(c) For1< p<∞,
|u(z)|pdσ ≤Cp
|f|p dt, f ∈ Lp. (d) For all f ∈L1(), we have the distribution function inequality
σ({z:|u(z)|> λ})≤(c1/λ)
|f(t)|dt, λ >0.
If(b)or(c)holds for one value of p,1< p<∞, then(a)holds.
The constants Cpdepend only on p and the constant N(σ)in(5.5). In fact, if(a)holds, we can take Cp = N(σ)Bp where Bp is the constant in Theorem 5.1withα=1. If(c)or(d)holds then(5.5)holds with N(σ)≤4pCp. Proof. If (a) holds, then by (5.6) and Theorem 5.1, (c) and (d) hold. Clearly, (c) implies (b), and if (b) holds for somep, then the closed graph theorem for Banach spaces shows that (c) holds for the same value ofp.
Now suppose that (d) holds or that (c) holds for some p,1< p<∞. As in the proof of Lemma 5.5, takeI = {x0 <t <x0+h}, and setu(z)= Py∗ f(x), f(t)=4χI(t). Thenfp =4h1/p, andu(z)>1 on Q =I ×(0,h).
Hence
σ(Q)≤σ({|u(z)|>1})≤Cp
|f|pdt =4pCph, and (5.5) holds.
at a point z0 for which f(z0)=0. If f(z0)=0, then log f(z) has a single- valued determination on some neighborhood ofz0, andv(z)=Re(log f(z)) is harmonic on this neighborhood. Hence (6.1) holds with equality if f(z0)=0.
Lemma 6.1. (Jensen’s Inequality). Let(X, μ)be a measure space such that μis a probability measure,μ(X)=1. Letv∈ L1(μ)be a real function, and letϕ(t)be a convex function on. Then
ϕ
vdμ
≤
ϕ(v)dμ.
Proof. The convexity of ϕ means that ϕ(t) is the supremum of the linear functions lying belowϕ:
ϕ(t0)=sup{at0+b:at+b≤ϕ(t),t ∈}.
Wheneverat+b≤ϕ(t), we have a
vdμ
+b=
(av+b)dμ≤
ϕ(v)dμ, and the supremum of the left sides of these inequalities isϕ(
vdμ).
Jensen’s inequality is also true if
vdμ= −∞, provided thatϕis defined att = −∞and increasing on [−∞,∞). The proof is trivial in that case.
Theorem 6.2. Letv(z)be a subharmonic function on, and letϕ(t)be an increasing convex function on[−∞,∞), continuous at t = −∞. Thenϕ◦ v is a subharmonic function on.
Proof. Since every convex function is continuous on, ϕis continuous on [−∞,∞). It follows immediately thatϕ◦vis upper semicontinuous. Ifz0∈ and ifr <r(z0), then becauseϕis increasing
ϕ(v(z0))≤ϕ
⎛
⎝ 1 πr2
(z0,r)
v(z)d x d y
⎞
⎠.
By Jensen’s inequality, then ϕ(v(z0))≤ 1
πr2
(z0,r)
ϕ(v)d x d y,
which is (6.1) forϕ(v).
For example, if f(z) is analytic on, then|f(z)|p =exp(plog|f(z)|) is a subharmonic function onif 0< p<∞, and
log+|f(z)| =max(log|f(z)|,0)
is also a subharmonic function on. Notice the contrast with the situation for harmonic functions, where we have|u|p subharmonic only for p≥1 (by H¨older’s inequality).
Theorem 6.3. Letv:→[−∞,∞]be an upper semicontinuous function.
Thenvis subharmonic onif and only if the following condition holds: If u(z)is a harmonic function on a bounded open subset W ofand if
Wlimz→ζ(v(z)−u(z))≤0 for allζ ∈∂W , then
v(z)≤u(z), z∈W.
Proof. Assumev(z) is subharmonic on. Letu(z) andWbe as in the above statement. ThenV(z)=v(z)−u(z) is subharmonic onW, and
Wlimz→ζV(z)≤0 for allζ ∈∂W.
Using a standard maximum principle argument we now showV ≤0 inW.
