Christ M., Kenig C.E., Sadosky C. Harmonic Analysis and Partial Differential Equations: Essays in Honor of Alberto P. Calderon

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104

Chapter 6: Some Extremal Problems in Martingale Theory

To prove (6.23), we shall show that there is a function U that majorizes V

on [ 0, oo) x liii such that, for all n > 1,

EU(fn, gn) < EU(fn-1, gn-1) < ... < EU(fo, go) < 0.

(6.24)

The existence of such a U gives (6.23) as an immediate consequence. Let U

be the function on [ 0, oo) x lFll defined as follows. If lyl < rx, then

U(x, Y) = V (X' y);

(6.25)

if rx < IyI < px, then

(2p - 1) (p -

1)v-1

_p+

Iy2p-, x) (IyI + x)-'-';

(6.26)

U(x, y) _

(p + 1)p-1

(i_P2

and if 0 < px < lyl, then

U(x, y) = (IyI + (p2 - p - 1)x)(Iyl -

x)p-1.

(6.27)

This function, which is continuous, has the desired properties.

(The

method leading to its discovery is briefly described in §6.6.) By Lemma

6.4 below, it majorizes V. Let n > 1. Then, by (6.31) in Lemma 6.5 below

and the condition of differential subordination,

U(fa, gn) < U(fn-1, g.-1) + V(fn-1, gn-1)dn +,O(.fn-1, g.-j) - en.

(6.28)

Using (6.32) in Lemma 6.5, we see that the last two terms in this expres-

sion are integrable and, by the martingale property, integrate to 0. The other

two terms are also integrable so the left side of (6.24) follows. The right side

follows from U(fo,go) < 0: Igol = Ieol < Idol = fo and U(x,y) < 0 if IyI < x,

which is implied by Lemma 6.5 or can be seen directly from the definition of

U. Therefore inequality (6.21) holds. It is sharp as we shall prove below, so

(6.20) also holds.

Lemma 6.4. If p > 2, then V < U on [ 0, oo) x H.

It will be clear from the proof that V (x, y) < U(x, y) if and only if

0 < rx < I yI. Also, notice that, if p = 2, then U = V.

Proof. The majorization of V by U will follow from the continuity of U and

V once we have shown that V < U on each of the domains Dk, 0 < k < 2,

where

Do = {(x, y) : 0 < I yl < rx},

D1 = {(x, y) : rx < I yI < px},

D2 = {(x,y):0<px<IyI}.

Burkholder 105

(i) In view of (6.25), the inequality V < U holds trivially on D0.

(ii)

Let (x, y) E D1. Then jyj + x > 0, so to prove V(x, y) < U(x, y) we can

assume by homogeneity that jyj + x = 1. Thus, jyj = 1- x and the inequality

V (x, y) < U(x, y) is equivalent to F(x) > 0 where

F(x) = (2p (p

)(p

-1)p-1

i l - x -

p22-

p i 1x)

- (1 - x)p +

prp-1xp.

(6.29)

The assumptions on (x, y). imply that x E I where I = (1/(p+ 1), 2/(p+ 1))

but it is convenient to view F as a function on (0, 1), in which case it is easy

to see that F(2/(p + 1)) = F'(2/(p + 1)) = 0, F"(2/(p + 1)) > 0, Fis

positive on (0, 1) so F" is increasing there, and

F"(1/(p+1))

?(p-

1) [rp-1 -pp-3]

(p + 1)p-2

Therefore, F"(1/(p+1)) > 0 if 2< p < 3; F"(1/4) = 0 and F"(1/(p+1)) < 0

if p > 3 as can be seen by checking that the map cp defined on [2, oo) by

<p(p) = (p - 1) log(p - 1) - (p - 1) log 2 - (p - 3) log p

is strictly concave and satisfies W(2) = o(3) = 0.

These properties of F imply that if 2 < p < 3, then F is strictly convex

and positive on I. Now assume that p > 3. Then there is a number zp E I

such that F is strictly concave on [1/(p + 1), zp] and is strictly convex on

[zp, 2/(p+ 1)]. Therefore, F is positive on [zp, 2/(p+ 1)). So F is positive at

both zp and 1/(p+ 1) where F(1/(p+1)) > 0 follows from p+l > (p/(p-1))p,

which holds for p > 3. The concavity implies that F is positive on the whole

interval [1/(p + 1), zp]. Accordingly, V < U on D1. (iii) Now suppose that

(x, y) belongs to D2. Then jyj - x > (p -1)x > 0. Using homogeneity again,

we can assume that IyI - x = 1.

