2. Analytic Continuation and Domain of Analyticity
Definition 1. Let , be domains in , such that . If , are analytic functions and in , then is an analytic continuation of in and is an analytic continuation of in .
Theorem 1. If has an analytic continuation in , then the continuation is unique.
Let and be domains, and be analytic functions, where is an analytic continuation of in . The function, , defined by is an analytic continuation of in .
Let the series, , be convergent in the disk, . Any point on the boundary circle K: that has a neighborhood in which f has an analytic continuation is called regular, otherwise, the point is deemed singular. If has a radius of convergence, r, then there is at least one singular point on the circle, . If all points on are singular, then the disk, , is termed the domain of analyticity for f.
3. Overconvergence of Power Series
Let
be the radius of convergence,
r. The sequence,
, of partial sums of the series diverges at any point outside
. However, it is possible for a suitably chosen subsequence,
, of the sequence of partial sums to turn out to be uniformly convergent for some
. In such a case, we say that the series
exhibits the property of overconvergence. Consider Porter’s example of a series demonstrating overconvergence [
9]:
Example 1. where is the coefficient with the maximal absolute value among the coefficients in the binomial .
We write down the considered functional series in terms of powers of z: . Considering that the lowest degree of z in is , the power series is obtained without rearranging the terms of the original series. Therefore, the original series is convergent in the disk of convergence of the power series derived from it. We have , and this equality is fulfilled for infinitely many n. Using Hadamard’s formula, we determine the radius of convergence of the considered power series to be . The sequence of partial sums of the grouped series is the subsequence of the sequence of partial sums of the resulting power series. Therefore, is convergent in the unit disk, and using , we find that the considered sequence is also convergent in the disk, , which means the overconvergence of . Another possible justification is the following: we have the inequality , then 2. Thus, the series is convergent in the circle, , which with respect to z, is the interior of the lemniscate . Therefore, the sequence is convergent inside the lemniscate and divergent outside of it. On the other hand, since is obtained from a grouped series without rearrangement, its circle of convergence is the largest circle centered at the origin contained in the lemniscate 2. Obviously, this is the unit circle, and we again show that the series possesses overconvergence.
Let us determine those
:
. We calculate
We define the sequences
and
by
. Then,
for all
and, therefore,
for each
k and every integer
. Thus, each sequence
trivially ensures the overconvergence of the power series for
. On the other hand,
The above calculations are motivated by the following definition:
Definition 2. We will say that the series admits Ostrowski gaps with respect to its coefficients if there exist sequences and , satisfying the following conditions:
- (a)
;
- (b)
;
- (c)
if , then .
5. Ostrowski’s Theorem and Its Consequences
Theorem 2. Let , and as there are infinitely many indices, s, we denote them by with the following: with . Then, in a neighborhood of every regular point of the boundary of the disk of convergence, the series possesses the property of overconvergence, i.e., the subsequence is uniformly convergent in a sufficiently small neighborhood of every regular point of .
An interesting consequence of Ostrowski’s result is the following Hadamard theorem:
Theorem 3. Let have a radius of convergence of 1, such that for all sufficiently large s, and is fixed. Then the series has no analytical continuation.
Proof. Assume that the series, , has an analytic continuation in a neighborhood, U, of a point, . Then, the assumptions of Ostrowski’s theorem are fulfilled; therefore, there exists a neighborhood of in which . But for all sufficiently large k since . We have that resulting in in V. The latter leads to a contradiction since the partial sums of the considered series cannot be convergent outside the circle of convergence. Thus, the assumption that the power series has analytic continuation is wrong, and the unit disk is a domain of analyticity for the series. □
Example 2. For example, the power series, , has a radius of convergence of 1 and Ostrowski gaps, but no overconvergence, because it is analytically noncontinuable, according to Hadamard’s theorem.
Every power series, , with a nonzero radius of convergence, , is uniformly convergent in every circle, . Therefore, the overconvergence property could be considered as an extension of uniform convergence in a region outside the convergence circle, . Now, we will prove another Ostrowski theorem, which is the opposite of the one proved above:
Theorem 4. If the power series has the overconvergence property, then it has Hadamard–Ostrowski gaps, i.e., can be represented as the sum of two power series, the first with a radius of convergence greater than the radius of convergence of the initial power series, and the second series allows Ostrowski gaps in terms of their coefficients.
Again, without loss of generality, we will assume that the initial power series has a radius of convergence of
. We will prove that there exist sequences
and
that satisfy the conditions of the definition for Hadamard–Ostrowski gaps. We will now write down the power series in the following form:
In such a case, the first power series on the right-hand side of the equality has a radius of convergence ; this is because , and the second power series, according to the choice of the sequence, and , exhibits Ostrowski gaps. We will prove the existence of and with the requested properties by the following:
Theorem 5. If has a subsequence of the sequence of partial sums, which converges uniformly in a neighborhood of a regular point, then there exist positive values and , such that for all sufficiently large k, we have the following: for .
