Extension Technique for Functions of Diffusion Operators: a stochastic approach

It has recently been shown that complete Bernstein functions of the Laplace operator map the Dirichlet boundary condition of a related elliptic PDE to the Neumann boundary condition. The importance of this mapping consists in being able to convert problems involving non-local operators, like fractional Laplacians, into ones that only involve differential operators. We generalise this result to diffusion operators associated with stochastic differential equations, using a method which is entirely based on stochastic analysis.


Introduction
This paper is strongly motivated by Kwaśnicki & Mucha's recent work, [15], where they look for functions ψ such that ψ(−∆) can be both well-defined and associated with a local PDE operator via the so-called Dirichlet-to-Neumann map, by extension method. Here, ∆ stands for the Laplace operator on R d , and they found that the above works for complete Bernstein functions.
Our contribution to this problem is two-fold. First, we show that, for the majority of complete Bernstein functions ψ, the above Dirichlet-to-Neumann map also works when ∆ is replaced by a diffusion operator of type where the coefficients are continuous functions on R d satisfying a linear growth condition. Second, our method of proof is based on a stochastic approach totally different to the purely analytic approach taken in [15]. Though the basic ideas for our stochastic approach have been around for a while in different contexts-see below for more details-it is technically quite demanding to push the method through for general complete Bernstein functions. In an abstract way, the mechanism connecting the stochastic and analytic approach used by us and the authors of [15], respectively, is the one-to-one relationship of local times of generalised diffusions and complete Bernstein functions; and this relationship was first verified in a rather hidden application given in [12].
In the remaining part of the introduction we put the Dirichlet-to-Neumann map into context, in Section 2 we formulate our results, in Section 3 we prove them, in Section 4 we discuss our conditions, and we finish the paper with some auxiliary results given in the Appendix.
Consider a bounded continuous function u on R d × [0, ∞) which is harmonic in R d × (0, ∞), i.e. ∆ x u(x, y) + ∂ 2 y u(x, y) = 0, (x, y) ∈ R d × (0, ∞), (1.1) where ∆ x denotes the Laplace operator acting on the d-dimensional x-component of u. Then it has been long 1 known that the square root of the Laplacian, −(−∆ x ) 1/2 , can map the Dirichlet boundary condition pointwise to the Neumann boundary condition of the elliptic PDE (1.1), i.e.
−(−∆ x ) 1/2 u(·, 0) = lim y↓0 ∂ y u(·, y), and this map is the earliest example of the so-called Dirichlet-to-Neumann map. This example of course triggered the question whether there are functions ψ, other than the square root, such that −ψ(−∆ x ) maps the Dirichlet boundary condition to the Neumann boundary condition of an associated PDE, and the first systematic answer to this question was given by Caffarelli & Silvestre in their 2007 paper [7]: they were able to construct a mapping for any fractional Laplacian, −(−∆ x ) α/2 , 0 < α < 2, their associated PDE read and they proved −(−∆ x ) α/2 u(·, 0) = c α lim y↓0 [y 1−α ∂ y u(·, y)].
So, they did not map the Dirichlet boundary condition to a proper Neumann boundary condition, they rather mapped it to a weighted Neumann boundary condition. However, the importance of their mapping consists in being able to convert problems involving non-local operators, like fractional Laplacians, into ones that only involve (maybe degenerate) differential operators.
Since then, in particular because of this link between non-local and local operators, there has been huge activity generalising the above extension method in many ways, mostly replacing ∆ x by more general diffusion operators, but also considering equations in more general function spaces. Avoiding to quote every of these many activities, the made progress can be seen best when looking into applications, and we concentrate on the two areas of application we would value most: analysing fractional obstacle problems (and related optimal stopping problems), see [21,6,4,5,2]; proving Harnack inequalities and maximum principles with respect to fractional operators, see [7,22,23,8,3,16].
However, in all this work, the authors restricted themselves to fractional powers, and the only work we are aware of, where functions ψ other than fractional powers were systematically treated, is Kwaśnicki & Mucha's recent paper [15]. There is also a follow-up paper, [14], on ArXiv, by Kwaśnicki, and we will compare the results in both papers with ours in Section 4. Below, we only add a few words on the origins of our stochastic approach.
The first paper to be mentioned is [17], by Molchanov & Ostrovskii, in 1969. From the analytic point of view, this paper deals with fractional Laplacians like the 2007-paper [7], by Caffarelli & Silvestre, and they applied a technique which in some way is the stochastic analogue to the extension method.
While the heuristic idea of the extension method is to study an operator acting on functions on R d via a PDE in R d × (0, ∞), the stochastic analogue of this idea would be to study the stochastic process generated by that operator via a pair of stochastic processes taking values in R d × (0, ∞). The process generated by the operator of interest would be the trace of the corresponding pair of processes on the boundary R d × {0} of R d × (0, ∞)-see Section 2 for precise definitions within our context. However, Molchanov & Ostrovskii did not make a connection between the generator of the trace process and the Neumann boundary condition of a solution to a PDE. The only paper we are aware of, where this connection has been made in a stochastic setting, is the 1986-paper [9], by Hsu. Formally, the connection is easily made by combining Itô's formula and random time change. The extra difficulty in Hsu's case was that, instead of R d ×(0, ∞), he considered bounded domains in R d+1 with sufficiently smooth boundary. In our case, because of the general setup, the difficulty comes from a lack of regularity, which has made it harder to apply Itô's formula.

Results
Our stochastic extension method is based on four ingredients.