So here jyj = 1 + x and the inequality

V (x, y) < U(x, y) is equivalent to G(x) > 0, where

G(x) = (1 + x + (p2 - p - 1)x) - (1 + x)p+ prp-'xp.

Here the assumptions on (x, y) imply that x E J where J is the interval

(0,1/(p-1)). By (ii), U > V on the intersection of the boundary of D1 with

that of D2. This implies that

G(1/(p - 1)) > 0. (6.30)

If2<p<3,then

G'(x) = p2 - p

- p(1 +

x)p-1 + p2rp-1xp-1

> p2 _ p _ p2p-2[1 +

xp-1] + p2rp-1xp-1

> p2 _ p _ p2p-2

> 0.

106 Chapter 6: Some Extremal Problems in Martingale Theory

The first inequality follows from a classical moment inequality; the second is

a consequence of

p(p -

1)P-1

> 22P-3, p>2,

which follows from 0(2) = 0, Vi'(2) > 0, and ?P"(p) > 0, p > 2, where

V,(p) = loge + (p - 1) log(p - 1) - (2p - 3) log 2;

and the third follows from i(2) = ii(3) = 0 and the concavity of rt on [2, 3],

where 77(p) = p - 1 - 2p-2. Because G(0) = 0 and G' is positive on J, we see

that G is positive on J. If p > 3, then G is concave on [ 0,1/(p - 1)], so by

(6.30) and G(0) = 0, G is positive on J in this case also. Therefore, V < U

on D2. This completes the proof of Lemma 6.4.

Lemma 6.5. There are continuous functions

co :

[ 0, oo) x H -* JR and 0 :

[0,oo)xH-*H such that, ifx>0,x+h>0,yEH, kE)Eli, andIkI5IhI,

then

U(x + h, y + k) < U(x, y) + cp(x, y)h + i(x, y) k

(6.31)

and

Wx,y)I + I

'(x,y)I <- CP(Iy1

+x)P-1.

(6.32)

Proof. The functions cp and 1' are defined on [ 0, oo) x lll( as follows. If IyI < rx,

then

(P(x, y) =

-p2rP-1XP-1

and V)(x, y) = pl yIP-2y;

if rx < IyI < px,

then

W(x, y) = P

(i)P_1

{(p - 2) y- (p2 - p + 1)x}(IyI + x)P2,

(Pi)P_'

(x, y)=p{(2p-1)IyI

-p(p-2)x}(IyI+x)n-2y',

where y' = y/Iyi; and if 0 < px < IyI, then

w(x,y) = p{(p - 2)lyi - (p2 - p - 1)x}(IyI

x)P-2,

&(x, y) = p{IyI +p(p - 2)x}(IyI

x)P-2y'.

It is easy to check that these functions are continuous and satisfy (6.32) for

some choice of c, that does not depend on x or y.

We show now that (6.31) holds. By the continuity of the right and left.

sides of (6.31), it is enough to show this for x > 0,

x + h > 0, and with

I ki 0 {0, r1 hl }. Fix x, y, h, k and define I and J by

Burkholder

107

I = {tE]R:x+ht>0},

J = {t E_ R : (x + ht, y + kt) E Do U D1 U D2 }.

Then I\J is finite: the coefficient of t2 in Iy + ktl2 - m2(x + ht)2 is different

from zero for m = 0, r, p. Define G on I by

G(t) = U(x + ht, y + kt).

Then G and its first derivative G' are continuous on I and

G'(t) = Us (x + ht, y + kt)h + Uy(x + ht, y + kt) k

implying that G'(0) = cp(x, y)h+,O(x, y) k. The inequality (6.31) is therefore

equivalent to

G(1) < G(0) + G'(0),

which follows from the concavity of G. We shall prove this concavity by

showing that if t E J, then G"(t) < 0 since if the continuous function G'

is nonincreasing on each component of J, then it is nonincreasing on I. By

translation, it is enough to prove that if (x, y) E DOUDl UD2, then G"(0) _< 0.

