9. H ARMONIC FUNCTIONS AND P OISSON INTEGRAL FORMULA
Let Ω ⊆ C be a region (connected open set). A function u : Ω "→ R is said to be harmonic if u ∈ C2 (Ω)
and ∆u(z) = 0 for all z ∈ Ω. Here ∆u(z) := ux,x (z) + uy,y (z) with z = x + iy. We recall some basic properties of
harmonic functions. We do not prove them here, but expect that you already know them11.
(1) (Mean value property). Let u ∈ C2 (Ω). Then u is harmonic if and only if it has the mean value
property (MVP). Recall that u is said to have MVP if u(z) =
1 R 2π
iθ
2π 0 u(z + re )dθ
for all z ∈ Ω and r > 0
such that D(z, r) ⊆ Ω. Here we have averaged u over a circle centered at z. One can also average over
R
disks. In other words, MVP is equivalent to saying u(z) = πr12 D(z,r) u(w)dm(w) for all z ∈ Ω and r > 0
such that D(z, r) ⊆ Ω.
(2) (Maximum principle). Let Ω be a bounded region, u ∈ C(Ω) and ∆u = 0 in Ω. Then max u = max u.
Ω
∂Ω
Further, if u(z) = max u for some z ∈ Ω, then u must be constant. Since −u is harmonic whenever u is,
Ω
the same holds for the minimum too.
(3) (Relationship to analytic functions). If f : Ω → C is holomorphic, then u = Re f is harmonic in
Ω. The converse is true if Ω is a disk, i.e., if Ω is a disk and u is harmonic on Ω, then there is a
holomorphic function on Ω whose real part is equal to u.
It is good to understand the need for various conditions in the above facts.
Exercise 1.
(1) What statements among the above change if Ω is not connected?
(2) Show that the maximum principle is false if Ω is not bounded. Take Ω = H, the upper half-plane.
(3) Give a region Ω and a harmonic function u : Ω "→ R that is not the real part of a holomorphic function.
A useful computation: The following integral will be very important to us.
Lemma 2. For all z ∈ C and any r > 0,
Z2π
log |z − reiθ |
0
dθ
= log(|z| ∨ r).
2π
Proof. The function z "→ log |z| is harmonic on C \ {0} (why?). If r < |z|, then D(z, r) ⊆ C \ {0} and hence by
MVP, we see that
R2π
0
log |z − reiθ | dθ
2π = log |z|.
Now suppose r > |z|. Write log |z − reiθ | = log |z| + log r + log | 1z − 1r e−iθ |. Integrate over θ to get
Z2π
0
log |z − reiθ |
dθ
= log |z| + log r +
2π
Z2π
0
!1 1
! dθ
!
!
log ! − e−iθ ! .
z r
2π
! !
! !
But ! 1z ! < 1r , hence by the previously considered case, the last integral is equal to log |1/z|. Adding the first
two terms we get
R2π
0
log |z − reiθ | dθ
2π = log r.
This completes the proof when |z| < r and |z| > r. When |z| = r, firstly convince yourself that the integral
is well defined. Second, show that r "→
holds for |z| = r.
R2π
0
log |z − reiθ | dθ
2π is continuous in r. Conclude that the formula also
!
11Rudin’s Real and complex analysis has a whole chapter on harmonic functions and contains all the facts we mention.
26
Poisson integral formula: By the maximum principle, if u and v are continuous in Ω, harmonic in Ω, and
equal to each other on ∂Ω, then u = v. In other words, u |∂Ω determines u. But, is there an explicit way to
recover u from its boundary values?
Let Ω = rD. Then, if u ∈ C(rD) and u is harmonic in rD, then u(z) =
P(z, reiθ ) =
r2 − |z|2
r2 − t 2
= 2 2
iθ
2
r + t − 2rt cos(α − θ)
|re − z|
1
2π
R2π
0
u(reiθ )P(z, reiθ )dθ, where
if z = teiα with t < r.
For fixed r, the function (z, θ) #→ P(z, reiθ ) is called the Poisson kernel for rD.