First, let X = [X(t), t ≥ 0] be a diffusion, taking values in R d , which is a global solution of the SDE dX i (t) = p k=1 σ ik (X(t)) dB k (t) + a i (X(t)) dt, i = 1, . . . , d, (2.1) where (B 1 , . . . , B p ) T is a Brownian motion, σ = (σ ik ) is a d × p − matrix of functions on R d , and a = (a 1 , . . . , a d ) T is a vector field on R d . Of course, the infinitesimal generator associated with this diffusion, in short 'generator of X', formally reads Assume that both coefficients a, σ are continuous but also satisfy some growth condition to allow for global solutions of (2.1). Second, letm be a Krein string, that is a monotone non-decreasing right-continuous function m : [0, ∞] → [0, ∞] such thatm(∞) = ∞. Eitherm ≡ ∞, orm is uniquely associated with a locally finite measure on ([0, R), B([0, R))), where R = sup{y ≥ 0 :m(y) < ∞} > 0. If R < ∞, set y 1 = R, and otherwise, if the measure is not identically zero, let y 1 > 0 be the right endpoint of its support in [0, ∞). This way, except for the trivial stringsm ≡ ∞, andm(y) = 0, ∀y ∈ [0, ∞), any string is also uniquely associated with a measure on ([0, y 1 ] ∩ R, B([0, y 1 ] ∩ R)), whose restriction to [0, y 1 ) is locally finite. In what follows, excluding the two trivial strings, we are going to use the same notation m for both string and measure because it will always be clear from the context which one of the two objects is meant. Unless stated otherwise, the measure-version ofm refers to the measure on Third, consider the elliptic PDE where the product L x u(x, y) ×m(dy) is understood in the sense of distributions with respect to y, for any fixed x. We are looking for solutions of the following type: • u(·, y) ∈ C 2 (R d ), for any y ∈ (0, y 1 ), for any x ∈ R d , and any smooth function g : (0, y 1 ) → R with compact support.
Remark 2.2. Existence of solutions to (2.3) in the sense of Definition 2.1 implicitly requires the coefficients of the operator L x to be 'good enough'. In this paper, the only explicit assumption on these coefficients is continuity. All other assumptions are made implicitly via properties of solutions to equations. First, we require existence of global solutions to the SDE (2.1), but all other implicit assumptions will be made via properties of solutions to the PDE (2.3).
Fourth, let Y = [Y (t), t ≥ 0] be another diffusion, which is independent of X, and which takes values in the interval [0, y 1 ] ∩ R, where y 1 > 0 can be finite or infinite. Suppose that this diffusion is regular in (0, y 1 ), right-singular but not absorbing at zero, and left-singular at y 1 .
Such a one -dimensional diffusion can be uniquely determined by a scale function and a speed measure-see [1] for a general account on this theory based on stochastic calculus. For the purpose of this paper, we choose the scale function to be the identity, and we want to construct Y as in the proof of Lemma A.1 in the Appendix, using a measurem on ([0, y 1 ] ∩ R, B([0, y 1 ] ∩ R)) which is associated with a non-trivial Krein string.
This construction is a special case of [1, Theorem (6.5)], and following the terminology given in [1], this theorem tells us in particular that the corresponding speed measure would not bem but where m is defined on B([0, y 1 )) in the first case, and on B([0, y 1 ]) in the second. But, according to [1,Proposition (5.34)], for our choice of regular and singular points, this speed measure would have to satisfy: (S1) m(K) < +∞, for any compact subset K ⊆ [0, y 1 ); 2 In what follows, ∂i and ∂ij is short notation for ∂/∂xi and ∂ 2 /∂xi∂xj, respectively. So, we should ask whether the wanted regularity of Y in terms of regular and singular points puts a constraint on the type of strings for which our method works. That said, by definition of y 1 , any non-trivial Krein stringm is locally finite on [0, y 1 ), so (S1) is always satisfied, at least. Property (S2) is not restrictive either: strings which do not lead to this property of m would be related to the case where y 1 is absorbing for Y . Property (S3) though, which is due to the wanted regularity of Y in (0, y 1 ), requires the string to be strictly increasing on (0, y 1 ).
In this paper, we therefore restrict ourselves to non-trivial Krein strings which are strictly increasing on (0, y 1 ). Note that, by [1,Theorem (6.5)], using such strings in the construction described in the proof of Lemma A.1 in the Appendix indeed gives diffusions Y of the wanted regularity.
Of course, any measure m satisfying (S1-3) admits a Lebesgue decomposition into the sum of two singular measures, one of which is absolutely continuous with respect to the Lebesgue measure. By (S3), the density of the absolutely continuous measure can be chosen to be strictly positive. Therefore, without restricting generality, using a measurable function b : [0, y 1 ] ∩ R → R, we can write the Lebesgue decompositions of both measures m andm in the form respectively, where m 0 =m({0}) =m(0), m 1 =m({y 1 }) =m(y 1 ) −m(y 1 −), and the measure n satisfies n([0, y 1 ] ∩ R \ N ) = 0, for some Borel set N ⊆ (0, y 1 ) of Lebesgue measure zero. Note that, since b maps into R, the density b −2 (y) is indeed positive, for all y ∈ [0, y 1 ] ∩ R. Furthermore, for technical reasons, we set b(y) = 0, for all y ∈ (N ∪ {0, y 1 }) ∩ R, where b −2 (y) = +∞, if b(y) = 0, as usual.
Eventually, if the Lebesgue decomposition of the speed measure m is written in the above form, then [1,Theorem (7.9)] yields that Y solves an SDE of type where B is a Brownian motion, and [L y t (Y ), t ≥ 0] stands for the symmetric local time of Y at y ∈ [0, y 1 ]. Further details on how this local time is defined can be found in the proof of Lemma A.1 in the Appendix. Remark 2.3. Though we write the Lebesgue decomposition of the speed measure in a slightly different way, [1, Theorem (7.9)] can still be applied in our context. Note that L y 1 t (Y ) = 0, t ≥ 0, if y 1 = ∞, and that one can assume that (B 1 , . . . , B p ) and B are independent, without loss of generality.
Our first goal is to establish a version of Itô's lemma for u(X(·), Y (·)), when u is a solution to (2.3). Solutions of (2.3) are jointly measurable, they are continuous in x, and also continuous in y (recall that ∂ ± y u do exist), but they might not be jointly continuous. This suggests that more regularity than stated in Definition 2.1 would be needed for Itô's lemma to hold true. We should nonetheless try to keep assumptions on the regularity of u as 'weak' as possible. Adding the following condition seems to be enough. Assumption 2.4. The functions u, ∂ i u, ∂ ij u, 1 ≤ i, j ≤ d, can be extended to locally bounded functions on R d × [0, y 1 ] ∩ R by taking the limits lim y↓0 u(x, y), lim y↓0 ∂ i u(x, y), lim y↓0 ∂ ij u(x, y), and lim y↑y 1 u(x, y), lim y↑y 1 ∂ i u(x, y), lim y↑y 1 ∂ ij u(x, y), at any fixed x ∈ R d , where the latter three limits are only taken when y 1 < ∞.
