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Previous set of notes: 246A Notes 5. Next set of notes: Notes 2.

— 1. Jensen’s formula —

Suppose ${f}$ is a non-zero rational function ${f =P/Q}$, then by the fundamental theorem of algebra one can write

$\displaystyle f(z) = c \frac{\prod_\rho (z-\rho)}{\prod_\zeta (z-\zeta)}$

for some non-zero constant ${c}$, where ${\rho}$ ranges over the zeroes of ${P}$ (counting multiplicity) and ${\zeta}$ ranges over the zeroes of ${Q}$ (counting multiplicity), and assuming ${z}$ avoids the zeroes of ${Q}$. Taking absolute values and then logarithms, we arrive at the formula

$\displaystyle \log |f(z)| = \log |c| + \sum_\rho \log|z-\rho| - \sum_\zeta \log |z-\zeta|, \ \ \ \ \ (1)$

as long as ${z}$ avoids the zeroes of both ${P}$ and ${Q}$. (In this set of notes we use ${\log}$ for the natural logarithm when applied to a positive real number, and ${\mathrm{Log}}$ for the standard branch of the complex logarithm (which extends ${\log}$); the multi-valued complex logarithm ${\log}$ will only be used in passing.) Alternatively, taking logarithmic derivatives, we arrive at the closely related formula

$\displaystyle \frac{f'(z)}{f(z)} = \sum_\rho \frac{1}{z-\rho} - \sum_\zeta \frac{1}{z-\zeta}, \ \ \ \ \ (2)$

again for ${z}$ avoiding the zeroes of both ${P}$ and ${Q}$. Thus we see that the zeroes and poles of a rational function ${f}$ describe the behaviour of that rational function, as well as close relatives of that function such as the log-magnitude ${\log|f|}$ and log-derivative ${\frac{f'}{f}}$. We have already seen these sorts of formulae arise in our treatment of the argument principle in 246A Notes 4.

Exercise 1 Let ${P(z)}$ be a complex polynomial of degree ${n \geq 1}$.
• (i) (Gauss-Lucas theorem) Show that the complex roots of ${P'(z)}$ are contained in the closed convex hull of the complex roots of ${P(z)}$.
• (ii) (Laguerre separation theorem) If all the complex roots of ${P(z)}$ are contained in a disk ${D(z_0,r)}$, and ${\zeta \not \in D(z_0,r)}$, then all the complex roots of ${nP(z) + (\zeta - z) P'(z)}$ are also contained in ${D(z_0,r)}$. (Hint: apply a suitable Möbius transformation to move ${\zeta}$ to infinity, and then apply part (i) to a polynomial that emerges after applying this transformation.)

There are a number of useful ways to extend these formulae to more general meromorphic functions than rational functions. Firstly there is a very handy “local” variant of (1) known as Jensen’s formula:

Theorem 2 (Jensen’s formula) Let ${f}$ be a meromorphic function on an open neighbourhood of a disk ${\overline{D(z_0,r)} = \{ z: |z-z_0| \leq r \}}$, with all removable singularities removed. Then, if ${z_0}$ is neither a zero nor a pole of ${f}$, we have

$\displaystyle \log |f(z_0)| = \int_0^1 \log |f(z_0+re^{2\pi i t})|\ dt + \sum_{\rho: |\rho-z_0| \leq r} \log \frac{|\rho-z_0|}{r} \ \ \ \ \ (3)$

$\displaystyle - \sum_{\zeta: |\zeta-z_0| \leq r} \log \frac{|\zeta-z_0|}{r}$

where ${\rho}$ and ${\zeta}$ range over the zeroes and poles of ${f}$ respectively (counting multiplicity) in the disk ${\overline{D(z_0,r)}}$.

One can view (3) as a truncated (or localised) variant of (1). Note also that the summands ${\log \frac{|\rho-z_0|}{r}, \log \frac{|\zeta-z_0|}{r}}$ are always non-positive.

