Natural boundaries and lacunary series

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Consider a power series

∑n=1∞zn!. \sum _{n=1}^{\infty } z^{n!}.

Show that the series converges to a holomorphic function on the open unit disk centered at the origin. Prove that the boundary of the disk is the natural boundary of the function.

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For ∣an∣<1\left|a_n\right|<1,

f(z)=∏n=1∞an−z1−aˉnz. f(z) = \prod _{n=1}^{\infty } \frac{a_n-z}{1-\bar a_n z}.
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(a)

Show that ff defines a holomorphic function on D={z:∣z∣<1}D = \left\{ z : \left|z\right|<1\right\}.

(b)

Prove that ff does not have an analytic continuation to any larger disk {z:∣z∣<r}\left\{ z:\left|z\right|<r\right\} where r>1r>1.

Prove that the function

f(z)=∑m,n=−∞∞1(m+nz)2,m2+n2>0, f(z) = \sum _{m,n=-\infty }^{\infty } \frac{1}{(m+nz)^{2}}, \quad m^{2}+n^{2} > 0,

considered in the upper half-plane Im(z)>0\mathrm{Im}\left(z\right) > 0, is regular there and has the real axis as an essential boundary of continuation.

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Prove that the function ∑n=1∞1n2⋅1z−e2πni2\displaystyle \sum_{n=1}^{\infty } \frac{1}{n^{2}}\cdot \frac{1}{z-e^{2\pi ni\sqrt{2}}}, regular for ∣z∣<1\left|z\right| < 1, cannot be continued beyond the circle ∣z∣=1\left|z\right| = 1.

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Prove that the same singularity (problem 5571, the absence of continuation beyond the boundary of the circle of convergence) is exhibited by the series:

∑n=0∞zn!;∑n=0∞12nz2n;∑n=0∞anz2n, \sum _{n=0}^{\infty } z^{n!}; \qquad \sum _{n=0}^{\infty } \frac{1}{2^{n}}z^{2^{n}}; \qquad \sum _{n=0}^{\infty } a_{n}z^{2^{n}},

if ∑n=0∞an\sum_{n=0}^{\infty } a_{n} diverges.

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Let aa, bb, cc be the numbers represented by the three vertices of a triangle, and xx, yy, zz proper fractions. The series:

∑m=0∞∑n=0∞∑p=0∞xmynzpma+nb+pcm+n+p−ζ,m+n+p>0, \sum _{m=0}^{\infty }\sum _{n=0}^{\infty }\sum _{p=0}^{\infty } \frac{x^{m}y^{n}z^{p}}{\dfrac {ma+nb+pc}{m+n+p}-\zeta }, \quad m+n+p > 0,

represents, outside the triangle, an analytic function of ζ\zeta that cannot be continued inside the triangle. Prove that, contrary to what is stated in Hermite's book Cours d’analyse, the series also converges inside the triangle on an everywhere-dense set of points.

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Let the coefficients of the polynomial

P(x)=a0xr+a1xr−1+⋯+ar−1x+ar,a0≠0,r≥1 P(x) = a_0x^r+a_1x^{r-1}+\cdots +a_{r-1}x+a_r , \qquad a_0\neq 0, \quad r\geq 1

be real. The analytic function defined by the power series

[P(0)]+[P(1)]z+[P(2)]z2+⋯+[P(n)]zn+⋯ [P(0)]+[P(1)]z+[P(2)]z^2+\cdots +[P(n)]z^n+\cdots

is

  1. a rational function, if the rr numbers a0,a1,…,ar−1a_0,a_1,\ldots ,a_{r-1} are all rational,

  2. not a rational function, if the rr numbers a0,a1,…,ar−1a_0,a_1,\ldots ,a_{r-1} are not all rational. (Cf. Problem 168 of Part II.) [Problem 85 of Part I.]

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Denote by QnQ_n the ``cross sum'' of the number nn, i.e. the sum of its digits. E.g. we have Q137=1+3+7=11Q_{137}=1+3+7=11. The power series

Q1z+Q2z2+Q3z3+⋯+Qnzn+⋯ Q_1z+Q_2z^2+Q_3z^3+\cdots +Q_nz^n+\cdots

has the unit circle as its natural boundary.

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In an infinite table of logarithms let the logarithms of all the positive integers 1,2,3,…1,2,3,\ldots be written underneath each other in their natural order, in such a fashion that decimal places of the same order appear in the same vertical column. If we interpret the digits in any vertical column as the coefficients of a power series, then we obtain a power series that cannot be analytically continued.

In a different formulation we are required to prove the following theorem: Let jj be an integer and in the decimal representation of log⁡1,log⁡2,log⁡3,…,log⁡n,…\log 1,\log 2,\log 3,\ldots ,\log n,\ldots let the jjth digit after the decimal point be denoted by d1,d2,d3,…,dn,…d_1,d_2,d_3,\ldots ,d_n,\ldots respectively. Then the analytic function defined by the power series

d1z+d2z2+d3z3+⋯+dnzn+⋯ d_1z+d_2z^2+d_3z^3+\cdots +d_nz^n+\cdots

has as its natural boundary the circle ∣z∣=1\left|z\right|=1.