Remark 2.5. Under the above assumption, the particular limit of u(x, y), y ↓ 0, exists, which we will also denote by f (x). So f : R d → R is a locally bounded measurable function which plays the role of a Dirichlet boundary condition for the solution u. In what follows, to emphasise this role, we will always write u f for the extension of u, when Assumption 2.4 is assumed. Using u f (x, 0) as an alternative notation for the limit of u f (x, y), y ↓ 0, leads to the wanted equality u f (x, 0) = f (x), x ∈ R d . The other limits taken under Assumption 2.4 are going to be denoted by ∂ i u f (x, 0), ∂ ij u f (x, 0) and, when and L x u f (·, y 1 −) used further below would refer to these limits, too.
Corollary 2.6. Let u f be the extension of a solution u to (2.3) satisfying Assumption 2.4. Then, the limit ∂ + y u f (x, 0) = lim y↓0 ∂ + y u(x, y) exists, and this limit extends ∂ + y u to a locally bounded function on R d × [0, y 1 ), which is càdlàg in y, for any fixed x ∈ R d .
If both y 1 < ∞ &m([0, y 1 )) < +∞, then the limit ∂ − y u f (x, y 1 −) = lim y↑y 1 ∂ + y u(x, y) also exists, extending ∂ − y u to a locally bounded function on R d × (0, y 1 ], which is càglàd in y, for any fixed x ∈ R d . Remark 2.7. This corollary is fairly standard, so we only sketch its proof in the Appendix. This sketch also makes clear that, once all x-direction second order partial derivatives ∂ ij u, 1 ≤ i, j ≤ d, are locally bounded in R d × (0, y 1 ), then u, ∂ i u, 1 ≤ i ≤ d, ∂ ± y u are necessarily locally bounded in R d × (0, y 1 ), too. Assumption 2.4 rather puts conditions on how the x-direction partial derivatives behave near the boundary of R d × (0, y 1 ). Now, for fixed x ∈ R d , let (Ω, F, P x ) be a complete probability space big enough to carry all random variables X, Y, (B 1 , . . . , B r ), B, as described above, with X, Y starting at X(0) a.s. = x, Y (0) a.s. = 0, respectively. Moreover, we choose a suitable 3 filtration, all processes are adapted to, and all stopping times refer to. The following stopping times, with respect to levels r ≥ 0, will be frequently used, with respect to Y , but also with respect to other one-dimensional processes, for example τ r ( X ) with respect to X etc.
Next, observe that the pair of random variables (X, Y ) describes a stochastic process on (Ω, F, P x ) taking values in R d × [0, y 1 ] ∩ R . This process is associated with a so-called trace process, Z = [Z(t), t ≥ 0], which is the trace of the process (X, Y ) when touching the hyperplane {(x, 0) : x ∈ R} ⊆ R d × [0, y 1 ] ∩ R , and which can be given by , and Z would be a killed process with finite life time L 0 ∞ (Y ). We denote its cemetery-state by †, which is added to R d in the usual way, and we also define X(∞) = †, to be consistent with (2.6). Any function f : The purpose of Lemma 2.8 is to be able to work out the limit of E x [f (Z(t)) − f (x)]/t, when t ↓ 0, for any fixed x ∈ R d , and any bounded function f : R d → R of a certain degree of regularity. For the corresponding result, which is the main result of this paper, we have to differ between infinite and finite y 1 , but also between y 1 is absorbing and not-absorbing in the case of finite y 1 . Moreover, because of P x ({τ y 1 (Y ) < +∞}) ∈ {0, 1}, see [1, Lemma (2.9)] for example, the case of y 1 is absorbing splits into two further cases.
We first describe the above cases analytically in terms of the Krein string defining the diffusion Y , and then we state the main result. Theorem 2.11. Choose a non-trivial Krein stringm which is strictly increasing on (0, y 1 ), and select a PDE operator L x according to (2.2) and Remark 2.2. Let u be a bounded solution to (2.3) satisfying Assumption 2.4, and suppose that the extension . Consider one of the following cases: Then, for any x ∈ R d , The question arises whether the representation of lim t↓0 E x [f (Z(t)) − f (x)]/t given in the above theorem has got anything to do with the generator of Z, if a generator exists for the process Z, and that's what we clarify, first.
To start with, under the made assumptions, both processes X and Y are strong Markov processes, the process X because of [11,Theorem 5.4.20], and Y by construction in [1,Chapter VI].
Of course, the generator of X is given by L x , being formally defined via (2.2), and a dense domain in some Banach space. In our setup, since the coefficients of the SDE (2.1) are supposed to be continuous, it is easy to see that X is a Feller process, and hence the natural choice of Banach space would be a space of continuous functions with a growth condition at infinity-let us denote this space by C 0 (R d ). If the coefficients of (2.1) satisfy the linear growth condition (4.1), then would be a possible choice. Here, the function x → |x| 2 plays the role of a Lyapunov function, and other Lyapunov functions ensuring existence of global solutions to (2.1) could be considered. Also, the above choice of C 0 (R d ) is separable and contains C b (R d ), so let us assume that the chosen Banach Then, L x generates a strongly continuous contraction semigroup on that Banach space, which we denote by e tLx , t ≥ 0, slightly abusing notation because this notation is usually reserved for semigroups generated by self-adjoint operators on Hilbert spaces.
By the strong Markov property of Y , the process [T t , t ≥ 0] is a (possibly killed) subordinator, see Section 68 of Chapter VIII in [19], for example. Let ψ(λ), λ > 0, denote the Laplace exponent of this subordinator, that is where γ t stands for the law of the random time T t . Then Phillips' theorem, see [18], yields that the family of Bochner integrals defines a strongly continuous contraction semigroup on C 0 (R d ), whose generator has a domain including the domain of L x . This semigroup obviously coincides with the transition semigroup of Z on , and hence Z is a strong Markov process whose generator coincides with the generator of the strongly Even more, formally substituting e −τ λ by e −τ (−Lx) in (2.8), gives suggesting that, formally, the generator of the semigroup [0,∞) e τ Lx γ t (dτ ), t ≥ 0, that is the generator of Z, could be identified with −ψ(−L x ), in some sense. This functional calculus based on subordination has indeed been established, see [20] for a good account on this theory. Furthermore, if L x was a selfadjoint operator on a Hilbert space, then the functional calculus for self-adjoint operators on a Hilbert space would produce the same operator −ψ(−L x ), assuming that ψ is an admissible function in this calculus, too. All in all, the well-established operator −ψ(−L x ) on C 0 (R d ), where ψ is the Laplace exponent of the subordinator [T t , t ≥ 0], plays the role of the generator of the trace process Z, and the domain of L x is even a core for −ψ(−L x ). Therefore, for any f in the domain of L x , and since uniform convergence implies pointwise convergence, we have got the following answer to our question below Theorem 2.11.
