Exercises
7.1 Functions of Bounded Mean Oscillation
7.1.1 Definition and Basic Properties of BMO
Definition 7.1.1. For f a complex-valued locally integrable function on Rn, set f
BMO=sup
Q
1
|Q|
Q
f(x)−Avg
Q
f dx,
where the supremum is taken over all cubes Q in Rn. The function f is called of bounded mean oscillation iff
BMO <∞and BMO(Rn)is the set of all locally integrable functions f on Rnwithf
BMO<∞.
Several remarks are in order. First it is a simple fact that BMO(Rn)is a linear space, that is, if f,g∈BMO(Rn)andλ∈C, then f+g andλf are also in BMO(Rn) and
f+g
BMO ≤ f
BMO+g
BMO, λf
BMO = |λ|f
BMO. But BMOis not a norm. The problem is that iff
BMO=0, this does not imply that f =0 but that f is a constant. See Proposition 7.1.2. Moreover, every constant function c satisfiesc
BMO=0. Consequently, functions f and f+c have the same BMO norms whenever c is a constant. In the sequel, we keep in mind that elements of BMO whose difference is a constant are identified. Although BMO is only a seminorm, we occasionally refer to it as a norm when there is no possibility of confusion.
We begin with a list of basic properties of BMO.
Proposition 7.1.2. The following properties of the space BMO(Rn)are valid:
(1) Iff
BMO=0, then f is a.e. equal to a constant.
(2) L∞(Rn)is contained in BMO(Rn)andf
BMO≤2f
L∞.
(3) Suppose that there exists an A>0 such that for all cubes Q in Rnthere exists a constant cQsuch that
sup
Q
1
|Q|
Q|f(x)−cQ|dx≤A. (7.1.1) Then f∈BMO(Rn)andf
BMO≤2A.
(4) For all f locally integrable we have 1
2f
BMO≤sup
Q
1
|Q|inf
cQ
Q|f(x)−cQ|dx≤f
BMO.
(5) If f∈BMO(Rn), h∈Rn, andτh(f)is given byτh(f)(x) =f(x−h), thenτh(f) is also in BMO(Rn)and
τh(f)
BMO=f
BMO.
(6) If f ∈BMO(Rn)andλ >0, then the functionδλ(f)defined byδλ(f)(x) = f(λx)is also in BMO(Rn)and
δλ(f)
BMO=f
BMO.
(7) If f∈BMO then so is|f|. Similarly, if f,g are real-valued BMO functions, then so are max(f,g), and min(f,g). In other words, BMO is a lattice. Moreover,
|f|
BMO ≤ 2f
BMO, max(f,g)
BMO ≤ 3 2
f
BMO+g
BMO
, min(f,g)
BMO ≤ 3 2
f
BMO+g
BMO
.
(8) For locally integrable functions f define f
BMOballs=sup
B
1
|B|
B
f(x)−Avg
B
f dx, (7.1.2) where the supremum is taken over all balls B in Rn. Then there are positive constants cn,Cnsuch that
cnf
BMO≤f
BMOballs≤Cnf
BMO.
Proof. To prove (1) note that f has to be a.e. equal to its average cNover every cube [−N,N]n. Since[−N,N]nis contained in[−N−1,N+1]n, it follows that cN=cN+1
for all N. This implies the required conclusion. To prove (2) observe that Avg
Q
f−Avg
Q
f ≤2 Avg
Q |f| ≤2f
L∞.
For part (3) note that
f−Avg
Q
f ≤ |f−cQ|+ Avg
Q
f−cQ ≤ |f−cQ|+ 1
|Q|
Q|f(t)−cQ|dt. Averaging over Q and using (7.1.1), we obtain thatf
BMO≤2A. The lower in- equality in (4) follows from (3) while the upper one is trivial. Property (5) is imme- diate. For (6) note that AvgQδλ(f) =AvgλQf and thus
1
|Q|
Q
f(λx)−Avg
Q
δλ(f) dx= 1
|λQ|
λQ
f(x)−Avg
λQ f dx.
