The Fokker-Planck Equation

Proof.

Step 1: an integral identity for the average.
Fix \(x\) and \(t \gt 0\). The deterministic time \(t\) is a stopping time for \(\{ \mathcal{F}^{(n)}_t \}\) with \(E^x[ t ] = t \lt \infty\), so Dynkin's formula applies to \(f\) and gives

\[ u( t, x ) = f( x ) + E^x \Bigl[ \int_0^t A f( X_s )\, ds \Bigr] . \]

The integrand is jointly measurable in \(( s, \omega )\). Each component of \(X\) is a continuous adapted real process, hence progressively measurable by the progressive measurability criterion, the components assemble into a measurable \(\mathbb{R}^m\)-valued map by component-wise measurability, and \(A f\) is continuous, so the composition \(( s, \omega ) \mapsto A f( X_s( \omega ) )\) is jointly measurable. It is also bounded. The function \(A f\) is continuous with support inside the support of \(f\), by the second-order formula of the generator theorem, so \(\| A f \|_\infty \lt \infty\). The product measure of Lebesgue measure on \([ 0, t ]\) with \(\mathbb{P}\) is finite, so the bounded integrand is integrable against it, and Fubini's theorem exchanges the expectation with the time integral:

\[ u( t, x ) = f( x ) + \int_0^t E^x \bigl[ A f( X_s ) \bigr]\, ds . \]

Step 2: the integrand is continuous in time.
Write \(g( s ) = E^x[ A f( X_s ) ]\). For almost every path the map \(s \mapsto A f( X_s )\) is continuous, because the path is continuous and \(A f\) is. The absolute value is dominated by the constant \(\| A f \|_\infty\), which is integrable with respect to \(\mathbb{P}\), so the dominated convergence theorem applies along any sequence \(s_k \to s\) and shows that \(g\) is continuous on \([ 0, \infty )\). The fundamental theorem of calculus then upgrades the integral identity of Step 1. The map \(t \mapsto u( t, x )\) is continuously differentiable, with

\[ \frac{ \partial u }{ \partial t }( t, x ) = E^x \bigl[ A f( X_t ) \bigr] . \]

Setting \(t = 0\) in the definition of \(u\) gives \(u( 0, x ) = E^x[ f( X_0 ) ] = f( x )\), since the diffusion starts at \(x\).

Step 3: membership in the domain.
Fix \(t \geq 0\). The Markov property, read in several components as declared, asserts two things about the average of the bounded Borel function \(f\): the map \(y \mapsto E^y[ f( X_t ) ] = u( t, y )\) is itself bounded and Borel, and for every \(r \geq 0\),

\[ \mathbb{E} \bigl[ f( X^x_{t+r} ) \bigm| \mathcal{F}_r \bigr] = u( t, X^x_r ) \quad \text{almost surely.} \]

Taking expectations on both sides and applying the tower property to the left side collapses the conditioning:

\[ E^x \bigl[ u( t, X_r ) \bigr] = E^x \bigl[ f( X_{t+r} ) \bigr] = u( t + r, x ) . \]

The difference quotient that defines the generator of \(u( t, \cdot )\) at \(x\) is therefore a difference quotient of \(u\) in its time slot:

\[ \frac{ E^x[ u( t, X_r ) ] - u( t, x ) }{ r } = \frac{ u( t + r, x ) - u( t, x ) }{ r } . \]

As \(r \downarrow 0\) the right side converges to the time derivative established in Step 2. The function \(u( t, \cdot )\) is bounded and Borel, so the integrability demanded by the definition of the generator holds, the limit of the difference quotient exists at every \(x\), and

\[ A \bigl( u( t, \cdot ) \bigr)( x ) = \frac{ \partial u }{ \partial t }( t, x ) . \]

This proves part (a).

