and has distribution $N(0,t)$. Thus $P_t f(x)=\mathbb{E}[f(x+B_t)]$. A common test-function class is $C_c^2(\mathbb{R})$: the functions with two continuous derivatives and compact support. The following result shows why this class is convenient.

Theorem: If $f\in C_c^2(\mathbb{R})$ and $P_t f(x)=\mathbb{E}[f(x+B_t)]$, then $f\in D_{\mathrm{pt}}(A)$ and $$ Af(x)=\frac12 f''(x)\qquad\text{for every }x\in\mathbb{R}. $$

Proof. Fix $x$, and let $Z$ have distribution $N(0,1)$, so $B_t$ has the same distribution as $\sqrt{t}Z$. Taylor's formula with an integral remainder gives $$ f(x+h)=f(x)+h f'(x)+\frac12h^2 f''(x)+h^2 r_x(h), $$ where $r_x(h)\to0$ as $h\to0$. Since $f''$ is bounded and uniformly continuous, the remainder can be bounded by a constant independent of $h$. Substitute $h=\sqrt{t}Z$, take expectations, and use $\mathbb{E}[Z]=0$ and $\mathbb{E}[Z^2]=1$: $$ \frac{P_t f(x)-f(x)}{t} =\frac12 f''(x)+\mathbb{E}\!\left[Z^2r_x(\sqrt{t}Z)\right]. $$ For each fixed $Z$, the remainder term tends to zero as $t\downarrow0$. Its absolute value inside the expectation is bounded by a constant times $Z^2$, which is integrable. The dominated convergence theorem therefore makes the expectation tend to zero. The displayed quotient converges to $\frac12 f''(x)$, as required. Compact support makes $f$ bounded, so the expectations used here are finite. $\square$

This proves a useful inclusion: $C_c^2(\mathbb{R})$ is a class of functions on which the Brownian pointwise generator can be calculated by differentiating twice. The theorem does not say that every function in the pointwise generator domain is in $C_c^2(\mathbb{R})$.

Worked Example: A Polynomial Outside the Test-Function Class

For Brownian motion started at $x$, compute the pointwise generator of $f(x)=x^2$, and compare its domain status with $C_c^2(\mathbb{R})$.

Since $B_t$ has mean zero and variance $t$, $$ P_t f(x)=\mathbb{E}[(x+B_t)^2] =x^2+2x\mathbb{E}[B_t]+\mathbb{E}[B_t^2] =x^2+t. $$ Thus $P_t|f|(x)$ is finite and $$ \frac{P_t f(x)-f(x)}{t} =\frac{x^2+t-x^2}{t}=1. $$ The limit exists at every $x$, so $Af(x)=1$. The calculation is checked directly by the Brownian transition's second moment: the extra $t$ is exactly its variance. Although $x^2$ is twice continuously differentiable, it is not compactly supported; it belongs to the pointwise domain without belonging to $C_c^2(\mathbb{R})$.

Worked Example: A Kink at the Starting State

Test whether $f(x)=|x|$ belongs to the pointwise Brownian generator domain at $x=0$. The transition expectation is finite because a normal random variable has finite first absolute moment. Since $B_t$ has distribution $\sqrt{t}Z$, $$ P_t f(0)=\mathbb{E}[|B_t|] =\sqrt{t}\,\mathbb{E}[|Z|]. $$ Using the standard normal density, $$ \mathbb{E}[|Z|] =\frac{2}{\sqrt{2\pi}}\int_0^\infty z e^{-z^2/2}\,dz =\sqrt{\frac{2}{\pi}}. $$ The integral equals $1$, since an antiderivative of $z e^{-z^2/2}$ is $-e^{-z^2/2}$. Consequently $$ \frac{P_t f(0)-f(0)}{t} =\frac{\sqrt{2/\pi}}{\sqrt{t}}\longrightarrow+\infty. $$ There is no finite generator value at zero, so $|x|\notin D_{\mathrm{pt}}(A)$. The divergence comes from increments of typical size $\sqrt{t}$, not $t$; dividing their expected absolute size by $t$ does not give a finite limit.

Check: at $t=1$, the quotient is $\sqrt{2/\pi}$, and at $t=1/4$, it is $2\sqrt{2/\pi}$. Halving the time multiplies the quotient by $1/\sqrt{1/2}$, consistent with its divergence as $t\downarrow0$.

Linearity and Dependence on the Domain

Proposition: The pointwise generator domain is a vector space, and $A$ is linear on it. In particular, if $f,g\in D_{\mathrm{pt}}(A)$ and $a,b\in\mathbb{R}$, then $af+bg\in D_{\mathrm{pt}}(A)$ and $A(af+bg)=aAf+bAg$.

