Solution to selected problems.

20. Applying Ito’s formula to u ∈ C 2 (R 3 − {0}) and B t ∈ R 3 \{0}, ∀t, a.s.
u(B t ) = u(x) +
∫ t
0
∇u(B s ) · dB s + 1 2
∫ t
0
△u(B s )ds = 1
‖x‖ + 3∑
i=1
∫ t
0
∂ i u(B s )dB i s
and M t = u(B t ) is a local martingale. This solves (a). Fix 1 ≤ α ≤ 3. Observe that E(u(B 0 ) α ) <
∞. Let p, q be a positive number such that 1/p + 1/q = 1. Then
∫
E x (u(B t ) α 1 1
‖x−y‖2
) =
R 3 ‖y‖ α e− 2t dy
(2πt) 3/2
∫
∫
1 1
‖x−y‖2
1 1
‖x−y‖2
=
{B(0;1)∩B c (x;δ)} ‖y‖ α e− 2t dy +
(2πt) 3/2
R 3 \{B(0;1)∩B c (x;δ)} ‖y‖ α e− 2t dy
(2πt) 3/2
∫
≤
sup
y∈{B(0;1)∩B c (x;δ)}
+
1
(2πt) 3 2
( ∫
R 3 \{B(0;1)∩B c (x;δ)}
e − ‖x−y‖2
2t ·
B(0;1)
1
‖y‖ α dy
1
‖y‖ αp dy ) 1/p ( ∫
R 3 \{B(0;1)∩B c (x;δ)}
(
1
‖x−y‖2
e− 2t
(2πt) 3/2
) q
dy) 1/q
Pick p > 3/α > 1. Then the first term goes to 0 as t → ∞. In particular it is finite for all t ≥ 0.
The first factor in the second term is finite while the second factor goes to 0 as t → ∞ since
∫
1 1
(2πt) 3/2 (2πt)
3(q−1)/2
e−
‖x−y‖2
2t/q
dy =
∫
1
(2πt) 3(q−1)/2
1 1
‖x−y‖2
q 3/2 e− 2t/q
dy =
(2πt/q) 3/2
1 1
(2πt) 3(q−1)/2 q 3/2
and the second factor is finite for all t ≥ 0. (b) is shown with α = 1. (c) is shown with α = 2. It
also shows that
lim
t→∞ Ex (u((B t ) 2 ) = 0
which in turn shows (d).
21. Applying integration by parts to (A ∞ − A·)(C ∞ − C . ),
(A ∞ − A·)(C ∞ − C . ) = A ∞ C ∞ −
∫ ·
Since A, C are processes with finite variation,
[A, C] t = ∑
△A s △C s
∫ t
0
(A ∞ − A s− )dC s =
Substituting these equations,
0