Proof: By perturbing ${r}$ slightly if necessary, we may assume that none of the zeroes or poles of ${f}$ (which form a discrete set) lie on the boundary circle ${\{ z: |z-z_0| = r \}}$. By translating and rescaling, we may then normalise ${z_0=0}$ and ${r=1}$, thus our task is now to show that

$\displaystyle \log |f(0)| = \int_0^1 \log |f(e^{2\pi i t})|\ dt + \sum_{\rho: |\rho| < 1} \log |\rho| - \sum_{\zeta: |\zeta| < 1} \log |\zeta|. \ \ \ \ \ (4)$

We may remove the poles and zeroes inside the disk ${D(0,1)}$ by the useful device of Blaschke products. Suppose for instance that ${f}$ has a zero ${\rho}$ inside the disk ${D(0,1)}$. Observe that the function

$\displaystyle B_\rho(z) := \frac{\rho - z}{1 - \overline{\rho} z} \ \ \ \ \ (5)$

has magnitude ${1}$ on the unit circle ${\{ z: |z| = 1\}}$, equals ${\rho}$ at the origin, has a simple zero at ${\rho}$, but has no other zeroes or poles inside the disk. Thus Jensen’s formula (4) already holds if ${f}$ is replaced by ${B_\rho}$. To prove (4) for ${f}$, it thus suffices to prove it for ${f/B_\rho}$, which effectively deletes a zero ${\rho}$ inside the disk ${D(0,1)}$ from ${f}$ (and replaces it instead with its inversion ${1/\overline{\rho}}$). Similarly we may remove all the poles inside the disk. As a meromorphic function only has finitely many poles and zeroes inside a compact set, we may thus reduce to the case when ${f}$ has no poles or zeroes on or inside the disk ${D(0,1)}$, at which point our goal is simply to show that

$\displaystyle \log |f(0)| = \int_0^1 \log |f(e^{2\pi i t})|\ dt.$

Since ${f}$ has no zeroes or poles inside the disk, it has a holomorphic logarithm ${F}$ (Exercise 46 of 246A Notes 4). In particular, ${\log |f|}$ is the real part of ${F}$. The claim now follows by applying the mean value property (Exercise 17 of 246A Notes 3) to ${\log |f|}$. $\Box$

An important special case of Jensen’s formula arises when ${f}$ is holomorphic in a neighborhood of ${\overline{D(z_0,r)}}$, in which case there are no contributions from poles and one simply has

$\displaystyle \int_0^1 \log |f(z_0+re^{2\pi i t})|\ dt = \log |f(z_0)| + \sum_{\rho: |\rho-z_0| \leq r} \log \frac{r}{|\rho-z_0|}. \ \ \ \ \ (6)$

This is quite a useful formula, mainly because the summands ${\log \frac{r}{|\rho-z_0|}}$ are non-negative; it can be viewed as a more precise assertion of the subharmonicity of ${\log |f|}$ (see Exercises 60(ix) and 61 of 246A Notes 5). Here are some quick applications of this formula:

Exercise 3 Use (6) to give another proof of Liouville’s theorem: a bounded holomorphic function ${f}$ on the entire complex plane is necessarily constant.

Exercise 4 Use Jensen’s formula to prove the fundamental theorem of algebra: a complex polynomial ${P(z)}$ of degree ${n}$ has exactly ${n}$ complex zeroes (counting multiplicity), and can thus be factored as ${P(z) = c (z-z_1) \dots (z-z_n)}$ for some complex numbers ${c,z_1,\dots,z_n}$ with ${c \neq 0}$. (Note that the fundamental theorem was invoked previously in this section, but only for motivational purposes, so the proof here is non-circular.)

