The Radon Inversion Problem for Holomorphic Functions in the Unit Disc

Abstract

This paper deals with the so-called Radon inversion problem formulated in the following way: Given a $p>0$ and a strictly positive function H continuous on the unit circle ${\partial {\mathbb {D}}}$, find a function f holomorphic in the unit disc ${\mathbb {D}}$ such that $\int _0^1|f(zt)|^pdt=H(z)$ for $z \in {\partial {\mathbb {D}}}$. We prove solvability of the problem under consideration. For $p=2$, a technical improvement of the main result related to convergence and divergence of certain series of Taylor coefficients is obtained.

1 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 ${\mathbb {D}}$ denote the unit disc of ${\mathbb {C}}$ and $\mathcal {O}({\mathbb {D}})$ be the space of functions holomorphic in ${\mathbb {D}}$. Given a $p>0$, the transform under consideration assigns to a function $f \in \mathcal {O}({\mathbb {D}})$ a function $\mathcal {R}^p(f):\overline{{\mathbb {D}}} \longmapsto {\mathbb {R}}_+$ for which

$$\begin{aligned} \mathcal {R}^p(f)(z):=\int _0^1 |f(zt)|^p dt,\quad \text {where }z \in \overline{{\mathbb {D}}}. \end{aligned}$$

Then one may pose the following problem: Given a $p>0$ and a strictly positive continuous function H on the unit circle ${\partial {\mathbb {D}}}$, find a holomorphic function $f \in \mathcal O({\mathbb {D}})$ such that

$$\begin{aligned} \mathcal {R}^p(f)(z)= H(z)\quad \text {for }z \in {\partial {\mathbb {D}}}. \end{aligned}$$

(1)

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 $\mathcal {R}^p(f)$ is well-defined at all points of ${\partial {\mathbb {D}}}$. Hence, our purpose is to find a function such that (1) not only holds almost everywhere, but for all $z \in {\partial {\mathbb {D}}}$.

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 $g \in \mathcal {O}({\mathbb {D}})$ such that

$$\begin{aligned} \lim _{t \rightarrow 1} |g(zt)| = G(z)\quad \text {for }z \in {\partial {\mathbb {D}}}\end{aligned}$$

(2)

given a strictly positive function G continuous on ${\partial {\mathbb {D}}}$. The role of radial limits is just played by values of the function $\mathcal {R}^p(f)(z)$. 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 $\sigma$ and $\mu$ stand for normalized Lebesgue measures on ${\partial {\mathbb {D}}}$ and $\overline{{\mathbb {D}}}$ respectively, i.e. $\sigma ({\partial {\mathbb {D}}})=1$ and $\mu ({\mathbb {D}})=1$. The supremum norm and $L_p$ norms on ${\partial {\mathbb {D}}}$ and ${\mathbb {D}}$ will be denoted by

$$\begin{aligned}&\Vert f\Vert =\sup _{z \in \overline{{\mathbb {D}}}}|f(z)|,\qquad \Vert f\Vert _{1,{\partial {\mathbb {D}}}}=\int _{{\partial {\mathbb {D}}}}|f(z)|d\sigma (z), \\&\Vert f\Vert ^2_{2,{\mathbb {D}}}=\int _{{\partial {\mathbb {D}}}}\int _0^1|f(zt)|^2dtd\sigma (z). \end{aligned}$$

Let $\mathcal {C}_L(\overline{{\mathbb {D}}})$ stand for the space of continuous functions in $\overline{{\mathbb {D}}}$ which are bounded by $L>0$.

For $p=2$, if $f=\sum _{n=0}^{\infty } a_n z^n$ satisfies (1), then we observe that

$$\begin{aligned} \Vert \mathcal {R}^2(f)\Vert _{1,{\partial {\mathbb {D}}}}= & {} \int _{{\partial {\mathbb {D}}}} \int _0^1 \left| \sum _{n=0}^{\infty } a_n z^nt^n \right| ^2 dt\,d\sigma (z)\\= & {} \sum _{n=0}^\infty \frac{|a_n|^2}{2n+1}=\Vert H\Vert _{1,{\partial {\mathbb {D}}}}<\infty . \end{aligned}$$

One may consider what happens with the series

$$\sum _{n=0}^\infty \left( \frac{|a_n|^2}{2n+1}\right) ^s,$$

when $s \in [0,1)$. It turns out that it is possible to construct a function $f=\sum _{n=0}^{\infty } a_n z^n$ such that (1) holds and $\sum _{n=0}^\infty \left( |a_n|^2/(2n+1)\right) ^s=\infty$ for $s\in [0,1)$. 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 $s \in [0,2)$ is presented. For analogues in several complex variables see [5, 6, 12].

2 Radon Inversion Problem for Holomorphic Functions in the Unit Disc

For $p>0$ and $q=\min \{p,1\}$ we may define a metric space

$$\begin{aligned} \mathcal {HR}^p({\mathbb {D}}):=\left\{ f \in \mathcal {O}({\mathbb {D}}):\sup _{z \in \partial {\mathbb {D}}} \mathcal {R}^p(f)(z) < \infty \right\} \end{aligned}$$

with metric

$$\begin{aligned} d(f,g):=\sup _{z \in \partial {\mathbb {D}}} \left( \mathcal {R}^p(f-g)(z)\right) ^{1/q}. \end{aligned}$$

Proposition 1

Space $(\mathcal {HR}^p({\mathbb {D}}),d)$ is complete for any $p>0$. Moreover, if $f_n {\mathop {\underset{n\rightarrow \infty }{\rightarrow }}\limits ^{d}} f$ in $\mathcal {HR}^p({\mathbb {D}})$, then $\mathcal {R}^p(f_n) \rightrightarrows \mathcal {R}^p(f)$ as ${n \rightarrow \infty }$ on ${\partial {\mathbb {D}}}$.

Proof

Let $\{f_n\}_{n \in {\mathbb {N}}}$ be a Cauchy sequence in $\mathcal {HR}^p({\mathbb {D}})$. Set $\varepsilon >0$. There exists $N \in {\mathbb {N}}$ such that

$$\begin{aligned} \left( \mathcal {R}^p(f_m-f_n)(z)\right) ^{{1/q}} < \varepsilon \quad \text {for all }m,n\ge N,\, z \in \partial {\mathbb {D}}\end{aligned}$$

and hence

$$\begin{aligned} \int _0^1|(f_m - f_n)(zt)|^pdt < \varepsilon ^q \quad \text {for all }m,n\ge N,\, z \in \partial {\mathbb {D}}. \end{aligned}$$

(3)

Integration over ${\partial {\mathbb {D}}}$ gives

$$\begin{aligned} \int _{\partial {\mathbb {D}}}\int _0^1|(f_m - f_n)(zt)|^pdt d\sigma (z) < \varepsilon ^q \quad \text {for all } m,n\ge N. \end{aligned}$$

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 ${\mathbb {D}}$, $w \in K$ and $r={\text {dist}}(K,{\partial {\mathbb {D}}})/2$, so that $\overline{{\mathbb {D}}}(w,r) \subset {\mathbb {D}}$. By applying the sub-mean-value theorem to the function $|f_m-f_n|^p$ and $\overline{{\mathbb {D}}}(w,r)$, we get for all $m,n \ge N$

$$\begin{aligned} \begin{array}{ll} |(f_m-f_n)(w)|^p &{}\le \displaystyle \frac{1}{\mu (\overline{{\mathbb {D}}}(w,r))}\int _{{\mathbb {D}}(w,r)}|(f_m - f_n)(\zeta )|^pd\mu (\zeta ) \\ &{}\le \displaystyle {} \frac{1}{r^2}\int _{\partial {\mathbb {D}}}\int _0^1|(f_m - f_n)(zt)|^pdt d\sigma (z) \\ &{}< \displaystyle \frac{\varepsilon ^q}{r^2}. \end{array} \end{aligned}$$

In particular,

$$\begin{aligned} \sup _{w \in K}|(f_m-f_n)(w)|^p < \frac{\varepsilon ^q}{r^2} \quad \text {for all }m,n\ge N. \end{aligned}$$

This implies that $\{f_n\}_{n\in {\mathbb {N}}}$ converges locally uniformly on compact subsets of ${\mathbb {D}}$. Let f be the limit function for $\{f_n\}_{n \in {\mathbb {N}}}$. Since $f_n$ are holomorphic, so is f.

Now we shall show that $f_n {\mathop {\longrightarrow }\limits ^{d}} f$ and $f \in \mathcal {HR}^p({\mathbb {D}})$. By (3),

$$\begin{aligned} \int _0^\delta |(f_m - f_n)(zt)|^pdt < \varepsilon ^q \quad \text {for all } \delta \in (0,1)\, m,n\ge N,\, z \in \partial {\mathbb {D}}. \end{aligned}$$

(4)

Since $f_m$ converges to f locally uniformly, we may pass with m to the limit in (4) to obtain

$$\begin{aligned} \int _0^\delta |(f - f_n)(zt)|^pdt \le \varepsilon ^q \quad \text {for all }\delta \in (0,1)\, n\ge N,\, z \in \partial {\mathbb {D}}. \end{aligned}$$

The above inequality holds for any $\delta \in (0,1)$, so

$$\begin{aligned} \int _0^1|(f - f_n)(zt)|^pdt \le \varepsilon ^q \quad \text {for all }n\ge N\, z \in \partial {\mathbb {D}}. \end{aligned}$$

Therefore, $f_n {\mathop {\underset{n\rightarrow \infty }{\rightarrow }}\limits ^{d}} f$. Moreover, by the triangle inequality, for all $n \ge N$

$$\begin{aligned} \sup _{z \in \partial {\mathbb {D}}} \left( \mathcal {R}^p(f)(z)\right) ^{1/q} \le \sup _{z \in \partial {\mathbb {D}}}\left( \mathcal {R}^p(f-f_n)(z)\right) ^{1/q}+\sup _{z \in \partial {\mathbb {D}}}\left( \mathcal {R}^p(f_n)(z)\right) ^{1/q} < \infty , \end{aligned}$$

which implies that $f \in \mathcal {HR}^p({\mathbb {D}})$.

