Errata
p. 2, line 11 down: Ai1 ∩ · · · ∩ Ain
p. 4, line 4 down: P(N < ∞) > 0 =⇒ EP [N ] < ∞
p. 6, line 13 up: ω ∈ [0, 1)
p. 8, line 13 down: Change [0, ∞)2 to [0, 1)2 .
p. 27, line 11 up: Change Ξ−1 to (Λ0 )−1
Exercise 1.3.19 on pp. 32–33: There are a number of annoying errors here.
In line 4 up on p. 32, etz should be e−tz . In line 6 down on p. 33, the last
2
2 − tδ2
expression should be tδ
e
. The most serious error is in the estimation of
E(t, δ) starting on line 8 on p. 33. One should write E(t, δ) as
Z
δ
e−
tz 2
2
etR(z) + etR(−z) − 2 dz.
0
One should then use a second order Taylor’s expansion to write etR(z) +etR(−z) −2
2
as t R(z) + R(−z) + F (t, z), where |F (t, z)| ≤ t2 R(z)2 + R(−z)2 etR(−|z|)| .
Now observe that
|R(z) + R(−z)| =
∞
X
z 2m
z4
≤
.
m
2(1 − z 2 )
m=2
Thus, for δ ≤ 21 , one can dominate |E(t, δ)| by a constant times the sum of
Z
δ
2
4 − tz2
|z| e
t
dz
and t
0
2
Z
δ
|z|6 e−
tz 2
2
dz,
0
3
and so there is a C < ∞ such that |E(t, δ)| ≤ Ct− 2 for all t ≥ 1 and 0 < δ ≤ 12 .
p
Finally, take δ = 3t−1 log t to arrive at
Γ(t + 1) r 2π 3
t+1 −t −
≤ Ct− 2
t e
t for t ≥ 9, and from this get (1.3.21).
p. 43, line 7 up: supn≥1 EP [Xn2 ] < ∞
1
2
Errata
p. 43, line 3 up: Change to
P
max |Sm | ≥ 2t
1≤m≤n
≤
P(|Sn | ≥ t)
.
1 − max1≤m≤n P(|Sn − Sm | ≥ t)
1
1
p. 45, line 6 down: 2− p ∨1 EP |X|p p
p. 50, line 6 down: Theorem 1.3.15
p. 55, line 4 down: Change to
1
1
P |S̃[β m ] | ≥ β 2 ≤ 3 exp −β 2 log(2) [β m ] .
p. 55, lines 13–14 down: Change to
lim S̃n − a ≤ inf lim S̃km − a
k≥2 m→∞
n→∞
Skm − Skm−1
= lim lim − a
Λkm
P-almost surely.
k→∞ m→∞
p. 55, line 2 up: P |S̃n − a|
p. 62, line 15 up: Change “it introduced” to “it was introduced”
p. 66, line 10 down: Change to “Show that (2.1.1) for ϕ ∈ Cc∞ (R; R)”
p. 83, line 1 down: Change C-valued to R-valued.
p. 83, line 18 up: Exercise 2.4.37
p. 88, line 1 up: Delete ±
p. 89, line 6 down: (e, X)RN
p. 91, line 6 down: ≤ 2−2
p. 90, lines 10 and 4 up: Change to ≤
2
ξ 2 σm
2Σ2n
and ≤
2
R 4 rn
4
p. 91, line 5 down: ≤ 2−2
−1
p. 95, lines 9–8 up: The hint should read (Πξ)(2) = C(22)
C(21) ξ(1) if ξ(2) = 0(2)
and Πξ = ξ if ξ(1) = 0.
p. 98, footnote: Change to “footnote in Exercise 2.1.13”
p. 103, line 9 up: Delete “In”
2
; R)
p. 111, line 5 down: converges in L2 (γ0,1
Errata
3
p. 116, line 1 up: Change to
`
X
ν` Γ ∩ B(0, `) =
νk Γ ∩ B(0, k) \ B(0, k − 1)
k=1
p. 117, line 2 down: Change µ` to ν` .
p. 122, line 2 down: Reduce to the case when µ is symmetric, µ({0}) > 0,
and therefore
p. 122, line 19 down: Change “for all ξ ∈ RN ” to “uniformly on compacts”
p. 133, line 10 down: Change +A (1 − ηR )ϕ to −A (1 − ηR )ϕ in final the
expression.
p. 134, line 5 up: Change the second expression to
Z
1 − η(y) 2
|y| M (dy)
|ξ|
|y|
B(0,r)
p. 136, line 3 down: Change `µ (R−1 ξ 0 − x) to `µ (R−1 ξ 0 − ξ) in final expression.
