Section 14.9

The Maximum Modulus Principle

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Problem 5645

Applying Cauchy's integral to the function (f(z))n\left(f(z)\right)^{n}, where nn is an integer tending to ∞\infty, prove the theorem: if f(z)f(z) is regular inside a contour CC, and on the contour the inequality ∣f(z)∣⩽M\left|f(z)\right| \leqslant M holds, then it remains valid inside the contour as well.

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Problem 5646

Prove the theorem: if f(z)f(z) is regular inside the disk ∣z∣⩽R\left|z\right| \leqslant R, and ∣f(z)∣<M\left|f(z)\right| < M, while f(0)=0f(0) = 0, then for ∣z∣<R\left|z\right| < R the inequality ∣f(z)∣<MR⋅∣z∣\left|f(z)\right| < \dfrac {M}{R}\cdot \left|z\right| holds, and equality at even one point is possible only when f(z)f(z) has the form f(z)=MReiαzf(z) = \dfrac {M}{R}e^{i\alpha }z. (Schwarz's Lemma.)

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Problem 5647

Prove the theorem: if f(z)f(z) is regular for r<∣z∣<Rr < \left|z\right| < R, then for r<ρ<Rr < \rho < R the identity

12π∫02π∣f(ρeφi)∣2 dφ=∑∣an∣2ρ2n, \frac{1}{2\pi }\int _{0}^{2\pi } \left|f(\rho e^{\varphi i})\right|^{2}\, d\varphi = \sum \left|a_{n}\right|^{2}\rho ^{2n},

holds, where ana_{n} is a coefficient of the Laurent series.

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Problem 5648

Prove the theorem: if f(z)=∑n=0∞anznf(z) = \sum_{n=0}^{\infty } a_{n}z^{n} converges for ∣z∣<R\left|z\right| < R, and moreover ∣f(z)∣⩽M\left|f(z)\right| \leqslant M, then ∑∣an∣2R2n⩽M2\sum \left|a_{n}\right|^{2}R^{2n} \leqslant M^{2}.

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Problem 5649

Prove that if the identity w=∑n=0∞anznw = \sum_{n=0}^{\infty } a_{n}z^{n} gives a mapping of the disk ∣z∣<R\left|z\right| < R onto an area SS, then S=π∑n=0∞n∣an∣2R2nS = \pi \sum_{n=0}^{\infty } n\left|a_{n}\right|^{2}R^{2n}.

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Problem 5650

Prove the theorem: if f(z)f(z) is regular in the disk ∣z∣<R\left|z\right| < R and is not constant, then the function

I(r)=∫02π∣f(reφi)∣2 dφ I(r) = \int _{0}^{2\pi } \left|f(re^{\varphi i})\right|^{2}\, d\varphi

increases together with rr for 0<r<R0 < r < R.

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Problem 5651

Prove the theorem: among all functions f(z)f(z), regular inside the disk ∣z∣<R\left|z\right| < R and satisfying the conditions ∫02π∣f(Reφi)∣2 dφ=M\int_{0}^{2\pi } \left|f(Re^{\varphi i})\right|^{2}\, d\varphi = M, the linear function gives the smallest area onto which the disk ∣z∣<R\left|z\right| < R is mapped.

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Problem 5652

Prove the theorem: for a function f(z)f(z), regular for r<∣z∣<Rr < \left|z\right| < R, the quantity

I(ρ)=∫02π∣f(ρeφi)∣2 dφ=K(ln⁡ρ) I(\rho ) = \int _{0}^{2\pi } \left|f(\rho e^{\varphi i})\right|^{2}\, d\varphi = K(\ln \rho )

is a convex function of ln⁡ρ\ln \rho, i.e., the inequality holds (for m>0m > 0, n>0n > 0):

K(mln⁡ρ1+nln⁡ρ2m+n)⩽mK(ln⁡ρ1)+nK(ln⁡ρ2)m+n. K\left(\frac{m\ln \rho _{1}+n\ln \rho _{2}}{m+n}\right) \leqslant \frac{mK(\ln \rho _{1})+nK(\ln \rho _{2})}{m+n}.
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Problem 5653

Prove the theorem: if f(z)f(z), besides the previous conditions (problem 5652), is also nonzero for r<∣z∣<Rr < \left|z\right| < R, then the quantity

Iσ(ρ)=∫02π∣f(ρeφi)∣σ dφ I_{\sigma }(\rho ) = \int _{0}^{2\pi } \left|f(\rho e^{\varphi i})\right|^{\sigma }\, d\varphi

is likewise, for any σ>0\sigma > 0, a convex function of ln⁡ρ\ln \rho.