]/t which is much weaker than convergence in a Banach space (note that pointwise convergence is not even metrizable). Thus, if the conditions of Theorem 2.11 are satisfied, then u f (·, 0) does not necessarily have to be in the domain of the generator of the trace process.
(b) Some generators L x , for example ∆ x , can also be considered in L p (R d )-spaces, 1 ≤ p < ∞, determined by a reference measure, which in case of ∆ x would simply be the Lebesgue measure on R d . Then, under the conditions of Theorem 2.11, if u f (·, 0) is in the L p (R d )-domain of the generator L x , the assertion of Corollary 2.12 would only hold true for almost every x ∈ R d with respect to the reference measure, by obvious reasons. Of course, this assertion would nevertheless imply the identity in the corresponding L p -space, though only the sum on the right-hand side, not each summand, has to be in that L p -space. (c) Subject to m 0 = 0, the conditions of this corollary translate into conditions on L x ,m, and the Dirichlet data, such that the operator −ψ(−L x ) acts as Dirichlet-to-Neumann map, too. For how these conditions compare to existing conditions, and for how the Laplace exponent ψ depends onm, the reader is referred to Section 4.
So far, we have considered (X, Y ) a stochastic process on some probability space (Ω, F, P x ), taking values in R d × [0, y 1 ] ∩ R , with X and Y almost surely starting at x and 0, respectively. The starting point Y (0) a.s. = 0 was crucial for showing that the generator associated with the transition semigroup of the trace process maps the Dirichlet boundary condition of a solution of (2.3) to its Neumann boundary condition.
But, we can also apply our Itô's lemma 2.8 to obtain stochastic representations for fundamental solutions of our PDE (2.3) with respect to different boundary conditions. For this purpose, we need the processes X and Y starting at x ∈ R d and arbitrary y ∈ (0, y 1 ), respectively, and we are going to write X x and Y y instead of X and Y in the remaining part of this section, in order to emphasise the specific choice of starting values. Also, slightly abusing notation, we will denote the underlying probability measure by P dropping the sub-index x, though P could in principle be different for different (x, y) ∈ R d × (0, y 1 ). Now, for arbitrary but fixed (x, y) ∈ R d × (0, y 1 ), let π(x, y, dx ′ ) denote the probability law determined by where f : R d → R can be any bounded measurable function.
Corollary 2.14. For any bounded solution u of (2.3) satisfying Assumption 2.4, if the extension and hence π(x, y, dx ′ ) acts as a fundamental solution in each of theses cases.

Proofs
Proof of Lemma 2.8. The following proof of (a) demonstrates the main idea used in all three cases, and hence we will restrict ourselves to what is different to this proof when proving (b),(c).
(a) To start with, we 'mollify' u f introducing where ̺ ε (y) = ̺(y/ε)/ε, y ∈ R, using a 'right-hand' mollifier ̺ ∈ C 2 (R) with compact support in (−1, 0) satisfying ̺ ≥ 0 and R ̺(y) dy = 1. Note that, because of Assumption 2.4, on the one hand, and because the support of ̺ is bounded away from zero, on the other, the mollified solution u ε f is an element of C 2 (R d × [0, ∞)). Recall that u f is not necessarily jointly continuous, so we do not know if [u f (X(t), Y (t)), t ≥ 0] is a continuous stochastic process. However, is predictable, too, because the stochastic integrals which are well-defined by (2.1), (2.5), and Assumption 2.4, give continuous processes in t. Thus, using [10, Prop. I 2.18 b)], to prove part (a) of the lemma, it is sufficient to show that P x ({Z(t ∧ τ ) = 0}) = 1, for any t ≥ 0, and any predictable stopping time τ . Next, set τ N = τ N ( X ) ∧ N , and choose a sequence S N , N = 1, 2, . . . , of finite stopping times which is L 2 (P x )-localising for the local martingale [ for any t ≥ 0, any N = 1, 2, . . . , and any predictable stopping time τ . Also, by Assumption 2.4, the First, for any ε > 0, the classical version of Itô's lemma gives We claim that, for any chosen sequence of ε-values converging to zero, there is a subsequence (ε n ) ∞ n=1 such that the above equation's left-hand side almost surely converges to when n → ∞. Indeed, the limit of u ε f (X(t N ), Y (t N )) − u ε f (x, 0), ε ↓ 0, is obvious, by Remark 2.5. Next, for 0 ≤ s ≤ t N , partial integration yields where, by Corollary 2.6, the supremum on the right-hand side is uniformly bounded in ε, for any chosen sequence of ε-values converging to zero. Therefore, follows by bounded convergence. Lastly, it is not hard to see that all stochastic integrals on the left-hand side of (3.1) converge duly in L 2 (P x ), when ε ↓ 0, proving the claim we made above. We only discuss the case of ) dB(s)-the other cases are easier because the coefficients σ ik are supposed to be continuous. Since ) ds ] < +∞. Then, again by dominated convergence, the ε-limit of the righthand side can be taken with respect to the integrand, and this limit is zero, for all s ∈ [0, t N ].
So, for P x ({Z(t N ) = 0}) = 1, the remaining step is to show that almost surely converges to To see this, we use the construction of [Y (t), t ≥ 0] given in the proof of Lemma A.1 in the Appendix. Also, recall the Lebesgue decomposition of the measures m andm, where m has got the properties (S1-3). Of course, the measurem satisfies the same properties, and (S1) implies that b −2 is locally integrable on [0, y 1 ).
