- #1
MathLearner123
- 17
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I need help please! So I'm reading 'Complex made simple' by David C. Ullrich. I made all the requirements for this proof but the author don't give the proof of this final theorem, instead it gives a similar proof for another set of theorems.
Let ##\mathbb{D}' = \mathbb{D} \setminus \{0\}##.
Theorem 20.3 (Great Picard Theorem) If ##f \in H(\mathbb{D}')## and ##f(\mathbb{D}') \subset \mathbb{C} \setminus \{0, 1\}## then ##f## has a pole or a removable singularity at ##0##.
Theorem A. If ##u: \mathbb{C} \rightarrow \mathbb{R}## is a bounded harmonic function then ##u## is constant.
Proof. Since ##\mathbb{C}## is simply connected, there exists a real-valued harmonic function ##v## such that ##u+i v## is holomorphic. Let ##h=e^{u+i v}##. Then ##h## is an entire function, and ##h## is bounded, since ##|h|=e^u##. Thus ##h## is constant, and hence ##u=\log |h|## is constant.
Theorem B. If ##f: \mathbb{D}^{\prime} \rightarrow \mathbb{R}## is harmonic and bounded then ##u## extends to a function harmonic in ##\mathbb{D}##.
The problem with Theorem B is that we cannot say that ##u## has a harmonic conjugate, since ##\mathbb{D}^{\prime}## is not simply connected. (This is precisely analogous to the problem with Theorem 20.3: We cannot say that there exists ##\tilde{f}: \mathbb{D}^{\prime} \rightarrow \Pi^{+}##with ##\lambda \circ \tilde{f}=f## because ##\mathbb{D}^{\prime}## is not simply connected.)
One can give a slightly informal proof of Theorem B as follows: There does exist a "multi-valued" harmonic conjugate ##v## (this is an informal way of saying that there is a holomorphic function ##u+i v## defined in some disk contained in ##\mathbb{D}^{\prime}## which admits unrestricted continuation in ##\mathbb{D}^{\prime}## ). Now it is not hard to see that in fact there exists a multi-valued harmonic conjugate ##v## and a real number ##c## such that any two branches of ##v## differ by a multiple of c. If ##c=0## we are done, so we assume that ##c \neq 0##, and define ##h=e^{2 \pi(u+i v) / c}##. The fact that any two branches of ##v## differ by a multiple of ##c## shows that ##h## is an actual single-valued holomorphic function. As in the proof of Theorem A we see that ##h## is bounded; hence ##h## has a removable singularity, and hence ##u## has a removable singularity at the origin as well.
If the "informal" parts of that proof bother you don't worry, we will give a more formal version soon. The corresponding informal proof of Theorem 20.3 is this: Although we cannot assert that there exists ##\tilde{f}: \mathbb{D}^{\prime} \rightarrow \Pi^{+}##with ##\lambda \circ \tilde{f}=f##, there does exist a "multi-valued function" ##\tilde{f}## with this property (that is, an ##\tilde{f}## defined in some disk contained in ##\mathbb{D}^{\prime}## which admits unrestricted continuation in ##\mathbb{D}^{\prime}## ). It turns out that there exists ##\phi \in \Gamma(2)## such that any two branches of ##\tilde{f}## differ by a power of ##\phi##, and, since ##\phi## cannot be elliptic, we know that there exists a nonconstant bounded function ##h \in H\left(\Pi^{+}\right)## such that ##h \circ \phi=h##. It follows that ##h \circ \tilde{f}## is single-valued, and hence has a removable singularity at the origin.
It is now quite plausible, and not hard to prove using the results in the previous chapter, that in fact ##f## has a (possibly infinite) limit at the origin, so that in particular ##f## does not have an essential singularity (for the details here see the formal proof below).