If $f_n {\mathop {\underset{n\rightarrow \infty }{\rightarrow }}\limits ^{d}} f$, then $d(f,f_n) \mathop {\longrightarrow }\limits _{n\rightarrow \infty } 0$. Finally, by the triangle inequality,

$$\begin{aligned} \sup _{z \in \partial {\mathbb {D}}}\left( \left( \mathcal {R}^p(f)(z)\right) ^{1/q} -\left( \mathcal {R}^p(f_n)(z)\right) ^{1/q}\right) \le \sup _{z \in \partial {\mathbb {D}}}\left( \mathcal {R}^p(f-f_n)(z)\right) ^{1/q} {\mathop {\longrightarrow }\limits _{n\rightarrow \infty }} 0. \end{aligned}$$

Thus, $\left( \mathcal {R}^p(f_n)\right) ^{1/q} \rightrightarrows \left( \mathcal {R}^p(f)\right) ^{1/q}$ as ${n \rightarrow \infty }$ and consequently $\mathcal {R}^p(f_n) \rightrightarrows \mathcal {R}^p(f)$ as ${n \rightarrow \infty }$ on ${\partial {\mathbb {D}}}$. $\square$

Theorem 2

Fix $p>0$. Let $f \in \mathcal {C}({\mathbb {D}})$. If $\mathcal {R}^p(f) \in \mathcal {C}({\partial {\mathbb {D}}})$, then $\mathcal {R}^p(f) \in \mathcal {C}(\overline{{\mathbb {D}}})$.

Proof

Consider a sequence of functions $\phi _n(z):=\int _0^{a_n} |f(zt)|^p dt$, where $\{a_n\}_{n \in {\mathbb {N}}}$ is a sequence increasing to 1 and $z \in {\partial {\mathbb {D}}}$. Since f is continuous in ${\mathbb {D}}$, $\{\phi _n(z)\}_{n \in {\mathbb {N}}}$ is an increasing sequence of continuous functions on ${\partial {\mathbb {D}}}$. Moreover,

$$\lim _{n \rightarrow \infty } \phi _n (z)= \mathcal {R}^p(f)(z)~ \text { for every } ~z \in \overline{{\mathbb {D}}}.$$

If $\mathcal {R}^p(f)$ is continuous on ${\partial {\mathbb {D}}}$, then by Dini’s theorem $\phi _n$ converges uniformly to $\mathcal {R}^p(f)$ on ${\partial {\mathbb {D}}}$.

Fix $\varepsilon >0$. By continuity of $\mathcal {R}^p(f)$ on ${\partial {\mathbb {D}}}$, there exists $\delta >0$ such that for any $z,\bar{z} \in {\partial {\mathbb {D}}}$ if $|z-\bar{z}|\le 2 \delta$, then

• $\left| \mathcal {R}^p(f)(z) -\mathcal {R}^p(f)(\bar{z})\right| \le \displaystyle \frac{\varepsilon }{3}$

• $\displaystyle \frac{1}{1-\delta }\int _{1-\delta }^1 |f(zt)|^p dt \le \frac{\varepsilon }{3}$ for $z \in {\partial {\mathbb {D}}}$

• $\displaystyle \left( \frac{1}{1-\delta } - 1\right) \int _0^1|f(zt)|^p dt\le \frac{\varepsilon }{3}$ for $z \in {\partial {\mathbb {D}}}$.

Let $|z-w|\le \delta$, where $w \in {\mathbb {D}}, z \in {\partial {\mathbb {D}}}$. Then there exists $w^* \in {\partial {\mathbb {D}}}$ such that $w=sw^*$ for some $s \in (1-\delta ,1)$. Hence we may estimate

$$\begin{aligned} \begin{array}{ll} &{}\displaystyle \left| \mathcal {R}^p(f)(w^*) -\mathcal {R}^p(f)(w)\right| \\ &{}\quad = \displaystyle \left| \int _0^1|f(w^*t)|^p dt - \int _0^1|f(wt)|^p dt \right| \\ &{}\quad = \displaystyle \left| \int _0^1|f(w^*t)|^p dt - \int _0^1|f(w^*st)|^p dt \right| \\ &{}\quad = \displaystyle \left| \int _0^1|f(w^*t)|^p dt - \displaystyle \frac{1}{s} \int _0^s|f(w^*t)|^p dt \right| \\ &{}\quad \le \displaystyle \left| \int _0^1|f(w^*t)|^p dt - \frac{1}{s}\int _0^1|f(w^*t)|^p dt \right| +\displaystyle \frac{1}{s}\int _s^1|f(w^*t)|^p dt \\ &{}\quad \le \displaystyle \left( \frac{1}{1-\delta } - 1\right) \int _0^1|f(w^*t)|^p dt+\displaystyle \frac{1}{1-\delta }\int _{1-\delta }^1|f(w^*t)|^p dt \\ &{}\quad \le \displaystyle \frac{2\varepsilon }{3}. \end{array} \end{aligned}$$

Finally, observe that by the triangle inequality, $|z-w^*| \le |z-w| + |w-w^*| \le 2 \delta$. Therefore,

$$\begin{aligned} \left| \mathcal {R}^p(f)(z) -\mathcal {R}^p(f)(w)\right|\le & {} \left| \mathcal {R}^p(f)(z) -\mathcal {R}^p(f)(w^*)\right| + \left| \mathcal {R}^p(f)(w^*) -\mathcal {R}^p(f)(w)\right| \\\le & {} \frac{\varepsilon }{3} + \frac{2\varepsilon }{3}= \varepsilon \end{aligned}$$

for $|z-w| \le \delta$. $\square$

This theorem implies immediately that for any solution g to the Radon inversion problem, the function $\mathcal {R}^p(g)$ is continuous in $\overline{{\mathbb {D}}}$.

Lemma 3

Fix $p>0$. Let H be a strictly positive continuous function on ${\partial {\mathbb {D}}}$. For every $\varepsilon _1,\varepsilon _2 \in (0,1)$ and compact set $K \subset {\mathbb {D}}$ there exist $N \in {\mathbb {N}}$ and a polynomial Q such that $Q\ne 0$ on ${\partial {\mathbb {D}}}$ and if $f_n(z)=(np+1)^{1/p}Q(z)z^{n}$, then for $n \ge N$ the following conditions are satisfied

• 1. $|f_n(z)| \le \varepsilon _1$ for $z \in K$

• 2. $\displaystyle H(z)(1 - \varepsilon _2)< \mathcal {R}^p(f_n)(z)< H(z)(1 + \varepsilon _2)$ for $z \in {\partial {\mathbb {D}}}$.

Proof

Due to the fact that every continuous function can be approximated by a function of class $\mathcal {C}^\infty$, there exists a strictly positive function $v \in {\mathcal {C}^{\infty } }({\partial {\mathbb {D}}})$ such that

$$\begin{aligned} v(z)\left( 1 -\frac{\varepsilon _2}{5}\right) \le H(z) \le v(z)\left( 1 + \frac{\varepsilon _2}{5}\right) \quad \text {for} ~~ z \in {\partial {\mathbb {D}}}. \end{aligned}$$

(5)

Also, there exists a function u such that $v(z)=e^{pu(z)}$ for $z \in {\partial {\mathbb {D}}}$. Since $v \in {\mathcal {C}^{\infty } }({\partial {\mathbb {D}}})$, u satisfies Hölder condition. Let $\mathcal {S}(u)$ denote the Schwarz integral of u. By its properties, ${\text {Re}}(\mathcal {S}(u))=u$ on ${\partial {\mathbb {D}}}$ and $\mathcal {S}(u)$ is continuous in $\overline{{\mathbb {D}}}$ (see [3, 7]). Hence, there exists $r\in (0,1)$ such that

$$\begin{aligned} e^{pu(z)}\left( 1 - \frac{\varepsilon _2}{5}\right) \le e^{p {\text {Re}}(\mathcal {S}(u)(tz))}\le e^{pu(z)}\left( 1 +\frac{\varepsilon _2}{5}\right) \quad \text {for }t\in [r,1]~~ \text {and}~~z\in {\partial {\mathbb {D}}}. \end{aligned}$$

(6)

There exists a polynomial Q(z) that approximates the function $e^{\mathcal {S}(u)(z)}$ in the following way

$$\begin{aligned} \left( 1 - \frac{\varepsilon _2}{5}\right) e^{p{\text {Re}}(\mathcal {S}(u)(z))} \le |Q(z)|^p\le \left( 1+ \frac{\varepsilon _2}{5}\right) e^{p{\text {Re}}(\mathcal {S}(u)(z))}\quad \text {for }z\in \overline{{\mathbb {D}}}. \end{aligned}$$

