One of the most interesting areas in complex analysis is the study of generalizations of polynomials. One avenue of generalization is the study of entire functions, which behave similarly to polynomials in many ways. Polynomials are also nice because for any choice of \(n\) complex numbers, one can easily construct a polynomial with precisely those \(n\) numbers as its roots. A natural question is: for some sequence of complex number \(\{a_n\}_{n\in\mathbb{N}}\), is there an analytic function whose set of roots is precisely \(\{a_n\}\)?
The answer is yes, as long as the \(a_n\) are subject to the mild condition that \(\{a_n\}_{n\in\mathbb{N}}\) has no limit points. This condition exists since any analytic function whose set of roots has a limit point must be the constant zero function as is shown by the identity theorem (this is an idea I used to struggle with; see here). Traditionally, this result is shown with a long technical argument that hinges on the Weierstrass factorization theorem, but if one imposes some more conditions on \(\{a_n\}_{n\in\mathbb{N}}\), it is much quicker to explicitly construct an analytic function with those roots. This construction is called a Blaschke product. First, we establish the following lemma that we will need later. Lemma: Let \(0<|a|<1\) and \(|z|\leq r<1\). Then, \[\left|\frac{a+|a|z}{(1-\overline{a}z)a}\right|\leq\frac{1+r}{1-r}.\] Proof: Put \[f(z)=\frac{a+|a|z}{(1-\overline{a}z)a}\] \[g(z)=f(z)-\frac{1}{1-\overline{a}z}=\frac{|a|z}{(1-\overline{a}z)a}.\] By the reverse triangle inequality, \[|1-\overline{a}z|\geq|1-|\overline{a}z||\geq1-|\overline{a}z|=1-|\overline{a}||z|=1-|a||z|.\] Since \(|a|<1\) and \(|z|\leq r\), the above gives us \(|1-\overline{a}z|\geq1-r\). Therefore, \(\frac{1}{|1-\overline{a}z|}\leq\frac{1}{1-r}\) and \[|g(z)|=\frac{|z|}{|1-\overline{a}z|}\leq\frac{r}{1-r}.\] Now by the reverse triangle inequality again, \[\left||f(z)|-\frac{1}{|1-\overline{a}z|}\right|\leq\left|f(z)-\frac{1}{1-\overline{a}z}\right|=|g(z)|\leq\frac{r}{1-r}.\] Hence, \[|f(z)|\leq\frac{1}{|1-\overline{a}z|}+\frac{r}{1-r}\leq\frac{1}{1-r}+\frac{r}{1-r}=\frac{1+r}{1-r}.\] \(\square\) Of course, to construct an infinite product, one must know how infinite products work. We will be invoking the following result from the theory of infinite products without proof. Theorem: Let \(G\) be a region in \(\mathbb{C}\) and let \(\{f_n\}_{n\in\mathbb{N}}\) be a sequence in \(H(G)\) such that no \(f_n\) is identically zero. If \(\sum_{n=1}^{\infty}{[f_n(z)-1]}\) converges absolutely and uniformly on compact subsets of \(G\) then \(\prod_{n=1}^{\infty}{f_n(z)}\) converges in \(H(G)\). This theorem gives us a sufficient condition for an infinite product of holomorphic functions to converge to a holomorphic function. These are the conditions that we will check when we construct our infinite product to confirm that the product is indeed a well-defined holomorphic function. Let \(\{a_n\}_{n\in\mathbb{N}}\) be a sequence of complex numbers with \(0<|a_n|<1\) for all \(n\in\mathbb{N}\) and \(\sum_{n=1}^{\infty}{(1-|a_n|)}<\infty\). We claim that \[B(z)=\prod_{n=1}^{\infty}{\frac{|a_n|}{a_n}\left(\frac{a_n-z}{1-\overline{a_n}z}\right)}\] converges in \(H(B_1(0))\) with \(|B(z)|\leq1\) for all \(z\in B_1(0)\). This is what is called a Blaschke product. The roots of \(B(z)\) are precisely the prescribed complex numbers \(\{a_n\}_{n\in\mathbb{N}}\). So if we can establish that this product converges to a holomorphic function, we will