Exercise 3. Show that if f : rT #→ R is continuous, then u : rD → R defined by u(z) =
is harmonic in rD and u(zn ) → u(z0 ) whenever z0 ∈ rT, zn ∈ rD and zn → z0 .
1 R 2π
iθ
iθ
2π 0 u(re )P(z, re )dθ
The exercise is not trivial, but it is very much like many other proofs we have seen, for example Fejér’s
theorem or the proof of Fourier inversion etc. For fixed r, z, it may be seen that dµ(θ) = P(z, reiθ ) dθ
2π is a
probability measure on T . If zn → reiα for some α, then this probability distribution concentrates closer and
closer to the point eiα .
Another useful computation: We shall also need the following integral.
Lemma 4. If a ∈ rD, then
Z2π
log |reiθ − a| P(a, reiθ )
0
dθ
= log(r2 − |a|2 ) − log r.
2π
Proof. Assume |a| < r (then use continuity argument to get the result for |a| = r). The function u(z) = log |z−a|
is harmonic on C \ {a} only, and hence Poisson integral cannot be applied directly (or we would get u(a) =
−∞!). We use a trick attributed to Kelvin, the so called reflection principle. The idea is to find a function v
harmonic in a neighbourhood of rD such that v = u on rT. Then, the integral we want may also be written
as
R 2π
0
v(reiθ ) dθ
2π to which Poisson integral formula applies to give v(a) as the answer!
What is v? Recall that if |ζ| < 1, then the Mobius transformation w #→
i.e.,
|w−ζ|
|1−ζw|
= 1 if |w| = 1 and |ζ| < 1. Apply this to w =
1=
z
r
| rz − ar |
1−
1
az
r2
w−ζ
1−ζw
with |z| = r and ζ =
=
a
r
maps the unit circle to itself,
to get
r|z − a|
.
r2 − az
2
Thus, if we set v(z) = log |r2 − az| − log r, then v(z) = u(z) whenever |z| = r. Further, v is harmonic on C \ { ra }
which contains rD. We have found the v with the properties we wanted. Hence
Z2π
0
log |reiθ − a|
dθ
= v(a) = log(r2 − |a|2 ) − log r.
2π
!
This is what we wanted to show.
A physicsly remark: From the inverse square law (Coulomb’s law), the electric potential due to a unit
1
charge at the origin of R3 , is ϕ(x) = %x%
(a scalar field). Both for mathematics and physics, it makes sense
to study the analogous laws in other dimensions. It turns out that in two dimensions, the correct analogue
of Coulomb’s law is inverse distance force, which means that the potential is ϕ(z) = log |z| (gradient of the
potential should give the force). This is a useful way to sometimes interpret various manipulations above.
27
For example, what we did in the proof of Lemma 4 is this. First we considered u(z) = log |z − a|, the
potential due to a unit positive charge located at a. We replaced it by v, which is essentially the potential
2
due to a unit charge placed at ra (the “reflection” of a in the unit circle). By symmetry it turns out to give
the same potential on the circle rT (This must be unclear since it is inaccurate, but think of a unit charge at
+i and a unit charge at −i. They both give the same potential on the real line. Our situation is analogous).
10. P OISSON -J ENSEN FORMULA
Lemma 5. [Jensen’s formula] Let f be a meromorphic function on the whole plane. If f (0) "= 0, ∞, then, for any r > 0
we have
log | f (0)| =
Z2π
0
log | f (reiθ )|
n
dθ m
r
r
− ∑ log
+ ∑ log
2π i=1
|ai | j=1
|b j |
where a1 , . . . , am are the zeros of f in rD and b1 , . . . , bn are the poles of f in rD (zeros and poles are always repeated
according to multiplicity, thus ai s need not be distinct and the same for b j s).
1
n
Proof. Write f (z) = g(z) ∏m
i=1 (z − ai ) ∏ j=1 z−b j , where g is a meromorphic function on the whole plane that
has no zeros or poles in rD. Then,
m
n
i=1
j=1
log | f (z)| = log |g(z)| + ∑ log |z − ai | − ∑ log |z − b j |.