R
p. 136, line 6 down: Insert limR→∞ before RN η̂R (ξ 0 ) dξ 0 .
p. 136, line 9 up: Insert y before µ n1 (dy).
p. 138, line 2 down: δmµ ? πM r (−∞, 0) = 0
R
p. 138, line 8 down: mη0 ≥ R η0 (y)y M (dy)
n
n
p. 139, line 6 down: Change 2 α to 2− α
1
p. 141, line 9 down: Change t = 2 to t = 2 α .
p. 142, line 7 down: Insert ϕ2 (|y|) before M (dy).
p. 143, line 12 down: Change righthand side to − Γ(1−α)
α
√ζ
−1
α
.
p. 143, lines 11–5 up: Change to:
In addition, for ξ > 0,
Z
Z
√
e−r − 1
ξα
Γ(1 − α) α
α
fα ( −1ξ) = ξ
dr = −
r−α e−r dr = −
ξ .
1+α
α (0,∞)
α
(0,∞) r
Hence, since − Γ(1−α)
α
√ζ
−1
α
is also analytic in C+ , fα (ζ) = − Γ(1−α)
α
√ζ
−1
α
for ζ ∈ C+ . By continuity, this equality extends to C+ .
p. 146, line 5 down: Change to “rotationally invariant, symmetric, α-stable”
4
Errata
p. 146, line 1 up Change to:
1-stable, one can use `µ (ξ) = `µ (−ξ) to see that m = 0 and that the ν in the
expression for `µ (ξ) can be taken to be symmetric.
p. 147, lines 3 & 7R down: Insert “symmetric” after “rotationally invariant”
in line 3, and delete SN −1 in line 7.
p. 148, line 9 up: Change “Exercise 3.3.11” to “Exercise 3.3.12”
p. 157, line 5 down: Change to ντ = δyτ
p. 160, lines 5–7 down: Although it is true that FD(RN ) ( BD(RN ) , the given
open set does not provide a proof. Instead, let ∆ be a subset of [0, 1], and set
A = {1[t,∞) : t ∈ ∆}. Show that A is a closed subset of D(R) and that the map
t
1[t,∞) ∈ D(R) is measurable as a map from (R, B[0,1] ) into D(R, FD(R) .
Conclude that if A were an element of FD(R) , then ∆ would have to be Borel
measurable. Hence, A ∈ BD(R) \ FD(R) if ∆ ∈
/ B[0,1] .
p. 160, line 15 up: Change C(RN ) to D(RN )
p. 162, line 6 up: Change the description of the set B to
K
\
nk−1 +1
(
X
(τ1 , . . . , τnK ) ∈ (0, ∞)NK :
m=1
k=1
τm > tk−1 &
nk
X
)
τm ≤ tk
m=1
The reduction to the case when 1 ≤ n1 < · · · < nK is not so simple as
indicated. Perhaps a better approach is to work by induction on K ≥ 1. The
case K = 1 is already covered. Thus, assume that K ≥ 2 and that the result
holds for K − 1. If nK = 0, there is nothing to do. If nK ≥ 1 and n1 = 0, set
K0 = min{k : nk ≥ 1}. Then 2 ≤ K0 ≤ K and, by the induction hypothesis,
P N (tk ) = nk , 1 ≤ k ≤ K = P τ1 > tK0 −1 & Tnk ≤ tk < Tnk +1 , K0 ≤ k ≤ K
= P τ1 ∈ (tK0 −1 , tK0 ] & Tnk − τ1 ≤ tk − τ1 < Tnk +1 − τ1 , K0 ≤ k ≤ K
Z tK0
=
e−t P Tnk −1 ≤ tk − t < Tnk , K0 ≤ k ≤ K dt
tK0 −1
=e
−tK
Y
K0 <k≤K
= e−tK
Y
1≤k≤K
(tk − tk−1 )nk −nk−1
(nk − nk−1 )!
(tk − tk−1 )nk −nk−1
.
(nk − nk−1 )!
Z
tK0
tK0 −1
(tK0 − t)nK0 −1
dt
(nK0 − 1)!