Exercise 5 (Shifted Jensen’s formula) Let ${f}$ be a meromorphic function on an open neighbourhood of a disk ${\{ z: |z-z_0| \leq r \}}$, with all removable singularities removed. Show that

$\displaystyle \log |f(z)| = \int_0^1 \log |f(z_0+re^{2\pi i t})| \mathrm{Re} \frac{r e^{2\pi i t} + (z-z_0)}{r e^{2\pi i t} - (z-z_0)}\ dt \ \ \ \ \ (7)$

$\displaystyle + \sum_{\rho: |\rho-z_0| \leq r} \log \frac{|\rho-z|}{|r - \rho^* (z-z_0)|}$

$\displaystyle - \sum_{\zeta: |\zeta-z_0| \leq r} \log \frac{|\zeta-z|}{|r - \zeta^* (z-z_0)|}$

for all ${z}$ in the open disk ${\{ z: |z-z_0| < r\}}$ that are not zeroes or poles of ${f}$, where ${\rho^* = \frac{\overline{\rho-z_0}}{r}}$ and ${\zeta^* = \frac{\overline{\zeta-z_0}}{r}}$. (The function ${\Re \frac{r e^{2\pi i t} + (z-z_0)}{r e^{2\pi i t} - (z-z_0)}}$ appearing in the integrand is sometimes known as the Poisson kernel, particularly if one normalises so that ${z_0=0}$ and ${r=1}$.)

Exercise 6 (Bounded type)
• (i) If ${f}$ is a holomorphic function on ${D(0,1)}$ that is not identically zero, show that ${\liminf_{r \rightarrow 1^-} \int_0^{2\pi} \log |f(re^{i\theta})|\ d\theta > -\infty}$.
• (ii) If ${f}$ is a meromorphic function on ${D(0,1)}$ that is the ratio of two bounded holomorphic functions that are not identically zero, show that ${\limsup_{r \rightarrow 1^-} \int_0^{2\pi} |\log |f(re^{i\theta})||\ d\theta < \infty}$. (Functions ${f}$ of this form are said to be of bounded type and lie in the Nevanlinna class for the unit disk ${D(0,1)}$.)

Exercise 7 (Smoothed out Jensen formula) Let ${f}$ be a meromorphic function on an open set ${U}$, and let ${\phi: U \rightarrow {\bf C}}$ be a smooth compactly supported function. Show that

$\displaystyle \sum_\rho \phi(\rho) - \sum_\zeta \phi(\zeta)$

$\displaystyle = \frac{-1}{2\pi} \int\int_U ((\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}) \phi(x+iy)) \frac{f'}{f}(x+iy)\ dx dy$

$\displaystyle = \frac{1}{2\pi} \int\int_U ((\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y}^2) \phi(x+iy)) \log |f(x+iy)|\ dx dy$

where ${\rho, \zeta}$ range over the zeroes and poles of ${f}$ (respectively) in the support of ${\phi}$. Informally argue why this identity is consistent with Jensen’s formula.

When applied to entire functions ${f}$, Jensen’s formula relates the order of growth of ${f}$ near infinity with the density of zeroes of ${f}$. Here is a typical result:

Proposition 8 Let ${f: {\bf C} \rightarrow {\bf C}}$ be an entire function, not identically zero, that obeys a growth bound ${|f(z)| \leq C \exp( C|z|^\alpha)}$ for some ${C, \alpha > 0}$ and all ${z}$. Then there exists a constant ${C'>0}$ such that ${D(0,R)}$ has at most ${C' R^\alpha}$ zeroes (counting multiplicity) for any ${R \geq 1}$.

Entire functions that obey a growth bound of the form ${|f(z)| \leq C_\varepsilon \exp( C_\varepsilon |z|^{\rho+\varepsilon})}$ for every ${\varepsilon>0}$ and ${z}$ (where ${C_\varepsilon}$ depends on ${\varepsilon}$) are said to be of order at most ${\rho}$. The above theorem shows that for such functions that are not identically zero, the number of zeroes in a disk of radius ${R}$ does not grow much faster than ${R^\rho}$. This is often a useful preliminary upper bound on the zeroes of entire functions, as the order of an entire function tends to be relatively easy to compute in practice.