(7)

Let $M:= \sup _{z\in {\partial {\mathbb {D}}}} H(z)$. The maximum principle implies that

$$\begin{aligned} \sup _{z \in \overline{{\mathbb {D}}}} e^{p{\text {Re}}(\mathcal {S}(u)(z))} =\sup _{z \in {\partial {\mathbb {D}}}} e^{p u(z)} =\sup _{z \in {\partial {\mathbb {D}}}} v(z){\mathop {\le }\limits ^{(5)}} \frac{5}{5-\varepsilon _2}\sup _{z \in {\partial {\mathbb {D}}}}H(z) = \frac{5M}{5-\varepsilon _2}. \end{aligned}$$

(8)

There exists a constant $C >0$ such that $5M/(5-\varepsilon _2)\le C H(z)$ for $z \in {\partial {\mathbb {D}}}$. Choose $N_ 1 \in {\mathbb {N}}$ such that $r^{np+1} \le \varepsilon _2/(5C)$ for all $n\ge N_1$. Then consider a polynomial $f_n(z)=(np+1)^{1/p}Q(z)z^{n}$, where $n \ge N_1$. For $z \in {\partial {\mathbb {D}}}$ and $n \ge N_1$ the above inequalities give the required estimation

$$\begin{aligned} \displaystyle \int _0^1 |f_n(zt)|^p dt&\ge (np+1)\displaystyle \int _{r}^1 |Q(zt)|^pt^{np}dt {\mathop {\ge }\limits ^{(7)}} \left( 1-\displaystyle \frac{\varepsilon _2}{5}\right) (np+1)\displaystyle \int _{r}^1 e^{p{\text {Re}}(\mathcal {S}(u)(zt))}t^{np}dt \\&{\mathop {\ge }\limits ^{(6)}}e^{p u(z)}\left( 1-\displaystyle \frac{\varepsilon _2}{5}\right) ^2(np+1)\displaystyle \int _{r}^1 t^{np}dt {\mathop {\ge }\limits ^{(5)}} H(z)\left( 1 - \displaystyle \frac{\varepsilon _2}{5}\right) ^3(1-r^{np+1}) \\&\ge H(z)\left( 1 - \displaystyle \frac{\varepsilon _2}{5}\right) ^4 \ge ~H(z)(1-\varepsilon _2) \end{aligned}$$

and

$$\begin{aligned} \displaystyle \int _0^1 |f_n(zt)|^p dt= & {} (np+1)\displaystyle \int _{0}^1 |Q(zt)|^pt^{np}dt \\&{\mathop {\le }\limits ^{(7)}}&\displaystyle \left( 1+\frac{\varepsilon _2}{5}\right) (np+1)\displaystyle \int _{0}^1 e^{p{\text {Re}}(\mathcal {S}(u)(zt))}t^{np}dt \\&{\mathop {\le }\limits ^{(6)}}&\displaystyle (np+1)\left( 1+\frac{\varepsilon _2}{5}\right) \left( \frac{5M}{5-\varepsilon _2}\displaystyle \int _{0}^rt^{np}dt + \displaystyle \left( 1+\frac{\varepsilon _2}{5}\right) e^{pu(z)}\displaystyle \int _{r}^1t^{np} dt \right) \\&{\mathop {\le }\limits ^{(5)}}&\displaystyle \frac{5M}{5-\varepsilon _2}\left( 1+\frac{\varepsilon _2}{5}\right) r^{np+1}+H(z)\left( 1+\frac{\varepsilon _2}{5}\right) ^3(1-r^{np+1})\\&{\mathop {\le }\limits ^{(8)}}&\displaystyle H(z) \left( 1+\frac{\varepsilon _2}{5}\right) C r^{np+1} +H(z)\left( 1+\frac{\varepsilon _2}{5}\right) ^3(1-r^{np+1})\\\le & {} \displaystyle H(z) \left( 1+\frac{\varepsilon _2}{5}\right) \left( C r^{np+1} +\left( 1+\frac{\varepsilon _2}{5}\right) ^2\right) \\\le & {} H(z)(1+\varepsilon _2). \end{aligned}$$

Now, to make $f_n$ satisfy the statement 1 it is enough to select a natural number $N_2 \ge N_1$ such that for all $n \ge N_2$

$$\begin{aligned} |f_n(z)| = (np+1)^{1/p}|Q(z)||z|^{n} \le \varepsilon _1 \quad \text {for}~~z \in K. \end{aligned}$$

Finally, observe that the function $f_n$ constructed above, where n is sufficiently large, has the required properties. $\square$

Remark 4

We can replace the statement 2 in Lemma 3 by one of the following

$2'$.:

$H(z)(1 - \varepsilon _2)< \mathcal {R}^p(f)(z)< H(z)$ for $z \in {\partial {\mathbb {D}}}$

$2''$.:

$H(z) - \varepsilon _2< \mathcal {R}^p(f)(z)< H(z)$ for $z \in {\partial {\mathbb {D}}}$.

Proof

For statement $2'$ it is enough to take $\tilde{H}(z) := H(z)/(1-\tilde{\varepsilon }_2)$ and $\tilde{\varepsilon }_2:= \varepsilon _2/(2-\varepsilon _2)$ in Lemma 3. To obtain conclusion $2''$ we apply the lemma above with $\hat{\varepsilon }_2:=\tilde{\varepsilon }_2/M$, where $M:=\sup _{z \in {\partial {\mathbb {D}}}} |H(z)|$. $\square$

Lemma 5

Fix $p>0$. Let $p>0$ and Q be a polynomial such that $Q(z) \ne 0$ for $z \in {\partial {\mathbb {D}}}$. For every $\varepsilon _3 \in (0,1)$ there exists $N \in {\mathbb {N}}$ such that for $n \ge N$ and all functions $g \in \mathcal {C}_L(\overline{{\mathbb {D}}})$ the function $f_n(z)=(np+1)^{1/p}Q(z)z^{n}$ satisfies

$$\begin{aligned} \mathcal {R}^p(f_n)(z) + \mathcal {R}^p(g)(z) - \varepsilon _3<\mathcal {R}^p(f_n+g)(z) < \mathcal {R}^p(f_n)(z) +\mathcal {R}^p(g)(z)+\varepsilon _3. \end{aligned}$$

Proof

Choose $k \in {\mathbb {N}}$ such that $k\ge p$. First, by triangle inequality, observe that

$$\begin{aligned} \mathcal {R}^{p/k}(f^k_n+g^k)(z)= & {} \int _{0}^{1} |(f_n^k+g^k)(zt)|^{p/k} dt \le \int _{0}^{1} |f_n(zt)|^pdt\\&+ \int _{0}^{1} |g(zt)|^pdt =\mathcal {R}^p(f_n)(z) +\mathcal {R}^p(g)(z). \end{aligned}$$

Let $\delta \in (0,1)$ be such that $\int _\delta ^1|g(zt)|^p < \varepsilon _3/8$. Since

$$\begin{aligned} \int _{0}^{\delta } |f_n(zt)|^p dt \le (np+1)\Vert Q\Vert ^p \int _{0}^{\delta } t^{np} dt = \Vert Q\Vert ^p \delta ^{np+1}, \end{aligned}$$

there exists $N_1 \in {\mathbb {N}}$ such that $\int _{0}^{\delta } |f_n(zt)|^p dt <\varepsilon _3/8$ for all $n\ge N_1$. Thus, the above inequalities imply the following

$$\begin{aligned} \displaystyle \mathcal {R}^{p/k}(f^k_n+g^k)(z)= & {} \displaystyle \int _{0}^{1} |(f_n^k+g^k)(zt)|^{p/k} dt \\= & {} \displaystyle \displaystyle \int _0^{\delta } |(f_n^k+g^k)(zt)|^{p/k} dt +\displaystyle \int _{\delta }^{1} |(f_n^k+g^k)(zt)|^{p/k} dt\\\ge & {} \displaystyle \int _{0}^{\delta } |g(zt)|^p dt - \displaystyle \int _{0}^{\delta } |f_n(zt)|^p dt + \displaystyle \int _{\delta }^{1} |f_n(zt)|^p dt - \displaystyle \int _{\delta }^{1} |g(zt)|^p dt \\= & {} \mathcal {R}^p(f_n)(z) +\mathcal {R}^p(g)(z) - 2 \displaystyle \int _{0}^{\delta } |f_n(zt)|^p dt - 2 \int _{\delta }^{1} |g(zt)|^p dt \\> & {} \displaystyle \mathcal {R}^p(f_n)(z) +\mathcal {R}^p(g)(z) - \frac{\varepsilon _3}{2}. \end{aligned}$$