Errata
5
If n1 ≥ 1, then, by the preceding case when n1 = 0,
P N (tk ) = nk , 1 ≤ k ≤ K
= P Tn1 ≤ t1 & Tn1 +1 − Tn1 > t1 − Tn1
& Tnk − Tn1 ≤ tk − Tn1 < Tnk +1 − Tn1 2 ≤ k ≤ K
1
=
(n1 − 1)!
Z
e−tK
(n1 − 1)!
Z
=
t1
tn1 −1 e−t P N (t1 − t) = 0 & N (tk − t) = nk − n1 , 2 ≤ k ≤ K dt
0
t1
tn1 −1
0
K
K
Y
Y
(tk − tk−1 )nk −nk−1
(tk − tk−1 )nk −nk−1
dt = e−tK
.
(nk − nk−1 )!
(nk − nk−1 )!
k=2
k=1
p. 170, line 17 down: Change 4.3.11 to 4.3.21
X (m+1)2−n −X(m2−n )
X (m+1)2−n +X(m2−n )
p. 182, line 5up: Change
to
2
2
h
i p1
p. 182, line 4 up: EP supt∈[0,1] kXn+1 (t) − Xn (t)kpB
p. 188, line 6 down: Change to
!
P
sup e, B(t) RN ≥ R
t∈[0.T ]
R2
≤ 2P e, B(T ) RN ≥ R ≤ 2e− 2T
R2
In order to get the second inequality, one has to show that γ0,1 ]R, ∞) ≤ 12 e− 2 .
q
R ∞ x2
R2
2
This follows trivially from R e− 2 dx ≤ R−1 e− 2 when R ≥
π . When
q
R R −x2
R2
R ≤ π2 , one has to show that 12 − √12π 0 e− 2 dx ≤ 21 e− 2 , or, equivalently,
q R
−x2
R2
R
that 1 ≤ e− 2 + π2 0 e− 2 dx. Finally, observe that the right hand side is
q
non-decreasin of 0 ≤ R ≤ π2 .
p. 190, lines 16 & 8 up: Change 16 up to
p  p1


EP 
sup
x,y∈[0,T ]ν
y6=x
and
QN
i=1
to
Qν
i=1
ν
kX̃(y) − X̃(x)kB  
+r−α
.
  ≤ K(ν, r, α)CT p
α
|y − x|
in line 8 up
p. 191, line 11 down: On B(1)2 righthand side should be replaced by
P∞
p. 192, line 7 up: n=m L(2−n−1 ) ≤ CL(2−m−1 )
B(1)2
2 .
6
Errata
p. 192, line 4 up:
P kB − Bn k[0,1] ≥ RL(2
−n−1
∞
X
) ≤
P Mm+1 ≥ C −1 RL(2−m−1 )
m=n
p. 192, line 2 up:
−1 2
P Mn ≥ RL(2−n ) ≤ 2n(1−2 R )+1
p. 199, line 14 up: Change hEP [X], x∗ i = EP [hX, x∗ i] to hEµ [X], x∗ i =
Eµ [hX, x∗ i].
RR
RR
p. 222, line 3 up: Change
to
B
p. 230, line 12 up: Replace P Fn by P Fn .
p. 237, line 4 down: Change the right hand side to
1
α
Z
{M(η) f ≥ α} ∩ B(0, R)|f (y) dy
p. 259, line 11 up: Change u : E × F −→ R to u : E 2 −→ R
y
y
,
h
to
,
h
p. 259, lines 4 up: Change kyk
kykE
F
F
E
p. 263, line 13 down:
h
1
i 2p
P
2p sup E kXn − Xn−1 kE Fn−1
n∈Z+
p. 268, lines 2 & 6 down: Insert
to ξ in line 6
L2p (µ;R)
1
2
before (ξ, Cξ)RN in line 2, and change x
p. 269, line 6 down: Change ω ∈ Θ to ω ∈ Ω
p. 273, line 10 down: Change A ∈ Bt × Ft to A ∈ B[0,t] × Ft .