have found an analytic function whose roots we have chosen. Define \(f_n(z)=\frac{|a_n|}{a_n}\left(\frac{a_n-z}{1-\overline{a_n}z}\right)\) for each \(n\in\mathbb{N}\). We must check that \(\sum_{n=1}^{\infty}{[f_n(z)-1]}\) converges absolutely and uniformly on any compact subset of \(B_1(0)\). So let \(K\subseteq B_1(0)\) be an arbitrary compact subset. We can pick \(0<r<1\) so that \(K\subseteq B_r(0)\). Now, observe \[\begin{split} f_n(z)-1&=\frac{|a_n|}{a_n}\left(\frac{a_n-z}{1-\overline{a_n}z}\right)-1\\ &=\frac{|a_n|a_n-|a_n|z}{(1-\overline{a_n}z)a_n}-\frac{a_n-|a_n|^2z}{(1-\overline{a_n}z)a_n}\\ &=\frac{|a_n|a_n-a_n+|a_n|^2z-|a_n|z}{(1-\overline{a_n}z)a_n}\\ &=\frac{a_n(|a_n|-1)+|a_n|z(|a_n|-1)}{(1-\overline{a_n}z)a_n}\\ &=(|a_n|-1)\left(\frac{a_n+|a_n|z}{1-\overline{a_n}z}\right). \end{split}\] Since \(|a_n|<1\), we have \(||a_n|-1|=1-|a_n|\). Moreover, we may recognize the second factor above as the one from the lemma we have proven. Therefore, by the lemma, for all \(z\in B_r(0)\), \[|f_n(z)-1|\leq(1-|a_n|)\left(\frac{1+r}{1-r}\right).\] So for all \(z\in B_r(0)\), we have \[\sum_{n=1}^{\infty}{|f_n(z)-1|}\leq\frac{1+r}{1-r}\sum_{n=1}^{\infty}{1-|a_n|}<\infty\] by assumption. Therefore, \(\sum_{n=1}^{\infty}{[f_n(z)-1]}\) converges absolutely on \(K\). Uniform convergence follows immediately. Letting \(M_n=(1-|a_n|)\left(\frac{1+r}{1-r}\right)\), the above inequality tells us that \(\sum_{n=1}^{\infty}{M_n}<\infty\), so \(\sum_{n=1}^{\infty}{[f_n(z)-1]}\) converges uniformly on \(K\) due to the Weierstrass \(M\)-test. Since none of the of the \(f_n\) are identically zero, the theorem we wrote earlier tells us that \(B(z)\) converges in \(H(B_1(0))\). Of course, the roots are as desired. \(|B(z)|\leq1\) follows as a nice application of the classification of automorphisms of the unit disk (which itself follows from Schwarz's lemma). The classification tells us that every automorphism of the unit disk (i.e. conformal maps from the unit disk to itself) is a Möbius transformation of the form \(c\varphi_a\) where \(|c|=1\), \(|a|<1\), and \[\varphi_a(z)=\frac{z-a}{1-\overline{a}z}.\] Notice that for every \(n\in\mathbb{N}\), we have \(f_n=\frac{|a_n|}{a_n}\varphi_{a_n}\). Therefore, each \(f_n\) is actually an automorphism of the unit disk, and it follows that their infinite product must also map into the unit disk. If one does not know the traditional result that if \(\{a_n\}_{n\in\mathbb{N}}\) is the zero set of some holomorphic function as long as \(\{a_n\}_{n\in\mathbb{N}}\) has no limit points, one may still use Blaschke products to answer the following very interesting question. Let \(G\subseteq\mathbb{C}\) be an open region and let \(f\colon G\to\mathbb{C}\) be a holomorphic function. Is it necessarily true that \(f\) extends continuously to the boundary of \(G\)? The answer is no, as the following argument with a Blaschke product shows. For each \(n\in\mathbb{N}\), define \(a_n=\left(1-\frac{1}{(n+1)^2}\right)\exp{\left(i\sum_{k=1}^{n}{\frac{1}{k}}\right)}\). By construction, \(0<|a_n|<1\) for all \(n\) and \(\sum_{n=1}^{\infty}{(1-|a_n|)}=\sum_{n=1}^{\infty}{\frac{1}{(n+1)^2}}=\frac{\pi^2}{6}-1<\infty\). So the Blaschke product with roots \(\{a_n\}_{n\in\mathbb{N}}\) is well-defined. Let this Blaschke product be \(f(z)\). Since \(\sum_{k=1}^{\infty}{\frac{1}{k}}=\infty\) but \(\frac{1}{k}\to0\), it is clear that for any angle \(\theta\), the argument of \(a_n\) will come arbitrarily close to \(\theta\) for infinitely many \(n\). The