If $0<p \le 1$, then $k=1$ does the job. For $p> 1$ we need a little bit more effort to accomplish this. Denote

$$\begin{aligned} \mathcal {I}_{k}^p(f_n,g)(z):=\int _{0}^{1} \left| \sum _{m=1}^{k-1}{ k \atopwithdelims ()m } (f_n(zt))^m (g(zt))^{k-m}\right| ^{p/k}dt \end{aligned}$$

and notice that

$$\begin{aligned} \begin{array}{ll} \mathcal {I}_{k}^p(f_n,g)(z) &{}\le \displaystyle \sum _{m=1}^{k-1}{ k \atopwithdelims ()m }^{p/k}\int _{0}^{1} |f_n(zt)|^{mp/k} |g(zt)|^{(k-m)p/k}dt \\ &{}\le \displaystyle \sum _{m=1}^{k-1}{ k \atopwithdelims ()m }^{p/k} \Vert Q\Vert ^{mp/k}L^{p(k-m)/k}(np+1)^{m/k}\int _0^1 t^{nmp/k}dt \\ &{}= \displaystyle \sum _{m=1}^{k-1}{ k \atopwithdelims ()m }^{p/k} \Vert Q\Vert ^{mp/k}L^{p(k-m)/k}\frac{k(np+1)^{m/k}}{nmp+k} \\ &{}\underset{{n \rightarrow \infty }}{\rightarrow } 0. \end{array} \end{aligned}$$

Therefore, there exists $N_2 \in {\mathbb {N}}$ such that $\mathcal {I}_{k}^p(f_n,g)(z)<\varepsilon _3/2$ for $n \ge N_2$. This yields the conclusion rather quickly

$$\begin{aligned} \begin{array}{ll} \mathcal {R}^p(f_n+g)(z) &{}=\displaystyle \int _{0}^{1} |(f_n+g)(zt)|^p dt \\ &{}=\displaystyle \int _{0}^{1} \left( |(f_n+g)(zt)|^k\right) ^{p/k}dt \\ &{}=\displaystyle \int _{0}^{1} \left| \sum _{m=0}^k{ k \atopwithdelims ()m }(f_n(zt))^m (g(zt))^{k-m}\right| ^{p/k}dt \\ &{}\le \displaystyle \int _{0}^{1} |f_n(zt)|^p dt + \displaystyle \int _{0}^{1} |g(zt)|^p dt + \mathcal {I}_{k}^p(f_n,g)(z)\\ &{}<\displaystyle \mathcal {R}^p(f_n)(z)+ \mathcal {R}^p(g)(z) +\frac{\varepsilon _3}{2} \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{ll} \mathcal {R}^p(f_n+g)(z) &{}=\displaystyle \int _{0}^{1} |(f_n+g)(zt)|^p dt \\ &{}=\displaystyle \int _{0}^{1} \left| \sum _{m=0}^{k}{ k \atopwithdelims ()m }(f_n(zt))^m (g(zt))^{k-m}\right| ^{p/k}dt \\ &{}\ge \displaystyle \int _{0}^{1} |(f_n^k+g^k)(zt)|^{p/k} dt -\mathcal {I}_{k}^p(f_n,g)(z) \\ &{}>\displaystyle \int _{0}^{1} |f_n(zt)|^p dt + \int _{0}^{1} |g(zt)|^p dt -\frac{\varepsilon _3}{2} - \frac{\varepsilon _3}{2} \\ &{}=\mathcal {R}^p(f_n)(z) + \mathcal {R}^p(g)(z) - \varepsilon _3. \end{array} \end{aligned}$$

$\square$

Theorem 6

Fix $p>0$. Let G be a strictly positive continuous function defined on ${\partial {\mathbb {D}}}$. There exists a function $f \in \mathcal {O}({\mathbb {D}})$ such that

$$\begin{aligned} \mathcal {R}^p(f)(z) = G(z) \quad \text {for}~ z \in {\partial {\mathbb {D}}}. \end{aligned}$$

Moreover, $\mathcal {R}^p(f) \in \mathcal {C}(\overline{{\mathbb {D}}})$.

Proof

Since G is strictly positive, there exists $m \in {\mathbb {N}}$ such that $G(z) - 2^{-m}>0$ for all $z \in {\partial {\mathbb {D}}}$. We will apply Lemmas  3 and 5 iteratively to construct a sequence $\{f_n\}_{n\in {\mathbb {N}}}$ of functions in the disc algebra and a sequence of real numbers $0<r_0\le r_1 \le \cdots \le r_n \le r_{n+1}\le \cdots \le 1$ satisfying the following conditions

• (a) $|f_n(z)| \le 2^{-n} \quad \text {for}~z \in \overline{{\mathbb {D}}}_{r_{n-1}}$

• (b) $\displaystyle \mathcal {R}^p\left( \sum _{k=1}^{n} f_k\right) (z)< G(z)- 2^{-(m+n+2)} \quad \text {for}~~z\in {\partial {\mathbb {D}}}$

• (c) $\displaystyle \int _0^{r_{n}} \left| \sum _{k=1}^{n} f_k(zt)\right| ^p dt >G(z) - 2^{-(m+n)} \quad \text {for}~z\in {\partial {\mathbb {D}}}$.

We begin by selecting $r_0=1/2$ and applying Lemma 3 in version $2''$ with

$$\begin{aligned} H:=G-2^{-(m+3)},\quad K:=\overline{{\mathbb {D}}}_{r_0},\quad \varepsilon _1:=\frac{1}{2},\quad \varepsilon _2:=2^{-(m+2)} \end{aligned}$$

to obtain a function $f_1 \in \mathcal {O}({\mathbb {D}}) \cap {\mathcal {C}^{\infty }(\overline{{\mathbb {D}}})}$ such that

• $|f_1(z)|\le 1/2$ for $z \in \overline{{\mathbb {D}}}_{1/2}$

• $G(z) -2^{-(m+1)}< \mathcal {R}^p(f_1)(z) < G(z)-2^{-(m+3)}$ for $z\in {\partial {\mathbb {D}}}$.

Then choose $1\ge r_1 \ge (1+r_0)/2$ such that

$$\begin{aligned} \int _0^{r_{1}} | f_1(zt)|^p dt >G(z) - 2^{-(m+1)}\quad \text {for } z\in {\partial {\mathbb {D}}}. \end{aligned}$$

Proceeding inductively, let us assume that numbers $r_0\le r_1\le r_2 \le \cdots r_n \le 1$ and functions $f_1,f_2,\ldots ,f_n\in \mathcal {O}({\mathbb {D}}) \cap {\mathcal {C}^{\infty }(\overline{{\mathbb {D}}})}$ with properties a)−c) have been found. We may apply Lemmas 3 and 5 with

$$\begin{aligned} H:= G- \mathcal {R}^p\left( \sum _{k=1}^n f_k\right) - 2^{-(m+n+2)},\quad g(z):=\sum _{k=1}^n f_k(z),\quad K:= \overline{{\mathbb {D}}}_{r_{n}}, \\ \varepsilon _1:= 2^{-(n+1)},\quad \varepsilon _2:= 2^{-(m+n+3)}, \quad \varepsilon _3:= 2^{-(m+n+3)} \end{aligned}$$

to produce a new function $f_{n+1}\in \mathcal {O}({\mathbb {D}}) \cap {\mathcal {C}^{\infty }(\overline{{\mathbb {D}}})}$ such that the following statements hold

• $|f_{n+1}(z)|\le 2^{-(n+1)} \quad \text {for}~~ z \in \overline{{\mathbb {D}}}_{r_{n}}$

• $\displaystyle \mathcal {R}^p(f_{n+1}) >G - \mathcal {R}^p\left( \sum _{k=1}^n f_k\right) - 2^{-(m+n+2)} - 2^{-(m+n+3)}$ on ${\partial {\mathbb {D}}}$

• $\displaystyle \mathcal {R}^p(f_{n+1})< G - \mathcal {R}^p\left( \sum _{k=1}^n f_k\right) - 2^{-(m+n+2)}$ on ${\partial {\mathbb {D}}}$

• $\displaystyle \mathcal {R}^p\left( \sum _{k=1}^{n+1} f_k\right) >\displaystyle \mathcal {R}^p(f_{n+1}) + \mathcal {R}^p\left( \sum _{k=1}^n f_k\right) - 2^{-(m+n+3)}$ on ${\partial {\mathbb {D}}}$

• $\displaystyle \mathcal {R}^p\left( \sum _{k=1}^{n+1} f_k\right) < \displaystyle \mathcal {R}^p(f_{n+1}) + \mathcal {R}^p\left( \sum _{k=1}^n f_k\right) + 2^{-(m+n+3)}$ on ${\partial {\mathbb {D}}}$.

This gives on ${\partial {\mathbb {D}}}$

$$\begin{aligned} \begin{array}{ll} \displaystyle \mathcal {R}^p\left( \sum _{k=1}^{n+1} f_k\right) &{}<\displaystyle \mathcal {R}^p\left( \sum _{k=1}^n f_k\right) + \mathcal {R}^p( f_{n+1}) + 2^{-(m+n+3)} \\ &{}< \displaystyle \mathcal {R}^p\left( \sum _{k=1}^n f_k\right) + G - \mathcal {R}^p\left( \sum _{k=1}^n f_k\right) - 2^{-(m+n+2)} + 2^{-(m+n+3)} \\ &{}=G - 2^{-(m+n+3)} \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{ll} \displaystyle \mathcal {R}^p\left( \sum _{k=1}^{n+1} f_k\right) &{}> \displaystyle \mathcal {R}^p\left( \sum _{k=1}^n f_k\right) + \mathcal {R}^p( f_{n+1}) - 2^{-(m+n+3)} \\ &{}> \displaystyle \mathcal {R}^p\left( \sum _{k=1}^n f_k\right) + G - \mathcal {R}^p\left( \sum _{k=1}^n f_k\right) - 2^{-(m+n+1)} \\ &{}=G - 2^{-(m+n+1)}. \end{array} \end{aligned}$$

Next select $1\ge r_{n+1}\ge (1+r_n)/2$ such that

$$\int _0^{r_{n+1}} \left| \sum _{k=1}^{n+1} f_k(zt)\right| ^p dt >G(z) - 2^{-(m+n+1)}$$

and the induction process is complete.