The actual proof is going to use a covering-map argument in place of the analytic continuation - I think this makes the details of the proof easier to follow, possibly at the price of making it harder to see where the proof comes from. So for the sake of illustrating how the covering-map argument corresponds to the analytic continuation argument, here's the "formal" proof of Theorem B:
Proof of Theorem B. Define a function ##U: \Pi^{+} \rightarrow \mathbb{R}## by ##U(z)=u\left(e^{2 \pi i z}\right)##. Then ##U## is harmonic, being the composition of a harmonic function and a holomorphic function. Since ##\Pi^{+}##is simply connected. there exists ##F \in## ##H\left(\Pi^{+}\right)##such that ##U=\mathbb{R e}(F)##. Now, although ##U(z+1)=U(z)## we need not have ##F(z+1)=F(z)##. But ##\mathbb{R e}(F(z+1)-F(z))=U(z+1)-U(z)=0##. and so ##F(z+1)-F(z)## must be an imaginary constant: There exists ##c \in \mathbb{R}## such that ##F(z+1)-F(z)=i c## for all ##z \in \Pi^{+}##. Choose a real number ##\alpha \neq 0## so that ##c \alpha## is a multiple of ##2 \pi## (you can take ##\alpha=2 \pi / c## unless ##c=0## ). Let #### E(z)=e^{\alpha F(z)} . ####
It follows that ##E(z+1)=E(z)##, and so there exists ##f \in H\left(\mathbb{D}^{\prime}\right)## with #### E(z)=f\left(e^{2 \pi i z}\right) \text
{. } ####
Now the fact that ##u## is bounded shows that ##f## is bounded, so that ##f## has a removable singularity at the origin. The fact that ##u## is bounded also shows that ##f## is bounded away from 0 , so in particular ##f(0) \neq 0##. Hence ##u## has a limit at 0 , since ##u=\log (|f|) / \alpha##. (Since ##f\left(e^{2 \pi i z}\right)=e^{\alpha F(z)}## it follows that #### \left.\log \left(\left|f\left(e^{2 \pi i z}\right)\right|\right)=\alpha \operatorname{Re}(F(z))=\alpha U(z)=\alpha u\left(e^{2 \pi i z}\right) \cdot\right) ####
If we do the same thing with the Big Picard Theorem, replacing functions in ##\mathbb{D}^{\prime}## by the corresponding periodic functions in the upper half-plane, we see that we need to prove this:
Theorem 20.4. Suppose that ##F: \Pi^{+} \rightarrow \mathbb{C} \backslash\{0,1\}## is holomorphic and satisfies ##F(z+1)=F(z)## for all
##z \in \Pi^{+}##. Then there exists ##L \in \mathbb{C}_{\infty}## such
that ##F\left(z_j\right) \rightarrow L## for every sequence
##\left(z_j\right)## in ##\Pi^{+}##with ##\operatorname{Im}\left(z_j\right)
\rightarrow \infty##.
So, I need to show that Theorem 20.4 implies Theorem 20.3
I've tried to do something similar, but don't get something useful.
What I've tried:
So, supposing that ##f \in H(\mathbb{D}')## and ##f(\mathbb{D}') \subset \mathbb{C} \setminus \{0, 1\}## we have that ##f : \mathbb{D}' \to \mathbb{C} \setminus \{0, 1\}## is holomorphic. So, we can define ##F : \Pi^+ \to \mathbb{C} \setminus \{0, 1\}## by ##F(z) = f(e^{2 \pi i z})##. From here we obtain that ##F## is holomorphic on ##\Pi^+## (which is simply connected) so, because ##\lambda : \Pi^+ \to \mathbb{C} \setminus \{0,1\}## is a covering map we have that there exists the lift of ##f##: ##\tilde{f} : \Pi^+ \to \Pi^+## with ##\lambda \circ \tilde{f} = F##, where ##\tilde{f}## is holomorphic. Because ##F(z) = F(z+1)## for all ##z \in \Pi^+## we obtain that, from Theorem 20.4 there exists ##L \in \mathbb{C}_\infty## such that ##F(z_j) \to L## for every sequence ##(z_j) \subset \Pi^+## with ##\Im(z_j) \to \infty##. So, we have that ##(\lambda \circ \tilde{f})(z_j) \to L## for every sequence ##(z_j)## with ##Im(z_j) \to \infty##.
How can I continue from here? I know that I've written so much, but I need these for my bachelor degree. Thanks for reading!
