Continuity of Diffusion Semigroups from SDE Stability

From Initial Conditions to Semigroup Continuity

A diffusion semigroup acts on a payoff $f$ by averaging it over the distribution of the process at a fixed time. To show that this average varies continuously with the starting point, it is useful to compare two solutions on the same probability space and drive them with the same Brownian motion. This is called a synchronous coupling. Under global Lipschitz assumptions, the solutions remain close in mean square when their initial states are close. Boundedness and continuity of $f$ then transfer that closeness to the expectations defining the semigroup.

Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carry an $m$-dimensional Brownian motion $\mathbf{B}_t$. Consider an $\mathbb{R}^d$-valued SDE with time-independent coefficients, $dX_t=b(X_t)\,dt+\sigma(X_t)\,d\mathbf{B}_t$, and deterministic initial state $X_0=x\in\mathbb{R}^d$. Here $b:\mathbb{R}^d\to\mathbb{R}^d$ is the drift, and $\sigma:\mathbb{R}^d\to\mathbb{R}^{d\times m}$ is the diffusion coefficient. Use Euclidean norms for vectors and the Frobenius norm for matrices.

Definition: The global Lipschitz condition is that, for some $L\ge0$ and every $z,w\in\mathbb{R}^d$, $|b(z)-b(w)|\le L|z-w|$ and $\|\sigma(z)-\sigma(w)\|\le L|z-w|$. The linear growth condition is that, for some $K\ge0$, $|b(z)|^2+\|\sigma(z)\|^2\le K^2(1+|z|^2)$ for every $z$.

Under these conditions the standard global Lipschitz existence and uniqueness theorem for SDEs gives a unique strong solution for each deterministic $x$, with $\mathbb{E}\sup_{0\le s\le T}|X_s^x|^2<\infty$ for every finite $T$. This is the standard Picard-iteration theorem; its proof is not needed here. A strong solution is constructed on the given space using the given Brownian motion. Pathwise uniqueness means that two solutions on the same space with the same Brownian motion and initial state agree almost surely at every time. These statements are distinct from weak existence, where the probability space and Brownian motion may be chosen as part of the solution, and uniqueness in law, which asserts equality of distributions rather than equality of coupled paths.

For each $x$, the integral form of the equation is $X_t^x=x+\int_0^t b(X_s^x)\,ds+\int_0^t \sigma(X_s^x)\,d\mathbf{B}_s$. The stochastic integral is the Ito integral: its integrand is progressively measurable and belongs to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T\|\sigma(X_s^x)\|^2\,ds<\infty$ on every finite interval. This integrability follows from linear growth and the stated second-moment bound. The drift integral is an ordinary time integral.

Definition: The diffusion semigroup at time $t\ge0$, acting on a bounded measurable function $f:\mathbb{R}^d\to\mathbb{R}$, is $P_t f(x)=\mathbb{E}[f(X_t^x)]$. The expectation is under $\mathbb{P}$, and the starting state $x$ is deterministic.

A Mean-Square Stability Estimate

Fix $x,y\in\mathbb{R}^d$, and construct $X^x$ and $X^y$ using the same Brownian motion. This shared noise matters: it makes the difference of their stochastic terms an Ito integral with integrand $\sigma(X_s^x)-\sigma(X_s^y)$. The Ito isometry applies because this progressively measurable difference is square-integrable on every finite interval, by the moment bounds and linear growth.

Proposition: Under the global Lipschitz and linear growth conditions above, for every finite $T$ and $0\le t\le T$, $$ \mathbb{E}|X_t^x-X_t^y|^2 \le 3\exp\!\big(3L^2(T+1)t\big)|x-y|^2. $$

Proof. Set $D_t=X_t^x-X_t^y$. Subtract the two integral equations. For vectors $u,v,w$, the inequality $|u+v+w|^2\le3(|u|^2+|v|^2+|w|^2)$ follows from Cauchy–Schwarz. Apply it to the initial difference, the drift integral, and the Ito integral. Cauchy–Schwarz in time gives $\left|\int_0^t(b(X_s^x)-b(X_s^y))\,ds\right|^2\le t\int_0^t|b(X_s^x)-b(X_s^y)|^2\,ds$. By the Ito isometry, the expected square of the stochastic integral is $\int_0^t\mathbb{E}\|\sigma(X_s^x)-\sigma(X_s^y)\|^2\,ds$. Consequently, using the Lipschitz bounds and $t\le T$,

\begin{aligned} \mathbb{E}|D_t|^2 &\le 3|x-y|^2+ 3tL^2\int_0^t\mathbb{E}|D_s|^2\,ds+ 3L^2\int_0^t\mathbb{E}|D_s|^2\,ds\\ &\le 3|x-y|^2+3L^2(T+1)\int_0^t\mathbb{E}|D_s|^2\,ds. \end{aligned}