By statement a), the sequence $\sum _{k=1}^{n}f_k$ converges uniformly to the function $f:=\sum _{k=1}^{\infty }f_k$ on compact subsets of ${\mathbb {D}}$ and $f \in \mathcal {O}({\mathbb {D}})$. Then notice that for all $r \in (0,1)$ and $z \in {\partial {\mathbb {D}}}$ we have

$$\begin{aligned} \begin{array}{ll} \displaystyle \int _0^r |f(zt)|^p dt &{}= \displaystyle \int _0^r \lim _{n \rightarrow \infty } \left| \sum _{k=1}^{n}f_k(zt)\right| ^p dt =\displaystyle \lim _{n \rightarrow \infty }\int _0^r \left| \sum _{k=1}^{n}f_k(zt)\right| ^p dt\\ &{}\le \displaystyle \lim _{n \rightarrow \infty }( G(z) -2^{-(m+n+2)} ) \\ &{}= G(z). \end{array} \end{aligned}$$

Hence,

$$\begin{aligned} \mathcal {R}^p(f)(z) \le G(z) \quad \text {for}~~ z \in {\partial {\mathbb {D}}}. \end{aligned}$$

Let $q=\max \{p,1\}$. For a given $n\in {\mathbb {N}}$ and all $z \in {\partial {\mathbb {D}}}$ from conditions a) and c) it follows that

$$\begin{aligned} \begin{array}{ll} \left( \mathcal {R}^p(f)(z) \right) ^{1/q} &{}\ge \displaystyle \left( \int _0^{r_n} \left| \sum _{k=1}^{\infty }f_k(zt)\right| ^p dt\right) ^{1/q} \\ &{}\ge \displaystyle \left( \int _0^{r_n} \left| \sum _{k=1}^{n}f_k(zt)\right| ^p dt \right) ^{1/q} - \displaystyle \left( \int _0^{r_n} \left| \sum _{k=n+1}^{\infty }f_k(zt)\right| ^p dt \right) ^{1/q} \\ &{}> \displaystyle \left( G(z) - 2^{-(m+n)}\right) ^{1/q} - \left( \sum _{k=n+1}^{\infty }2^{-k}\right) ^{p/q} \\ &{}=\displaystyle \left( G(z) - 2^{-(m+n)}\right) ^{1/q}- 2^{-np/q} \mathop {\longrightarrow }\limits _{n\rightarrow \infty } (G(z))^{1/q}. \end{array} \end{aligned}$$

Therefore,

$$\begin{aligned} \mathcal {R}^p(f)(z)\ge G(z) ~~\text {for all}~~ z \in {\partial {\mathbb {D}}}. \end{aligned}$$

These inequalities establish the conclusion $\mathcal {R}^p(f)= G ~~ \text {on}~~{\partial {\mathbb {D}}}$. Finally, it follows from Theorem 2 that $\mathcal {R}^p(f)\in \mathcal {C}(\overline{{\mathbb {D}}})$. $\square$

3 Divergent Taylor Series

Lemma 7

Let $h \in \mathcal {C}({\partial {\mathbb {D}}}) ~\text {be a strictly positive function},~\phi \in \mathcal {C}(\overline{{\mathbb {D}}})$ be a bounded function and $a,\theta$, $\varepsilon \in (0,1)$. There exist a natural number $N \in {\mathbb {N}}$ and a sequence of orthogonal polynomials $\{P_n\}_{n \in {\mathbb {N}}}$ such that on ${\partial {\mathbb {D}}}$ the following statements hold

(s1):

$\mathcal {R}^2(P_k)<ah ,~~ k=0,1,\ldots$

(s2):

$\displaystyle \mathcal {R}^2\left( \sum _{j=0}^k P_j\right) <h,~~ k=0,1,\ldots$

(s3):

$\mathcal {R}^2\left( \displaystyle \sum _{j=0}^{k}P_{j}\right) > \displaystyle \frac{1}{2}a\left( h- \mathcal {R}^2\left( \sum _{j=0}^{k-1} P_j\right) \right) + \mathcal {R}^2\left( \displaystyle \sum _{j=0}^{k-1}P_{j}\right) ,~~k=1,2,\ldots$

(s4):

$\displaystyle \left| \mathcal {R}^2\left( \sum _{j=0}^{k} P_j\right) -\displaystyle \sum _{j=0}^{k}\mathcal {R}^2(P_j)\right| <\varepsilon ,~~ k=1,2,\ldots$

(s5):

$\displaystyle \left| \mathcal {R}^2\left( \phi +\sum _{j=0}^{k} P_j\right) -\sum _{j=0}^{k}\displaystyle \mathcal {R}^2(P_j)-\displaystyle \mathcal {R}^2(\phi )\right| <\varepsilon ,~~ k=1,2,\ldots$

(s6):

$\displaystyle \mathcal {R}^2\left( \sum _{j=0}^N P_j\right) >\theta h$.

Proof

First notice that $P_0 \equiv 0$ satisfies conditions $(s1) - (s2)$. Then Lemmas 3 and 5 applied to

$$ H:= \frac{5}{6}ah, \quad g:=\phi , \quad \varepsilon _2:= \frac{1}{5}, \quad \varepsilon _3:=\varepsilon $$

produces a polynomial $P_1$ with the following properties

• $\displaystyle \frac{1}{2}ah<\frac{4}{5}H<\displaystyle \mathcal {R}^2(P_{1})<H<ah$ on ${\partial {\mathbb {D}}}$

• $\displaystyle \mathcal {R}^2(P_{1}) + \mathcal {R}^2(\phi ) -\varepsilon< \mathcal {R}^2(\phi +P_{1})<\mathcal {R}^2(P_{1}) + \mathcal {R}^2(\phi ) -\varepsilon$ on ${\partial {\mathbb {D}}}$.

Of course, the above statements imply $(s1){-}(s5)$ immediately.

We then proceed inductively as follows: Suppose that orthogonal polynomials $P_0,P_1,\ldots , P_k$ have been found such that conditions $(s1){-}(s5)$ hold. By (s4) and (s5), there exist $\rho _1, \rho _2 \in (0,1)$ such that

• $\displaystyle \left| \mathcal {R}^2\left( \sum _{j=0}^{k} P_j\right) -\displaystyle \sum _{j=0}^{k}\mathcal {R}^2(P_j)\right|<\rho _1<\varepsilon$ on ${\partial {\mathbb {D}}}$

• $\displaystyle \left| \mathcal {R}^2\left( \phi +\sum _{j=0}^{k} P_j\right) -\sum _{j=0}^{k}\displaystyle \mathcal {R}^2(P_j)-\displaystyle \mathcal {R}^2(\phi )\right|<\rho _2<\varepsilon$ on ${\partial {\mathbb {D}}}$.

Select $N_1 \in {\mathbb {N}}$ such that $N_1 > \deg (P_k)$. For our inductive step, again we may apply Lemmas 3 and 5 to

$$\begin{aligned} H:= & {} \frac{5}{6}a\left( h - \mathcal {R}^2\left( \sum _{j=0}^k P_j\right) \right) , \quad g_1:=\sum _{j=0}^k P_j, \quad g_2:=\sum _{j=0}^k P_j +\phi , \quad \varepsilon _2:= \frac{1}{5}, \\&\varepsilon _3:= \min \left\{ \frac{1}{6}\inf _{z \in {\partial {\mathbb {D}}}}a\left( h(z)- \mathcal {R}^2\left( \sum _{j=0}^k P_j\right) (z)\right) ,\varepsilon - \rho _1, \varepsilon - \rho _2 \right\} \end{aligned}$$

to obtain a polynomial $P_{k+1}(z):=\left( 2n_{k+1}+1\right) ^{1/2}Q_{k+1}(z)z^{n_{k+1}}$, where $n_{k+1} \ge N_1$ and $Q_{k+1}\ne 0$ on ${\partial {\mathbb {D}}}$. Therefore, the polynomial $P_{k+1}$ is orthogonal to $P_0,P_1,\ldots ,P_k$ and satisfies the following statements

• (a) $\displaystyle (1-\varepsilon _2)H<\displaystyle \mathcal {R}^2(P_{k+1})<H$ on ${\partial {\mathbb {D}}}$

• (b) $\displaystyle \mathcal {R}^2(g_1) +\displaystyle \mathcal {R}^2(P_{k+1})-\varepsilon _3<\displaystyle \mathcal {R}^2(g_1 + P_{k+1}) < \displaystyle \mathcal {R}^2(g_1) +\displaystyle \mathcal {R}^2(P_{k+1})+\varepsilon _3$ on ${\partial {\mathbb {D}}}$

• (c) $\displaystyle \mathcal {R}^2(g_2) +\displaystyle \mathcal {R}^2(P_{k+1})-\varepsilon _3<\displaystyle \mathcal {R}^2(g_2 + P_{k+1}) < \displaystyle \mathcal {R}^2(g_2) +\displaystyle \mathcal {R}^2(P_{k+1})+\varepsilon _3$ on ${\partial {\mathbb {D}}}$.

