Brownian Motion and Potential Theory

Abstract

Diffusion can be understood on several levels. The study of diffusion on a macroscopic level, of a substance such as heat, involves the notion of the flux of the quantity.

A The Trotter product formula

It is often of use to analyze the solution operator to an evolution equation of the form

$$\frac{\partial u} {\partial t} = Au + Bu$$

in terms of the solution operators e^tA and e^tB, which individually might have fairly simple behavior. The case where A is the Laplace operator and B is multiplication by a function is used in §2 to establish the Feynman–Kac formula, as a consequence of Proposition A.4 below.

The following result, known as the Trotter product formula, was established in [Tro].

Theorem A.1.

Let A and B generate contraction semigroups e ^tA and e ^tB , on a Banach space X. If $\overline{A + B}$ is the generator of a contraction semigroup R(t), then

$$R(t)f {=\lim _{n\rightarrow \infty }}{\bigl ({e}^{(t/n)A}\,{e{}^{(t/n)B}}\bigr )}^{n}f, $$

(A.1)

for all f ∈ X.

Here, $\overline{A + B}$ denotes the closure of A + B. A simplified proof in the case where A + B itself is the generator of R(t) is given in an appendix to [Nel2]. We will give that proof.

Proposition A.2.

Assume that A, B, and A + B generate contraction semigroups P(t), Q(t), and R(t) on X, respectively, where $\mathcal{D}(A + B) = \mathcal{D}(A) \cap \mathcal{D}(B)$. Then (A.1) holds for all f ∈ X.

Proof.

It suffices to prove (A.1) for $f \in \mathcal{D} = \mathcal{D}(A + B)$. In such a case, we have

$$P(h)Q(h)f - f = h(A + B)f + o(h), $$

(A.2)

since P(h)Q(h)f − f = (P(h)f − f) + P(h)(Q(h)f − f). Also, R(h)f − f = h(A + B) + o(h), so

$$P(h)Q(h)f - R(h)f = o(h)\text{ in }X,\ \text{ for }f \in \mathcal{D}.$$

Since A + B is a closed operator, $\mathcal{D}$ is a Banach space in the norm $\|{f\|}_{\mathcal{D}} =\| (A + B)f\| +\| f\|$. For each $f \in \mathcal{D},\ {h}^{-1}{\bigl (P(h)Q(h) - R(h)\bigr )}f$ is a bounded set in X. By the uniform boundedness principle, there is a constant C such that

$$\frac{1} {h}{\bigl \|P(h)Q(h)f - R(h)f\bigr \|} \leq C\|{f\|}_{\mathcal{D}},$$

for all h > 0 and $f \in \mathcal{D}$. In other words, {h^{− 1}(P(h)Q(h) − R(h)) : h > 0 } is bounded in $\mathcal{L}(\mathcal{D},X)$, and the family tends strongly to 0 as h → 0. Consequently,

$$\frac{1} {h}{\bigl \|P(h)Q(h)f - R(h)f\bigr \|}\rightarrow 0$$

uniformly for f is a compact subset of $\mathcal{D}$.

Now, with t ≥ 0 fixed, for any $f \in \mathcal{D},\ \{R(s)f : 0 \leq s \leq t\}$ is a compact subset of $\mathcal{D}$, so

$${\bigl \|{\bigl (P(h)Q(h) - R(h)\bigr )}R(s)f\bigr \|} = o(h), $$

(A.3)

uniformly for 0 ≤ s ≤ t. Set h = t ∕ n. We need to show that (P(h)Q(h))ⁿ f − R(hn)f → 0, as n → ∞. Indeed, adding and subtracting terms of the form (P(h)Q(h))^j R(hn − hj), and using $\|P(h)Q(h)\| \leq 1$, we have

$$\begin{array}{rcl} &{\bigl \|{\bigl (P(h)Q {(h)}\bigr )}^{n}f - R(hn)f\bigr \|} & \\ &\leq {\bigl \| {\bigl (P(h)Q(h) - R(h)\bigr )}R(h(n - 1))f\bigr \|} & \\ & +{\bigl \|{\bigl (P(h)Q(h) - R(h)\bigr )}R(h(n - 2))f\bigr \|}& \\ & \text{ }\quad \text{ }\quad +\ \cdots \ +\ {\bigl \|{\bigl ( P(h)Q(h) - R(h)\bigr )}f\bigr \|}.\end{array}$$

(A.4)

This is a sum of n terms that are uniformly o(t ∕ n), by (A.3), so the proof is done.