Proof. ([
9]) Let us also prove this statement for
n:
, the case
is similar. Let
V be a neighborhood of a regular point belonging to
, such that
converges uniformly in
V. We denote
and let
be a one-sheeted map that transforms the unit disk,
D, inside a circle of radius,
, as we can consider that
. Let
,
be single-valued analytic functions, such that
and
. Therefore,
has the form
. Since
h is a one-sheeted map in
D, the same is true for
, and we have
, at
. Since
for
, then according to Cauchy’s inequality for the coefficients of
, we will have the following:
Since
and
, then
Let us consider the limit
as
n and
p tend to infinity, such that the quotient
is less than a suitably chosen positive number,
, independent of
p and
n. From
, we have the following:
For , we can redefine g by continuity at point : . Since , there exists and , such that for all . Hence, .
Let
be the analytic continuation of
f in
V. Let
and we can replace
with
because for sufficiently large
k, according to the overconvergence condition, we have
. Then the inequality
is written in the following form:
. Taking into account that
, we obtain
. Thus, the inequality
holds for all
and
.
We obtain for all .
We can define the sequences
and
as follows:
,
. The sequences defined in this way satisfy the conditions of Definition 3, which we check immediately:
□
Second shorter proof (R. Kovacheva). The main idea is to employ a harmonic majorant for a sequence of analytic functions, each having a unique modulus. □
Definition 4. Let be a harmonic function in a region, R, of the complex plane, and let be a sequence of analytic functions in R having a unique modulus in R. If for every connected compact , the conditionis satisfied, then we will say that is the harmonic majorant of the sequence in R.
We will use the following theorem [
10]:
Theorem 6. Let be a harmonic majorant of in R. Let there exist a connected compact consisting of more than one point for whichthen the strict inequality holds for every connected compact in R.
Let be a power series with a radius of convergence of 1, and V be a neighborhood of a regular point of . Let be a compact subset of V such that . Let be uniformly convergent in . We will prove that for all sufficiently large k, there exists : , and is a fixed number that we will determine. Putting , we obtain the sequences, defining the Hadamard–Ostrowski gaps. We will first prove and, subsequently, determine .
We fix
:
, and applying the Corollary 1, we have the following:
Let
F be the analytic continuation of
f in
. Due to
in the compact subsets of
D, then for all sufficiently large
k, we will have the following:
According to (*), zero is the harmonic majorant of the sequence
in
, and from (**), it is established that on the compact
U, the strict inequality
holds. According to Theorem 6, the strict inequality holds for every compact
, i.e.,
. We fix
:
, consider the compact
. Then, there exists
, depending on
, such that
and, hence, we obtain the following:
Now, let us define
in the following form:
:
where
, and we can choose it in such a way to satisfy
and define
, such that we have the following:
The sequences and are correctly defined because the constants q and are chosen such that , and the proof is complete.
6. Decomposition of Analytic Functions in Series by Orthogonal Polynomials: Overconvergence and Examples
6.1. Regular Compacts and Green’s Function
Let E be a singly connected compact in , and be a domain in the extended complex plane , and suppose there exists a function with the following properties:
- (1)
is a harmonic function in G except at point ;
- (2)
The function is harmonic in a neighborhood of ;
- (3)
.
We will denote by the Green’s function for the domain, G, with a pole at the infinity point. It is unique. If we assume that satisfies the conditions (1)–(3), then is harmonic in G and vanishes on . According to the maximum principle, . Due to (2), we have and, therefore, it is nonnegative on the boundary of and is a harmonic function inside. From the maximum principle, we obtain in G.
Definition 5. We will call a simply connected compact E regular if its complement, , in the extended complex plane possesses the Green’s function with a pole at the infinite point.
Consider a regular and convex compact,
E, with its complement,
G. Let
represent the Green’s function with a pole at the infinity point for
G and let
represent the harmonic conjugate of
in
G. Then, the function
is analytic in
G, and we consider the function
. We will show that
conformally maps
G to the exterior of the unit circle. Indeed, if
, then
and taking into account
, we have that
maps
G to the exterior of the unit circle. It suffices to show that
in
G, which is equivalent to
If we assume that for some , then we have , which combined with the Cauchy-Riemann equations and , this gives . Thus, we establish that . We will show that this is impossible for points in G, referencing the following:
Theorem 7. All points, , for which the partial derivatives and simultaneously vanish, belong to the convex hull of E.
Since E is a convex compact, it coincides with its convex hull and, therefore, it contains all (if any) zeros of , meaning that there are no points in G annihilating . Thus, we establish that the map is conformal; moreover, according to the inverse mapping theorem, is locally invertible on G.