Step 4: the space-time bridge.
For part (b), fix \(s \gt 0\) and \(x \in \mathbb{R}^m\). The case \(s = 0\) is the shared initial condition. The space-time extension with sign \(\varepsilon = -1\) and offset \(\tau_0 = s\), read in several components as declared, makes

\[ \widehat{S}_t = \bigl( s - t , X^x_t \bigr) , \quad t \geq 0 , \]

an Itô diffusion in \(\mathbb{R}^{1+m}\) with drift \(( -1, \mathbf{b} )\), noise coefficient the \(( 1 + m ) \times n\) matrix whose first row vanishes and whose remaining rows are \(\sigma\), and starting point \(( s, x )\). Writing \(\tau\) for the first coordinate of the enlarged state, the differential operator of \(\widehat{S}\) has drift part \(- \partial_\tau + \sum_i b_i \partial_i\) and second-order part governed by the enlarged squared coefficient, whose first row and first column vanish because the first row of the noise coefficient does. Hence for \(v \in C^2( \mathbb{R}^{1+m} )\),

\[ \widehat{L} v( \tau, x ) = - \frac{ \partial v }{ \partial \tau }( \tau, x ) + L \bigl( v( \tau, \cdot ) \bigr)( x ) . \]

Step 5: a bounded region the clock must leave.
Fix \(\delta \in ( 0, s )\) and \(R \gt | x |\), and set

\[ U = ( \delta, s + 1 ) \times B_R \subseteq \mathbb{R}^{1+m} , \]

an open bounded set containing the starting point \(( s, x )\). Let \(\tau_U\) be the first exit time of \(\widehat{S}\) from \(U\). The first coordinate of \(\widehat{S}_t\) is \(s - t\), which equals \(\delta\) at time \(t = s - \delta\), and a point whose first coordinate is \(\delta\) lies outside \(U\). Hence \(\tau_U \leq s - \delta\) on every path, and in particular \(E^{( s, x )}[ \tau_U ] \leq s - \delta \lt \infty\). The clock coordinate, moving at unit speed, guarantees for free the finite mean that a state coordinate alone could not.

Step 6: a compactly supported twin of the competitor.
The competitor \(w\) lives on \([ 0, \infty ) \times \mathbb{R}^m\) and cannot be handed directly to a formula that expects a function on all of \(\mathbb{R}^{1+m}\). The smooth bump function repairs this. In one dimension, with radii \(r_1 = \tfrac{1}{2}( s + 1 - \delta )\) and \(r_2 = r_1 + \tfrac{\delta}{2}\), translation of the bump by \(c = \tfrac{1}{2}( \delta + s + 1 )\) produces a smooth \(\chi : \mathbb{R} \to \mathbb{R}\) equal to \(1\) on \([ \delta, s + 1 ]\) and vanishing outside \(( \tfrac{\delta}{2}, s + 1 + \tfrac{\delta}{2} )\), and translation preserves smoothness. In \(m\) dimensions, radii \(R\) and \(R + 1\) produce a smooth \(\varphi\) equal to \(1\) on \(\overline{B_R}\) and vanishing outside \(B_{R+1}\). The product \(\psi( \tau, x ) = \chi( \tau )\, \varphi( x )\) is smooth, equals \(1\) on \([ \delta, s + 1 ] \times \overline{B_R}\), and is supported inside the compact set \([ \tfrac{\delta}{2},\, s + 1 + \tfrac{\delta}{2} ] \times \overline{B_{R+1}}\). Define

\[ \widetilde{w}( \tau, x ) = \begin{cases} \psi( \tau, x )\, w( \tau, x ) , & \tau \gt 0 , \\ 0 & \tau \leq 0 . \end{cases} \]

On the open set \(\{ \tau \gt \tfrac{\delta}{2} \}\) both factors are twice continuously differentiable, and on the open set \(\{ \tau \lt \tfrac{\delta}{2} \} \cup \{ | x | \gt R + 1 \}\) the function vanishes identically because \(\psi\) does. These two open sets cover \(\mathbb{R}^{1+m}\), so \(\widetilde{w} \in C^2( \mathbb{R}^{1+m} )\), and it is bounded by \(\| w \|_\infty\).

