The Radon Inversion Problem for Holomorphic Functions in the Unit Disc

This paper deals with the so-called Radon inversion problem formulated in the following way: Given a p>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$p>0$$\end{document} and a strictly positive function H continuous on the unit circle ∂D\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\partial {\mathbb {D}}}$$\end{document}, find a function f holomorphic in the unit disc D\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {D}}$$\end{document} such that ∫01|f(zt)|pdt=H(z)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\int _0^1|f(zt)|^pdt=H(z)$$\end{document} for z∈∂D\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$z \in {\partial {\mathbb {D}}}$$\end{document}. We prove solvability of the problem under consideration. For p=2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$p=2$$\end{document}, a technical improvement of the main result related to convergence and divergence of certain series of Taylor coefficients is obtained.


Introduction
Generally, if f is a function defined on a manifold M, the Radon transform assigns to f its integrals over a given family of submanifolds of M. In 1917 such transformations on hyperplanes were first studied by Radon [10]. The present paper is devoted to the Radon-type transform defined on functions holomorphic in the complex unit disc that evaluates integrals of p−th power of the modulus of the function along radii of the unit disc. Let D denote the unit disc of C and O(D) be the space of functions holomorphic in D. Given a p > 0, the transform under consideration assigns to a function f ∈ O(D) Communicated Then one may pose the following problem: Given a p > 0 and a strictly positive continuous function H on the unit circle ∂D, find a holomorphic function f ∈ O(D) such that It is worth mentioning that an almost everywhere version of the problem above was already solved by the first author in [8] even in several complex variables. However, the function R p ( f ) is well-defined at all points of ∂D. Hence, our purpose is to find a function such that (1) not only holds almost everywhere, but for all z ∈ ∂D.
As was observed in [8], the problem under consideration is similar to the generalized inner function problem which in one variable is to find a function g ∈ O(D) such that lim t→1 |g(zt)| = G(z) for z ∈ ∂D (2) given a strictly positive function G continuous on ∂D. The role of radial limits is just played by values of the function R p ( f ) (z). Similarly as in [8], we intend to find a solution to (1) by constructive methods that were applied in proving the existence of a non-constant inner function in several complex variables. The main result of this paper is Theorem 6 in which we prove solvability of the Radon inversion problem. For details of the first two independent constructions of inner functions in the unit ball we refer the reader to [1,2,9]. Throughout the article we shall use the following notation. Let σ and μ stand for normalized Lebesgue measures on ∂D and D respectively, i.e. σ (∂D) = 1 and μ(D) = 1. The supremum norm and L p norms on ∂D and D will be denoted by Let C L (D) stand for the space of continuous functions in D which are bounded by L > 0.
For p = 2, if f = ∞ n=0 a n z n satisfies (1), then we observe that One may consider what happens with the series ∞ n=0 |a n | 2 2n + 1 s , when s ∈ [0, 1). It turns out that it is possible to construct a function f = ∞ n=0 a n z n such that (1) holds and ∞ n=0 |a n | 2 /(2n + 1) s = ∞ for s ∈ [0, 1). This is the second main result of this paper given in Theorem 9. Our motivation to consider the above Taylor series is [4], where a construction of an analytic function in the disc algebra with a divergent series of Taylor coefficients with every power s ∈ [0, 2) is presented. For analogues in several complex variables see [5,6,12].