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Problem 5654

Show that the condition f(z)≠0f(z) \neq 0 in the previous problem (problem 5653) can be dispensed with.

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Problem 5655

Prove the theorem: if f(z)f(z) is regular for ∣z∣<R\left|z\right| < R, then the quantity Iσ(ρ)I_{\sigma }(\rho ) (problem 5653) increases together with ρ\rho.

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Problem 5656

Prove the theorem: if f(z)f(z) is regular for r<∣z∣<Rr < \left|z\right| < R, and M(ρ)M(\rho ) is the maximum modulus of f(ρeφi)f(\rho e^{\varphi i}), then for r<ρ<Rr < \rho < R the quantity M(ρ)M(\rho ) is a convex function of ln⁡ρ\ln \rho (Hadamard's Three-Circles Theorem).

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Remark.
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Hint. It is useful to consider the function znf(z)=F(z)z^{n}f(z) = F(z), choosing nn so that the quantities M(r)M(r) and M(R)M(R) for F(z)F(z) are equal.

Problem 5657

Prove the theorem: if a polynomial f(z)f(z) of degree nn does not exceed MM in modulus on the interval −1⩽z⩽1-1 \leqslant z \leqslant 1, then for any zz the inequality ∣f(z)∣⩽M(a+b)n\left|f(z)\right| \leqslant M(a+b)^{n} holds, where aa and bb are the semi-axes of the ellipse passing through the point zz and having its foci at the points z=±1z = \pm 1.

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Remark.
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Hint. Set z=12(ζ+1ζ)z = \dfrac {1}{2}\left(\zeta +\dfrac {1}{\zeta }\right) and apply the theorem on the modulus to the function

ζ−nf(ζ2+12ζ). \zeta ^{-n}f\left(\frac{\zeta ^{2}+1}{2\zeta }\right).
Problem 5658

Prove the theorem: if for ∣z∣<1\left|z\right| < 1 the function f(z)f(z) is regular and ∣f(z)∣<M\left|f(z)\right| < M, while f(α)=0f(\alpha ) = 0, where ∣α∣<1\left|\alpha \right| < 1, then for ∣z∣<1\left|z\right| < 1 the inequality

∣f(z)∣⩽M⋅∣z−α1−α‾z∣ \left|f(z)\right| \leqslant M\cdot \left|\frac{z-\alpha }{1-\overline{\alpha }z}\right|

holds.

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Problem 5659

Prove the theorem: if f(z)f(z) is regular for ∣z∣<1\left|z\right| < 1; f(0)≠0f(0) \neq 0; ∣f(z)∣<M\left|f(z)\right| < M, and f(z)=0f(z) = 0 at z=z1,z2,z3,…z = z_{1}, z_{2}, z_{3}, \ldots, then the infinite product ∏ν=1∞∣zν∣\prod_{\nu =1}^{\infty } \left|z_{\nu }\right| converges.

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Remark.
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Hint. Consider the inequality

∣f(z)∣⩽M∣z−z11−z1‾z⋅z−z21−z2‾z⋯z−zn1−zn‾z∣. \left|f(z)\right| \leqslant M\left|\frac{z-z_{1}}{1-\overline{z_{1}}z}\cdot \frac{z-z_{2}}{1-\overline{z_{2}}z}\cdots \frac{z-z_{n}}{1-\overline{z_{n}}z}\right|.
Problem 5660

Prove the theorem: if for ∣z∣<1\left|z\right| < 1 the function f(z)f(z) is regular and bounded, and for a<φ<ba < \varphi < b the quantity f(reφi)f(re^{\varphi i}) tends uniformly to zero as r→1r \to 1, then f(z)f(z) is identically zero.

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Remark.
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Hint. Consider the function f(z)⋅f(ωz)…f(ωn−1z)f(z)\cdot f(\omega z)\ldots f(\omega^{n-1}z), where ω=e2πi/n\omega = e^{2\pi i/n}, and nn is sufficiently large.

Problem 5661

Prove the theorem: if the sequence fn(z)f_{n}(z), for ∣z∣⩽1\left|z\right| \leqslant 1, is uniformly bounded and tends to 00 as n→∞n \to \infty on a set of points having the origin as a limit point, then fn(z)→0f_{n}(z) \to 0 as n→∞n \to \infty at any interior point of the disk ∣z∣<1\left|z\right| < 1.