Step 7: Dynkin on the bounded region.
The set \(U\) is open and bounded, the exit time has finite mean by Step 5, and \(\widetilde{w}\) is \(C^2\) on \(\mathbb{R}^{1+m}\), so Dynkin's formula on a bounded region applies to \(\widehat{S}\), \(U\), and \(\widetilde{w}\):

\[ E^{( s, x )} \bigl[ \widetilde{w}( \widehat{S}_{\tau_U} ) \bigr] = \widetilde{w}( s, x ) + E^{( s, x )} \Bigl[ \int_0^{\tau_U} \widehat{L} \widetilde{w}( \widehat{S}_r )\, dr \Bigr] . \]

Each ingredient now simplifies. The set \(U\) is open and \(\psi \equiv 1\) on the larger set \([ \delta, s + 1 ] \times \overline{B_R}\), so on \(U\) the functions \(\widetilde{w}\) and \(w\) agree on an open neighborhood of every point, and their derivatives agree there too. For \(r \lt \tau_U\) the point \(\widehat{S}_r\) lies in \(U\), where the first coordinate lies in \(( \delta, s + 1 ) \subseteq ( 0, \infty )\), so the equation satisfied by \(w\) applies and

\[ \widehat{L} \widetilde{w}( \widehat{S}_r ) = \widehat{L} w( \widehat{S}_r ) = 0 . \]

The integral term vanishes. At the starting point, \(( s, x ) \in U\) gives \(\psi( s, x ) = 1\) and \(\widetilde{w}( s, x ) = w( s, x )\). At the exit, path continuity places \(\widehat{S}_{\tau_U}\) in \(\overline{U} \subseteq [ \delta, s + 1 ] \times \overline{B_R}\), where \(\psi \equiv 1\), so \(\widetilde{w}( \widehat{S}_{\tau_U} ) = w( \widehat{S}_{\tau_U} )\). The formula collapses to

\[ w( s, x ) = E^{( s, x )} \bigl[ w( \widehat{S}_{\tau_U} ) \bigr] . \]

Step 8: send the ball to infinity, then the floor to zero.
Hold \(\delta\) and let \(R \to \infty\) through integers. For almost every path the supremum \(\sup_{ 0 \leq r \leq s } | X_r |\) is finite, because the path is continuous on a compact interval. Once \(R\) exceeds that supremum the state coordinate cannot exit \(B_R\) before time \(s - \delta\), so the exit is through the clock face: \(\tau_U = s - \delta\) and \(\widehat{S}_{\tau_U} = ( \delta, X_{s-\delta} )\). The integrand \(w( \widehat{S}_{\tau_U} )\) is dominated by the constant \(\| w \|_\infty\), so the dominated convergence theorem gives, the expectation now written under \(E^x\) because the enlarged process is built pathwise from \(X^x\) on one probability space,

\[ w( s, x ) = E^x \bigl[ w( \delta, X_{s-\delta} ) \bigr] . \]

Now let \(\delta \downarrow 0\). Path continuity gives \(X_{s-\delta} \to X_s\) almost surely, continuity of \(w\) on \([ 0, \infty ) \times \mathbb{R}^m\) gives \(w( \delta, X_{s-\delta} ) \to w( 0, X_s ) = f( X_s )\) almost surely, and the same domination by \(\| w \|_\infty\) lets dominated convergence pass the limit through the expectation:

\[ w( s, x ) = E^x \bigl[ f( X_s ) \bigr] = u( s, x ) . \]

This proves part (b).

Proof.

Step 1: complete derivatives integrate to zero.
Let \(h \in C^1( \mathbb{R}^m )\) have compact support, and fix a coordinate \(i\). The function \(\partial_i h\) is continuous with compact support, hence bounded and integrable, so Fubini's theorem writes its integral as an iterated one, integrating first along the \(i\)-th coordinate line. Reading Lebesgue measure on \(\mathbb{R}^m\) as the product of the \(i\)-th line factor with the remaining coordinates rests on the same monotone-class machinery that the track has already taken on faith for product measures, and we extend that act of trust to this one identification. On each such line the map \(y_i \mapsto h( y )\) is continuously differentiable and vanishes outside a compact interval, so the fundamental theorem of calculus gives

\[ \int_{\mathbb{R}} \partial_i h( y )\, dy_i = 0 , \]

and integrating this zero over the remaining coordinates gives \(\int_{\mathbb{R}^m} \partial_i h( y )\, dy = 0\).