Radon Inversion Problem for Holomorphic Functions in the Unit Disc
For p > 0 and q = min{ p, 1} we may define a metric space Proof Let { f n } n∈N be a Cauchy sequence in HR p (D). Set ε > 0. There exists N ∈ N such that Integration over ∂D gives Since f m − f n is a holomorphic function, on the basis of [11,Prop. 1.5.4.], | f m − f n | p is a subharmonic function. Let K be a compact subset of D, w ∈ K and r = dist(K , ∂D)/2, so that D(w, r ) ⊂ D. By applying the sub-mean-value theorem to the function | f m − f n | p and D(w, r ), we get for all m, n ≥ N In particular, This implies that { f n } n∈N converges locally uniformly on compact subsets of D. Let f be the limit function for { f n } n∈N . Since f n are holomorphic, so is f .
Since f m converges to f locally uniformly, we may pass with m to the limit in (4) to obtain The above inequality holds for any δ ∈ (0, 1), so Therefore, f n d → n→∞ f . Moreover, by the triangle inequality, for all n ≥ N which implies that f ∈ HR p (D).
Finally, by the triangle inequality, Thus, (R p ( f n )) 1/q ⇒ (R p ( f )) 1/q as n → ∞ and consequently R p ( f n ) ⇒ R p ( f ) as n → ∞ on ∂D.
Proof Consider a sequence of functions φ n (z) := a n 0 | f (zt)| p dt, where {a n } n∈N is a sequence increasing to 1 and z ∈ ∂D. Since f is continuous in D, {φ n (z)} n∈N is an increasing sequence of continuous functions on ∂D. Moreover, If R p ( f ) is continuous on ∂D, then by Dini's theorem φ n converges uniformly to Fix ε > 0. By continuity of R p ( f ) on ∂D, there exists δ > 0 such that for any z,z ∈ ∂D if |z −z| ≤ 2δ, then Finally, observe that by the triangle inequality, |z − w * | ≤ |z − w| + |w − w * | ≤ 2δ. Therefore, This theorem implies immediately that for any solution g to the Radon inversion problem, the function R p (g) is continuous in D.

Lemma 3 Fix p > 0. Let H be a strictly positive continuous function on ∂D.
For every ε 1 , ε 2 ∈ (0, 1) and compact set K ⊂ D there exist N ∈ N and a polynomial Q such that Q = 0 on ∂D and if f n (z) = (np + 1) 1/ p Q(z)z n , then for n ≥ N the following conditions are satisfied Proof Due to the fact that every continuous function can be approximated by a function of class C ∞ , there exists a strictly positive function Also, there exists a function u such that v(z) = e pu(z) for z ∈ ∂D.
Since v ∈ C ∞ (∂D), u satisfies Hölder condition. Let S(u) denote the Schwarz integral of u. By its properties, Re(S(u)) = u on ∂D and S(u) is continuous in D (see [3,7]). Hence, there exists r ∈ (0, 1) such that for t ∈ [r , 1] and z ∈ ∂D. (6) There exists a polynomial Q(z) that approximates the function e S(u)(z) in the following way Let M := sup z∈∂D H (z). The maximum principle implies that For z ∈ ∂D and n ≥ N 1 the above inequalities give the required estimation Now, to make f n satisfy the statement 1 it is enough to select a natural number Finally, observe that the function f n constructed above, where n is sufficiently large, has the required properties.

Remark 4
We can replace the statement 2 in Lemma 3 by one of the following Proof For statement 2 it is enough to takeH (z) : To obtain conclusion 2 we apply the lemma above withε 2 :=ε 2 /M, where M := sup z∈∂D |H (z)|.
Lemma 5 Fix p > 0. Let p > 0 and Q be a polynomial such that Q(z) = 0 for z ∈ ∂D. For every ε 3 ∈ (0, 1) there exists N ∈ N such that for n ≥ N and all functions g ∈ C L (D) the function f n (z) = (np + 1) 1/ p Q(z)z n satisfies Proof Choose k ∈ N such that k ≥ p. First, by triangle inequality, observe that Let δ ∈ (0, 1) be such that Thus, the above inequalities imply the following If 0 < p ≤ 1, then k = 1 does the job. For p > 1 we need a little bit more effort to accomplish this. Denote Therefore, there exists N 2 ∈ N such that I p k ( f n , g)(z) < ε 3 /2 for n ≥ N 2 . This yields the conclusion rather quickly and

