Patrick Corn
Math 185, 7/21/05
Solution Set 8
Problem 1: Suppose such an fexists. Let g(z) = z2; then fand gagree on a set with a limit
point, namely the set {1
2k:k∈N}, and so they must agree everywhere by the identity theorem,
which is absurd.
Problem 2: We’ve done exercise II.8.2. It shows that the function g(z) = f(z) is holomorphic
on {z:z∈G}. In this case, that set equals G. At any rate, fand gare holomorphic functions
that, by assumption, agree on G∩R. Clearly G∩Ris open in R, and if it were empty, then G∩H+
and G∩H−would be disjoint open sets whose union was G; since Gis symmetric with respect to
the real axis, they would both have to be nonempty, which would contradict the connectedness of
G. So G∩Rcontains some open interval in R, which certainly is a set with a limit point. Thus
f=gon G, which gives us what we want.
Problem 3: Suppose that |f|attains a local minimum in Gat z0, but f(z0)6= 0. Then find
a neighborhood Nof z0in Gsuch that |f(z)| ≥ |f(z0)|for all z∈N. Of course fcannot vanish
on N, so the function g(z) = 1/f(z) is holomorphic on N, and it has a local maximum at z0∈N,
which implies, by the maximum modulus principle, that g, and hence f, is constant on all of N.
Then fis constant on all of G, by the identity theorem (it agrees with the zero function on an open
disk N, which certainly has lots of limit points). So there you go.
Problem 4: Fix w∈D, where Dis the open unit disk. Then define
φ1(z) = z−w
1−zw , φ2(z) = z−f(w)
1−zf (w)
Both φ1and φ2are LFTs sending Dto D(you can see that the unit circle maps to itself by
multiplying the denominator by zand taking absolute values; see also exercise III.9.4 and III.9.5).
Now then, the function g=φ2◦f◦φ−1
1is a holomorphic map from Dto itself, and g(0) = 0. Then
by Schwarz’s lemma, we have |g(s)| ≤ |s|for all s∈D. Set s=φ1(z); then we get
|φ2(f(z))| ≤ |φ1(z)|
which is precisely the statement of Pick’s lemma. (Note that the part about |g′(0)| ≤ 1 is exactly
what you need to do exercise VII.17.3, which I did not assign.)
Lastly, equality holds in Schwarz’s lemma if and only if g(z) = cz, and in this case f=φ−1
2◦g◦φ1
is a composition of three LFTs, so it’s an LFT sending Dto itself. Also note that if fis an LFT
sending Dto itself, then gis an LFT sending Dto itself with g(0) = 0; it is clear that the only
possibility for gis that it is a dilation g(z) = cz. (Again, see exercise III.9.4 or III.9.5.) So equality
holds if and only if fis an LFT sending Dto itself.
Problem 5: There are a couple of ways to do this; you can use the result of the previous
exercise, or think about adapting its proof. I think I’ll do the latter. Suppose that zand ware two
distinct fixed points. Then define φ1and φ2as above. But, of course, in this situation φ1=φ2.
At any rate, we see that g=φ1◦f◦φ−1
1is a holomorphic function from Dto itself, and f(w) = w
implies that g(0) = 0, just as in the last problem. Note also that g(φ1(z)) = φ1(f(z)) = φ1(z). We
can apply Schwarz’s lemma to see that |g(s)| ≤ |s|for all s∈D, and equality actually holds for
s=φ1(z). Therefore g(z) = cz for some constant c. But g(z) = cz cannot have a nonzero fixed
point, unless c= 1. In this case, f=φ−1
1◦g◦φis also the identity function, because gis.
Problem 6: Suppose uis a positive harmonic function on C. We can find a holomorphic function
g(z) with real part equal to u. Let φbe an LFT sending the right half plane to the open unit disk,
for instance φ(z) = z−1
z+1 . Then φ◦gis an entire function, because φis holomorphic on the image
of g. And the image of φ◦gis contained in the open unit disk. So we have ourselves a bounded
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