Step 2: the first-order term moves with one sign change.
Fix \(i\) and apply Step 1 to \(h = b_i\, \varphi\, \psi\), which is continuously differentiable because each factor is, and compactly supported because \(\varphi\) is. The product rule expands \(\partial_i h\) into three terms, and the vanishing of the total integral rearranges into

\[ \int_{\mathbb{R}^m} b_i\, ( \partial_i \varphi )\, \psi\, dy = - \int_{\mathbb{R}^m} \varphi\, \partial_i \bigl( b_i\, \psi \bigr)\, dy . \]

Step 3: the second-order term moves with two.
Fix \(i\) and \(l\). First apply Step 1 to \(h = ( a_{il}\, \psi )\, \partial_l \varphi\), which is \(C^1\) with compact support because \(a_{il} \psi\) is \(C^2\) and \(\partial_l \varphi\) is \(C^1\) with compact support. Expanding \(\partial_i h\) by the product rule and rearranging,

\[ \int_{\mathbb{R}^m} a_{il}\, \psi\, ( \partial_{il} \varphi )\, dy = - \int_{\mathbb{R}^m} \partial_i \bigl( a_{il}\, \psi \bigr)\, ( \partial_l \varphi )\, dy . \]

Then apply Step 1 in the coordinate \(l\) to \(h = \varphi\, \partial_i ( a_{il} \psi )\), again \(C^1\) with compact support, and rearrange once more:

\[ - \int_{\mathbb{R}^m} \partial_i \bigl( a_{il}\, \psi \bigr)\, ( \partial_l \varphi )\, dy = \int_{\mathbb{R}^m} \varphi\, \partial_{li} \bigl( a_{il}\, \psi \bigr)\, dy . \]

The function \(a_{il} \psi\) is twice continuously differentiable, so its mixed partial derivatives in the coordinates \(i\) and \(l\) are equal, and the right side equals \(\int \varphi\, \partial_{il}( a_{il} \psi )\, dy\).

Step 4: assemble.
Multiply the conclusion of Step 3 by \(\tfrac{1}{2}\), sum over \(i\) and \(l\), sum the conclusion of Step 2 over \(i\), and add. The left sides assemble to \(\langle L \varphi, \psi \rangle\) by the formula for \(L\), and the right sides assemble to \(\langle \varphi, L^* \psi \rangle\) by the formula for \(L^*\). Every integrand along the way is continuous and vanishes off the support of \(\varphi\), so every integral converges absolutely.

Proof.

Step 1: integrals against the density.
For \(t \gt 0\) and a non-negative Borel \(g : \mathbb{R}^m \to \mathbb{R}\), the identity \(E^x[ g( X_t ) ] = \int g( y )\, p_t( x, y )\, dy\) holds. It is the hypothesis when \(g\) is an indicator, extends to simple functions by linearity, and passes to arbitrary non-negative \(g\) by monotone limits of simple functions on both sides. Splitting a bounded Borel \(g\) into positive and negative parts extends it to all bounded Borel functions, both sides being finite because the density integrates to one.

Step 2: the average evolves.
Fix \(\varphi \in C_0^2( \mathbb{R}^m )\). Steps 1 and 2 of the proof of the backward theorem, which used only Dynkin's formula, the joint measurability of the integrand, Fubini, and dominated convergence, established that \(t \mapsto E^x[ \varphi( X_t ) ]\) is continuously differentiable with

\[ \frac{ d }{ dt }\, E^x \bigl[ \varphi( X_t ) \bigr] = E^x \bigl[ A \varphi( X_t ) \bigr] , \]

and we take that conclusion verbatim. On \(C_0^2\) the generator theorem identifies \(A \varphi\) with \(L \varphi\), a continuous function supported inside the support of \(\varphi\).