P.S. I see some random '##>##' in text. Ignore it please. Is from cite I think.
Let ##\mathbb{D}' = \mathbb{D} \setminus \{0\}##.
Theorem 20.3 (Great Picard Theorem) If ##f \in H(\mathbb{D}')## and ##f(\mathbb{D}') \subset \mathbb{C} \setminus \{0, 1\}## then ##f## has a pole or a removable singularity at ##0##.
Theorem A. If ##u: \mathbb{C} \rightarrow \mathbb{R}## is a bounded harmonic function then ##u## is constant.
Proof. Since ##\mathbb{C}## is simply connected, there exists a real-valued harmonic function ##v## such that ##u+i v## is holomorphic. Let ##h=e^{u+i v}##. Then ##h## is an entire function, and ##h## is bounded, since ##|h|=e^u##. Thus ##h## is constant, and hence ##u=\log |h|## is constant.
Theorem B. If ##f: \mathbb{D}^{\prime} \rightarrow \mathbb{R}## is harmonic and bounded then ##u## extends to a function harmonic in ##\mathbb{D}##.
The problem with Theorem B is that we cannot say that ##u## has a harmonic conjugate, since ##\mathbb{D}^{\prime}## is not simply connected. (This is precisely analogous to the problem with Theorem 20.3: We cannot say that there exists ##\tilde{f}: \mathbb{D}^{\prime} \rightarrow \Pi^{+}##with ##\lambda \circ \tilde{f}=f## because ##\mathbb{D}^{\prime}## is not simply connected.)
One can give a slightly informal proof of Theorem B as follows: There does exist a "multi-valued" harmonic conjugate ##v## (this is an informal way of saying that there is a holomorphic function ##u+i v## defined in some disk contained in ##\mathbb{D}^{\prime}## which admits unrestricted continuation in ##\mathbb{D}^{\prime}## ). Now it is not hard to see that in fact there exists a multi-valued harmonic conjugate ##v## and a real number ##c## such that any two branches of ##v## differ by a multiple of c. If ##c=0## we are done, so we assume that ##c \neq 0##, and define ##h=e^{2 \pi(u+i v) / c}##. The fact that any two branches of ##v## differ by a multiple of ##c## shows that ##h## is an actual single-valued holomorphic function. As in the proof of Theorem A we see that ##h## is bounded; hence ##h## has a removable singularity, and hence ##u## has a removable singularity at the origin as well.
If the "informal" parts of that proof bother you don't worry, we will give a more formal version soon. The corresponding informal proof of Theorem 20.3 is this: Although we cannot assert that there exists ##\tilde{f}: \mathbb{D}^{\prime} \rightarrow \Pi^{+}##with ##\lambda \circ \tilde{f}=f##, there does exist a "multi-valued function" ##\tilde{f}## with this property (that is, an ##\tilde{f}## defined in some disk contained in ##\mathbb{D}^{\prime}## which admits unrestricted continuation in ##\mathbb{D}^{\prime}## ). It turns out that there exists ##\phi \in \Gamma(2)## such that any two branches of ##\tilde{f}## differ by a power of ##\phi##, and, since ##\phi## cannot be elliptic, we know that there exists a nonconstant bounded function ##h \in H\left(\Pi^{+}\right)## such that ##h \circ \phi=h##. It follows that ##h \circ \tilde{f}## is single-valued, and hence has a removable singularity at the origin.
It is now quite plausible, and not hard to prove using the results in the previous chapter, that in fact ##f## has a (possibly infinite) limit at the origin, so that in particular ##f## does not have an essential singularity (for the details here see the formal proof below).