Thus, we assume that ] is a standard one-dimensional Wiener process, given on (Ω, F, P x ), and Note that A ∞ = +∞, a.s., as stated in [1, Lemma (6.15)], and t → A t is continuous because the measurem is finite on compact subsets of [0, ∞). Therefore, s = A A −1 s , for all s ≥ 0, a.s., and time change yields , the set {b 2 = 0} has Lebesgue measure zero, and hence the last integral becomes using the PDE (2.3) to identify the limit of the measures ∂ 2 y u ε f (X(A s ), y) dy, when ε ↓ 0. But, which almost surely is the ε-limit of the second summand in (3.2). Note the subtle point that the indicator 1 (0,∞) is due to the fact that the PDE only holds in the open half plane. Finally, as ) ds is the ε-limit of the first summand in (3.2), the ε-limit of (3.2) can almost surely be given by Observe that, different to the proof of (a), it is technically more demanding to work with the mollified version of u f (x, y), when y is close to y 1 < ∞. We therefore choose h ∈ (0, y 1 ) and build a function u f,h on the whole half-space by Let u ε f,h denote the mollified version of u f,h using the same mollifier ̺ as in the proof of (a), and let Y h denote the process [Y (t ∧ τ y 1 −h (Y )), t ≥ 0]. Note that, for any such h, the stopping time τ y 1 −h (Y ) is almost surely finite, because m([0, y 1 − h]) < +∞.
We first apply the classical version of Itô's lemma to u ε f,h (X(t N ), Y h (t N )) and let ε go to zero. We then prove our claim by letting h go to zero, too.
Recall that changing b on a set of Lebesgue measure zero would not change the law of Y , thus it would not change the law of Y h , either. Without loss of generality, we can therefore assume that b(y 1 − h) = 0. As a consequence, Y h satisfies the SDE, and hence applying Itô's formula to u ε f,h (X(t N ), Y h (t N )) gives an equation identical to (3.1) with u ε f and Y replaced by u ε f,h and Y h , respectively. Then, as in the proof of part (a), for any sequence of ε-values converging to zero, there exists a subsequence (ε n ) ∞ n=1 such that, when n → ∞, the left-hand side of this equation converges almost surely to The next step is to find the ε-limit of the right-hand side, i.e.
the second integral of which can be written as since Y h is absorbing at y 1 − h, and b(y 1 − h) = 0. But s = A A −1 s , for all s < τ y 1 −h (Y ), a.s., and hence, when ε ↓ 0, the second integral converges almost surely to applying time change and partial integration as in the proof of (a).
By dominated convergence, when ε ↓ 0, the first integral of (3.4) converges to so that summing up gives the following almost sure limit of (3.4), when ε ↓ 0. Of course, being the limit of a left-hand and a right-hand side of Itô's formula, respectively, (3.3) almost surely equals the ε-limit of (3.4), and hence 0 a.s.
Eventually, we choose a whole sequence of h-values converging to zero. Since countably many h-values still form a set of Lebesgue measure zero, b(y 1 − h) can be assumed to be zero, for any h in this countable set, and we have to show that the h-limit of the above equation's right-hand side almost surely equals Z(t N ). First when h ↓ 0, and Z(t N ) equals the right-hand side of (3.5), on each {τ y 1 −h (Y ) > t N }. Therefore, without loss of generality, we only show that the h-limit of this right-hand side almost surely equals Z(t N ) under the assumption that τ y 1 (Y ) is finite.
Under this assumption, τ y 1 (Y ) − τ y 1 −h (Y ) → 0, almost surely, when h ↓ 0, and hence, using dominated convergence, all summands on the right-hand side of (3.5), except the last one, can be shown to converge in L 2 (P x ) to their Z(t N )-counterparts, when h ↓ 0, in a straight forward way (recall that t N satisfies t N ≤ τ N (L 0 · (Y )), by definition). Identifying the limit of the last summand is more involved because ∂ + y u f (·, y)1 { · ≤N } may become unbounded, when y approaches y 1 . Recall that ∂ + y u f (·, 0)1 { · ≤N } is bounded by Corollary 2.6. However, as the left-hand side of (3.5) is zero, if all other summands converge in L 2 (P x ), then the last summand does, too, so that where the last line follows by monotone convergence. The above justifies that and hence by dominated convergence, proving part (b) of the lemma.
(c) Choose h ∈ (0, y 1 ), and define t N , u f,h , u ε f,h as in the proof of (b). Since Y is not absorbing at y 1 < ∞, the local time L y 1 · (Y ) does not vanish, and hence the classical version of Itô's lemma gives for all x ∈ R d , the term involving L y 1 · (Y ) vanishes, so that the ε-limit of the above left-hand side almost surely equals by the same arguments used in the proof of (a). Furthermore, unlike in case (b) where y 1 is absorbing, it now must hold thatm([0, y 1 ]) < +∞ (see the beginning of Remark 2.9), and hence s = A A −1 s , for all s ≥ 0, a.s. Therefore, the ε-limit of the second integral on the right-hand side of (3.6) can almost surely be given by again applying time change and partial integration as in the proof of (a). Here, the 'artificial' jump of ∂ + y u f,h (X(A s ), ·) at y 1 − h/2, which we created when extending u f to half-space, equals All in all, letting ε go to zero on both sides of (3.6) yields 0 a.s.
and it needs to be shown that the h-limit of the above right-hand side almost surely coincides with Z(t N ), where Z(t) is the case-(c)-version of what has been defined in the proofs of (a),(b). By Corollary 2.6, ∂ + y u f always behaves well near the boundary of R d × (0, y 1 ) at zero. In case (c), Corollary 2.6 also implies that ∂ + y u f behaves well near the boundary of R d × (0, y 1 ) at y 1 . So, all terms can be shown to converge almost surely or in L 2 (P x ) to their Z(t N )-counterparts in a straight forward way, when h ↓ 0. Below, we verify that when h ↓ 0, finishing the proof of part (c) and the lemma. Since [1, Lemma (6.34)(i)] gives we can almost surely bound and the task is to show that both summands vanish a.s., when h ↓ 0.
First, observe that X(A s ) ≤ N , for s ≤ A −1 t N , a.s., and that 0 < h/2 < y 1 /2 by our choice of h. Therefore, the first summand is bounded by where it follows from the proof of Corollary 2.6 that Thus the limit of the first summand almost surely vanishes. For the second summand, note that A −1 t N is almost surely finite, and hence, for almost every ω ∈ Ω, and each y ∈ R, s → L y s (W )(ω) can be considered a continuous distribution function of a finite measure Since L y s (W )(ω) is continuous in y, too, the measures ν y 1 −h/2 (ω) converge weakly to ν y 1 (ω), when h goes to zero. As a consequence, when h ↓ 0, the second summand converges almost surely to zero, because s → A s is continuous, on the one hand, and because x → ∂ − y u f (x, y 1 −) is by assumption a bounded continuous function, for x ≤ N , on the other.