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Remark.
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Hint. Let z1,z2,…,zmz_{1}, z_{2}, \ldots , z_{m} be points lying inside a fixed disk ∣z∣>ρ\left|z\right| > \rho, at which fn(z)→0f_{n}(z) \to 0. Considering the formula

fn(z)=∑ν=1nfn(zν)ω′(zν)⋅ω(z)z−zν+ω(z)2πi∫∣t∣=1fn(t)ω(t)(t−z) dt;ω(z)=∏ν=1m(z−zν), f_{n}(z) = \sum _{\nu =1}^{n} \frac{f_{n}(z_{\nu })}{\omega '(z_{\nu })}\cdot \frac{\omega (z)}{z-z_{\nu }}+\frac{\omega (z)}{2\pi i}\int _{\left|t\right|=1} \frac{f_{n}(t)}{\omega (t)(t-z)}\, dt; \qquad \omega (z) = \prod _{\nu =1}^{m} (z-z_{\nu }),

fix, for a given ε>0\varepsilon > 0, a sufficiently small ρ\rho, then a correspondingly large mm, and then let nn tend to ∞\infty.

Problem 5662

Prove the theorem: if the functions fn(z)f_{n}(z) are regular inside a contour CC and uniformly bounded on it, and inside the contour there is a limit point of points at which fn(z)→0f_{n}(z) \to 0 as n→∞n \to \infty, then fn(z)→0f_{n}(z) \to 0 at every point inside the contour as well. (Vitali's Theorem.)

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Problem 5663

Prove the theorem: if the functions fn(z)f_{n}(z) are regular in the disk ∣z∣<1\left|z\right| < 1, have no roots there, and are bounded in modulus by 11, and, moreover, fn(0)→0f_{n}(0) \to 0 as n→∞n \to \infty, then throughout the interior of the disk ∣z∣<1\left|z\right| < 1 the functions fn(z)→0f_{n}(z) \to 0 as n→∞n \to \infty.

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Remark.
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Hint. Use the Poisson integral for ln⁡∣fn(z)∣\ln \left|f_{n}(z)\right|.

Problem 5664

Let zνz_{\nu } be the roots of an entire function f(z)f(z), and

0<∣z1∣⩽∣z2∣⩽∣z3∣⩽⋯⩽∣zN−1∣⩽R<∣zN∣, 0 < \left|z_{1}\right| \leqslant \left|z_{2}\right| \leqslant \left|z_{3}\right| \leqslant \cdots \leqslant \left|z_{N-1}\right| \leqslant R < \left|z_{N}\right|,

while for the function f(z)f(z) the inequality

∣f(z)∣<Ce∣z∣σ \left|f(z)\right| < Ce^{\left|z\right|^{\sigma }}

holds. Prove that for ∣z∣⩽3R\left|z\right| \leqslant 3R

∣f(0)∣<∣f(z)∏ν=1N−1(1−zzν)∣<Ce(3R)σ2N−1. \left|f(0)\right| < \left|\frac{f(z)}{\prod _{\nu =1}^{N-1} \left(1-\dfrac {z}{z_{\nu }}\right)}\right| < \frac{Ce^{(3R)^{\sigma }}}{2^{N-1}}.
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Problem 5665

Prove that, under the same conditions (problem 5664), the series ∑ν=1∞1∣zν∣n\displaystyle \sum_{\nu =1}^{\infty } \frac{1}{\left|z_{\nu }\right|^{n}} converges for n>σn > \sigma.

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Problem 5666

Under the same conditions (problem 5664), prove the Weierstrass infinite-product expansion:

f(z)=eg(z)∏ν=1∞φν(z), f(z) = e^{g(z)}\prod _{\nu =1}^{\infty } \varphi _{\nu }(z),

where g(z)g(z) is some entire function, and

φν(z)=(1−zzν)ezzν+z22zν2+…+zn−1(n−1)zνn−1. \varphi _{\nu }(z) = \left(1-\frac{z}{z_{\nu }}\right)e^{\frac{z}{z_{\nu }}+\frac{z^{2}}{2z_{\nu }^{2}}+\ldots +\frac{z^{n-1}}{(n-1)z_{\nu }^{n-1}}}.
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Problem 5667

Let

f(z)=∑n=0∞anzn f(z) = \sum _{n=0}^{\infty } a_{n}z^{n}

satisfy, in the disk ∣z∣⩽R\left|z\right| \leqslant R, the condition: Re(f(z))⩽M\mathrm{Re}\left(f(z)\right) \leqslant M, where M>0M > 0. Then

∣a1∣<2∣M−Re(a0)∣R. \left|a_{1}\right| < \frac{2\left|M-\mathrm{Re}\left(a_{0}\right)\right|}{R}.
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Remark.
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Hint. Consider first the case a0=0a_{0} = 0, using the expansion f(z)2M−f(z)=a12Mz+bz2+⋯\dfrac {f(z)}{2M-f(z)} = \dfrac {a_{1}}{2M}z+bz^{2}+\cdots and the relation ∣f(z)∣<∣2M−f(z)∣\left|f(z)\right| < \left|2M-f(z)\right|.