Theorem 6 Fix p > 0. Let G be a strictly positive continuous function defined on ∂D.
There exists a function f ∈ O(D) such that Proof Since G is strictly positive, there exists m ∈ N such that G(z) − 2 −m > 0 for all z ∈ ∂D. We will apply Lemmas 3 and 5 iteratively to construct a sequence { f n } n∈N of functions in the disc algebra and a sequence of real numbers 0 < r 0 ≤ r 1 ≤ · · · ≤ r n ≤ r n+1 ≤ · · · ≤ 1 satisfying the following conditions We begin by selecting r 0 = 1/2 and applying Lemma 3 in version 2 with Then choose 1 ≥ r 1 ≥ (1 + r 0 )/2 such that Proceeding inductively, let us assume that numbers r 0 ≤ r 1 ≤ r 2 ≤ · · · r n ≤ 1 and with properties a)−c) have been found. We may apply Lemmas 3 and 5 with to produce a new function f n+1 ∈ O(D) ∩ C ∞ (D) such that the following statements hold This gives on ∂D and the induction process is complete.
By statement a), the sequence n k=1 f k converges uniformly to the function f := ∞ k=1 f k on compact subsets of D and f ∈ O(D). Then notice that for all r ∈ (0, 1) and z ∈ ∂D we have Hence, Let q = max{ p, 1}. For a given n ∈ N and all z ∈ ∂D from conditions a) and c) it follows that Therefore, These inequalities establish the conclusion R p ( f ) = G on ∂D. Finally, it follows from Theorem 2 that R p ( f ) ∈ C(D).

Furthermore, combining (a)-(c) and the induction hypothesis gives conditions
Similarly, The induction is now complete. Observe that statements (s2) and (s3) imply that is an increasing sequence. Hence, by Dini's theorem, R 2 n j=0 P j converges uniformly to h. Therefore, there exists N ∈ N such that Theorem 8 There exists a holomorphic function f (z) = ∞ n=0 a n z n such that Proof For every j ∈ N we may apply the preceding lemma with φ ≡ 0, h ≡ 1, a := 2 − j , θ := 1/2 to obtain N j ∈ N and a sequence of orthogonal polynomials Observe that if n > m, then It follows from the above that is a Cauchy sequence in HR 2 (D) which by Proposition 1 is convergent in HR 2 (D). Therefore, we may define a function Denote R 2 n j=1 1/ j 2 N j k=0 P jk by P n , for the moment. If n > m, then on ∂D Observe that there exists a constant C > 0 such that sup z∈∂D |P n (z)| < C for all n ∈ N. Hence, is a Cauchy sequence in the space of continuous functions, so R 2 ( f ) ∈ C(∂D) and consequently by Theorem 2, the function R 2 ( f ) is also continuous in D. Now we turn to conclusion 2. Let P jk (z) = n∈I jk p jkn (z) = n∈I jk a jkn z n be the homogeneous expansion of P jk , where I jk is the set of monomials' degrees of the polynomial P jk . By construction, I jm ∩ I ik = ∅ for ( j, m) = (i, k). Set s ∈ [0, 1). Notice that by statement b), which implies that Finally, we conclude from all this that j,k,n for s ∈ [0, 1). The proof of the theorem is now complete.
Observe that the above theorem does not depend on all the statements from Lemma 7. However, we may improve Theorem 8 so that the function f satisfies also the Radon inversion problem. This can be done by applying the full version of Lemma 7 with suitable arguments.