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Here is another construction of such a function. Let {nk}\left\{ n_{k}\right\} be a sequence of integers such that n1>1n_{1} > 1 and nk+1>2knkn_{k+1} > 2 k n_{k}. Define

h(z)=∑k=1∞5kznk h(z) = \sum _{k = 1}^{\infty } 5^{k} z^{n_{k}}

Prove that the series converges if ∣z∣<1\left|z\right| < 1 and prove that there is a constant c>0c > 0 such that ∣h(z)∣>c⋅5m\left|h(z)\right| > c \cdot 5^{m} for all zz with ∣z∣=1−(1/nm)\left|z\right| = 1-\left(1 / n_{m}\right).

Hence hh has no finite radial limits. Prove also that hh must have infinitely many zeros in UU. (Compare with Exercise 15, Chap. 12). In fact, prove that to every complex number α\alpha there correspond infinitely many z∈Uz \in U at which h(z)=αh(z) = \alpha.

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Suppose f(z)=Σanzn,an≥0f(z) = \Sigma a_{n} z^{n}, a_{n} \geq 0, and the radius of convergence of the series is 1. Prove that ff has a singularity at z=1z = 1. Hint: Expand ff in powers of z−12z-\frac{1}{2}. If 1 were a regular point of ff, the new series would converge at some x>1x > 1. What would this imply about the original series?

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Suppose

f(z)=∑k=1∞(z−z22)3k=∑n=1∞anzn f(z) = \sum _{k = 1}^{\infty }\left(\frac{z-z^{2}}{2}\right)^{3^{k}} = \sum _{n = 1}^{\infty } a_{n} z^{n}

Find the regions in which the two series converge. Show that this illustrates Theorem 16.5. Find the singular point of ff which is nearest to the origin.

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Prove that the following statements are correct if {nk}\left\{ n_{k}\right\} is a sequence of positive integers which tends to ∞\infty sufficiently rapidly. If

f(z)=∑k=1∞znkk f(z) = \sum _{k = 1}^{\infty } \frac{z^{n_{k}}}{k}

then ∣f′(z)∣>nk/(10k)\left|f^{\prime }(z)\right| > n_{k} /(10 k) for all zz such that

1−1nk<∣z∣<1−12nk 1-\frac{1}{n_{k}} < \left|z\right| < 1-\frac{1}{2 n_{k}}

Hence

∫01∣f′(reiθ)∣dr=∞ \int _{0}^{1}\left|f^{\prime }\left(r e^{i \theta }\right)\right| d r = \infty

for every θ\theta, although

lim⁡R→1∫0Rf′(reiθ)dr \lim _{R \rightarrow 1} \int _{0}^{R} f^{\prime }\left(r e^{i \theta }\right) d r

exists (and is finite) for almost all θ\theta. Interpret this geometrically, in terms of the lengths of the images under ff of the radii in UU.

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Prove that the sum of the power series ∑n=0∞z2n\sum_{n = 0}^{\infty } z^{2^{n}} is a complete analytic function in the disk of convergence B1(0)={z:∣z∣<1}B_{1}(0) = \left\{ z:\left|z\right| < 1\right\}.

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Prove that when the radius of convergence of the power series ∑n=0+∞anzn\sum_{n = 0}^{+\infty } a_{n} z^{n} equals one and all an⩾0a_{n} \geqslant 0, then such a series cannot be continued to the point z=1z = 1 (Pringsheim's theorem).

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Let f(z)=∑n=0∞cnzn,lim‾⁡n→∞∣cn∣1/n=1f(z) = \sum_{n = 0}^{\infty } c_{n} z^{n}, \varlimsup_{n \rightarrow \infty }\left|c_{n}\right|^{1 / n} = 1, and let all coefficients cnc_{n} be nonnegative. Prove that the point z=1z = 1 is a singular point of the function f(z)f(z).

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Prove that the statement of Problem 43 remains valid also under the weaker assumptions:

Re(cn)⩾0(n>n0),lim‾⁡n→∞(Re(cn))1/n=1 \mathrm{Re}\left(c_{n}\right) \geqslant 0 \quad \left(n > n_{0}\right), \quad \varlimsup _{n \rightarrow \infty }\left(\mathrm{Re}\left(c_{n}\right)\right)^{1 / n} = 1
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Prove that every point of the circle ∣z∣=1\left|z\right| = 1 is a singular point for the functions

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(1)

f(z)=∑n=0∞z2nf(z) = \sum_{n = 0}^{\infty } z^{2^{n}}.

(2)

f(z)=∑n=0∞zn!f(z) = \sum_{n = 0}^{\infty } z^{n!}.

References

[CJP+11] Chen J., Jiang G., Pan Y., Qin T., Tong Y., Wu Q., Xu S., Li T. Problems and Solutions in Mathematics. 2nd ed., World Scientific, 2011.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.
[Rud87] Rudin W. Real and Complex Analysis. 3rd ed., McGraw-Hill, 1987.
[ШПК15] Шабунин М.И., Половинкин Е.С., Карлов М.И. Сборник задач по теории функций комплексного переменного. 4-е изд., БИНОМ. Лаборатория знаний, 2015.