Step 3: transpose.
Fix \(t \gt 0\). Both \(\varphi\) and \(L \varphi\) are bounded and Borel, so Step 1 rewrites both sides of Step 2 as integrals against the density, and the duality theorem with \(\psi = p_t( x, \cdot ) \in C^2( \mathbb{R}^m )\) moves the operator across the pairing:

\[ \begin{align*} E^x \bigl[ L \varphi( X_t ) \bigr] &= \bigl\langle L \varphi,\, p_t( x, \cdot ) \bigr\rangle \\\\ &= \bigl\langle \varphi,\, L^* \bigl( p_t( x, \cdot ) \bigr) \bigr\rangle . \end{align*} \]

Step 4: differentiate under the integral.
Fix \(t \gt 0\) and choose \(0 \lt t_0 \lt t \lt t_1\). On the compact set \([ t_0, t_1 ] \times \operatorname{supp} \varphi\) the derivative \(\partial_t p\) is continuous, hence bounded by some \(M\). For \(0 \lt | h |\) small enough that \(t + h \in [ t_0, t_1 ]\), the mean value theorem bounds the difference quotient of \(p\) in its time slot by \(M\) at every \(y \in \operatorname{supp} \varphi\). The quotient converges pointwise to \(\partial_t p_t( x, y )\), the product with \(\varphi\) is dominated by \(M \| \varphi \|_\infty\) times the indicator of the compact support, which is Lebesgue integrable, so the dominated convergence theorem passes the time derivative through the pairing:

\[ \frac{ d }{ dt } \bigl\langle \varphi,\, p_t( x, \cdot ) \bigr\rangle = \bigl\langle \varphi,\, \partial_t\, p_t( x, \cdot ) \bigr\rangle . \]

The left side is \(\tfrac{d}{dt} E^x[ \varphi( X_t ) ]\) by Step 1, which Steps 2 and 3 computed. Combining,

\[ \bigl\langle \varphi,\, \partial_t\, p_t( x, \cdot ) - L^* \bigl( p_t( x, \cdot ) \bigr) \bigr\rangle = 0 \quad \text{for every } \varphi \in C_0^2( \mathbb{R}^m ) . \]

Step 5: localize.
Fix \(t \gt 0\) and write \(g( y ) = \partial_t\, p_t( x, y ) - L^* ( p_t( x, \cdot ) )( y )\), a continuous function because \(p\) is \(C^2\) in the pair and the coefficients carry the standing regularity. Suppose \(g( y_0 ) \neq 0\) for some \(y_0\), without loss of generality \(g( y_0 ) \gt 0\). Continuity provides \(r \gt 0\) with \(g \gt \tfrac{1}{2}\, g( y_0 )\) on the closed ball \(\overline{ B_r( y_0 ) }\). The smooth bump function with radii \(\tfrac{r}{2}\) and \(r\), translated to be centered at \(y_0\) as in Step 6 of the uniqueness proof, is a smooth \(\varphi\) with \(0 \leq \varphi \leq 1\), equal to \(1\) on \(\overline{ B_{r/2}( y_0 ) }\), and supported in \(\overline{ B_r( y_0 ) }\). Then \(\varphi \in C_0^2\) and

\[ \langle \varphi, g \rangle \geq \tfrac{1}{2}\, g( y_0 )\, \bigl| B_{r/2}( y_0 ) \bigr| \gt 0 , \]

where \(| B_{r/2}( y_0 ) |\) is the Lebesgue measure of the smaller ball, positive because the ball contains a cube of positive side. This contradicts Step 4, so \(g \equiv 0\) on \(\mathbb{R}^m\), for every \(t \gt 0\), and the asserted equation holds.

Proof.

Step 1: absolute convergence.
The integrand \(( \partial_i f )\, p\) is bounded in absolute value by \(\| \partial_i f \|_\infty\, p\), whose integral is \(\| \partial_i f \|_\infty\), and \(f\, \partial_i p\) is bounded in absolute value by \(\| f \|_\infty\, | \partial_i p |\), whose integral is finite by hypothesis. Both integrals converge absolutely.

Step 2: a cutoff whose gradient dies.
Let \(H\) be the smooth bump function with radii \(1\) and \(2\), and for \(R \geq 1\) set \(\chi_R( y ) = H( y / R )\). Then \(\chi_R\) is smooth, equals \(1\) on \(\overline{B_R}\), vanishes outside \(B_{2R}\), and takes values in \([ 0, 1 ]\). Its gradient contracts with the scaling. The chain rule gives \(\partial_i \chi_R( y ) = \tfrac{1}{R}\, ( \partial_i H )( y / R )\), and \(\partial_i H\), being continuous with compact support, is bounded, so \(\| \partial_i \chi_R \|_\infty \leq \tfrac{1}{R}\, \| \partial_i H \|_\infty\).