Property (7) is a consequence of the easy fact that |f| −Avg
Q
|f| ≤ f−Avg
Q
f +Avg
Q
f−Avg
Q
f .
Also, the maximum and the minimum of two functions can be expressed in terms of the absolute value of their difference. We now turn to (8). Given any cube Q in Rn, we let B be the smallest ball that contains it. Then|B|/|Q|=2−nvn
√nn, where vnis the volume of the unit ball, and
1
|Q|
Q
f(x)−Avg
B
f dx≤|B|
|Q|
1
|B|
B
f(x)−Avg
B
f dx≤vn
√nn 2n f
BMOballs.
It follows from (3) that f
BMO≤21−nvn
√nnf
BMOballs. To obtain the reverse conclusion, given any ball B find the smallest cube Q that contains it and argue similarly using a version of (3) for the space BMOballs. Example 7.1.3. We indicate why L∞(Rn)is a proper subspace of BMO(Rn). We claim that the function log|x|is in BMO(Rn)but not in L∞(Rn). To prove that it is in BMO(Rn), for every x0∈Rnand R>0, we must find a constant Cx0,Rsuch that the average of|log|x|−Cx0,R|over the ball{x : |x−x0| ≤R}is uniformly bounded.
Since 1 vnRn
|x−x0|≤R
log|x| −Cx0,R dx= 1 vn
|z−R−1x0|≤1
log|z| −Cx0,R+log R dz, we may take Cx0,R=CR−1x0,1+logR, and things reduce to the case that R=1 and x0is arbitrary. If R=1 and|x0| ≤2, take Cx0,1=0 and observe that
|x−x0|≤1
log|x| dx≤
|x|≤3
log|x| dx=C.
When R=1 and|x0| ≥2, take Cx0,1=log|x0|. In this case notice that 1
vn
|x−x0|≤1
log|x| −log|x0| dx= 1 vn
|x−x0|≤1
log |x|
|x0|
dx≤log 2,
since when|x−x0| ≤1 and|x0| ≥2 we have that log||x|x
0|≤log|x|0x|+1
0| ≤log32 and log|x|x|0|≤log|x|x0|
0|−1≤log 2. Thus log|x|is in BMO.
The function log|x|turns out to be a typical element of BMO, but we make this statement a bit more precise later. It is interesting to observe that an abrupt cutoff of a BMO function may not give a function in the same space.
Example 7.1.4. The function h(x) =χx>0log1xis not in BMO(R). Indeed, the prob- lem is at the origin. Consider the intervals(−ε,ε), where 0<ε<12. We have that
(−ε,ε)Avgh= 1 2ε
+ε
−ε h(x)dx= 1 2ε
ε 0 log1
xdx=1+log1ε
2 .
But then 1 2ε
+ε
−ε h(x)−Avg
(−ε,ε)h dx≥ 1 2ε
0
−ε Avg
(−ε,ε)h dx=1+log1ε
4 ,
and the latter is clearly unbounded asε→0.
Let us now look at some basic properties of BMO functions. Observe that if a cube Q1is contained in a cube Q2, then
Avg
Q1
f−Avg
Q2
f ≤ 1
|Q1|
Q1
f−Avg
Q2
f dx
≤ 1
|Q1|
Q2
f−Avg
Q2
f dx
≤ |Q2|
|Q1|f
BMO.
(7.1.3)
The same estimate holds if the sets Q1and Q2are balls.
A version of this inequality is the first statement in the following proposition.
For simplicity, we denote byf
BMOthe expression given byf
BMOballsin (7.1.2), since these quantities are comparable. For a ball B and a>0, aB denotes the ball that is concentric with B and whose radius is a times the radius of B.
Proposition 7.1.5. (i) Let f be in BMO(Rn). Given a ball B and a positive integer m, we have
Avg
B
f−Avg
2mB
f ≤2nmf
BMO. (7.1.4)
(ii) For anyδ >0 there is a constant Cn,δ such that for any ball B(x0,R)we have Rδ
Rn
f(x)−AvgB(x0,R)f
(R+|x−x0|)n+δ dx≤Cn,δf
BMO. (7.1.5)
An analogous estimate holds for cubes with center x0and side length R.