Note that the proof of Proposition A.2 used the contractivity of P(t) and of Q(t), but not that of R(t). On the other hand, the contractivity of R(t) follows from (A.1). Furthermore, the hypothesis that P(t) and Q(t) are contraction semigroups can be generalized to $\|P(t)\| \leq {e}^{at},\ \|Q(t)\| \leq {e}^{bt}$. If C = A + B generates a semigroup R(t), we conclude that $\|R(t)\| \leq {e}^{(a+b)t}$.

We also note that only certain properties of S(h) = P(h)Q(h) play a role in the proof of Proposition A.2. We use

$$S(h)f - f = hCf + o(h),\quad f \in \mathcal{D} = \mathcal{D}(C),$$

(A.5)

where C is the generator of the semigroup R(h), to get

$$S(h)f - R(h)f = o(h),\quad f \in \mathcal{D}.$$

(A.6)

As above, we have ${h}^{-1}\|S(h)f - R(h)f\| \leq C\|{f\|}_{\mathcal{D}}$ in this case, and consequently ${h}^{-1}\|S(h)f - R(h)f\| \rightarrow 0$ uniformly for f in a compact subset of $\mathcal{D}$, such as {R(s)f: 0 ≤ s ≤ t}. Thus we have analogues of (A.3) and (A.4), with P(h)Q(h) everywhere replaced by S(h), proving the following.

Proposition A.3.

Let S(t) be a strongly continuous, operator-valued function of t ∈ [0,∞), such that the strong derivative S′(0)f = Cf exists, for $f \in \mathcal{D} = \mathcal{D}(C)$, where C generates a semigroup on a Banach space X. Assume $\|S(t)\| \leq 1$ or, more generally, $\|S(t)\| \leq {e}^{ct}$. Then, for all f ∈ X,

$${e}^{tC}f {=\lim_{ n\rightarrow \infty }S}{({n}^{-1}t)}^{n}\ f.$$

(A.7)

This result was established in [Chf], in the more general case where S′(0) has closureC, generating a semigroup.

Proposition A.2 applies to the following important family of examples. Let $X = {L}^{p}({\mathbb{R}}^{n}),\ 1 \leq p < \infty$, or let $X = {C}_{o}({\mathbb{R}}^{n})$, the space of continuous functions vanishing at infinity. Let A = Δ, the Laplace operator, and B = − M_V, that is, Bf(x) = − V (x)f(x). If V is bounded and continuous on ${\mathbb{R}}^{n}$, then B is bounded on X, so Δ − V, with domain $\mathcal{D}(\Delta )$, generates a semigroup, as shown in Proposition 9.12 of Appendix A. Thus Proposition A.2 applies, and we have the following:

Proposition A.4.

If $X = {L}^{p}({\mathbb{R}}^{n}),\ 1 \leq p < \infty$, or $X = {C}_{o}({\mathbb{R}}^{n})$, and if V is bounded and continuous on ${\mathbb{R}}^{n}$, then, for all f ∈ X,

$${e}^{t(\Delta -V )}f {=\lim _{ n\rightarrow \infty }}{\Bigl ({e}^{(t/n)\Delta }{e{}^{-(t/n)V }}\Bigr )}^{n}f. $$

(A.8)

This is the result used in §2. If $X = {L}^{p}({\mathbb{R}}^{n}),\ p < \infty$, we can in fact take $V \in {L}^{\infty }({\mathbb{R}}^{n})$. See the exercises for other extensions of this proposition.

It will be useful to extend Proposition A.2 to solution operators for time-dependent evolution equations:

$$\frac{\partial u} {\partial t} = Au + B(t)u,\quad u(0) = f.$$

(A.9)

We will restrict attention to the special case that A generates a contraction semigroup and B(t) is a continuous family of bounded operators on a Banach space X. The solution operator S(t, s) to (A.9), satisfying S(t, s)u(s) = u(t), can be constructed via the integral equation

$$u(t) = {e}^{tA}f +{ \int \limits \limits }_{0}^{t}{e}^{(t-s)A}B(s)u(s)\ ds, $$

(A.10)

parallel to the proof of Proposition 9.12 in Appendix A on functional analysis. We have the following result.

Proposition A.5.

If A generates a contraction semigroup and B(t) is a continuous family of bounded operators on X, then the solution operator to (A.9) satisfies

$$S(t,0)f {=\lim }_{t\rightarrow \infty }\ {\Bigl ({e}^{(t/n)A}\,{e}^{(t/n)B((n-1)t/n)}\Bigr )}\cdots {\Bigl ({e}^{(t/n)A}\,{e}^{(t/n)B(0)}\Bigr )}f, $$

(A.11)

for each f ∈ X.

There are n factors in parentheses on the right side of (A.11), the jth from the right being e^{(t ∕ n)A} e^{(t ∕ n)B((j − 1)t ∕ n)}.