Let
be a real number and consider the image of the circle
by
, i.e.,
. We put
and have
,
which means that the image is an analytical Jordan curve, whose support we will denote by
. Thus, every level curve
of the Green’s function of
G is an analytic Jordan curve, which is the image of a circle with a radius greater than 1. Using this result, we immediately obtain the following properties:
- (1)
If , then the curves and do not intersect; moreover, the first one is contained inside the bounded region enclosed by .
- (2)
Through each point, passes a unique level curve , namely .
Based on these properties, an analyticity parameter is defined for a function analytic in a regular and convex compact. Denote by , the bounded region in is enclosed by , i.e., and , . Let f be analytic in E and suppose that it is not an entire function, which means that there are singular points in the finite plane, . If any singular points lie on , then we set the analyticity parameter, . Otherwise, the singularities of f lie in G. Since they are isolated, by using properties (1) and (2), we prove the existence of , such that on there is at least one singularity, while there are no singularities of f in , and we have . Indeed, we may set .
6.2. Capacity of Compact Set, Regular Measure, and Orthogonal Polynomials
Let
E be a regular convex compact with complement
. There are three fundamentally different approaches to defining the capacity of a compact set [
12]:
- (1)
Fekete’s method:
Given an arbitrary set of
distinct points
, belonging to
E, we define the following:
The sequence is monotonically decreasing and bounded below; hence, it is convergent. The number is called the transfinite Fekete diameter of E.
- (2)
Chebyshev’s constant:
Let
n be a natural number. We consider all polynomials
of degree
, with leading coefficients equal to 1. The minimax problem
We define the sequence by . It has been proved that this series is convergent and the limit is called the Chebyshev’s constant for E.
For each natural, n, let us denote by the polynomial with leading coefficient 1 that has the smallest maximum of the modulus over E, and by , the polynomial with zeros in E that has the same min–max property on E. According to the introduced notations, and . Similar to , is defined, with the equality, , being valid.
- (3)
Robben’s constant:
Since the function,
, is harmonic in a neighborhood of
, there exists the limit,
; let us denote it by
. A capacity of
E is defined as
. Hence,
Theorem 8. For each compact, K, in the complex plane, the equality holds.
By the -algebra generated by the compact subsets of the complex plane, we refer to a non-empty set that is closed with respect to countable unions and additions of compacts in . Its elements are called Borel sets. A nonnegative real-valued function defined on the considered -algebra of subsets of —which is monotone under set-theoretic inclusion, cancels on the empty set, and is countably additive—is called a Borel measure on . In addition, we require that the measure under consideration be finite, i.e., on each compact to take a finite value. The procedure for constructing an orthogonal system of polynomials is standard, by introducing a scalar product in the vector space of analytic functions over a regular compact set, E, by means of . If represents a finite Borel measure, it can be chosen to have compact support matching E. The considered vector space with the introduced scalar product becomes a Hermitian space, in which we can separate an orthonormal basis, for example, by the Gram–Schmidt orthogonalization of the series of polynomials , and subsequent normalization. This shows the existence of a series of polynomials, , with .
We consider all polynomials,
, of degree
, and with leading coefficients equal to 1. The minimization problem,
, has a unique solution about
, namely
, where
belongs to the orthogonal system
.
We will use the following standard notations: and .
Definition 6. We will call the measure, μ, regular if its support is a regular compact, and completely regular if it is regular with Since
, the inequality in the definition becomes an equality:
Example 3. Fully regular Borel measures: ([13]) If R is a bounded region, whose boundary is a Jordan curve, the measure, μ, defined by is completely regular at and , for the fixed .
Let .
Theorem 9. If the measure, μ, is completely regular, then Proof. ([
14]). Let
and
be the polynomials solving the corresponding minimization problems on the
norm and Chebyshev max norm. Therefore,
Since
is completely regular, we have
; therefore,
□
The following important theorem establishes a relationship between the asymptotics of the orthogonal system, , and the conformal map, .
Lemma 1. The following equality holds: .
Theorem 10. For an arbitrary polynomial of degree n, the following inequality holds: Proof. Theorem 16 in
Section 7. As a consequence of the orthogonal system,
, at
, we have the following:
□
6.3. Fourier Series for Analytic Functions Overconvergence
Let
f be an analytic function within a regular convex compact,
E, and let
be the orthogonal system of polynomials defined above. Regarding
f, we can write the formal orthogonal series as follows:
, with partial sums
, where
We will assume that f is not an entire function and possesses an analyticity parameter, .
Lemma 2. Under the assumptions above, the following holds: , and the sequence of partial sums, , is uniformly convergent in the compact subsets of to f.
The series
is called the Fourier series of
f on the orthogonal system,
, with a radius of convergence of
Definition 7. Let be a Fourier series with a finite radius of convergence . We say that a Fourier series has the overconvergence property if there exists a subsequence, , of the sequence of partial sums, , which is uniformly convergent in a domain located outside .