We can assumeW is connected. Leta=supWV(z) and supposea>0. Let {zn} be a sequence inW such that V(zn)→a. Sincea>0, the zn cannot accumulate on∂W, and there is a limit point z∈W. By the semicontinuity, V(z)=a, and the set
E= {z∈W :V(z)=a}
is not empty. The setEis closed becauseVis upper semicontinuous and has maximum valuea.
If z0∈ E, then because V(z)≤a on W, the mean value inequality (6.1) shows V(z)=a almost everywhere on (z0,r), for some r >0. HenceE is dense in(z0,r). BecauseEis closed this means(z0,r)⊂E, and Eis open. SinceWwas assumed to be connected, we have a contradiction and we conclude thata≤0.
Conversely, letz0 ∈and let(z0,r)⊂. Sincevis upper semicontinuous there are continuous functionsun(z) decreasing tov(z) on∂(z0,r) asn→ ∞.
LetUn(z) be the harmonic function on(z0,r) with boundary valuesun(z).
After a suitable change of scale,Unis obtained fromunby the Poisson integral formula for the unit disc. From Section 3 we know thatUn is continuous on (z0,r). By hypothesis we havev(z0)≤Un(z0), and hence
v(z0)≤lim
n
1 2π
un(z0+r eiθ)dθ = 1 2π
v(z0+r eiθ)dθ by monotone convergence. Averaging these inequalities againstr drthen gives (6.1), and sov(z) is subharmonic.
The proof just given shows that ifv(z) is subharmonic on, then (6.1) holds for anyr >0 such that(z0,r)⊂. It also shows that we can replace area means by circular means in (6.1). The condition
v(z0)≤ 1 2π
v(z0+r eiθ)dθ, 0<r <r(z0), is therefore equivalent to (6.1).
Corollary 6.4. If is a connected open set and if v(z) is a subharmonic function onsuch thatv(z)≡ −∞, then whenever(z0,r)⊂,
1 2π
v(z0+r eiθ)dθ = −∞.
Proof. Letun(z) be continuous functions decreasing tov(z) on∂(z0,r), and letUn(z) denote the harmonic extension ofun to(z0,r). If
1 2π
v(z0+r eiθ)dθ= −∞,
then since v is bounded above and since Poisson kernels are bounded and positive, we have
1 2π
Pz(θ)v(z0+r eiθ)dθ = −∞, |z|<1.
ConsequentlyUn(z)→ −∞for eachz∈(z0,r), and by Theorem 6.3v=
−∞on(z0,r). The nonempty set
{z∈:v(z)≡ −∞on a neighborhood ofz}
is then open and closed, and we again have a contradiction.
Theorem 6.5. Letv(z)be a subharmonic function in the unit disc D. Assume v(z)≡ −∞. For0<r <1, let
vr(z)=
⎧⎨
⎩
v(z), |z| ≤r,
1 2π
Pz/r(θ)v(r eiθ)dθ, |z|<r.
Then vr(z) is a subharmonic function in D, vr(z) is harmonic on |z|<
r, vr(z)≥v(z),z∈D, andvr(z)is an increasing function of r.
Proof. By Corollary 6.4 and by Section 3 we knowvr(z) is finite and harmonic on(0,r)= {|z|<r}. To see thatvr(z) is upper semicontinuous at a point z0∈∂(0,r) we must show
v(z0)≥ limz→z
|z|<r0
vr(z).
This follows from the approximate identity properties of the Poisson kernel and from the semicontinuity ofv. Writez0=r eiθ0. Forε >0 there isδ >0
such thatv(r eiθ)< v(z0)+εif|θ−θ0|< δ. Then if|z|<rand if|z−z0|is small,
vr(z)≤ 1 2π
|θ−θ0|≤δPz/r(θ)(v(z0)+ε)dθ + 1
2π
sup
θ v(r eiθ)
|θ−θ0|>δPz/r(θ)dθ
≤v(z0)+2ε.
Hencevr is upper semicontinuous.
If we again take continuous functionsun(z) decreasing tov(z) on∂(0,r), then as in the proof of Corollary 6.4 we have
v(z)≤vr(z).
Becausevis subharmonic, this inequality shows thatvr(z) satisfies the mean value inequality (6.1) at each pointz0 with|z0| =r. Consequentlyvr(z) is a subharmonic function onD.