Problem 5668

Considering, under the same conditions (problem 5667), the function

F(z)=1n∑ν=1n(f(ωνz)−a0), F(z) = \frac{1}{n}\sum _{\nu =1}^{n} \left(f(\omega ^{\nu }z)-a_{0}\right),

where ω=e2πin\omega = e^{\frac{2\pi i}{n}}, prove the inequality:

∣an∣⩽2∣M−Re(a0)∣Rn. \left|a_{n}\right| \leqslant \frac{2\left|M-\mathrm{Re}\left(a_{0}\right)\right|}{R^{n}}.
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Problem 5669

Under the same condition on the growth of f(z)f(z) as before (see problem 5664), prove that for ∣z∣⩽2R\left|z\right| \leqslant 2R, and hence also for ∣z∣⩽R\left|z\right| \leqslant R, we have:

∣ψ(z)∣=∣f(z)∏ν=1N(1−zzν)∣<Ce(2R)σ, \left|\psi (z)\right| = \left|\frac{f(z)}{\prod _{\nu =1}^{N} \left(1-\dfrac {z}{z_{\nu }}\right)}\right| < Ce^{(2R)^{\sigma }},

where in the product ∣zν∣⩽R\left|z_{\nu }\right| \leqslant R.

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Problem 5670

Using the fact that ln⁡ψ(z)\ln \psi (z) is a regular function of zz for ∣z∣⩽R\left|z\right| \leqslant R (problem 5669), prove that at z=0z = 0 the quantity

(ln⁡f(z))(n)−∑ν=1N(ln⁡φν(z))(n) \left(\ln f(z)\right)^{(n)}-\sum _{\nu =1}^{N} \left(\ln \varphi _{\nu }(z)\right)^{(n)}

is, in absolute value, less than cRn−σ\dfrac {c}{R^{n-\sigma }}, where cc is a corresponding constant.

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Problem 5671

Prove Hadamard's theorem: if f(z)f(z) satisfies, for every zz, the condition ∣f(z)∣<Ce∣z∣σ\left|f(z)\right| < Ce^{\left|z\right|^{\sigma }}, where σ<n\sigma < n and f(0)≠0f(0) \neq 0, then the Weierstrass expansion

f(z)=eg(z)⋅∏ν=1∞φν(z), f(z) = e^{g(z)}\cdot \prod _{\nu =1}^{\infty } \varphi _{\nu }(z),

holds, where g(z)g(z) is a polynomial whose degree is less than nn.

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Problem 5672

Prove Carathéodory's theorem: if f(z)=∑n=0∞an(z−a)nf(z) = \sum_{n=0}^{\infty } a_{n}(z-a)^{n} is regular in the disk ∣z−a∣⩽R\left|z-a\right| \leqslant R and there Re(f(z))<M\mathrm{Re}\left(f(z)\right) < M, where M>0M > 0, then for ∣z−a∣=r<R\left|z-a\right| = r < R the inequality

∣f(z)∣<∣a0∣+2∣M−Re(a0)∣⋅rR−r \left|f(z)\right| < \left|a_{0}\right|+\frac{2\left|M-\mathrm{Re}\left(a_{0}\right)\right|\cdot r}{R-r}

holds.

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Problem 5673

Prove Lindelöf's theorem: if ∣f(z)∣⩽C\left|f(z)\right| \leqslant C on the edges of the region −a<x<a-a < x < a, and inside this region ∣f(z)∣<eλ∣y∣\left|f(z)\right| < e^{\lambda \left|y\right|}, where λ\lambda is some constant, then also inside the region ∣f(z)∣⩽C\left|f(z)\right| \leqslant C, and moreover equality at an interior point is possible only when f(z)≡Cf(z) \equiv C.

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Remark.
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Hint. It is useful to study the function f(z)eεz2f(z)e^{\varepsilon z^{2}} for ε>0\varepsilon > 0.