The actual proof is going to use a covering-map argument in place of the analytic continuation - I think this makes the details of the proof easier to follow, possibly at the price of making it harder to see where the proof comes from. So for the sake of illustrating how the covering-map argument corresponds to the analytic continuation argument, here's the "formal" proof of Theorem B:
Proof of Theorem B. Define a function ##U: \Pi^{+} \rightarrow \mathbb{R}## by ##U(z)=u\left(e^{2 \pi i z}\right)##. Then ##U## is harmonic, being the composition of a harmonic function and a holomorphic function. Since ##\Pi^{+}##is simply connected. there exists ##F \in## ##H\left(\Pi^{+}\right)##such that ##U=\mathbb{R e}(F)##. Now, although ##U(z+1)=U(z)## we need not have ##F(z+1)=F(z)##. But ##\mathbb{R e}(F(z+1)-F(z))=U(z+1)-U(z)=0##. and so ##F(z+1)-F(z)## must be an imaginary constant: There exists ##c \in \mathbb{R}## such that ##F(z+1)-F(z)=i c## for all ##z \in \Pi^{+}##. Choose a real number ##\alpha \neq 0## so that ##c \alpha## is a multiple of ##2 \pi## (you can take ##\alpha=2 \pi / c## unless ##c=0## ). Let #### E(z)=e^{\alpha F(z)} . ####
It follows that ##E(z+1)=E(z)##, and so there exists ##f \in H\left(\mathbb{D}^{\prime}\right)## with #### E(z)=f\left(e^{2 \pi i z}\right) \text
{. } ####
Now the fact that ##u## is bounded shows that ##f## is bounded, so that ##f## has a removable singularity at the origin. The fact that ##u## is bounded also shows that ##f## is bounded away from 0 , so in particular ##f(0) \neq 0##. Hence ##u## has a limit at 0 , since ##u=\log (|f|) / \alpha##. (Since ##f\left(e^{2 \pi i z}\right)=e^{\alpha F(z)}## it follows that #### \left.\log \left(\left|f\left(e^{2 \pi i z}\right)\right|\right)=\alpha \operatorname{Re}(F(z))=\alpha U(z)=\alpha u\left(e^{2 \pi i z}\right) \cdot\right) ####
If we do the same thing with the Big Picard Theorem, replacing functions in ##\mathbb{D}^{\prime}## by the corresponding periodic functions in the upper half-plane, we see that we need to prove this:
Theorem 20.4. Suppose that ##F: \Pi^{+} \rightarrow \mathbb{C} \backslash\{0,1\}## is holomorphic and satisfies ##F(z+1)=F(z)## for all
##z \in \Pi^{+}##. Then there exists ##L \in \mathbb{C}_{\infty}## such
that ##F\left(z_j\right) \rightarrow L## for every sequence
##\left(z_j\right)## in ##\Pi^{+}##with ##\operatorname{Im}\left(z_j\right)
\rightarrow \infty##.
So, I need to show that Theorem 20.4 implies Theorem 20.3
I've tried to do something similar, but don't get something useful.
What I've tried:
So, supposing that ##f \in H(\mathbb{D}')## and ##f(\mathbb{D}') \subset \mathbb{C} \setminus \{0, 1\}## we have that ##f : \mathbb{D}' \to \mathbb{C} \setminus \{0, 1\}## is holomorphic. So, we can define ##F : \Pi^+ \to \mathbb{C} \setminus \{0, 1\}## by ##F(z) = f(e^{2 \pi i z})##. From here we obtain that ##F## is holomorphic on ##\Pi^+## (which is simply connected) so, because ##\lambda : \Pi^+ \to \mathbb{C} \setminus \{0,1\}## is a covering map we have that there exists the lift of ##f##: ##\tilde{f} : \Pi^+ \to \Pi^+## with ##\lambda \circ \tilde{f} = F##, where ##\tilde{f}## is holomorphic. Because ##F(z) = F(z+1)## for all ##z \in \Pi^+## we obtain that, from Theorem 20.4 there exists ##L \in \mathbb{C}_\infty## such that ##F(z_j) \to L## for every sequence ##(z_j) \subset \Pi^+## with ##\Im(z_j) \to \infty##. So, we have that ##(\lambda \circ \tilde{f})(z_j) \to L## for every sequence ##(z_j)## with ##Im(z_j) \to \infty##.
How can I continue from here? I know that I've written so much, but I need these for my bachelor degree. Thanks for reading!
P.S. I see some random '##>##' in text. Ignore it please. Is from cite I think.