Proof: First suppose that ${f(0)}$ is non-zero. From (6) applied with ${r=2R}$ and ${z_0=0}$ one has

$\displaystyle \int_0^1 \log(C \exp( C (2R)^\alpha ) )\ dt \geq \log |f(0)| + \sum_{\rho: |\rho| \leq 2R} \log \frac{2R}{|\rho|}.$

Every zero in ${D(0,R)}$ contribute at least ${\log 2}$ to a summand on the right-hand side, while all other zeroes contribute a non-negative quantity, thus

$\displaystyle \log C + C (2R)^\alpha \geq \log |f(0)| + N_R \log 2$

where ${N_R}$ denotes the number of zeroes in ${D(0,R)}$. This gives the claim for ${f(0) \neq 0}$. When ${f(0)=0}$, one can shift ${f}$ by a small amount to make ${f}$ non-zero at the origin (using the fact that zeroes of holomorphic functions not identically zero are isolated), modifying ${C}$ in the process, and then repeating the previous arguments. $\Box$

Just as (3) and (7) give truncated variants of (1), we can create truncated versions of (2). The following crude truncation is adequate for many applications:

Theorem 9 (Truncated formula for log-derivative) Let ${f}$ be a holomorphic function on an open neighbourhood of a disk ${\{ z: |z-z_0| \leq r \}}$ that is not identically zero on this disk. Suppose that one has a bound of the form ${|f(z)| \leq M^{O_{c_1,c_2}(1)} |f(z_0)|}$ for some ${M \geq 1}$ and all ${z}$ on the circle ${\{ z: |z-z_0| = r\}}$. Let ${0 < c_2 < c_1 < 1}$ be constants. Then one has the approximate formula

$\displaystyle \frac{f'(z)}{f(z)} = \sum_{\rho: |\rho - z_0| \leq c_1 r} \frac{1}{z-\rho} + O_{c_1,c_2}( \frac{\log M}{r} )$

for all ${z}$ in the disk ${\{ z: |z-z_0| < c_2 r \}}$ other than zeroes of ${f}$. Furthermore, the number of zeroes ${\rho}$ in the above sum is ${O_{c_1,c_2}(\log M)}$.

Proof: To abbreviate notation, we allow all implied constants in this proof to depend on ${c_1,c_2}$.

We mimic the proof of Jensen’s formula. Firstly, we may translate and rescale so that ${z_0=0}$ and ${r=1}$, so we have ${|f(z)| \leq M^{O(1)} |f(0)|}$ when ${|z|=1}$, and our main task is to show that

$\displaystyle \frac{f'(z)}{f(z)} - \sum_{\rho: |\rho| \leq c_1} \frac{1}{z-\rho} = O( \log M ) \ \ \ \ \ (8)$

for ${|z| \leq c_2}$. Note that if ${f(0)=0}$ then ${f}$ vanishes on the unit circle and hence (by the maximum principle) vanishes identically on the disk, a contradiction, so we may assume ${f(0) \neq 0}$. From hypothesis we then have

$\displaystyle \log |f(z)| \leq \log |f(0)| + O(\log M)$

on the unit circle, and so from Jensen’s formula (3) we see that

$\displaystyle \sum_{\rho: |\rho| \leq 1} \log \frac{1}{|\rho|} = O(\log M). \ \ \ \ \ (9)$

In particular we see that the number of zeroes with ${|\rho| \leq c_1}$ is ${O(\log M)}$, as claimed.

Suppose ${f}$ has a zero ${\rho}$ with ${c_1 < |\rho| \leq 1}$. If we factor ${f = B_\rho g}$, where ${B_\rho}$ is the Blaschke product (5), then

$\displaystyle \frac{f'}{f} = \frac{B'_\rho}{B_\rho} + \frac{g'}{g}$

$\displaystyle = \frac{g'}{g} + \frac{1}{z-\rho} - \frac{1}{z-1/\overline{\rho}}.$

Observe from Taylor expansion that the distance between ${\rho}$ and ${1/\overline{\rho}}$ is ${O( \log \frac{1}{|\rho|} )}$, and hence ${\frac{1}{z-\rho} - \frac{1}{z-1/\overline{\rho}} = O( \log \frac{1}{|\rho|} )}$ for ${|z| \leq c_2}$. Thus we see from (9) that we may use Blaschke products to remove all the zeroes in the annulus ${c_1 < |\rho| \leq 1}$ while only affecting the left-hand side of (8) by ${O( \log M)}$; also, removing the Blaschke products does not affect ${|f(z)|}$ on the unit circle, and only affects ${\log |f(0)|}$ by ${O(\log M)}$ thanks to (9). Thus we may assume without loss of generality that there are no zeroes in this annulus.