Proof of Lemma 2.10. Recall that y 1 is assumed to be finite.
(i) It has already pointed out at the beginning of Remark 2.9 that if Y is not absorbing at y 1 thenm([0, y 1 ]) < +∞. Vice versa, ifm([0, y 1 ]) < +∞, then the diffusion Y constructed in the proof of Lemma A.1 in the Appendix can never be absorbing at y 1 , because A t < +∞, a.s., for all t ≥ 0. Thus, Y is not absorbing at y 1 if and only ifm([0, y 1 ]) < +∞, which is equivalent to the statement to be proven under (i). Proof of Theorem 2.11. (a) Fix x ∈ R d , and choose an arbitrary but small t > 0. Define the stopping times τ N , S N as at the beginning of the proof of Lemma 2.8, and set Then, for fixed N ≥ 1, Lemma 2.8(a) yields and since t N ≤ τ N ∧ S ′ N , all stochastic integrals above have zero expectation, so that where we also used (3.10) which is an easy consequence of [1,Theorem (5.27)], when the scale function is the identity. Now, recall that Y (t) = W (A −1 t ), t ≥ 0, where A t is given in the proof of Lemma A.1 in the Appendix. Then, since [1, Lemma (6.34)(i)] yields = +∞, for part (a) of this proof. As a consequence, since τ N ∧ S ′ N grows to infinity, N → ∞, we have that, for almost every ω ∈ Ω, there exists N (ω) such that t N (ω) = T t (ω), for all N ≥ N (ω), and hence by dominated convergence, even if f was not continuous.
On the other hand, since t N ≤ T t , N ≥ 1, and since ∂ + All in all, to finish the proof of part (a) of the theorem, it remains to show that The proofs in the cases (b1) and (b2) are identical for a large part, and that's why we name this large part of the proof (b), and we only go into the differences between (b1) & (b2) at the end. By Lemma 2.10, the point y 1 will be absorbing in this case.
Note that the above reasoning also implies that ψ(0) has to be positive in case (b), though we obviously had ψ(0) = 0 in case (a). Now, fix x ∈ R d , t > 0, and define t N by (3.9), but using S ′ N = S N ∧ τ y 1 −1/N (Y h ), instead, where Y h again denotes the process [Y (t ∧ τ y 1 −h (Y )), t ≥ 0], for some h ∈ (0, y 1 ). Of course, if Y does hit y 1 then it would hit it after hitting y 1 − h, that is N = 1, 2, . . . , (3.12) though τ y 1 −h (Y ) converges to τ y 1 (Y ), when h ↓ 0, whether τ y 1 (Y ) is finite or not. Furthermore, if y 1 < ∞ is absorbing then L 0 τy 1 (Y ) (Y ) always equals L 0 ∞ (Y ) almost surely, because Y can only be absorbed at y 1 > 0 after hitting zero for the last time.
All in all, for fixed N ≥ 1, Lemma 2.8(b) yields where mart(t N ) denotes the sum of the stochastic integrals, at stopping time t N . Taking expectations on both sides, we therefore obtain that using (3.10) as in the proof of part (a). First, we deal with the limit of the above right-hand side. Time change yields where t N grows to T t ∧ τ y 1 −h (Y ), when N → ∞. So, since ∂ + y u f (·, 0) + m 0 L x u f (·, 0) is bounded, by dominated convergence, the limit of the right-hand side equals when h ↓ 0. Next, it is easy to see that, for almost every ω ∈ {T t < τ y 1 −h (Y )}, there exists N (ω) such that t N (ω) = T t (ω), for all N ≥ N (ω). Also, sincem([0, y 1 − h]) < +∞, we know that τ y 1 −h (Y ) is almost surely finite, and therefore {T t ≥ τ y 1 −h (Y )} a.s. = {T t > τ y 1 −h (Y )}, because the process Y cannot be at zero and y 1 − h at the same time. As a consequence, for almost every ω ∈ {T t ≥ τ y 1 −h (Y )}, there exists N (ω) such that t N (ω) = τ y 1 −h (Y )(ω), for all N ≥ N (ω), and we obtain that by dominated convergence, only using boundedness of u f , but not continuity.
On the whole, we have justified that by the same arguments as in the proof of part (a), only taking into account that, when t ↓ 0, for almost every ω ∈ Ω, it will eventually happen that t < L 0 ∞ (Y )(ω). Since f (X(T t ))1 {Tt=∞} = f ( †)1 {Tt=∞} = 0, the proof is complete in both cases (b1) and (b2). (c) By Lemma 2.10, the point y 1 < ∞ is not absorbing, andm([0, a.s. = +∞ as well as ψ(0) = 0, as in the proof of part (a). Now, fix x ∈ R d , t > 0, and define t N by (3.9), but using S ′ N = S N , instead-there is no localisation needed w.r.t. Y because of Corollary 2.6. Since ∂ − y u f (·, y 1 −) is continuous, Lemma 2.8(c) yields again using mart(t N ) for the sum of the stochastic integrals, at stopping time t N . As in part (a) of the proof, for almost every ω ∈ Ω, there exists N (ω) such that t N (ω) = T t (ω), for all N ≥ N (ω), and hence because Y cannot be at zero and y 1 at the same time. Furthermore, again by [1, Theorem (5.27)], and the last integral vanishes by assumption, in case (c). All in all, we obtain that and the rest is identical to the proof of (a).
Proof of Corollary 2.14. Fix (x, y) ∈ R d × (0, y 1 ), and write X x , Y y , u(x, y) instead of X, Y, f (x) whenever the notation X, Y, f (x) is used in the formulation of Lemma 2.8. It is readily checked that our proof of Lemma 2.8 works for these starting values, as well. Next, recall the stopping times τ N , S N introduced at the beginning of the proof of Lemma 2.8, and define By Lemma 2.10, the phrase 'y 1 < ∞ absorbing' refers to the cases (b1) & (b2) in Theorem 2.11, with (a) & (c) being the remaining cases. We will always choose N big enough ensuring 1/N < y < y 1 −1/N . The idea of proof, in all four cases, consists of the following steps: apply Lemma 2.8 to u f (X x (t N ), Y y (t N )), take expectations on both sides, and eventually take the limit N → ∞.
