Difficulties for d Results for d
Lemma 2.3.1 Let
K
D be the set of continuous probability kernels on D, equip- ped with the topology of uniform convergence, and let
K
′
D be the space of all positive linear operators K :
C
D →
C
D satisfying K 1 = 1, equipped with
the strong operator topology. Then a homeomorfism between
K
D and
K
′
D is given by
K f x = K
x
| f f
∈
C
D. 2.3.5
In particular, K
n
converges to K in the topology on
K
D if and only if K
n
f − K f → 0
∀ f ∈
C
D. 2.3.6
Proof of Lemma 2.3.1: Let K ∈
K
D, and for f ∈
C
D define K
′
f x : =
K
x
| f . By the continuity of K , the function x → K
x
| f is continuous. It is obvious that the map f
→ K
′
f is positive and linear, and therefore continuous, and satisfies K
′
1 = 1. Conversely, by the Riesz-Markov theorem [30], Theorem
IV.14, each such K
′
defines a probability measure K
x
for each x ∈ D. Since
K
′
f ∈
C
D for each f ∈
C
D, the map x → K
x
is continuous in the weak topology. Finally,
P
D is continuously imbedded in
C
D
∗
, the dual of
C
D, so
K
D is continuously imbedded in
C
D,
C
D
∗
, if we equip the latter with the topology of uniform convergence, where the uniform structure on
C
D
∗
is given by the semi-norms l
→ |l| f |. The topology of uniform convergence in
C
D,
C
D
∗
is then defined by the semi-norms p
f f
∈
C
D
given by p
f
F = sup
x ∈D
|Fx| f | F
∈
C
D,
C
D
∗
. 2.3.7
If K ∈
K
D, then p
f
K = sup
x ∈D
|K
x
| f | = K f , so uniform convergence of probability kernels corresponds to convergence of the associated operators in the
strong operator topology. From now on we identify kernels with linear operators as in 2.3.5.
Since D is convex, the Dirichlet problem on D always has a solution. We shall be interested in harmonic functions and functions of constant Laplacian.
Lemma 2.3.2 a For every φ
∈
C
∂ D there exists a unique f
∈
C
D ∩
C
2
D that solves
f = 0
on D f
= φ on ∂ D.
2.3.8 The solution is given by
f = H φ,
2.3.9 where H
∈
K
D is the probability kernel given by H
x
d y = P[B
x τ
∈ dy], 2.3.10
where B
x t
t ≥0
is Brownian motion starting at x and τ
: = inf{t ≥ 0 : B
x t
∈ ∂ D}. 2.3.11
b There exists a unique g
∗
∈
C
D ∩
C
2
D that solves −
1 2
g
∗
= 1 on D
g
∗
= 0 on ∂ D.
2.3.12 The solution is given with τ as in 2.3.11 by
g
∗
x = E
x
[τ ] 2.3.13
and satisfies g
∗
0 on D. There exists an L ∞ such that
g
∗
x ≤ L|x − y|
∀x ∈ D, y ∈ ∂ D. 2.3.14
Proof of Lemma 2.3.2: Formulas 2.3.9 and 2.3.10 can be found in [22], Propo- sition 4.2.7 and Theorems 4.2.12 and 4.2.19. For 2.3.13 see [22], Problem 4.2.25.
The fact that g
∗
0 on D can easily be deduced from the representation 2.3.13, but alternatively one may consult [29], Theorem 2.5. To prove 2.3.14, we assume
without loss of generality that y = 0 and x
1
∀x ∈ D, where for any x ∈
R
d
we write x = x
1
, . . . , x
d
. Now choose L such that |x − ˜x| ≤ L for all x, ˜x ∈ D.
Define a stopping time ˜τ by
˜τ := inf{t ≥ 0 : B
1 t
∈ {0, L}}, 2.3.15
where B
t
= B
1 t
, . . . , B
d t
is d-dimensional Brownian motion. By [22], Prob- lem 4.2.25, we have
g
∗
x = E
x
[τ ] ≤ E
x
[ ˜τ] = x
1
L − x
1
≤ Lx
1
≤ L|x − y|. 2.3.16