The proof has two parts. First, in close parallel to the derivation of (A.4), we have, for any $f \in \mathcal{D}(A)$, that the difference between the right side of (A.11) and

$${e}^{(t/n)(A+B((n-1)t/n))}\cdots {e}^{(t/n)(A+B(0))}f $$

(A.12)

has norm ≤ n ⋅o(1 ∕ n), tending to zero as n → ∞, for t in any bounded interval [0, T]. Second, we must compare (A.12) with S(t, 0)f. Now, for any fixed t > 0, define v(s) on 0 ≤ s ≤ t by

$$\frac{\partial v} {\partial s} = Av + B{\Bigl (\frac{j - 1} {n} t\Bigr )}v,\quad \frac{j - 1} {n} t \leq s < \frac{j} {n}t;\quad v(0) = f. $$

(A.13)

Thus (A.12) is equal to v(t). Now we can write

$$\frac{\partial v} {\partial s} = Av + B(s)v + R(s)v,\quad v(0) = f,$$

(A.14)

where, for n large enough, $\|R(s)\| \leq \epsilon$, for 0 ≤ s ≤ t. Thus

$$v(t) = S(t,0)f +{ \int \limits \limits }_{0}^{t}S(t,s)R(s)v(s)\ ds,$$

(A.15)

and the last term in (A.15) is small. This establishes (A.11).

Thus we have the following extension of Proposition A.4. Denote by $BC({\mathbb{R}}^{n})$ the space of bounded, continuous functions on ${\mathbb{R}}^{n}$, with the sup norm.

Proposition A.6.

If $X = {L}^{p}({\mathbb{R}}^{n}),\ 1 \leq p < \infty$, or $X = {C}_{o}({\mathbb{R}}^{n})$, and if V (t) belongs to $C{\bigl ([0,\infty ),BC({\mathbb{R}}^{n})\bigr )}$, then the solution operator S(t,0) to

$$\frac{\partial u} {\partial t} = \Delta u - V (t)u$$

satisfies

$$S(t,0)f {=\lim _{n\rightarrow \infty }}\ {\Bigl ({e}^{(t/n)\Delta }{e}^{-(t/n)V ((n-1)t/n)}\Bigr )}\cdots {\Bigl ({e}^{(t/n)\Delta }{e}^{-(t/n)V (0)}\Bigr )}f,$$

(A.16)

for all f ∈ X.

To end this appendix, we give an alternative proof of the Trotter product formula when Au = Δu and Bu(x) = V (x)u(x), which, while valid for a more restricted class of functions V (x) than the proof of Proposition A.4, has some desirable features. Here, we define v_k = (e^{(1 ∕ n)Δ} e^{− (1 ∕ n)V})^k f and set

$$v(t) = {e}^{s\Delta }{e}^{-sV }{v}_{ k},\ \text{ for }t = \frac{k} {n} + s,\ 0 \leq s \leq \frac{1} {n}.$$

(A.17)

We use Duhamel’s principle to compare v(t) with u(t) = e^{t(Δ − V)} f. Note that v(t) → v_{k + 1} as t ↗ (k + 1) ∕ n, and for k ∕ n < t < (k + 1) ∕ n,

$$\begin{array}{rcl} \frac{\partial v} {\partial t} & = \Delta v - {e}^{s\Delta }V {e}^{-sV }{v}_{k} & \\ & = (\Delta - V )v + [V,{e}^{s\Delta }]{e}^{-sV }{v}_{k}.\end{array}$$

(A.18)

Thus, by Duhamel’s principle,

$$v(t) = {e}^{t(\Delta -V )}f +{ \int \limits \limits }_{0}^{t}{e}^{(t-s)(\Delta -V )}R(s)\ ds,$$

(A.19)

where

$$R(s) = [V,{e}^{\sigma \Delta }]{e}^{-\sigma V }{v}_{ k},\ \text{ for }s = \frac{k} {n} + \sigma,\ 0 \leq \sigma < \frac{1} {n}.$$

(A.20)

We can write [V, e^σΔ] = [V, e^σΔ − 1], and hence

$$R(s) = V ({e}^{\sigma \Delta } - 1){e}^{-\sigma V }{v}_{ k} - ({e}^{\sigma \Delta } - 1)V {e}^{-\sigma V }{v}_{ k}. $$

(A.21)

Now, as long as

$$\mathcal{D}(\Delta - V ) = \mathcal{D}(\Delta ) = {H}^{2}({\mathbb{R}}^{n}), $$

(A.22)

we have, for 0 ≤ γ ≤ 1,

$${\bigl \|{e{}^{t(\Delta -V )}}\bigr \|}_{ \mathcal{L}({H}^{-2\gamma },{L}^{2})} ={\bigl \| {e{}^{t(\Delta -V )}}\bigr \|}_{ \mathcal{L}({L}^{2},{H}^{2\gamma })} \leq C(T){t}^{-\gamma }, $$