(iii) There exists a constant Cnsuch that for all f∈BMO(Rn)we have
sup
y∈Rn
sup
t>0
Rn|f(x)−(Pt∗f)(y)|Pt(x−y)dx≤Cnf
BMO. (7.1.6)
Here Pt denotes the Poisson kernel introduced in Chapter 2.
(iv) Conversely, there is a constant Cnsuch that for all f∈L1loc(Rn)for which
Rn
|f(x)|
(1+|x|)n+1dx<∞ we have
Cnf
BMO≤sup
y∈Rn
sup
t>0
Rn|f(x)−(Pt∗f)(y)|Pt(x−y)dx. (7.1.7) Proof. (i) We have
Avg
B
f−Avg
2B
f = 1
|B| B
f(t)−Avg
2B
f dt
≤ 2n
|2B|
2B
f(t)−Avg
2B
f dt
≤ 2nf
BMO.
Using this inequality, we derive (7.1.4) by adding and subtracting the terms Avg
2B
f, Avg
22B
f, . . . , Avg
2m−1B
f.
(ii) In the proof below we take B(x0,R)to be the ball B=B(0,1)with radius 1 centered at the origin. Once this case is known, given a ball B(x0,R)we replace the function f by the function f(Rx+x0). When B=B(0,1)we have
Rn
f(x)−Avg
B
f (1+|x|)n+δ dx
≤
B
f(x)−Avg
B
f (1+|x|)n+δ dx+
∑
∞ k=02k+1B\2kB
f(x)−Avg
2k+1B
f + Avg
2k+1B
f−Avg
B
f (1+|x|)n+δ dx
≤
B
f(x)−Avg
B
f dx +
∑
∞ k=02−k(n+δ) 2k+1B
f(x)−Avg
2k+1B
f + Avg
2k+1B
f−Avg
B
f dx
≤vnf
BMO+
∑
∞ k=02−k(n+δ)
1+2n(k+1)
(2k+1)nvnf
BMO
=Cn,δf
BMO.
(iii) The proof of (7.1.6) is a reprise of the argument given in (ii). Set Bt=B(y,t).
We first prove a version of (7.1.6) in which the expression(Pt∗f)(y)is replaced by AvgBt f . For fixed y,t we have
Rn
t f(x)−Avg
Bt
f (t2+|x−y|2)n+12
dx
≤
Bt
t f(x)−Avg
Bt
f (t2+|x−y|2)n+12 dx
+
∑
∞ k=0
2k+1Bt\2kBt
t f(x)−Avg
2k+1Bt
f + Avg
2k+1Bt
f−Avg
Bt
f (t2+|x−y|2)n+12 dx
≤
Bt
f(x)−Avg
Bt
f
tn dx
+
∑
∞ k=02−k(n+1) tn
2k+1Bt
f(x)− Avg
2k+1Bt
f + Avg
2k+1Bt
f−Avg
Bt
f dx
≤vnf
BMO+
∑
∞ k=02−k(n+1)
1+2n(k+1)
(2k+1)nvnf
BMO
=Cnf
BMO.
(7.1.8)
Using the inequality just proved, we also obtain
Rn
(Pt∗f)(y)−Avg
Bt
f Pt(x−y)dx = (Pt∗f)(y)−Avg
Bt
f
≤
Rn
Pt(x−y) f(x)−Avg
Bt
f dx
≤Cnf
BMO,
which, combined with the inequality in (7.1.8), yields (7.1.6) with constant 2Cn. (iv) Conversely, let A be the expression on the right in (7.1.7). For|x−y| ≤t we have Pt(x−y)≥cnt(2t2)−n+12 =cnt−n, which gives
A≥
Rn|f(x)−(Pt∗f)(y)|Pt(x−y)dx≥cn tn
|x−y|≤t|f(x)−(Pt∗f)(y)|dx.
Proposition 7.1.2 (3) now implies that f
BMO≤2A/(vncn).
This concludes the proof of the proposition.