details of this are not hard; one may phrase this in terms of the pigeonhole principle. Moreover, \(1-\frac{1}{(n+1)^2}\to 1\). So it is clear that every point on the unit circle is a limit point of the sequence \(\{a_n\}_{n\in\mathbb{N}}\). That is, every point of \(\partial B_1(0)\) is the limit point of the roots of the function \(f\) which is holomorphic on \(B_1(0)\). Thus, if \(f\) is to extend continuously to the boundary of its domain, this extension must be zero on the unit circle \(|z|=1\) due to continuity. But now the maximum modulus principle implies that \(f\) is the constant zero function, a contradiction. It turns out that there is a pretty developed theory of the space of functions which are holomorphic on the unit disk and can be continuously extended to the boundary (see here). Indeed, the fact that this space is properly contained in the space of bounded holomorphic functions on the disk (which is the Hardy space \(H^{\infty}\)) demonstrates why many statements of the maximum modulus principle state a hypothesis of the function being holomorphic in some open set \(G\) and continuous in the closure \(\overline{G}\).
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Holomorphic functions are very "rigid". One sense in which this is true is shown via Cauchy's estimate, which essentially states that a holomorphic function that is bounded must have derivatives that are not too large. Intuitively, this makes quite a bit of sense and one may ask if this is true outside of the setting of complex analysis. The answer is in the negative. Consider the family of real functions \(f_k\colon\mathbb{R}\to\mathbb{R}\) defined by \(f_k(x)=\sin{kx}\) where \(k\in\mathbb{R}\) is arbitrary. Clearly each \(f_k\) is bounded and analytic in some neighborhood of \(0\). Nonetheless, there is no \(M\) for which \(|f'_k(0)|<M\) for all \(k\in\mathbb{R}\) since \(f'_k(0)=k\) for every \(k\in\mathbb{R}\). So holomorphic functions are truly "rigid" in a sense that cannot be paralleled by real-analytic functions.
To show Cauchy's estimate, one needs to first develop the basic form of the Cauchy integral formula, use that to prove that holomorphic functions are analytic, and then study the coefficients of the power series expansion of holomorphic functions. Before we do this, we need the following lemma. Lemma: If \(|z|<1\), \[\int_{0}^{2\pi}{\frac{e^{is}}{e^{is}-z}\ \mathrm{d}s}=2\pi.\] Proof: We do this by the "Feynman trick" (differentiating under the integral sign, i.e. the Leibniz rule). Let \(\varphi(s,t)=\frac{e^{is}}{e^{is}-z}\) on \([0,2\pi]\times[0,1]\). Since we make the restrictions \(|z|<1\) and \(|t|<1\), we have \(|tz|<1\), so \(\varphi\) is continuously differentiable since it is the product of \(e^{is}\) and the closed form of the infinite geometric series with first term \(1\) and common ratio \(tz\). Define \[g(t)=\int_{0}^{2\pi}{\varphi(s,t)\ \mathrm{d}s}.\] Since \(\varphi\) is continuously differentiable, the fundamental theorem of calculus implies that \(g\) is continuously differentiable. So by the Leibniz rule, \[g'(t)=\int_{0}^{2\pi}{\frac{\partial}{\partial t}\varphi(s, t)\ \mathrm{d}s}=\int_{0}^{2\pi}{\frac{ze^{is}}{(e^{is}-tz)^2}\ \mathrm{d}s}.