Similarly, given a zero ${\rho}$ with ${|\rho| \leq c_1}$, we have ${\frac{1}{z-1/\overline{\rho}} = O(1)}$, so using Blaschke products to remove all of these zeroes also only affects the left-hand side of (8) by ${O(\log M)}$ (since the number of zeroes here is ${O(\log M)}$), with ${\log |f(0)|}$ also modified by at most ${O(\log M)}$. Thus we may assume in fact that ${f}$ has no zeroes whatsoever within the unit disk. We may then also normalise ${f(0) = 1}$, then ${\log |f(e^{2\pi i t})| \leq O(\log M)}$ for all ${t \in [0,1]}$. By Jensen’s formula again, we have

$\displaystyle \int_0^1 \log |f(e^{2\pi i t})|\ dt = 0$

and thus (by using the identity ${|x| = 2 \max(x,0) - x}$ for any real ${x}$)

$\displaystyle \int_0^1 \log |f(e^{2\pi i t})|\ dt \ll \log M. \ \ \ \ \ (10)$

On the other hand, from (7) we have

$\displaystyle \log |f(z)| = \int_0^1 \log |f(e^{2\pi i t})| \mathrm{Re} \frac{e^{2\pi i t} + z}{e^{2\pi i t} - z}\ dt$

which implies from (10) that ${\log |f(z)|}$ and its first derivatives are ${O( \log M )}$ on the disk ${\{ z: |z| \leq c_2 \}}$. But recall from the proof of Jensen’s formula that ${\frac{f'}{f}}$ is the derivative of a logarithm ${\log f}$ of ${f}$, whose real part is ${\log |f|}$. By the Cauchy-Riemann equations for ${\log f}$, we conclude that ${\frac{f'}{f} = O(\log M)}$ on the disk ${\{ z: |z| \leq c_2 \}}$, as required. $\Box$

Exercise 10
• (i) (Borel-Carathéodory theorem) If ${f: U \rightarrow {\bf C}}$ is analytic on an open neighborhood of a disk ${\overline{D(z_0,R)}}$, show that

$\displaystyle \sup_{z \in D(z_0,r)} |f(z)| \leq \frac{2r}{R-r} \sup_{z \in \overline{D(z_0,R)}} \mathrm{Re} f(z) + \frac{R+r}{R-r} |f(z_0)|.$

(Hint: one can normalise ${z_0=0}$, ${R=1}$, ${f(0)=0}$, and ${\sup_{|z-z_0| \leq R} \mathrm{Re} f(z)=1}$. Now ${f}$ maps the unit disk to the half-plane ${\{ \mathrm{Re} z \leq 1 \}}$. Use a Möbius transformation to map the half-plane to the unit disk and then use the Schwarz lemma.)
• (ii) Use (i) to give an alternate way to conclude the proof of Theorem 9.

A variant of the above argument allows one to make precise the heuristic that holomorphic functions locally look like polynomials:

Exercise 11 (Local Weierstrass factorisation) Let the notation and hypotheses be as in Theorem 9. Then show that

$\displaystyle f(z) = P(z) \exp( g(z) )$

for all ${z}$ in the disk ${\{ z: |z-z_0| < c_2 r \}}$, where ${P}$ is a polynomial whose zeroes are precisely the zeroes of ${f}$ in ${\{ z: |z-z_0| \leq c_1r \}}$ (counting multiplicity) and ${g}$ is a holomorphic function on ${\{ z: |z-z_0| < c_2 r \}}$ of magnitude ${O_{c_1,c_2}( \log M )}$ and first derivative ${O_{c_1,c_2}( \log M / r )}$ on this disk. Furthermore, show that the degree of ${P}$ is ${O_{c_1,c_2}(\log M)}$.