pp. 283 & 284, lines 1 up & 3, 6, 8 down: Change e
|ξ|2
2
to e
1+|ξ|2
2
p. 287, line 6 up: Change Fk−1,n to Fζk−1,n
p. 296, lines 11 down, 8 up, and 5 up: Replace p(−1,1) r−2 , r−1 (x −
c), r−1 (y − c) by p(−1,1) r−2 t, r−1 (x − c), r−1 (y − c) in 11 down, change {x +
ψ(s) + δs ψ(τ ), τ ∈ [0, t − s]} to {x + ψ(s) + δs ψ(τ ) ∈ I, τ ∈ [0, t − s]} in 8 up,
and change (x − x) to (c − x) in 5 up
Errata
7
p. 298, line 10 down:
h
i
−EP γ0,(1−ζxG )I Γ − x − ψ(ζxG ), ζxG ≤ t .
p. 301, Lemma 8.1.2: The statement of this lemma should be amended to
say that µ̂ is a continuous function of weak* convergence on bounded subsets
of E ∗ . The point is that Lebesgue’s Dominated Convergence Theorem applies
to sequences but not to nets. By Exercise 5.1.9, weak* convergence of bounded
sets admits a metric and therefore continuity can be checked with sequences.
p. 306, line 3 up: Replace P(kXkE ≤ r) ≥
3
4
by P(kXkE ≤ r) ≥
9
10 .
p. 307, lines 14 down & 7 up: In 14 down, replace P(kXkE ≤ R) ≥ 34 by
n+1
9
P(kXkE ≤ R) ≥ 10
. In line 7 up, replace P kXkE ≥ 32 2 R by P kXkE ≥
n
32 2 R .
p. 308, line 9 down: Replace kx∗ kE ∗ by khx∗ kH .
P
P∞
p. 309, line 14 down: Replace g = m=0 by g = m=0 .
p. 310, line 2 down: Replace khx∗ k2H by (hx∗ , hy∗ )H .
p. 311, Theorem 8.2.6: Like the analogous one in Lemma 3.1.2, the initial
statement of continuity in this result should be changed to x∗ ∈ E ∗ 7−→ hx∗ ∈ H
continuous from the weak* topology to the strong topology on bounded subsets
of E ∗ . In fact, only when H is finite dimensional does this continuity hold for
the whole space.
p. 311, line 17 down: Replace h ∈ H 7−→ h ∈ E by h ∈ BH (0, R) 7−→ h ∈ E.
p. 311, line 11-9 up: Change these lines to
|hhk − h, x∗ i| ≤ + min hk − h, hx∗ − hx∗m H + 2R ≤ (2R + 1),
1≤m≤n
which proves that khk − hkH ≤ (2R + 1) for k ≥ `.
p. 314, line 11 & 10 up: Change n + 13 to (n + 1)3 and remove {X n : n ≥ 0}
p. 315, line 11 & 9 up: Replace 72 by 180.
p. 319, line 1 up: Change first sum to
∞
X
(t − k)+ ∧ 1Xk,0
k=0
p. 323, 11 up: Replace
P`m
i=1
kg̃m,i − gm,i kH by
P`m
p. 325, line 5 down: Change kQn xk k to kQn xk kE
i=1
kg̃m,i − ΠLn gm,i kH .
8
Errata
p. 328, lines 2 & 15 down: Change “rotation” to “orthognal”
p. 331, line 4–9 down:
Hint: Begin by showing that the inequality is trivial when either n = 1 or
kAA> kop ≥ 2. To prove it when n ≥ 2 and kAA> kop ≤ 2, set ∆ = IRn − AA> ,
show that

 21
`−1
X
1
2 
a2jj 
|a`` an` | ≤ |∆n` | + (AA> )nn
for 1 ≤ ` < n
j=1
|1 − ann | ≤ |1 − a2nn | ≤ ∆nn +
n−1
X
a2n` ,
`=1
and proceed by induction on n.
√
2
m+1
P∞
p. 332, line 6 up: m=1 Xm (θ) +(−1) m2 8X0 (θ)Xm (θ)
Q∞ 2
p. 332, line 1 up: sinh z = z m=1 1 + mz2 π2 .
p. 334, line 1 up: θT [0, T ] ∈ ΘT (RN ).
p. 335, lines 18, 11, & 7 up: Change H(RN ) to H1 (RN )
p. 337, line 3 up: Change W BE (hx∗ , r) to W BE (h, r) .