If a > 0, then U(ax, ay) = a&U(x, y). Therefore G"(0) does not change sign

if (x, y) E Dj is replaced by (x/IyI, y/IyI) E D1. So it is enough to prove that

G"(0) < 0 under the assumption that IyI = 1. If (x, y) E Do and IyI = 1,

then 1 < rx and

G"(0) = p(p- 2)(y k)2

+plkI2 - p2(p - 1)rP-ix-2h2

< p(p - 2)h2 + ph2 - p2(p - 1)rh2 (since rx > 1)

= -p(p - 1)(pr - 1)h2

If (x, y) E D1 and IyI = 1, then rx < 1 < px and

G"(0)

= -p

(.4)P_I

x)r-3[(p

- 2)A1 + (h2 - Ik12)Bl],

where, using the positivity of px -1 and px2 - p(p - 3)x + 2p -1, we see that

Al = h2[px2 + p(p + 1)x + 1] + 2h(y k)(p - 1)(px - 1)

-(y k)2[px2 - p(p - 3)x + 2p - 11

> h2[px2 + p(p + 1)x + 1] - 2h2(p - 1)(px - 1)

-h2[px2

- p(p -

3)x + 2p - 1]

108

Chapter 6: Some Extremal Problems in Martingale Theory

and, since x < 1/r, that

B1 = (1 + x)[2p - 1 - p(p - 2)x]

is also nonnegative. Therefore, G"(0) < 0 on Dl.

If (x, y) E D2 and IyI = 1, then 0 < px < 1 and

G"(0) = -p(1 - x)'-'[(p

- 2)A2 +

(h2 - Ik12) B2].

Here

A2 = h2[px2 - p(p + 1)x + 2p - 1] -

1)(1 - px)

-(y.k)2(px2+p(p-2)x+1 -px]

h2[px2 - p(p+1)x+2p- 1] -2h2(p- 1)(1 - px)

-h2[px2+p(p-2)x+1-px]

= 0

in which we have used the positivity of 1 - px and px2 + p(p - 2)x + 1 - px.

Furthermore,

B2 = (1 - x)(1 + p(p - 2)x]

is also nonnegative, so G"(0) < 0 on D2. This completes the proof of (6.31)

and (6.21).

Sharpness. To complete the proof of Theorem 6.3, we shall show that the

inequality (6.21) is sharp by constructing an example. Let op be the positive

number satisfying

NPP = prP-1.

If p > 2, as we continue to assume, then aP is the best constant in the

inequality (6.21).'In fact, if 0 <,6 < /3P, then there is a ±1-transform g of a

martingale f such that

II9IIP>/311f11P

There are several steps in the construction of such a pair f and g. Through-

out, let 0 < 4r5 < 1 and xn = 1 + nb for all nonnegative integers n.

(i) As a first step, let H = (Hn)n>o be a Markov chain with values in the

closed right half-plane such that Ho = (1, p) and

P[H2n+1 = (xn - r(, pxn + rb)

IH2n =

(xn, pxn)]

P[H2n+1 = (2xn,2rxn)IH2n = (xn,pxn)]

P[H2n+2 = (xn+1,pxn+l)IH2n+1 = (xn - r6,pxn+rb)] _

P[H2n+2 = (0, (p - 1)xn+1)IH2n+1 = (xn - rb, pxn + rb)] _

Burkholder

109

Assume further that (2xn,2rxn) and (0,2rxn+i) are absorbing states.

Let

1r0 = 1 and, for n > 1,

()

1 n-I

2r8

¶n=7rn

6 =

1+fSk=0

1+(k+r)S)

If n > 1, then

P[H2n = (xn, pxn)] = irn,

P[H2n = (2xr-1 i 2rxn-1)] =

1rn-11

+ (n - 1 + r)S'

P[H2n = (0, 2rxn)] = 7rn-1

(r + 1)6

1 + (n

1)S

1 + (n - 1 + r)S

I + nS

Let F and G be the martingales defined by Hn = (Fn, Gn), and D the

difference sequence of F. Then F0 = 1, Go = p, and, for n > 1,

Fn=1+EDk,

k=1

Gn = p + 1(-1)kDk.

k=1

(ii) The second step is to show that, if n > 0, then

7rn

_ exp R(n, r, S) (6.33)

(1 + nS)P

where'R(n, r, S) < 2nr62 + lOnr2S2. To see this, note first that, if 0 < t < 1,

then

exp(-t - t2/(1 - t)) = exp(-t/(1 - t)) < 1 - t < exp(-t).

So if 0 < t < 1/2, then 1- t = exp(-t - Rt (t)) where 0 < R1(t) < 2t2. Also,

2rS 2rS

z 2

0 <

1+k5 - 1+(k+r)S

<2r S .