Direct estimation on ${\partial {\mathbb {D}}}$ gives (s2)

$$\begin{aligned} \begin{array}{ll} \displaystyle \mathcal {R}^2\left( \sum _{j=0}^{k+1} P_j\right) &{}< \displaystyle \mathcal {R}^2\left( \sum _{j=0}^k P_j\right) +\displaystyle \mathcal {R}^2(P_{k+1})+\varepsilon _3\\ &{}< \displaystyle \mathcal {R}^2\left( \sum _{j=0}^k P_j\right) + \displaystyle \frac{5}{6}a\left( h - \mathcal {R}^2\left( \sum _{j=0}^k P_j\right) \right) \\ &{}\quad +\frac{1}{6}a\left( h- \mathcal {R}^2\left( \sum _{j=0}^k P_j\right) \right) \\ &{}\le h \end{array} \end{aligned}$$

and (s3)

$$\begin{aligned} \begin{array}{ll} \displaystyle \mathcal {R}^2\left( \sum _{j=0}^{k+1} P_j\right) &{}> \displaystyle \mathcal {R}^2\left( \sum _{j=0}^k P_j \right) +\displaystyle \mathcal {R}^2(P_{k+1})-\varepsilon _3\\ &{}> \displaystyle \mathcal {R}^2\left( \sum _{j=0}^k P_j\right) + \displaystyle \frac{2}{3}a\left( h - \mathcal {R}^2\left( \sum _{j=0}^k P_j\right) \right) \\ &{}\quad -\frac{1}{6}a\left( h- \mathcal {R}^2\left( \sum _{j=0}^k P_j\right) \right) \\ &{}\ge \displaystyle \mathcal {R}^2\left( \sum _{j=0}^k P_j \right) + \displaystyle \frac{1}{2}a\left( h- \mathcal {R}^2\left( \sum _{j=0}^k P_j\right) \right) . \end{array} \end{aligned}$$

Furthermore, combining (a)–(c) and the induction hypothesis gives conditions $(s4)-(s5)$

$$\begin{aligned} \begin{array}{ll} \displaystyle \mathcal {R}^2\left( \sum _{j=0}^{k+1}P_{j}\right) &{}< \displaystyle \mathcal {R}^2\left( \sum _{j=0}^{k}P_{j}\right) +\displaystyle \mathcal {R}^2(P_{k+1})+\varepsilon _3<\displaystyle \sum _{j=0}^{k+1}\mathcal {R}^2(P_{j}) + \rho _1 + \varepsilon _3 \\ &{}=\displaystyle \sum _{j=0}^{k+1}\mathcal {R}^2(P_{j}) + \varepsilon \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{ll} \displaystyle \mathcal {R}^2\left( \sum _{j=0}^{k+1} P_j +\phi \right) &{}< \displaystyle \mathcal {R}^2\left( \sum _{j=0}^k P_j +\phi \right) \\ &{}\quad +\displaystyle \mathcal {R}^2(P_{k+1})+\varepsilon _3<\displaystyle \mathcal {R}^2(\phi ) +\displaystyle \sum _{j=0}^{k+1}\mathcal {R}^2(P_{j})+\rho _2+\varepsilon _3 \\ &{}<\displaystyle \mathcal {R}^2(\phi ) +\displaystyle \mathcal {R}^2\left( \sum _{j=0}^{k+1}P_{j}\right) +\varepsilon . \end{array} \end{aligned}$$

Similarly,

$$\begin{aligned} \mathcal {R}^2\left( \sum _{j=0}^{k+1}P_{j}\right) >\sum _{j=0}^{k+1}\mathcal {R}^2(P_{j}) - \varepsilon \end{aligned}$$

and

$$\begin{aligned} \mathcal {R}^2\left( \sum _{j=0}^{k+1} P_j +\phi \right) > \mathcal {R}^2(\phi ) +\mathcal {R}^2\left( \sum _{j=0}^{k+1}P_{j}\right) -\varepsilon . \end{aligned}$$

The induction is now complete.

Observe that statements (s2) and (s3) imply that

$$\begin{aligned} \lim _{n\rightarrow \infty } \mathcal {R}^2\left( \sum _{j=0}^n P_j \right) (z) = h(z),\quad z \in {\partial {\mathbb {D}}}\end{aligned}$$

and

$$\begin{aligned} \left\{ \mathcal {R}^2\left( \sum _{j=0}^n P_j \right) \right\} _{n \in {\mathbb {N}}} \end{aligned}$$

is an increasing sequence. Hence, by Dini’s theorem, $\mathcal {R}^2\left( \sum _{j=0}^n P_j \right)$ converges uniformly to h. Therefore, there exists $N \in {\mathbb {N}}$ such that

$$\begin{aligned} \mathcal {R}^2\left( \sum _{j=0}^N P_j\right) >\theta h\quad \text {on } {\partial {\mathbb {D}}}. \end{aligned}$$

$\square$

Theorem 8

There exists a holomorphic function $f(z)= \sum _{n=0}^{\infty }a_n z^n$ such that

• 1. $\mathcal {R}^2(f) \in \mathcal {C}(\overline{{\mathbb {D}}})$

• 2. $\displaystyle \sum _{n=1}^{\infty } \left( \frac{|a_n|^2}{2n+1}\right) ^s =\infty , \quad s \in [0,1)$.

Proof

For every $j \in {\mathbb {N}}$ we may apply the preceding lemma with $\phi \equiv 0,~ h \equiv 1,~ a:=2^{-j}, ~ \theta := 1/2$ to obtain $N_j \in {\mathbb {N}}$ and a sequence of orthogonal polynomials $\{P_{jk}\}_{0\le k \le N_j,~j \in {\mathbb {N}}}$ such that

• (a) $\mathcal {R}^2(P_{jk})<2^{-j},~~ k=0,\ldots ,N_j$

• (b) $\displaystyle \frac{1}{2}<\mathcal {R}^2\left( \sum _{k=0}^{N_j} P_{jk}\right) <1$.

Observe that if $n>m$, then

$$\begin{aligned} \begin{array}{ll} &{} \displaystyle \sup _{z \in {\partial {\mathbb {D}}}}\left( \mathcal {R}^2\left( \sum _{j=1}^{n} \frac{1}{j^2} \sum _{k=0}^{N_j}P_{jk}-\sum _{j=1}^{m} \frac{1}{j^2} \sum _{k=0}^{N_j}P_{jk}\right) (z)\right) ^{1/2} \\ &{}\quad = \displaystyle \sup _{z \in {\partial {\mathbb {D}}}}\left( \mathcal {R}^2\left( \sum _{j=m+1}^{n} \frac{1}{j^2} \sum _{k=0}^{N_j}P_{jk}\right) (z)\right) ^{1/2}\\ &{}\quad \le \displaystyle \sup _{z \in {\partial {\mathbb {D}}}}\left( \sum _{j=m+1}^{n} \frac{1}{j^2}\left( \mathcal {R}^2\left( \sum _{k=0}^{N_j}P_{jk}\right) (z)\right) ^{1/2}\right) \\ &{}\quad \le \displaystyle \sum _{j=m+1}^{n} \frac{1}{j^2} \underset{{{{n,m \rightarrow \infty }}}}{\rightarrow } 0. \end{array} \end{aligned}$$

It follows from the above that

$$\begin{aligned} \left\{ \sum _{j=1}^{n} \frac{1}{j^2} \sum _{k=0}^{N_j}P_{jk}\right\} _{n\in {\mathbb {N}}} \end{aligned}$$

is a Cauchy sequence in $\mathcal {HR}^2({\mathbb {D}})$ which by Proposition 1 is convergent in $\mathcal {HR}^2({\mathbb {D}})$. Therefore, we may define a function

$$\begin{aligned} f:= \sum _{j=1}^{\infty } \frac{1}{j^2} \sum _{k=0}^{N_j}P_{jk} \in \mathcal {O}({\mathbb {D}}). \end{aligned}$$

Denote $\mathcal {R}^2\left( \sum _{j=1}^{n} 1/j^2 \sum _{k=0}^{N_j}P_{jk}\right)$ by $\mathcal {P}_n$, for the moment. If $n > m$, then on ${\partial {\mathbb {D}}}$

$$\begin{aligned} \begin{array}{ll} &{}\displaystyle \left| (\mathcal {P}_n)^{1/2} - (\mathcal {P}_m)^{1/2}\right| \\ &{}\quad =\left| \displaystyle \left( \mathcal {R}^2\left( \sum _{j=1}^{n} \frac{1}{j^2} \sum _{k=0}^{N_j}P_{jk}\right) \right) ^{1/2}-\left( \mathcal {R}^2\left( \sum _{j=1}^{m} \frac{1}{j^2} \sum _{k=0}^{N_j}P_{jk}\right) \right) ^{1/2} \right| \\ &{}\quad \le \displaystyle \left( \mathcal {R}^2\left( \sum _{j=m+1}^{n}\frac{1}{j^2} \sum _{k=0}^{N_j}P_{jk}\right) \right) ^{1/2}\le \displaystyle \sum _{j=m+1}^{n}\left( \mathcal {R}^2\left( \frac{1}{j^2} \sum _{k=0}^{N_j}P_{jk}\right) \right) ^{1/2} \\ &{}\quad = \displaystyle \sum _{j=m+1}^{n} \frac{1}{j^2}\left( \mathcal {R}^2\left( \sum _{k=0}^{N_j}P_{jk}\right) \right) ^{1/2}\le \displaystyle \sum _{j=m+1}^{n} \frac{1}{j^2} \mathop {\longrightarrow }\limits _{n,m \rightarrow \infty } 0. \end{array} \end{aligned}$$