Of course, taking expectations removes all stochastic integrals, by our choice of t N . Furthermore, all integrals involving ∂ + y u f (X x (s), 0) or L x u f (X x (s), 0) disappear, because Y y (s) > 0, for all s ≤ t N . In both cases (b1) & (b2), all terms involving 'y 1 −' disappear, because s < τ y 1 (Y y ), for all s ≤ t N . In case (c), the integrals involving ∂ − y u f (X x (s), y 1 −) or L x u f (X x (s), y 1 −) disappear as in the proof of Theorem 2.11, using the assumed boundary condition.
Here, it follows from the construction of Y given in [1, Chapter VI] that both Thus, since τ N ∧ S N ∧ τ N (Y ) grows to infinity, N → ∞, we also have that, for almost every ω ∈ Ω, there exists N (ω) such that t N (ω) = τ 1 N (Y y (ω)), for all N ≥ N (ω). So, when taking the limit N → ∞ on the right-hand side of (3.14), at first, one can interchange limit and expectation, since u f is bounded. Then, in case (c), the limit of the term involving 'y 1 −' must vanish, so that in both cases (a) & (c) the limit will be E[ f (X x (τ 0 (Y y ))) ], because of the continuity of u f at the boundary R d × {0}.
It remains to prove the corollary in case of (b1) & (b2), and it obviously suffices to show that Here, the construction of Y given in [1, Chapter VI] implies almost-sure-convergence of when N → ∞, so that (3.15) follows from the behaviour of u f at the boundary R d × {0, y 1 }.

Discussion of our Conditions
In this section we mainly verify the conditions of Theorem 2.11 in the case considered by Kwaśnicki & Mucha in [15], i.e. L x = ∆ x , showing that applying our stochastic method more or less covers the most general result on the extension technique for functions of the Laplace operator. We will nevertheless comment on the situation where L x is a PDE operator more general than ∆ x , too. However, a detailed analysis of solutions to the corresponding PDE (2.3) is rather involved and would go beyond the scope of this paper. First, Theorem 2.11 is based on the existence of a solution to (2.3), which is an elliptic PDE with coefficients determined by L x andm.
Here, the operator L x is determined by the coefficients of the SDE (2.1), and what we need in the proof of Lemma 2.8, among other things, is that (2.1) has a global solution, and that (2.3) has a solution u such that u(·, y) ∈ C 2 (R d ), for all y ∈ (0, y 1 ). So, as a minimum-condition, we require the coefficients of (2.1) to be continuous knowing that for the wanted level of smoothness of u in the x-variable one would probably need smoother coefficients, even. We did not specify any further conditions ensuring existence of a global solution to (2.1), leaving the reader with choice. The standard condition would be linear growth, i.e. Remark 4.1. That we exclude the two trivial strings is not restrictive because they are not important and could be covered by other means. Forcing the string to be strictly increasing on (0, y 1 ), though, excludes a class of non-trivial strings. It is possible to construct diffusions, Y , whose speed measures are associated with non-trivial strings which are constant on subintervals of (0, y 1 ), see [12], and we think that our method would work for these less regular diffusions Y , too. But, our proofs rely upon deep results proven in [1], and for these less regular diffusions there are simply no such results available in the literature, that's why we restricted ourselves to strings strictly increasing on (0, y 1 ).
Second, in order to be able to apply Theorem 2.11, for given L x and non-trivial Krein stringm strictly increasing on (0, y 1 ), one would have to understand both regularity, and behaviour near the boundary of R d × (0, y 1 ), of solutions to (2.3). One way of looking into this problem would be via fundamental solutions, and Corollary 2.14 suggests the following approach.
Suppose the diffusion X has got absolutely continuous transition probabilities p t (x, x ′ ), i.e.
where f : R d → R can be any bounded measurable function. Then, Corollary 2.14 would yield where γ y (dt) is the law of τ 0 (Y y ), and γ y (dt, dt ′ ) is the joint law of (τ 0 (Y y ), τ y 1 (Y y )), respectively. Here, the laws γ y (dt) and γ y (dt, dt ′ ) are fully determined bym and do not depend on p t (x, x ′ ). Therefore, if there are special transition probabilities p ⋆ t (x, x ′ ) such that the solutions to (2.3) determined by the corresponding PDE operator satisfy the assumptions of Theorem 2.11, for all non-trivial Krein strings strictly increasing on (0, y 1 ), then the same would hold true for any PDE operator L x whose associated transition probabilities p t (x, x ′ ) 'behave' like p ⋆ t (x, x ′ ). In the remaining part of this section, the special PDE operator will be the Laplace operator ∆ x , and hence the associated special transition probabilities p ⋆ t (x, x ′ ) are given by the heat kernel. The above principle simply tells that our Theorem 2.11 should at least hold true in those cases where p t (x, x ′ ) associated with L x is sufficiently close to the heat kernel, in some sense.
In the case of L x = ∆ x it is rather easy to verify the conditions of Theorem 2.11. Indeed, taking the Fourier transformû (ξ, y) = We at first introduce the solutions to the PDE considered in [15]. Letm be a non-trivial Krein string identified with a locally finite measure on ([0, R), B([0, R))), as discussed at the beginning of Section 2. Denote by f N (λ, y) and f D (λ, y) the solutions to the integral equations respectively, for any fixed λ > 0, and define Then, for any f ∈ L 2 (R d ),û gives the Fourier transform of a weak solution to in the sense of distributions. Note that there is a typo in their definition of φ(λ, y) given in [15,Subsection A.3], which is 'φ λ (s)' in their notation: they wrote (ψ(λ)) −1 instead of ψ(λ). Another typo affects the expression for φ(0, y) given in [15,Subsection A.3]: they wrote ψ(λ) instead of ψ(0). Of course, ψ(0) is short for lim λ↓0 ψ(λ). (b) By Krein's correspondence, see [13,12], any ψ defined above is a complete Bernstein function, that is a function which admits a representation as given in Lemma A.1 with respect to an arbitrary Borel measure µ on [0, ∞) satisfying [0,∞) µ(dλ ′ )/(1 + λ ′ ) < +∞. Vice versa, any complete Bernstein function, except ψ ≡ 0, can be given via (4.2), using a non-trivial Krein string.
(c) We are aware of other definitions of complete Bernstein functions, but they are of course equivalent to ours.
Both ∆ x and ψ(−∆ x ) are considered operators on L 2 (R d ), with respect to the Lebesgue measure, and the domain of ψ(−∆ x ) consists of all f ∈ L 2 (R d ) such that ψ(| · | 2 )f ∈ L 2 (R d ).