Now use these inequalities and the definition of lr,,, to see that, for n > 1,

1 p

1 n-1

2rb

1 .: nS ex +`- 1 + kS

+ R2(n, r, S)/

k=0

where (R2 (n, r, S)I < lOnr2S2. Then use the inequality

n-1

0 < E 1 it A - 2r log(1 + nb) < 2nrS2

k=0

110

Chapter 6: Some Extremal Problems in Martingale Theory

to obtain (6.33).

(iii) Let b > 0. Choose bn so that 0 < 4rbn < 1 and limn,

nbn = b. Let

(Ff,k)k>o and (Gf,k)k>o be the martingales F and Gas above with b replaced

everywhere by 5n. Then

lim EFp,2n = 1 + 2pr log(1 + b),

(6.34)

n-ioo

lim EGn,2n = p' + 2prpp log(1 + b).

(6.35)

n- 00

To prove (6.34), observe that if n > 1, then

EFn,2n

(1 + nbn)pirn(bn)

n-1

+2p E(1 + k5n)p7rk(bn)

r6,,

k=0

1 + (k + r)5n

exp(R(n, r, bn))

n-1

+2p E exp(R(k, r,

bn))rbn

1 + kbn

k=0

n-1

+2p E exp(R(k, r, bn))R3(k, r, 5n),

k=0

where IR3(k, r, bn)I < r26n2. Consequently, (6.34) holds. Similarly, if n _> 1,

then

EGn,2n

r n

+ 2prp E(1 + kbn)p7rk(6n)1

+ (k + r)bn

k=0

n-1

+ 2PrP E(1 + (k + l)bn)p7rk(bn)1

1)brr)&

1 +

kbn

k=0

1)bn'

pn(1 + n5n)pirn(bn)

n-1

which converges to pp + 2PrP[r + (r + 1)]log(1 + 6). Therefore, (6.35) also

holds.

(iv) Now choose b so that

1 + 1[pp + 2prpplog(1 + b)]

2pp + i [1 + 2pr log(1 + b)] >

gyp'

{6.36}

This is possible because /3p > Q and the limit of the left side of (6.36) as

b -+ oo is /3p. After this choice of b, choose n so that

'-+ IEGP

n,2n

> ,Bp. (6.37)

5 +

EFn,2n

Burkholder

111

This is possible by (6.34), (6.35), and (6.36).

(v) Suppose that A E F is independent of the Markov chain H and

satisfies P(A) = 1/2.

(If necessary, the original probability space can be

replaced by a slightly larger one to make such a choice possible.) With b and

n as in (iv), let fo = go = (p + 1)/2. For 0 < k < 2n, let

(fk+1,9k+1)

= (r'n,k, Gn,k) on A,

(p,1)offA.

For k > 2n, let (fk+1, 9k+1) = (fen+1, 92n+1) From (i), it follows that f is a

martingale and g is a ±1-transform of f. Furthermore, the ratio

II91ip/Ilf11p

is given by the left side of (6.37). Therefore, I I9I

Ip >

011f 11p. This shows that

P. gives the best constant and completes the proof of Theorem 6.3.

6.4 A Sharp Norm Inequality for an Integral

with Respect to a Nonnegative

Martingale

Suppose that X = (Xt)t>o is a nonnegative martingale on a complete prob-

ability space (1, F, P) filtered by (.Ft)t>o, a nondecreasing right-continuous

family of sub-v-algebras of F where FO contains all A E .T with P(A) = 0.

So each Xt is integrable and the conditional expectation E(XtIF,) = X, if

0 < s < t. Let Y be the Ito integral of H with respect to X where H is a

predictable process with values in the closed unit ball of the Hilbert space

E[:

Yt = HoXo +

H. dX,.

(0.t]

Both X and Y are adapted to the filtration (.Ft)t>o and are right-continuous

on [ 0, oo) with limits from the left on (0, oo). Set IIX IIp = suPt>o IIX=IIp

Theorem 6.6. Let 2 < p < oo and r = (p - 1)/2. Then

IIYII, <_ PrP-11UXIIp

(6.38)

and pry-1 is the best possible constant.

The inequality (6.38) follows from (6.21) in the same way that the analo-

gous inequality for nonnegative submartingales in [7] follows from the corre-

sponding discrete-parameter version. The fact that pry-1 is the best possible

constant in (6.38) follows from the fact that it is the best possible constant

112

Chapter 6: Some Extremal Problems in Martingale Theory

in (6.21) even in the special case that g is a ±1-transform of f as proved in

§6.3.