Gronwall’s inequality says that a nonnegative function $u(t)$ satisfying $u(t)\le a+c\int_0^t u(s)\,ds$, for constants $a,c\ge0$, satisfies $u(t)\le ae^{ct}$. Apply it with $u(t)=\mathbb{E}|D_t|^2$, $a=3|x-y|^2$, and $c=3L^2(T+1)$. This gives the claimed estimate, including the case $x=y$, when both sides are zero. $\square$

The constants in this estimate are convenient bounds, not generally sharp. Its main consequence is that, for fixed $t$, $X_t^y$ approaches $X_t^x$ in $L^2(\mathbb{P})$ as $y\to x$, under the synchronous coupling. In particular it approaches in probability, since $\mathbb{P}(|X_t^y-X_t^x|>\varepsilon)\le \mathbb{E}|X_t^y-X_t^x|^2/\varepsilon^2$ for each $\varepsilon>0$.

Continuity of the Semigroup

Theorem: Suppose the coefficients satisfy the global Lipschitz and linear growth conditions. For every fixed $t\ge0$ and every bounded continuous $f:\mathbb{R}^d\to\mathbb{R}$, the function $x\mapsto P_t f(x)$ is continuous.

Proof. Let $x_n\to x$. Couple all the solutions with the same Brownian motion. The proposition implies $\mathbb{E}|X_t^{x_n}-X_t^x|^2\to0$, so $X_t^{x_n}\to X_t^x$ in probability. Continuity of $f$ implies $f(X_t^{x_n})\to f(X_t^x)$ in probability. One way to justify this implication is the subsequence criterion: every subsequence of variables converging in probability has a further subsequence converging almost surely; continuity of $f$ gives almost-sure convergence of the transformed further subsequence. This entails convergence in probability of the full transformed sequence.

Write $M=\sup_z|f(z)|<\infty$. For each $\varepsilon>0$, split the expectation according to whether $|f(X_t^{x_n})-f(X_t^x)|\le\varepsilon$. On that event the absolute difference is at most $\varepsilon$; on its complement it is at most $2M$. Thus

\mathbb{E}|f(X_t^{x_n})-f(X_t^x)| \le \varepsilon+ 2M\,\mathbb{P}\big(|f(X_t^{x_n})-f(X_t^x)|>\varepsilon\big) \longrightarrow \varepsilon

in the sense that the limit superior is at most $\varepsilon$. Letting $\varepsilon\downarrow0$ proves that this expectation tends to zero. Therefore $|P_t f(x_n)-P_t f(x)|\le\mathbb{E}|f(X_t^{x_n})-f(X_t^x)|\to0$. Since this holds for every sequence $x_n\to x$, $P_t f$ is continuous. At $t=0$, $X_0^x=x$, so $P_0f=f$, which is continuous by assumption. $\square$

The argument does not require $f$ to be uniformly continuous or Lipschitz. Boundedness controls the contribution from events where the payoff difference is large; continuity controls the difference when the coupled states are close. This is why the conclusion applies to bounded continuous payoffs that can oscillate rapidly far from the starting point.

Worked Applications

Worked Example: Brownian Motion and the Nonsharp Bound

Take the one-dimensional SDE $dX_t=dB_t$ with $X_0=x$. Here $b=0$, $\sigma=1$, and the Lipschitz constant can be chosen as $L=0$. The stability proposition gives $\mathbb{E}|X_t^x-X_t^y|^2\le3|x-y|^2$. It does not give equality with $|x-y|^2$.

Directly, $X_t^x=x+B_t$ and $X_t^y=y+B_t$, so $X_t^x-X_t^y=x-y$ for every outcome. Hence the exact calculation is $\mathbb{E}|X_t^x-X_t^y|^2=|x-y|^2$, which is sharper than the proposition’s estimate. For $f(z)=\sin z$, the normal characteristic function gives $P_t f(x)=\mathbb{E}[\sin(x+B_t)]=e^{-t/2}\sin x$. At $t=0$ this is $\sin x$, and for each fixed $t$ it is continuous in $x$, as the theorem predicts.

Check: Subtracting the explicit solutions cancels the entire Brownian term, leaving exactly $x-y$. The proposition’s factor of $3$ is an upper-bound constant from the three-term inequality, not an exact formula.