Definition 8. We state that admits Hadamard–Ostrowski gaps with respect to the coefficients, , if there exist sequences and , which satisfy the following conditions:
- (a)
;
- (b)
; and
- (c)
for .
The following sections contain the information needed to state the main result of the paper: the refined and complete proof of Theorem 11.
Theorem 11. Let f be analytic in E with Fourier series, , and a radius of convergence, , such that the sequence of coefficients, , has Hadamard–Ostrowski gaps. Then, in a sufficiently small neighborhood of every regular point on , the considered series possesses overconvergence.
Proof. The main idea is contained in [
3]; here, we clarify the details and remove some inaccuracies. Let
, and
be a regular point, with
being the analytic extension of
f in the neighborhood,
U, of
. We define
. In
, the following representation holds:
The first series on the right-hand side of the equality has a radius of convergence
, which gives a reason to consider only the series
since
is located outside
and, therefore, intersects
U and the part,
, outside
. Thus, the considerations are reduced to the series,
, for which we have
. We keep the old notation, but by
f, we will mean
.
Let us again
fix
:
, and we choose
such that
and put
We define the curves
, which are images of circles through
. Since
, then
is contained in
. Analogously,
, then
intersects
U in the part outside
. The same is true for the curve,
, because
The curves do not intersect since their images through are non-intersecting circles. We can choose a small enough so that the outermost curve lies in .
We will use Theorem 10, and for brevity, put
and
. Then,
, and we can choose
such that
. Therefore,
. We find the following estimates:
The function
is decreasing at
; therefore,
such that for
and
,
holds. With
fixed, we can choose
:
and
. At
, we have the following:
We again use Theorem 10 and keep the notations of
. On
, we have the following:
The function
is increasing; therefore,
such that for
and
, it holds that
. With
fixed, we can choose
:
and
. When
, we have the following:
Let
and put
. The function
is analytic in the ring between circles
and
, as well as on them. By applying Hadamard’s three-circle theorem [
15] and using
, we obtain the following:
For all sufficiently large
k and those
, for which
, we have the following:
We will prove that for all sufficiently small
, we have
. We calculate
Therefore, , and since it is a continuous function of t, there exists a neighborhood in which remains negative. Therefore, is decreasing in the interval , but at . Therefore, for any fixed .
Let
:
. Finally, if we fix
, then for every
, we have the following:
We denote by
the compact set in
with boundary
and let
. Since
is a series of analytic functions in
, then from the maximum principle, we obtain
□
7. Auxiliary Results
Here, we will state the additional results that are used in the proofs of Ostrowski’s theorems. Analytical functions that conformally map a region onto another region of the complex plane form a special class of maps that preserve the angles between curves in magnitude and direction. Single-valued analytic functions realize conformal mapping at all points where their derivatives do not vanish. Single-sheet analytic functions perform conformal and bijective correspondence. Writing them in the Taylor series provides estimates for the modulus of their coefficients.
Definition 9. We will state that the power series with positive coefficients, , dominates the series , and we will write it as follows: if .
The following properties follow from the definition of the sum and multiplication of the power series, as follows: if , then , also . As a consequence, if and , then .
Theorem 12. Let be analytic and single-sheeted in , i.e., bijective and conformally mapping the open unit disk onto a region of the complex plane. Then and when .
Theorem 13. Let be analytic and single-sheeted in , i.e., bijective and conformally mapping the open unit circle in a region of the complex plane. Then, and , where , and is a positive number.
Proof. ([
9]) Applying the previous Theorem 12 and Cauchy’s inequality for arbitrary
, we have
. Let us fix
n and define
so that the value of
is minimal. That is, we will determine the maximum of
, and directly find
from where it follows that a desired value is reached at
. Therefore,
for
. From Theorem 12, it follows that
. For
, we obtain
such that
. Consequently,
□
Theorem 14. If n and k are positive integers, then .
Proof. We carry out the proof by induction on
k. For
, we have
Suppose that for
it is valid that
and consider
:
with the latter being true because the sequence
is strictly increasing. □
Theorem 15. The coefficient of in the development of is equal to .
Proof. Assume that
and set
. We obtain
□
Theorem 16. For an arbitrary polynomial, , of degree n, the following inequality holds: Proof. The function,
, is analytic and single-valued in
G, is not constant, and therefore reaches its maximum on
. But,
and
; therefore,
Theorem 10 immediately follows from here; for example:
If is an arbitrary point, distinct from ∞, then there is a single level line; we denote it by , which passes through . Then, . □
Corollary 1. If E is the disk , then the Green’s function for G has the form . Therefore, and the inequality from Theorem 10 takes the following form: In particular, if
E is the unit disk, then
, and putting
, the inequality takes the following form:
for an arbitrary polynomial
of degree
n.