Observe that there exists a constant $C>0$ such that $\sup _{z \in {\partial {\mathbb {D}}}}|\mathcal {P}_n(z)|<C$ for all $n \in {\mathbb {N}}$. Hence,

$$\begin{aligned} \displaystyle \sup _{z \in {\partial {\mathbb {D}}}}\left| \mathcal {P}_n(z)- \mathcal {P}_m(z)\right|&=\displaystyle \sup _{z \in {\partial {\mathbb {D}}}}\left| \left( (\mathcal {P}_n(z))^{1/2} + (\mathcal {P}_m(z))^{1/2}\right) \left( (\mathcal {P}_n(z))^{1/2} - (\mathcal {P}_m(z))^{1/2}\right) \right| \\&\le 2C \displaystyle \sup _{z \in {\partial {\mathbb {D}}}}\left| (\mathcal {P}_n(z))^{1/2} - (\mathcal {P}_m(z))^{1/2}\right| \mathop {\longrightarrow }\limits _{n,m \rightarrow \infty } 0. \end{aligned}$$

This implies that

$$\left\{ \mathcal {R}^2\left( \sum _{j=1}^{n} \frac{1}{j^2} \sum _{k=0}^{N_j}P_{jk}\right) \right\} _{n\in {\mathbb {N}}}$$

is a Cauchy sequence in the space of continuous functions, so $\mathcal {R}^2(f) \in \mathcal {C}({\partial {\mathbb {D}}})$ and consequently by Theorem 2, the function $\mathcal {R}^2(f)$ is also continuous in $\overline{{\mathbb {D}}}$.

Now we turn to conclusion 2. Let $P_{jk} (z)= \sum _{n\in I_{jk}} p_{jkn}(z)= \sum _{n\in I_{jk}} a_{jkn}z^n$ be the homogeneous expansion of $P_{jk}$, where $I_{jk}$ is the set of monomials’ degrees of the polynomial $P_{jk}$. By construction, $I_{jm} \cap I_{ik} = \emptyset$ for $(j,m) \ne (i,k)$. Set $s \in [0,1)$. Notice that by statement b),

$$\begin{aligned} \Vert P_{jk}\Vert _{2,{\mathbb {D}}}^{2}=\Vert \mathcal {R}^2(P_{jk})\Vert _{1,{\partial {\mathbb {D}}}}<2^{-j},~~k=0,\ldots ,N_j, \end{aligned}$$

which implies that

$$\begin{aligned} \Vert P_{jk}\Vert _{2,{\mathbb {D}}}^{2s} \ge \Vert P_{jk}\Vert _{2,{\mathbb {D}}}^{2s}\left( \Vert P_{jk}\Vert _{2,{\mathbb {D}}}^{2} 2^{j}\right) ^{1-s} =\Vert P_{jk}\Vert _{2,{\mathbb {D}}}^{2}2^{j(1-s)},~~k=0,\ldots ,N_j. \end{aligned}$$

Due to the orthogonality of the polynomials $p_{jkn}$, a simple calculation shows that

$$\begin{aligned} \Vert P_{jk}\Vert _{2,{\mathbb {D}}}^2= & {} \displaystyle \left| \right| \sum _{n\in I_{jk}} p_{jkn}\left| \right| _{2,{\mathbb {D}}}^2 = \displaystyle \sum _{n\in I_{jk}} \Vert p_{jkn}\Vert _{2,{\mathbb {D}}}^2 = \displaystyle \sum _{n\in I_{jk}} \displaystyle \int _{{\partial {\mathbb {D}}}} \int _0^1 |a_{jkn}z^nt^n|^2 dt d\sigma (z) \\= & {} \displaystyle \sum _{n\in I_{jk}} \int _0^1 | a_{jkn}|^2 t^{2n}dt =\displaystyle \sum _{n\in I_{jk}} \frac{|a_{jkn}|^2}{2n+1}. \end{aligned}$$

Finally, we conclude from all this that

$$\begin{aligned} \displaystyle \sum _{j,k,n}\left( \frac{1}{j^4}\frac{|a_{jkn}|^2}{2n+1}\right) ^s\ge & {} \displaystyle \sum _{j,k}\left( \frac{1}{j^4}\right) ^s \left( \sum _{n \in I_{jk}}\frac{|a_{jkn}|^2}{2n+1}\right) ^s =\displaystyle \sum _{j,k}\frac{1}{j^{4s}}\Vert P_{jk}\Vert _{2,{\mathbb {D}}}^{2s} \\= & {} \displaystyle \sum _{j=1}^{\infty }\frac{1}{j^{4s}} \sum _{k=0}^{N_j} \Vert P_{jk}\Vert _{2,{\mathbb {D}}}^{2s} \ge \displaystyle \sum _{j=1}^{\infty }\frac{1}{j^{4s}}\sum _{k=0}^{N_j} \Vert P_{jk}\Vert _{2,{\mathbb {D}}}^{2}2^{j(1-s)}\\= & {} \displaystyle \sum _{j=1}^{\infty }\frac{1}{j^{4s}}\left\| \sum _{k=0}^{N_j} P_{jk}\right\| _{2,{\mathbb {D}}}^{2}2^{j(1-s)} > \displaystyle \sum _{j=1}^{\infty }\frac{1}{2j^{4s}} 2^{j(1-s)} \\= & {} \infty \end{aligned}$$

for $s \in [0,1)$. The proof of the theorem is now complete. $\square$

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 $\varPhi \in \mathcal {C}({\partial {\mathbb {D}}})$ be a strictly positive function. There exists a holomorphic function $f(z)= \sum _{n=0}^{\infty }a_n z^n$ such that

• 1. $\mathcal {R}^2(f) = \varPhi$   on  ${\partial {\mathbb {D}}}$

• 2. $\mathcal {R}^2(f) \in \mathcal {C}(\overline{{\mathbb {D}}})$

• 3. $\displaystyle \sum _{n=0}^{\infty } \left( \frac{|a_n|^2}{2n+1}\right) ^s =\infty , \quad s \in [0,1)$.

Proof

Without loss of generality let us assume that $\sup _{z \in {\partial {\mathbb {D}}}}|\varPhi (z)|<1$. We shall construct a sequence of orthogonal polynomials $\{Q_k\}_{k=0}^\infty$ with the following properties

(p1):

each $Q_k = \displaystyle \sum _{j=0}^{N_k} P_{kj}$ is a sum of orthogonal polynomials $P_{kj}$

(p2):

$\displaystyle \mathcal {R}^2(P_{kj})<\left\| \varPhi -\mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right\| ^{k}_{1, {\partial {\mathbb {D}}}}~~\text {on}~{\partial {\mathbb {D}}},~~ j=0,1,\ldots ,N_k$

(p3):

$\displaystyle \mathcal {R}^2\left( \sum _{m=0}^k Q_m\right) <\varPhi ~~\text {on}~{\partial {\mathbb {D}}}$

(p4):

$\displaystyle \mathcal {R}^2\left( \sum _{m=0}^k Q_m\right) >\displaystyle \frac{1}{4}\left( \varPhi - \mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right) + \mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) ~~\text {on}~{\partial {\mathbb {D}}}$

(p5):

$\displaystyle \Vert Q_k\Vert _{2, {\mathbb {D}}}^2>\frac{1}{2}\left\| \varPhi -\mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right\| _{1,{\partial {\mathbb {D}}}}$

We will apply Lemma 7 iteratively, with new $a,h, \phi , \theta , \varepsilon$ at each iteration. Let us begin our process by selecting $Q_0\equiv 0$. Next apply Lemma 7 to

$$\begin{aligned} a:=\inf _{z\in {\partial {\mathbb {D}}}}|\varPhi (z)|,\quad h:=\varPhi , \quad \phi \equiv 0, \quad \theta :=\frac{1}{2}. \end{aligned}$$

This produces $N_1 \in {\mathbb {N}}$ and orthogonal polynomials $P_{10},P_{11},\ldots ,P_{1N_1}$ such that

• $\displaystyle \mathcal {R}^2(P_{1j})<\Vert \varPhi \Vert _{1,{\partial {\mathbb {D}}}}~~\text {on}~{\partial {\mathbb {D}}},~~ j=0,1,\ldots ,N_1$

• $\displaystyle \mathcal {R}^2\left( \sum _{j=0}^{N_1}P_{1j}\right) <\varPhi ~~\text {on}~{\partial {\mathbb {D}}}$

• $\displaystyle \mathcal {R}^2\left( \sum _{j=0}^{N_1}P_{1j}\right)>\displaystyle \frac{1}{2}\varPhi > \frac{1}{4}\varPhi ~~\text {on}~{\partial {\mathbb {D}}}$.