Ifr >s, thenvr =(vs)r, and since for any subharmonic functionv, vr(z)≥ v(z), the functionsvr(z) increase withr.
Corollary 6.6. Ifv(z)is a subharmonic function on D, then m(r)= 1
2π
v(r eiθ)dθ is an increasing function of r.
The subharmonic function v(z) on has a harmonic majorant if there is a harmonic function U(z) such that v(z)≤U(z) throughout . If is connected, ifv(z)≡ −∞in, and ifv(z) has a harmonic majorant, then the Perron process for solving the Dirichlet problem produces theleast harmonic majorant u(z), which is a harmonic function majorizing v(z) and satisfying u(z)≤U(z) for every other harmonic majorant U(z) of v(z) (see Ahlfors [1966] or Tsuji [1959]). Since we are interested only in simply connected domains, we shall not need the beautiful Perron process to obtain harmonic majorants. We can use the Poisson kernel instead.
Theorem 6.7. Letv(z)be a subharmonic function in the unit disc D. Then v has a harmonic majorant if and only if
sup
r
1 2π
v(r eiθ)dθ =sup
r vr(0)<∞.
The least harmonic majorant ofv(z)is then u(z)=lim
r→1
Pz/r(θ)v(r eiθ)dθ/2π =lim
r→1vr(z).
Proof. If suprvr(0) is finite, then by Harnack’s theorem the functionsvr(z) increase to a finite harmonic functionu(z) on D. Sincev(z)≤vr(z),u(z) is a harmonic majorant ofv(z). Conversely, if U(z) is harmonic on D, and if U(z)≥v(z) on D, then by Theorem 6.3, U(z)≥vr(z) for each r. Conse- quently, suprvr(0)<∞, and againu(z)=limrvr(z) is finite and harmonic.
Sincevr(z)≤U(z), we have u(z)≤U(z), and sou(z) is the least harmonic majorant.
Since by continuity u(z)=limr→1u(r z), the least harmonic majorant of v(z) can also be written
u(z)= lim
r→1
Pz(θ)v(r eiθ)dθ/2π.
In particular, ifv(z)≥0 and ifv(z) has a harmonic majorant, then its least harmonic majorant is the Poisson integral of the weak-star limit of the bounded positive measuresv(r eiθ)dθ/2π.
Theorem 6.8. Letv(z)be a subharmonic function in the upper half plane H . If
sup
y
|v(x+i y)|d x =M <∞, thenv(z)has a harmonic majorant inH of the form
u(z)=
Py(x−t)dμ(t), whereμis a finite signed measure on.
Proof. The inequality v(z)≤ 2
πysup
η
|v(ξ+iη)|dξ, z=x+i y, y>0, (6.2)
is proved in the same way that the similar inequality (3.9) was proved to begin the proof of Theorem 3.5.
Fixy0 >0 and consider the harmonic function u(z)=uy0(z)=
Py−y0(x−t)v(t,y0)dt,
defined on the half plane{y> y0}. We claimv(z)≤u(z) ony> y0. To see this, letε >0 and let A>0 be large. Letun(t) be continuous functions decreasing tov(t +i y0) on [−A,A], and let
Un(z)= A
−APy−y0(x−t)un(t)dt, y> y0,
be the Poisson integral ofun. The function
V(z)=v(z)−εlog|z+i| −Un(z)
is subharmonic on y> y0. With ε fixed we have limz→∞V(z)= −∞, by (6.2), and ifAis large we have
z→(t,ylim0)V(z)≤0
for |t| ≥ A, again by (6.2). If |t|< A, then limz→(t,y0)V(z)≤v(t,y0)
−un(t,y0)≤0. It follows from Theorem 6.3 and a conformal mapping that V(z)≤0 on y >y0. Sending n→ ∞, then A→ ∞, and then ε→0, we obtainv(z)≤u(z) on y>y0. The measuresv(t,y0) dt remain bounded as y0 →0, and ifdμ(t) is a weak-star cluster point, then
ylim0→0uy0(z)=
Py(x−t)dμ(t) is a harmonic majorant ofv(z).
The functionu(z) is actually the least harmonic majorant ofv(z), but we shall not use this fact.