p. 340, lines 3 & 9 down: Change kxkE ∗ = 1 to kx∗ kE ∗ = 1 in line 3 and
x∗ ∈ E ∗ to x∗ ∈ BE ∗ (0, 1) in line 9
p. 346, line 6 up: Replace h ∈ H(RN ) by h ∈ H1 (RN )
p. 347, lines 3 & 13 down: Change kθk to kθkΘU (RN ) in line 3 and kη U ku to
kη U kΘU (RN ) in line 13
p. 348, line 8 down: Change to
Z
[0,∞)
∂τ u0 (s, τ )∂τ u0 (t, τ ) + 14 u0 (s, τ )u0 (t, τ ) dτ = u0 (s, t)
p. 350, line 13 up: Change S (RN ; R) to S 0 (RN ; R)
p. 357, line 11 down: Change “translation invariant” to “Euclidean invariant”
p. 359, line 12 up: Change 2n for n
p. 364, lines 4 & 5 down: Change
β
m
2
δ
Λ[β m−1 ]
to
δΛ[β m−1 ]
β
m
2
.
p. 381, line 5 down: Change L(µk , (Yk,n )∗ ) to L(µk , (Yk,n )∗ P)
Errata
9
p. 383, line 11 down: change the statements of parts (ii) and (ii) of Exercise
9.1.17 as follows.
(ii) For ` ∈ Z+ , let π` be the natural projection map from E onto E` , and
show that K ⊂⊂ E if
\
K=
π`−1 (K` ), where K` ⊂⊂ E` for each ` ∈ Z+ .
`∈Z+
Conclude from this that A ⊆ M1 (E) is tight if and only if (π` )∗ µ : µ ∈ A ⊆
M1 (E` ) is tight for every ` ∈ Z+ .
Q`
Next, set E` = k=1 Ek , and let π` denote the natural projection map from
E onto E` . Show that for each ϕ ∈ UbR (E; R) and > 0 there is an ` ≥ 1 such
that |ϕ(y) − ϕ(x)| < for all x, y ∈ E with π` (x) = π` (y). Use this to show
that µn =⇒ µ in M1 (E) if and only if hϕ ◦ π` , µn i −→ hϕ ◦ π` , µi for every ` ≥ 1
and ϕ ∈ Cb (E` ; R).
(iii) Let µ[1,`] be an element of M1 E` , and assume that the µ[1,`] ’s are consistent in the sense that, for every ` ∈ Z+ ,
µ[1,`+1] Γ × E`+1 = µ[1,`] (Γ)
for all Γ ∈ BE` .
Show that there is a unique µ ∈ M1 (E) such that µ[1,`] = (π` )∗ µ for every
` ∈ Z+ .
p. 396, line 2 down: Change to “In order to remove the assumption on the
fourth moment”
p. 399, line 6 up: Change § 4.2.2 to § 4.2.1
p. 406, lines 1 down and 1 up: Change “Exercise 8.2.16” to “Exercise 8.3.19”
p. 415, line 10 down: Change the RHS to
h(T,x,y)
g (N ) (T,ψ(T )−x)
p. 434, line 3 up: Change ϕ ∈ Cb2 (G; R) to “bounded ϕ ∈ C 2 (G; R)”
p. 450, line 2 down & 14 up: Change e
p. 452, line 8 down: ≤
π2
4
t
to e
π2
8
t
and (ii) to (iii)
tλM
e− 2
N
(πt) 2
p. 464, line 8 up: Change to Cb (∂G; R)
p. 472, line 5 up: The assertion in Exercise 11.1.36 is false and so this exercise
should be ignored.
p. 473, lines 1 & 10 down: Change (ii) to (i) and (iii) to (ii).
(N )
p. 473, line 2 up: Wy ∃t ∈ [0, ∞) ψ(t) = ϕ(t)
10
Errata
p. 474, line 7 down: Change RHS to supy∈R |u(r, y)| ∨ |u(−r, y)|
p. 475, line 3 down: Change uρ (z) = to uρ (z) ≥
Rs R
p. 477, line 5 down: Delete lims&0 0 G ϕ(y)pG (t, x, y) dy) dt on RHS
p. 479, line 2 up: Change Cc (G; R) to Cc1 (G; R)
p. 481, line 10 up: Replace B by B 2
p. 491, line 5 up: Change {Bt : t ∈ [0, ∞)} to {Ft : t ≥ 0}
p. 502, line 1 down: Theorem 11.4.6
p. 508, line 11 down: Change (ii) to (iii)
p. 515, line 1 down: Let G be a connected open subset