(A.23)

for 0 < t ≤ T. Thus, if we take γ ∈ (0, 1) and t ∈ (0, T], we have for

$$F(t) ={ \int \limits \limits }_{0}^{t}{e}^{(t-s)(\Delta -V )}R(s)\ ds, $$

(A.24)

the estimate

$$\|F{(t)\|}_{{L}^{2}} \leq C{\int \limits \limits }_{0}^{t}{(t - s)}^{-\gamma }\ \|R{(s)\|}_{{ H}^{-2\gamma }}\ ds.$$

(A.25)

We can estimate $\|R{(s)\|}_{{H}^{-2\gamma }}$ using (A.21), together with the estimate

$${\bigl \|{e}^{\sigma \Delta } - {1}\bigr \|}_{ \mathcal{L}({L}^{2},{H}^{-2\gamma })} \leq C\ {\sigma }^{\gamma },\quad 0 \leq \gamma \leq 1. $$

(A.26)

Since σ ∈ [0, 1 ∕ n] in (A.21), we have

$$\begin{array}{rcl} \|R{(s)\|}_{{H}^{-2\gamma }}& \leq C{n}^{-\gamma }\phi (V )\|{f\|}_{{L}^{2}}, & \\ \phi (V )& ={\Bigl (\| {V \|}_{\mathcal{L}({H}^{2\gamma })} +\| {V \|}_{{L}^{\infty }}\Bigr )}{e}^{s\|{V \|}_{{L}^{\infty }} }.\end{array}$$

(A.27)

Thus, estimating v(t) = u(t) at t = 1, we have

$${\Bigl \|{\Bigl ({e}^{(1/n)\Delta }{e{}^{-(1/n)V }}\Bigr )}^{n}f - {e}^{(\Delta -V )}{f}\Bigr \|}_{{ L}^{2}} \leq {C}_{\gamma }\phi (V )\|{f\|}_{{L}^{2}} \cdot {n}^{-\gamma }, $$

(A.28)

for 0 < γ < 1, provided multiplication by V is a bounded operator on ${H}^{2\gamma }({\mathbb{R}}^{n})$. Note that this holds if ${D}^{\alpha }V \in {L}^{\infty }({\mathbb{R}}^{n})$ for |α| ≤ 2, and

$$\|{V \|}_{\mathcal{L}({H}^{2\gamma })} \leq {C\ \sup }_{\vert \alpha \vert \leq 2}\ \|{D}^{\alpha }{V \|}_{{ L}^{\infty }}. $$

(A.29)

One can similarly establish the estimate

$${\Bigl \|{\Bigl ({e}^{(t/n)\Delta }{e{}^{-(t/n)V }}\Bigr )}^{n}f - {e}^{t(\Delta -V )}{f}\Bigr \|}_{{ L}^{2}} \leq C(t)\phi (V )\|{f\|}_{{L}^{2}} \cdot {n}^{-\gamma }. $$

(A.30)

Exercises

• 1. Looking at Exercises 2–4 of §2, Chap. 8, extend Proposition A.4 to any V, continuous on ${\mathbb{R}}^{n}$, such that Re V (x) is bounded from below and | ImV (x)| is bounded. (Hint: First apply those exercises directly to the case where V is smooth, real-valued, and bounded from below.)

• 2. Let $H = {L}^{2}(\mathbb{R}),\ Af = df/dx,\ Bf = ixf(x)$, so e^tA f(x) = f(x + t), e^tB f(x) = e^itx f(x). Show that Theorem A.1 applies to this case, but not Proposition A.2. Compute both sides of

  $${e}^{pA+qB}f {=\lim _{n\rightarrow \infty }}{\bigl ({e}^{(p/n)A}{e{}^{(q/n)B}}\bigr )}^{n}f,$$

  and verify this identity directly.

  Compare with the discussion of the Heisenberg group, in §14 of Chap. 7.

• 3. Suppose A and B are bounded operators. Show that

  $${\bigl \|{e}^{t(A+B)} -{\bigl ( {e}^{(t/n)A}{e{}^{(t/n)B}}\bigr )}^{n}\bigr \|} \leq \frac{Ct} {n}$$

  and that

  $${\bigl \|{e}^{t(A+B)} -{\bigl ( {e}^{(t/2n)A}{e}^{(t/n)B}{e{}^{(t/2n)A}}\bigr )}^{n}\bigr \|} \leq \frac{ct} {{n}^{2}}.$$

  (Hint: Use the power series expansions for e^{(t ∕ n)A}, and so forth.)