Worked Example: Ornstein–Uhlenbeck Stability

Consider $dX_t=-X_t\,dt+2\,dB_t$, with $X_0=x$. The coefficients are globally Lipschitz with $L=1$, and have linear growth. The explicit solution is $X_t^x=e^{-t}x+2\int_0^t e^{-(t-s)}\,dB_s$. Indeed, applying Ito’s product rule to $e^tX_t$ gives drift terms $e^tX_t\,dt-e^tX_t\,dt=0$ and diffusion term $2e^t\,dB_t$, yielding that expression. In this application the function is $g(t,z)=e^tz$, with $g_t=e^tz$, $g_z=e^t$, $g_{zz}=0$; the Ito correction $\tfrac12 g_{zz}(2)^2dt$ is zero.

The same stochastic integral occurs for both starting states, so $X_t^x-X_t^y=e^{-t}(x-y)$. Therefore $\mathbb{E}|X_t^x-X_t^y|^2=e^{-2t}|x-y|^2$. At $t=\log 2$ and $x-y=3$, the exact value is $e^{-2\log 2}\cdot9=9/4$. The general proposition, with $T=t=\log 2$ and $L=1$, gives the upper bound $3e^{3(\log 2+1)\log 2}\cdot9$; it is valid but much larger than the exact value.

Check: At $t=0$, the exact difference is $x-y$, and its squared value is $|x-y|^2$. As $t$ increases, the exact squared difference is multiplied by $e^{-2t}$, consistent with mean reversion.

Worked Example: A Bounded Continuous Payoff That Is Not Uniformly Continuous

For Brownian motion started at $x$, take $f(z)=\sin(z^2)$. This function is bounded and continuous, but its derivative $2z\cos(z^2)$ is unbounded, so a global Lipschitz estimate for $f$ is unavailable. Fix $t=1$, and compare $x,y\in[-1,1]$. On the event $|B_1|\le R$, both $x+B_1$ and $y+B_1$ lie in $[-1-R,1+R]$. The mean value theorem and $|f'(z)|\le2|z|$ give $|f(x+B_1)-f(y+B_1)|\le2(1+R)|x-y|$. On the complement the difference is at most $2$, since $|f|\le1$. Also, Markov’s inequality and $\mathbb{E}B_1^2=1$ give $\mathbb{P}(|B_1|>R)\le1/R^2$ for $R>0$. Combining these bounds, \[ |P_1f(x)-P_1f(y)| \le2(1+R)|x-y|+\frac{2}{R^2}. \] For any desired error, first choose $R$ large enough to make the second term small; then choose $|x-y|$ small enough to make the first term small. Thus $P_1f$ is continuous on $[-1,1]$, illustrating the localization behind the theorem.

Check: With $R=2$, the displayed bound is $6|x-y|+\tfrac12$, since $2/R^2=2/4=1/2$. This particular bound is not by itself small for very close $x,y$; choosing $R$ larger first makes the tail term smaller, after which $x$ and $y$ can be taken sufficiently close.

Scope and Common Pitfalls

The proof is a fixed-time statement for bounded continuous functions. The stability estimate gives a quantitative bound on the coupled states, but it does not by itself give a global Lipschitz bound for $P_tf$ unless additional regularity of $f$ is available. For example, if $f$ is Lipschitz with constant $L_f$, then Cauchy–Schwarz and the proposition yield $|P_tf(x)-P_tf(y)|\le L_f\sqrt{3\exp(3L^2(T+1)t)}\,|x-y|$ for $t\le T$. For a merely continuous bounded $f$, the theorem gives continuity without this rate.

Global Lipschitz assumptions are sufficient conditions for the stability argument, not necessary conditions for every diffusion semigroup to be continuous. If the coefficients are not globally Lipschitz, the proposition cannot simply be invoked: existence, uniqueness, moment bounds, or continuous dependence may require different hypotheses and proofs. The conclusion concerns the transition semigroup under $\mathbb{P}$; no change to a risk-neutral measure $\mathbb{Q}$ is involved.

Check Your Understanding

Use the synchronous coupling and identify which assumption justifies each step.

In the stability proof, where is the common Brownian motion used?
Why does the Ito isometry apply to the difference of the two stochastic integrands?
Which part of the semigroup-continuity proof uses boundedness of $f$, and which part uses continuity?
For $dX_t=-X_t\,dt+2\,dB_t$, compute $X_t^x-X_t^y$ and its expected square.
Does the theorem assert continuity for every bounded measurable $f$? Explain which step fails without continuity.