Proof. Fix $x$. By the domain assumptions, $P_t|f|(x)$ and $P_t|g|(x)$ are finite for sufficiently small positive $t$. The inequality $|af+bg|\le |a||f|+|b||g|$ shows that $P_t|af+bg|(x)$ is finite as well. Linearity of expectation gives $$ \frac{P_t(af+bg)(x)-(af+bg)(x)}{t} =a\frac{P_t f(x)-f(x)}{t} +b\frac{P_t g(x)-g(x)}{t}. $$ Both limits on the right exist and are finite. Taking their linear combination proves existence of the limit on the left and gives the asserted formula at each $x$. This includes $a=0$, $b=0$, and the zero function. $\square$

Even with the same process, a domain changes when the function-space requirement changes. The function $x^2$ above has a finite pointwise generator but is not eligible for the strong generator on $C_0(\mathbb{R})$, because it does not belong to $C_0(\mathbb{R})$. Conversely, a pointwise limit that exists at every state need not converge uniformly over the state space. One should therefore name the domain being used rather than treating “the generator domain” as an unqualified synonym for $C_c^2$.

An Ornstein–Uhlenbeck Example

For parameters $\kappa>0$ and $\sigma>0$, the Ornstein–Uhlenbeck transition from state $x$ at time $t$ is Gaussian with mean $e^{-\kappa t}x$ and variance $\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa t})$. This transition law lets us calculate pointwise generator limits directly, without assuming that the test function is compactly supported.

Worked Example: Linear and Quadratic Ornstein–Uhlenbeck Tests

Compute the pointwise generators of $f_1(x)=x$ and $f_2(x)=x^2$. Both expectations are finite under the Gaussian transition. For $f_1$, $$ P_t f_1(x)=e^{-\kappa t}x, \qquad \frac{P_t f_1(x)-f_1(x)}{t} =x\frac{e^{-\kappa t}-1}{t}\longrightarrow-\kappa x, $$ using $\lim_{t\downarrow0}(e^{-\kappa t}-1)/t=-\kappa$. Hence $Af_1(x)=-\kappa x$.

For $f_2$, the second moment is variance plus squared mean: $$ P_t f_2(x)=e^{-2\kappa t}x^2+\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa t}). $$ Therefore $$ \frac{P_t f_2(x)-f_2(x)}{t} =x^2\frac{e^{-2\kappa t}-1}{t} +\frac{\sigma^2}{2\kappa}\frac{1-e^{-2\kappa t}}{t} \longrightarrow-2\kappa x^2+\sigma^2. $$ So $Af_2(x)=-2\kappa x^2+\sigma^2$. To check the coefficients, use $(e^{-2\kappa t}-1)/t\to-2\kappa$ and $(1-e^{-2\kappa t})/t\to2\kappa$; the variance term consequently tends to $\sigma^2$. The result for $f_2$ also agrees with the identity $P_t f_2=(P_t f_1)^2+\operatorname{Var}(X_t\mid X_0=x)$.

These examples illustrate why the domain is part of the generator's specification. The same transition rule can be tested on compactly supported smooth functions, on polynomials whose transition expectations are finite, or on a normed function space with a uniform convergence requirement. The resulting operator is meaningful only together with the chosen class of functions and the convergence notion.

Common Mistake: Replacing an Expectation by a Square

A tempting but incorrect calculation for Brownian motion is to write $P_t(x^2)=(\mathbb{E}[x+B_t])^2=x^2$. This replaces the expectation of a square by the square of an expectation and discards the variance. The correct identity is $$ \mathbb{E}[(x+B_t)^2] =\big(\mathbb{E}[x+B_t]\big)^2+\operatorname{Var}(B_t) =x^2+t. $$ The incorrect calculation would give generator zero, while the correct transition calculation gives $A(x^2)=1$. In stochastic calculus this discrepancy is related to Brownian quadratic variation, but the generator limit here follows directly from the transition expectation.

Takeaway: Define the generator by its short-time transition limit, and state the domain and convergence mode. For Brownian motion, $C_c^2(\mathbb{R})$ is a useful class on which $Af=\tfrac12f''$, but the pointwise domain can include noncompactly supported functions and exclude nonsmooth ones.

Check Your Understanding

For each question, specify whether it concerns a pointwise limit or a norm limit.

For Brownian motion started at $x$, compute $P_t f(x)$ and $Af(x)$ when $f(x)=x$.
Why does $f(x)=x^2$ belong to the pointwise Brownian generator domain but not to $C_c^2(\mathbb{R})$?
At which state does the pointwise generator limit for $f(x)=|x|$ fail, and what is the order of its quotient as $t\downarrow0$?
For the Ornstein–Uhlenbeck transition, use its mean and variance to check the generator value for $f(x)=x^2$.
Explain why pointwise convergence of $(P_t f-f)/t$ does not, by itself, establish convergence in the supremum norm.