Remark 4.3.
Recall that the coefficients of the PDE (2.3) are determined by L x , and the restriction ofm to (0, y 1 ). Thus, the value of m 0 cannot be recovered by taking right-hand limits at zero of partial derivatives of solutions to (2.3). This observation is important because the complete Bernstein function ψ used in Corollary 2.12 actually depends on m 0 , so that −ψ(−L x ) might not map the Dirichlet boundary condition to the Neumann boundary condition, if m 0 > 0, and we indeed see an m 0 -related correction-term in the conclusion of both Theorem 2.11 and Corollary 2.12.
By the same reasoning, (4.4) should only be true if the representation of the complete Bernstein function ψ does not involve a positive m 0 . The minor mistake leading to this issue can be traced back to [15,Subsection A.3]: in their notation, the authors claim that −φ ′ λ (0) = ψ(λ), which is not always true. To correct their statement, a limit corresponding to our term m 0 L x u f (x, 0) in Corollary 2.12 should be added on the right-hand side of (4.4).
Otherwise, if m 0 = 0 and L x = ∆ x , our result under Remark 2.13(b) looks very similar to (4.4), though there are differences. The obvious one is that the limit on the right-hand side of (4.4) is in L 2 (R d ), while our limit ∂ + y u f (x, 0) = lim y↓0 ∂ + y u(x, y) is pointwise in x ∈ R d . Furthermore, the solution u used in (4.4) solves the PDE in R d ×(0, R), while our solution solves the PDE in R d ×(0, y 1 ), and our notion of solution is stronger. Last but not least, equality (4.4) holds true whenever f is in the L 2 (R d )-domain of ψ(−∆ x ), while we require a whole range of assumptions for our result of the same kind. However, we will see below that these are only two different pictures of the same result, if the Dirichlet data f = u f (·, 0) is regular enough.
Let S(R d ) be the Schwartz space of rapidly decreasing functions, which is a core for ∆ x on L 2 (R d ). Fix a non-trivial Krein stringm, choose f ∈ S(R d ), and define ψ &û via (4.2) & (4.3), respectively.
We now verify that induces a solution to (2.3) in the sense of Definition 2.1 satisfying the conditions of Corollary 2.12.
For those properties of φ(λ, y) which we are going to use without proof or further reference, the reader is referred to [15,Subsection A.3].
First, since f ∈ S(R d ), and since φ(λ, y) is bounded but also continuous in y, all partial derivatives ∂ i,j u, 1 ≤ i, j ≤ d, exist and are jointly continuous on R d × (0, R), by dominated convergence. Therefore, the restriction of u to R d × (0, y 1 ) is a solution to (2.3) in the sense of Definition 2.1, and this solution is bounded.
Remark 4.4. (a) The above five steps, which are valid for arbitrary Krein stringsm, show that, if the Dirichlet data f is regular enough, then the solution u constructed in [15] satisfies both the condition for (4.4) and all our conditions for Theorem 2.11 & Corollary 2.12. For the purpose of demonstration, we have chosen Dirichlet data f from S(R d ) which is a popular core for the generator ∆ x in several Banach spaces. However, all arguments used in the above five steps would also work only assuming (1 + |ξ| 2 )f ∈ L 1 (R d ), and such f are in the C 0 (R d )-domain of ∆ x , when C 0 (R d ) is given by (2.7). (b) The only disadvantage of our method is that, despite covering many of them, we do not cover all complete Bernstein functions ψ ≡ 0-see Remark 4.1.
The main difference to our work is that the coefficients of such PDEs only depend on y. However, it would be very interesting to study the Dirichlet-to-Neumann map associated with this type of PDE via trace processes.
for any Borel set Γ ⊆ [0, y 1 ] ∩ R, and this equality specifies the local time of Y used in [1], based on the speed measure m. But, if l y t (Y ) denotes the local time of Y used in [12], then t 0 1 Γ (Y (s)) ds = Γ l y t (Y )m(dy), t ≥ 0, a.s., for any Borel set Γ ⊆ [0, y 1 ] ∩ R. Therefore, since m({0}) =m({0}), if m 0 > 0, then L 0 t (Y ) = l 0 t (Y ), t ≥ 0, a.s., is obvious. However, being an easy consequence of [1, Lemma (6.34)], this equality of the two local times at zero also holds true when m 0 = 0. Note that, L y t (Y ) = 2 l y t (Y ), t ≥ 0, a.s., for all y ∈ (0, y 1 ). The case y = 0 is different because zero is a special singular point for Y . If y 1 was finite, then it would be a special singular point, too.
All in all, the Laplace exponent obtained for the right-inverse of l 0 t (Y ) in [12] is indeed identical to the Laplace exponent of our right-inverse local time L 0 t (Y ) −1 .
Proof of Corollary 2.6. We first show that ∂ + y u(·, y ⋆ ) is locally bounded on R d , for an arbitrary but fixed y ⋆ ∈ (0, y 1 ).
Of course, sup n≥0 |u(x n , y ⋆ )| < +∞ since u(·, y ⋆ ) is continuous and x n → x ∈ R d , n → ∞, so that lim sup n→∞ u(x n , y ⋆ ) = −∞, which contradicts the continuity of u(·, y ⋆ ). All in all ∂ + y u(·, y ⋆ ) is indeed locally bounded on R d . Therefore, ∂ + y u f (x, 0) = lim y↓0 ∂ + y u(x, y) = lim y↓0 ∂ + y u(x, y ⋆ ) + (y,y ⋆ ] L x u(x, y ′ )m(dy ′ ) , x ∈ R d , is a locally bounded function on R d , too, since L x u f is locally bounded on R d × [0, y 1 ). Obviously, for fixed x ∈ R d , the extended version ∂ + y u f (x, ·) defined this way is right-continuous at zero, and thus it is càdlàg on [0, y 1 ) because u(x, ·) is a difference of two convex functions on the interior (0, y 1 ), finishing the proof of the corollary in the case of ∂ + y u f . In the case of ∂ − y u f , the extension can be given by ∂ − y u f (x, y 1 −) = lim y↑y 1 ∂ + y u(x, y ⋆ ) − (y ⋆ ,y] L x u(x, y ′ )m(dy ′ ) , x ∈ R d , though we omit the proof.