Theorem 9
Let Φ ∈ C(∂D) be a strictly positive function. There exists a holomorphic function f (z) = ∞ n=0 a n z n such that Proof Without loss of generality let us assume that sup z∈∂D |Φ(z)| < 1. We shall construct a sequence of orthogonal polynomials {Q k } ∞ k=0 with the following properties We will apply Lemma 7 iteratively, with new a, h, φ, θ, ε at each iteration. Let us begin our process by selecting Q 0 ≡ 0. Next apply Lemma 7 to This produces N 1 ∈ N and orthogonal polynomials P 10 , P 11 , . . . , P 1N 1 such that Now it is enough to define Q 1 := N 1 j=0 P 1 j . Proceeding inductively, let us assume that orthogonal polynomials Q 0 , Q 1 , . . . , Q k with the required properties ( p1)−( p5) have been constructed. We then may apply Lemma 7 with to obtain N k+1 ∈ N and orthogonal polynomials P k+1,0 , P k+1,1 , . . . , P k+1,N k+1 that are also orthogonal to the polynomials Q 0 , Q 1 , . . . , Q k and satisfy the following conditions on ∂D.
If we define Q k+1 := N k+1 j=0 P k+1, j , then statements ( p1) − ( p2) and ( p5) immediately hold. Other properties ( p3) − ( p4) follow from simple calculations That completes the induction. Combining statements ( p3)−( p4) iteratively gives the following This leads to the conclusion that lim n→∞ R 2 Hence, n k=0 Q k n∈N is a Cauchy sequence in HR 2 (D) and by Proposition 1 is convergent in HR 2 (D). It follows that the function f := ∞ k=0 Q k is well-defined and holomorphic in D. Then Proposition 1 also tells us that R 2 ( f ) = lim n→∞ R 2 n k=0 Q k = Φ on ∂D. Furthermore, due to Theorem 2 the function R 2 ( f ) is continuous in D.
Now we prove the second part of the theorem. Taking into account that Q 0 ≡ 0, we may which is equivalent to If s ∈ [0, 1), then we may apply above inequality to obtain that Let P k j (z) = n∈I k j p k jn (z) = n∈I k j a k jn z n be the homogenous expansion of P k j , where I k j is the set of monomial's degrees of the polynomial P k j . By construction, I m j ∩ I ki = ∅ for (m, j) = (k, i). Orthogonality of the polynomials p k jn gives the equality We conclude with the following estimation k, j,n |a k jn | 2 for s ∈ [0, 1). This completes the proof.
It is worth mentioning that in contrast to Theorem 6, in the above theorem we have applied Proposition 1 to show that the function that has been constructed in Theorem 9 satisfies the Radon inversion problem. This provides two different approaches in proving solvability of the problem under consideration.

Final Remarks
Herein, we shall give some remarks about the comparison of the Radon inversion problem we have solved in this paper and the generalization of the inner function problem formulated in (2). The second problem in one variable can be easily solved by applying the Schwarz integral. There exists a continuous real valued function φ such that G = e φ . Solving a standard Dirichlet problem for φ we obtain a harmonic function u φ ∈ C(D) such that u φ = φ on ∂D. Then, the Schwarz integral for u φ produces a function S(u φ ) ∈ O(D) such that Re(S(u φ )) ∈ C(D) and Re(S(u φ )) = φ on ∂D (see e.g. [3,7]). Now it is enough to take g := e S(u φ ) . Here we observe that if f , g ∈ O(D) are solutions of (1) and (2) respectively, then both functions |g| and R p ( f ) are continuous up to the boundary ∂D as has been shown in Theorem 2. However, there is a significant difference between the problems under consideration regarding zero sets of their possible solutions. On the one hand, notice that if g is a solution of the generalized inner function problem (2), then continuity of |g| up to the boundary ∂D implies that g cannot have infinitely many zeros. On the other hand, there exists a solution of the Radon inversion problem that has infinitely many zeros. In Theorem 6 we have constructed a function f (z) := ∞ k=1 (n k p + 1) 1/ p Q k (z)z n k , where Q k is a polynomial with certain properties. The function f is a solution to (1). Observe that it may be modified suitably to have infinitely many zeros in the following way: For m ∈ N there exist R m ∈ (0, 1) and n m ∈ N sufficiently large such that m−1 k=1 (n k p + 1) 1/ p Q k (z)z n k < (n m p + 1) 1/ p R n m m |Q m (z)| for z ∈ ∂D R m .
Then by Rouché's theorem, polynomial m k=1 (n k p + 1) 1/ p Q k (z)z n k has the same number of zeros as (n m p+1) 1/ p Q m (z)z n m in the disc D R m . Choosing suitably increas-ing sequences {R m } m∈N ⊂ (0, 1) and {n m } m∈N ⊂ N we may guarantee that f has infinitely many zeros in D.
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