Exercise 12 (Preliminary Beurling factorisation) Let ${H^\infty(D(0,1))}$ denote the space of bounded analytic functions ${f: D(0,1) \rightarrow {\bf C}}$ on the unit disk; this is a normed vector space with norm

$\displaystyle \|f\|_{H^\infty(D(0,1))} := \sup_{z \in D(0,1)} |f(z)|.$

• (i) If ${f \in H^\infty(D(0,1))}$ is not identically zero, and ${z_n}$ denote the zeroes of ${f}$ in ${D(0,1)}$ counting multiplicity, show that

$\displaystyle \sum_n (1-|z_n|) < \infty$

and

$\displaystyle \sup_{1/2 < r < 1} \int_0^{2\pi} | \log |f(re^{i\theta})| |\ d\theta < \infty.$

• (ii) Let the notation be as in (i). If we define the Blaschke product

$\displaystyle B(z) := z^m \prod_{|z_n| \neq 0} \frac{|z_n|}{z_n} \frac{z_n-z}{1-\overline{z_n} z}$

where ${m}$ is the order of vanishing of ${f}$ at zero, show that this product converges absolutely to a holomorphic function on ${D(0,1)}$, and that ${|f(z)| \leq \|f\|_{H^\infty(D(0,1)} |B(z)|}$ for all ${z \in D(0,1)}$. (It may be easier to work with finite Blaschke products first to obtain this bound.)
• (iii) Continuing the notation from (i), establish a factorisation ${f(z) = B(z) \exp(g(z))}$ for some holomorphic function ${g: D(0,1) \rightarrow {\bf C}}$ with ${\mathrm{Re}(g(z)) \leq \log \|f\|_{H^\infty(D(0,1)}}$ for all ${z\in D(0,1)}$.
• (iv) (Theorem of F. and M. Riesz, special case) If ${f \in H^\infty(D(0,1))}$ extends continuously to the boundary ${\{e^{i\theta}: 0 \leq \theta < 2\pi\}}$, show that the set ${\{ 0 \leq \theta < 2\pi: f(e^{i\theta})=0 \}}$ has zero measure.

Remark 13 The factorisation (iii) can be refined further, with ${g}$ being the Poisson integral of some finite measure on the unit circle. Using the Lebesgue decomposition of this finite measure into absolutely continuous parts one ends up factorising ${H^\infty(D(0,1))}$ functions into “outer functions” and “inner functions”, giving the Beurling factorisation of ${H^\infty}$. There are also extensions to larger spaces ${H^p(D(0,1))}$ than ${H^\infty(D(0,1))}$ (which are to ${H^\infty}$ as ${L^p}$ is to ${L^\infty}$), known as Hardy spaces. We will not discuss this topic further here, but see for instance this text of Garnett for a treatment.

Exercise 14 (Littlewood’s lemma) Let ${f}$ be holomorphic on an open neighbourhood of a rectangle ${R = \{ \sigma+it: \sigma_0 \leq \sigma \leq \sigma_1; 0 \leq t \leq T \}}$ for some ${\sigma_0 < \sigma_1}$ and ${T>0}$, with ${f}$ non-vanishing on the boundary of the rectangle. Show that

$\displaystyle 2\pi \sum_\rho (\mathrm{Re}(\rho)-\sigma_0) = \int_0^T \log |f(\sigma_0+it)|\ dt - \int_0^T \log |f(\sigma_1+it)|\ dt$

$\displaystyle + \int_{\sigma_0}^{\sigma_1} \mathrm{arg} f(\sigma+iT)\ d\sigma - \int_{\sigma_0}^{\sigma_1} \mathrm{arg} f(\sigma)\ d\sigma$

where ${\rho}$ ranges over the zeroes of ${f}$ inside ${R}$ (counting multiplicity) and one uses a branch of ${\mathrm{arg} f}$ which is continuous on the upper, lower, and right edges of ${C}$. (This lemma is a popular tool to explore the zeroes of Dirichlet series such as the Riemann zeta function.)