\] Now, one notices that \(\frac{zi}{e^{is}-tz}\) is an antiderivative of the integrand, and this function takes on the same value at \(s=0\) and \(s=2\pi\), hence \(g'(t)=0\) so \(g\) is a constant function. In particular, \[g(1)=g(0)=\int_{0}^{2\pi}{\mathrm{d}s}=2\pi,\] which is precisely what we wanted to show. \(\square\) We are now in a position to prove the basic form of the Cauchy integral formula. This is one of the cornerstone results in all of complex analysis. Cauchy's Integral Formula (Baby Version): Let \(G\subseteq\mathbb{C}\) be a region and let \(f\colon G\to\mathbb{C}\) be continuously differentiable. Suppose that \(r>0\) and \(\overline{B_r(a)}\subseteq G\). If \(\gamma(t)=a+re^{it}\), \(t\in[0,2\pi]\), then for all \(z\in B_r(a)\), \[f(z)=\frac{1}{2\pi i}\int_{\gamma}{\frac{f(w)}{w-z}\ \mathrm{d}w}.\] Proof: We will assume without loss of generality that \(a=0\) and \(r=1\). Pick \(z\in\mathbb{C}\) with \(|z|<1\). We use a technique similar to the one used in proving the lemma. Define \[\varphi(s,t)=\frac{f(z(1-t)+te^{is})e^{is}}{e^{is}-z}-f(z),\] where \((s,t)\in[0,2\pi]\times[0,1]\). Since \(\varphi\) is continuously differentiable, so is \(g\), so the Leibniz rule gives us \[g'(t)=\int_{0}^{2\pi}{e^{is}f'(z(1-t)+te^{is})\ \mathrm{d}s}.\] Now we notice that \(-\frac{i}{t}f'(z(1-t)+te^{is})\) is an antiderivative of the integrand that takes on the same value at \(s=0\) and \(s=2\pi\). Hence, \(g'(t)=0\) and \(g\) is constant. In particular, \[g(1)=g(0)=f(z)\int_{0}^{2\pi}{\frac{e^{is}}{e^{is}-z}\ \mathrm{d}s}-2\pi f(z)=0,\] where we apply the lemma that we have proven. But \(g(1)=0\) is equivalent to \[\int_{0}^{2\pi}{\frac{f(e^{is})e^{is}}{e^{is}-z}-f(z)\ \mathrm{d}s}=0.\] Rearranging this gives us \[f(z)=\frac{1}{2\pi}\int_{0}^{2\pi}{\frac{f(e^{is})e^{is}}{e^{is}-z}\ \mathrm{d}s}.\] But of course, we can realize the integral on the RHS as simply \(\frac{1}{i}\int_{\gamma}{\frac{f(w)}{w-z}\ \mathrm{d}w}\). \(\square\) We made an assumption of continuous differentiability in this theorem, but it is a classical result that holomorphy is equivalent to this (which I will not show here). Notice that the Cauchy integral formula is similar in philosophy to the mean value property of harmonic functions, which states that a harmonic function in a ball takes on its average value over the boundary of the ball at the center of the ball. For harmonic functions, the value of the function at the center of a ball is determined by the values of the function on the boundary of the ball. The Cauchy integral formula shows that holomorphic functions abide by the same principle in an even stronger sense: the value of a holomorphic function at any point in a ball is completely determined by the values of the function on the boundary of the ball. The next step is to establish that holomorphic functions are analytic by explicitly constructing a power series and showing that the power series will converge. Then, the coefficients of the power series we have constructed can be studied to control the derivatives i.e., to obtain Cauchy's estimate. By now, it should be clear that there are many parallels between holomorphic and harmonic functions, so one may try to find inspiration for the proof of the analyticity of holomorphic functions from the proof of the analyticity of harmonic functions. But a similar approach will not be possible. In the proof of the analyticity of harmonic functions, we essentially use a Cauchy-type estimate to control the derivatives of the harmonic function, and then use Taylor's theorem to establish the convergence of the power series. But now, we do not have access to a priori the Cauchy estimate, so we need a different approach to establish the convergence of the power series. The solution is to realize that part of the integrand in the Cauchy integral formula can be expanded as a geometric series. Analyticity of Holomorphic Functions: Let \(f\) be continuously differentiable in \(B_R(a)\). Then, \[f(z)=\sum_{n=0}^{\infty}{\frac{f^{(n)}(a)}{n!