Step 3: the identity with the cutoff in place.
The function \(h = f\, \chi_R\, p\) is continuously differentiable with compact support, so Step 1 of the duality proof applies to it, and the product rule expands the vanishing integral of \(\partial_i h\) into three pieces:

\[ \begin{align*} \int_{\mathbb{R}^m} ( \partial_i f )\, \chi_R\, p\, dy &= - \int_{\mathbb{R}^m} f\, \chi_R\, ( \partial_i p )\, dy \\\\ &\quad - \int_{\mathbb{R}^m} f\, ( \partial_i \chi_R )\, p\, dy . \end{align*} \]

Step 4: remove the cutoff.
As \(R \to \infty\) through integers, \(\chi_R \to 1\) pointwise, and the first two integrals converge to their cutoff-free versions by the dominated convergence theorem, with the dominating functions of Step 1. The third integral dies outright:

\[ \Bigl| \int_{\mathbb{R}^m} f\, ( \partial_i \chi_R )\, p\, dy \Bigr| \leq \frac{ \| f \|_\infty\, \| \partial_i H \|_\infty }{ R } \int_{\mathbb{R}^m} p\, dy \to 0 , \]

since \(p\) integrates to one. The limit is the first asserted identity.

Step 5: the probabilistic form.
Suppose \(p \gt 0\) and \(Y\) has density \(p\). Then \(\partial_i p = s_i\, p\) by the chain rule, so the right side of the identity is \(- \int f\, s_i\, p\, dy\). Integrals of functions against the density convert to expectations of the same functions of \(Y\) exactly as in Step 1 of the Fokker-Planck proof, applied separately to positive and negative parts, the requisite finiteness being Step 1 here. The converted identity is the displayed expectation.

Proof.

Step 1: the score carries the gradient.
Fix \(t\) and \(i\). Since \(p_t \gt 0\), the chain rule gives \(s_{t, i} = \partial_i \log p_t = ( \partial_i p_t ) / p_t\), so

\[ p_t\, \mathbf{s}_t = \nabla p_t \]

as vector fields on \(\mathbb{R}^m\). This is the single exchange on which the whole section turns. Multiplied by the density, the score forgets the logarithm.

Step 2: the correction is half a Laplacian.
The factor \(\sigma_t^2\) is constant in \(y\), so it passes through spatial derivatives, and by Step 1,

\[ \operatorname{div} \Bigl( p_t\, \tfrac{ \sigma_t^2 }{ 2 }\, \mathbf{s}_t \Bigr) = \tfrac{ \sigma_t^2 }{ 2 }\, \operatorname{div} \bigl( \nabla p_t \bigr) = \tfrac{ \sigma_t^2 }{ 2 }\, \Delta p_t , \]

the second derivatives existing and being continuous by (P1).

Step 3: cancel and conclude.
The divergence is additive over the two summands of the drift, so the right side of the asserted equation is

\[ \begin{align*} \tfrac{1}{2}\, \sigma_t^2\, \Delta p_t - \operatorname{div} \bigl( p_t\, \mathbf{u}_t \bigr) - \operatorname{div} \Bigl( p_t\, \tfrac{ \sigma_t^2 }{ 2 }\, \mathbf{s}_t \Bigr) \\\\ &= \tfrac{1}{2}\, \sigma_t^2\, \Delta p_t - \operatorname{div} \bigl( p_t\, \mathbf{u}_t \bigr) - \tfrac{1}{2}\, \sigma_t^2\, \Delta p_t \\\\ &= - \operatorname{div} \bigl( p_t\, \mathbf{u}_t \bigr) , \end{align*} \]

and the continuity equation (P3) identifies the last line with \(\partial_t p_t\). The noise scale entered twice, once in the Laplacian and once in the drift correction, and the two entrances cancel exactly, for every choice of \(\sigma\), with no regularity demanded of \(\sigma\) at all.