Therefore,
dθ
log | f (re )|
=
2π
Z2π
=
Z2π
Z2π
0
iθ
0
0
dθ m
log |g(re )| + ∑
2π i=1
iθ
log |g(reiθ )|
Z2π
0
dθ
log |re − ai | −
2π
iθ
Z2π
0
log |reiθ − b j |
dθ
2π
dθ
+ m log r − n log r
2π
by Lemma 2. Subtract this from log | f (0)| to get
log | f (0)| −
Z2π
0
dθ
log | f (re )|
= log |g(0)| −
2π
Z2π
n
dθ m
+ ∑ log |ai | − m log r − ∑ log |b j | + n log r
2π i=1
j=1
= log |g(0)| −
Z2π
n
dθ m
r
r
− ∑ log
+ ∑ log
.
2π i=1
|ai | j=1
|b j |
iθ
0
0
log |g(reiθ )|
log |g(reiθ )|
As g is holomorphic in a neighbourhood of rD and has no zeros there, z → log |g(z)| is harmonic in a neighbourhood of rD. By MVP, log |g(0)| =
R2π
0
log |g(reiθ )| dθ
2π . Jensen’s formula follows.
!
We next prove a generalization of the Jensen’s formula.
Lemma 6. [Poisson-Jensen formula] Let f be a meromorphic function on the whole plane. If r > 0, w ∈ rD, f (w) "=
0, ∞, then,
log | f (w)| =
Z2π
0
log | f (reiθ )|P(w, reiθ )
! r2 − a w !
! r2 − b w !
n
dθ m
!
!
j !
i !
− ∑ log !
! + ∑ log !
!
2π i=1
r(w − ai )
r(w
−
b
j)
j=1
where a1 , . . . , am are the zeros of f in rD and b1 , . . . , bn are the poles of f in rD.
28
1
n
Proof. As before, write f (z) = g(z) ∏m
i=1 (z − ai ) ∏ j=1 z−b j , where g is a meromorphic function on the whole
plane that has no zeros or poles in rD. Then,
m
n
i=1
j=1
log | f (z)| = log |g(z)| + ∑ log |z − ai | − ∑ log |z − b j |.
Now put z = reiθ and integrate again P(w, reiθ ) over θ to get
Z2π
0
=
log | f (reiθ )|P(w, reiθ )
dθ
2π
Z2π
iθ
0
dθ m
log |g(re )|P(w, re ) + ∑
2π i=1
iθ
Z2π
0
dθ
log |re − ai |P(w, re ) −
2π
iθ
iθ
Z2π
0
log |reiθ − b j |
dθ
2π
By Lemma 4 we find the integrals in the two summands. Further, log |g| is harmonic in a neighbourhood of
rD, hence by the Poisson integral formula,
Z2π
0
log | f (reiθ )|P(w, reiθ )
R2π
0
log |g(reiθ )|P(w, reiθ ) dθ
2π = log |g(w)|. We conclude
m
n
dθ
= log |g(w)| + ∑ [log(r2 − |ai |2 ) − log r] − ∑ [log(r2 − |b j |2 ) − log r].
2π
i=1
j=1
n
Subtract from log | f (w)| = log |g(w)| + ∑m
i=1 log |w − ai | − ∑ j=1 log |w − b j | to get
log | f (0)| −
Z2π
0
log | f (reiθ )|P(w, reiθ )
m
n
r2 − |b j |2
dθ
r2 − |ai |2
= − ∑ log
+ ∑ log
2π
r|w − ai | j=1
r|w − b j |
i=1
!
which is what we wanted to show.
Even if f has a zero or pole at the origin, show that Jensen’s formula can be modified as follows.
Exercise 7. Let f be meromorphic on the whole plane and suppose that f (z) ∼ c f zd as z → 0 for some d ∈ Z
and c f ∈ C. Then, show that
log |c f | =
Z2π
0
log | f (reiθ )|
n
dθ m
r
r
− ∑ log
+ ∑ log
2π i=1
|ai | j=1
|b j |
where a1 , . . . , am are the zeros of f in rD and b1 , . . . , bn are the poles of f in rD (if d > 0 then d many of the ai s
are zero and if d < 0 then d many of the b j s are zero).
29