}(z-a)^n},\] and this power series has a radius of convergence of at least \(R\). Proof: Pick \(r\in(0,R)\). Let \(\gamma(t)=a+re^{it}\). By the Cauchy integral formula, for all \(z\in B_r(a)\), \[f(z)=\frac{1}{2\pi i}\int_{\gamma}{\frac{f(w)}{w-z}\ \mathrm{d}w}=\frac{1}{2\pi i}\int_{\gamma}{\frac{f(w)}{w-a}\left(\frac{1}{1-\frac{z-a}{w-a}}\right)\ \mathrm{d}w}.\] Observe that \(|z-a|<r=|w-a|\) since \(w\in\{\gamma\}\) and \(z\in B_r(a)\). Therefore, the factor of the integrand in parentheses can be realized as an infinite geometric series: \(\frac{1}{1-\frac{z-a}{w-a}}=\sum_{n=0}^{\infty}{\left(\frac{z-a}{w-a}\right)^n}\), and thus \[f(z)=\frac{1}{2\pi i}\int_{\gamma}{\sum_{n=0}^{\infty}{\frac{f(w)}{(w-a)^{n+1}}(z-a)^n}\ \mathrm{d}w}.\] We would like to interchange the sum and the integral. Observe that \(\{\gamma\}\) is compact and \(|f|\) is continuous, so \(|f|\) attains some maximum \(M\) on \(\{\gamma\}\). Moreover, \(|w-a|=r\). Hence, \[\frac{|f(w)||z-a|^n}{|w-a|^{n+1}}\leq\frac{M|z-a|^n}{r^{n+1}}=\frac{M}{r}\left(\frac{|z-a|}{r}\right)^n.\] Of course, since \(|z-a|<r\), we have that \(\sum_{n=0}^{\infty}{\frac{|z-a|}{r}}\) converges as a geometric series. Therefore, the Weierstrass \(M\)-test implies that the partial sums of \(\sum_{n=0}^{\infty}{\frac{f(w)}{(w-a)^{n+1}}(z-a)^n}\) converge uniformly. Since integration commutes with uniform convergence, we have \[f(z)=\sum_{n=0}^{\infty}{\frac{(z-a)^n}{2\pi i}\int_{\gamma}{\frac{f(w)}{(w-a)^{n+1}}\ \mathrm{d}w}}.\] Indeed, this is a power series that converges for \(|z-a|<r\). The theory of analytic functions tell us that the coefficients must actually be \(\frac{1}{n!}f^{(n)}(a)\). In particular, the coefficients are independent of \(r\). Therefore, since \(0<r<R\) was arbitrary, the power series above converges at least within \(B_R(a)\). \(\square\) The key point here was that \(\frac{1}{z}\) is the closed form of an infinite geometric series when \(z\) is in the unit disk. Indeed, a lot of complex analysis revolves around the quirks of the function \(z\mapsto\frac{1}{z}\). Cauchy's estimate now follows easily. Observe that the proof above establishes that if \(f\) is holomorphic in the region \(G\) with \(\overline{B_r(a)}\subseteq G\) and \(\gamma(t)=a+re^{it}\), then \(f^{(n)}(a)=\frac{n!}{2\pi i}\int_{\gamma}{\frac{f(w)}{(w-a)^{n+1}}\ \mathrm{d}w}\). Suppose \(G=B_R(a)\), and suppose \(|f(z)|\leq M\) on this domain. Then for all \(r<R\) we have that \[\begin{split} |f^{(n)}(a)|&=\frac{n!}{2\pi}\left|\int_{\gamma}{\frac{f(w)}{(w-a)^{n+1}}\ \mathrm{d}w}\right|\\ &\leq\frac{n!}{2\pi}\int_{\gamma}{\left|\frac{f(w)}{(w-a)^{n+1}}\right|\ \mathrm{d}w}\\ &\leq\frac{n!}{2\pi}\int_{\gamma}{\frac{M}{r^{n+1}}\ \mathrm{d}w}\\ &=\frac{n!}{2\pi}\cdot\frac{M}{r^{n+1}}\cdot2\pi r\\ &=\frac{n!M}{r^n}. \end{split}\] Now, we can take the limit \(r\to R^-\) to obtain Cauchy's estimate \[|f^{(n)}(a)|\leq\frac{n!M}{R^n}.\] As we mentioned before, this estimate is quite similar to the one we used to prove the analyticity of harmonic functions. Cauchy's estimate is a powerful statement. Among other things it is used to prove Liouville's theorem, of which the fundamental theorem of algebra is a simple consequence. |
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