Let us now modify the definition of 8p(posmar) that is given in §6.2. Here

let fp(posmar) be the least extended real number 0 such that I]YII, < 0[[X IIp

for all X and Y as above. By Theorem 6.6 and earlier results in [3] and (7],

we obtain the same value as in the discrete-parameter case:

i3p(posmar) = p` - 1 = 1/(p - 1) if 1 < p:5 2,

= prp-1 if 2 < p < oo.

If the definitions of /3p(mar) and /3p(possub) are modified in the same way,

then (6.13), (6.14), and (6.15) hold also in this case.

We note here that the sentence near the top of page 996 of [7] beginning

with "In fact, the constant . . . " is not clear in its context. It should read

"In fact, if 1 < p < 2, then the constant p' - 1 is already the best possible

for the smaller class of nonnegative martingales."

6.5 Some Open Questions and Remarks

(i) The following question from [4] is still open even for H = R. Let M and N

be right-continuous martingales on [ 0, oo) with limits from the left on (0, oo).

Suppose they are martingales relative to the same filtration, are ]Hl-valued,

and have respective quadratic variation processes [M, M] and [N, N]. If

[N, N]t < [M, M]t for all t > 0,

(6.39)

then does IINIIp < (p`

- 1)IIMIIp

for all p c (1,oo)? If so, is this inequality

true under the less restrictive condition

[M, M],,, < [N, N],,,,?

(6.40)

There are analogous questions for weak-type, exponential, and other mar-

tingale inequalities. Wang [14] and Baiiuelos and Wang [2] consider the

following condition:

[M, M]t - [N, N]t is nonnegative and nondecreasing in

t. (6.41)

Under this more restrictive condition, Wang [14] proves the inequality

I I NI Ip < (p - 1) I ] M I l p for all p e (1, oo). He also proves similar extensions

of some of the other inequalities in [4] and [7].

In [2], where the proof of

IINIIp < (p* - 1)IIMI[p is given under the condition (6.41) in the special case

that M and N are continuous, Bahuelos and Wang give applications of this

inequality to the Beurling-Ahlfors transform.

Burkholder

113

(ii) Let a > 0. What is the effect of replacing condition (6.10) by

fOvj < alDul,

(6.42)

and condition (6.18) by

for all n > 1?

(6.43)

Changsun Choi [9] has proved that if 0 < a < 1, then (6.12) holds if p" is

replaced by max{(a+I)p, q}, and that the same replacement in the analogous

inequality for nonnegative submartingales (inequality (2.1) of [7]) gives the

best constant there. Earlier, Hammack [11] studied the effect of dropping the

condition that the submartingale be nonnegative while keeping the original

condition (6.18). In this case, the weak-type inequality (inequality (4.1) of

[7]) continues to hold but with 6, rather than 3, being the best constant. On

the other hand, the LP-inequalities fail to hold for 1 < p < oo.

6.6 A Note on Method

Here are the steps that can lead in a natural way to the function U of §6.3.

Start with H = R and

> 0, and as in §6.3, let p > 2. Let S = [ 0, oo) x R

and define V,6 : S -a R by VV (x, y) = lyIP - /3pxp. Then attempt to find the

least majorant U of Vfi on -S such that the mapping

x ti U(x, y + ex)

is concave on [ 0, oo) for all real y and for both e = 1 and c _ -1. There is

no such U if $ is too small. However, it seems reasonable to expect that the

least $ for which there is such a majorant of VV will lead to a U that has

some smoothness, perhaps continuous derivatives of the first order on the

interior of S. Furthermore, if this U is least possible and the second-order

derivatives exist in a neighborhood of (x, y) E S, then either

U:x+2Uy+Uyy=0 or Uxx-2Uy+Uvv=0

at (x, y). Otherwise, U would not be extremal. With the use of all of this,

the right value of $ and the right function U are not hard to find for the

case H = H. For a general Hilbert space H, the right function U has the

same formula (interpret the absolute value jyj as the norm of y E HH) as can

be seen from the work in §6.3. Perhaps further insight may be gained from

Theorem 1.1 of [3] and §2 of [6]. The extremal problems for martingales that

we have considered here focus mostly on the comparison of the sizes of two

martingales related in some way, such as one being differentially subordinate

to the other.

These problems lead to interesting (upper) boundary value

problems.

If instead, the focus is on the comparison of the sizes of their

maximal functions, then the boundary value problems become even more

interesting; see [8].