Now it is enough to define $Q_1:=\sum _{j=0}^{N_1}P_{1j}$. Proceeding inductively, let us assume that orthogonal polynomials $Q_0,Q_1,\ldots , Q_k$ with the required properties $(p1)-(p5)$ have been constructed. We then may apply Lemma 7 with

$$\begin{aligned} a:= & {} \left\| \varPhi - \mathcal {R}^2\left( \sum _{m=0}^k Q_m\right) \right\| _{1,{\partial {\mathbb {D}}}}^{k+1}, \quad h:=\frac{3}{4}\left( \varPhi -\mathcal {R}^2\left( \sum _{m=0}^{k} Q_m\right) \right) , \\ \phi:= & {} \sum _{m=0}^kQ_m, \quad \theta :=\frac{2}{3},\quad \varepsilon :=\frac{1}{8}\inf _{z \in {\partial {\mathbb {D}}}}\left( \varPhi (z)-\mathcal {R}^2\left( \sum _{m=0}^{k} Q_m\right) (z)\right) \end{aligned}$$

to obtain $N_{k+1} \in {\mathbb {N}}$ and orthogonal polynomials $P_{k+1,0},P_{k+1,1},\ldots ,P_{k+1,N_{k+1}}$ that are also orthogonal to the polynomials $Q_0,Q_1,\ldots , Q_k$ and satisfy the following conditions

• (a) $\displaystyle \mathcal {R}^2(P_{k+1,j})<\left\| \varPhi - \mathcal {R}^2\left( \sum _{m=0}^k Q_m\right) \right\| _{1,{\partial {\mathbb {D}}}}^{k+1}~~\text {on}~{\partial {\mathbb {D}}},~~j=0,\ldots ,N_{k+1}$

• (b) $\displaystyle \mathcal {R}^2\left( \sum _{j=0}^{N_{k+1}}P_{k+1,j}\right) <h=\displaystyle \frac{3}{4}\left( \varPhi -\mathcal {R}^2\left( \sum _{m=0}^{k} Q_m\right) \right) ~~\text {on}~{\partial {\mathbb {D}}}$

• (c) $\displaystyle \mathcal {R}^2\left( \sum _{j=0}^{N_{k+1}}P_{k+1,j}\right) >\theta h=\displaystyle \frac{1}{2}\left( \varPhi -\mathcal {R}^2\left( \sum _{m=0}^{k} Q_m\right) \right) ~~\text {on}~{\partial {\mathbb {D}}}$

• (d) $\displaystyle \left| \displaystyle \mathcal {R}^2\left( \sum _{m=0}^{k}Q_m + \sum _{j=0}^{N_{k+1}}P_{k+1,j}\right) - \displaystyle \mathcal {R}^2\left( \sum _{m=0}^{k}Q_m \right) - \displaystyle \mathcal {R}^2\left( \sum _{j=0}^{N_{k+1}}P_{k+1,j}\right) \right| <2\varepsilon ~\text {on}~{\partial {\mathbb {D}}}$.

If we define $Q_{k+1}:=\sum _{j=0}^{N_{k+1}}P_{k+1,j}$, then statements $(p1)-(p2)$ and (p5) immediately hold. Other properties $(p3)-(p4)$ follow from simple calculations

$$\begin{aligned} \mathcal {R}^2\left( \sum _{m=0}^{k+1} Q_m\right)< & {} \mathcal {R}^2\left( \sum _{m=0}^k Q_m\right) + \mathcal {R}^2(Q_{k+1}) + 2\varepsilon <\mathcal {R}^2\left( \sum _{m=0}^k Q_m\right) \\&+\varPhi -\mathcal {R}^2\left( \sum _{m=0}^{k} Q_m\right) =\varPhi \end{aligned}$$

and

$$\begin{aligned} \mathcal {R}^2\left( \sum _{m=0}^{k+1} Q_m\right)> & {} \mathcal {R}^2\left( \sum _{m=0}^k Q_m\right) + \mathcal {R}^2(Q_{k+1}) - 2\varepsilon \ge \mathcal {R}^2\left( \sum _{m=0}^k Q_m\right) \\&+\frac{1}{4}\left( \varPhi -\mathcal {R}^2\left( \sum _{m=0}^{k} Q_m\right) \right) . \end{aligned}$$

That completes the induction. Combining statements $(p3){-}(p4)$ iteratively gives the following

$$\begin{aligned} \begin{array}{ll} \displaystyle \varPhi -\mathcal {R}^2\left( \sum _{m=0}^{k} Q_m\right) &{}<\displaystyle \varPhi - \frac{1}{4}\left( \varPhi - \mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right) - \mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \\ &{} =\displaystyle \frac{3}{4}\left( \varPhi -\mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right) \\ &{}<\displaystyle \left( \frac{3}{4}\right) ^k\varPhi . \end{array} \end{aligned}$$

This leads to the conclusion that $\lim _{n\rightarrow \infty }\mathcal {R}^2\left( \sum _{k=0}^n Q_k \right) = \varPhi$ on ${\partial {\mathbb {D}}}$. Moreover, if $n > m$, then

$$\begin{aligned} \displaystyle \sup _{z \in {\partial {\mathbb {D}}}}\left( \mathcal {R}^2\left( \sum _{k=0}^{n}Q_{k}- \sum _{k=0}^{m}Q_{jk}\right) (z)\right) ^{1/2}= & {} \displaystyle \sup _{z \in {\partial {\mathbb {D}}}}\left( \mathcal {R}^2\left( \sum _{k=m+1}^{n}Q_{k}\right) (z)\right) ^{1/2}\\\le & {} \displaystyle \sup _{z \in {\partial {\mathbb {D}}}}\left( \sum _{k=m+1}^{n}\left( \mathcal {R}^2( Q_{k})(z)\right) ^{1/2}\right) \\\le & {} \displaystyle \sup _{z \in {\partial {\mathbb {D}}}} \left( \sum _{k=m+1}^n \displaystyle \frac{\sqrt{3}}{2}\left( \varPhi -\mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right) ^{1/2}\right) \\< & {} \displaystyle \sum _{j=m+1}^{n} \left( \frac{\sqrt{3}}{2}\right) ^k \Vert \varPhi \Vert ^{1/2}\\&{\mathop {\longrightarrow }\limits _{n,m\rightarrow \infty }} 0. \end{aligned}$$

Hence, $\left\{ \sum _{k=0}^{n}Q_{k}\right\} _{n\in {\mathbb {N}}}$ is a Cauchy sequence in $\mathcal {HR}^2({\mathbb {D}})$ and by Proposition 1 is convergent in $\mathcal {HR}^2({\mathbb {D}})$. It follows that the function $f:= \sum _{k=0}^\infty Q_k$ is well-defined and holomorphic in ${\mathbb {D}}$. Then Proposition 1 also tells us that $\mathcal {R}^2(f)=\lim _{n\rightarrow \infty }\mathcal {R}^2\left( \sum _{k=0}^n Q_k \right) = \varPhi$ on ${\partial {\mathbb {D}}}$. Furthermore, due to Theorem 2 the function $\mathcal {R}^2(f)$ is continuous in $\overline{{\mathbb {D}}}$.

Now we prove the second part of the theorem. Taking into account that $Q_0\equiv 0$, we may observe that $f=\sum _{k=1}^\infty Q_k= \sum _{k=1}^\infty \sum _{j=0}^{N_k}P_{kj}$. Property (p2) implies that

$$\begin{aligned} \Vert P_{kj}\Vert _{2,{\mathbb {D}}}^{2}=\left\| \mathcal {R}^2(P_{kj})\right\| _{1,{\partial {\mathbb {D}}}}< \left\| \varPhi - \mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right\| _{1,{\partial {\mathbb {D}}}}^{k}, \quad j=0,\ldots ,N_k, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \left\| \varPhi - \mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right\| _{1,{\partial {\mathbb {D}}}}^{-k}\Vert P_{kj}\Vert _{2,{\mathbb {D}}}^{2} <1, \quad j=0,\ldots ,N_k. \end{aligned}$$

If $s \in [0,1)$, then we may apply above inequality to obtain that

$$\begin{aligned} \begin{array}{ll} \Vert P_{kj}\Vert _{2,{\mathbb {D}}}^{2s} &{}>\displaystyle \Vert P_{kj}\Vert _{2,{\mathbb {D}}}^{2s}\left( \displaystyle \left\| \varPhi - \mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right\| _{1,{\partial {\mathbb {D}}}}^{-k}\Vert P_{kj}\Vert _{2,{\mathbb {D}}}^{2} \right) ^{1-s} \\ &{}=\displaystyle \left\| \varPhi - \mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right\| _{1,{\partial {\mathbb {D}}}}^{k(s-1)}\Vert P_{kj}\Vert _{2,{\mathbb {D}}}^{2} \end{array} ,~~j=0,\ldots ,N_k. \end{aligned}$$

(9)

Let

$$\begin{aligned} P_{kj} (z)= \sum _{n\in I_{kj}} p_{kjn}(z)= \sum _{n\in I_{kj}} a_{kjn}z^n \end{aligned}$$

be the homogenous expansion of $P_{kj}$, where $I_{kj}$ is the set of monomial’s degrees of the polynomial $P_{kj}$. By construction, $I_{mj} \cap I_{ki} = \emptyset$ for $(m,j) \ne (k,i)$. Orthogonality of the polynomials $p_{kjn}$ gives the equality

$$\begin{aligned} \Vert P_{kj}\Vert _{2,{\mathbb {D}}}^2= \sum _{n\in I_{kj}} \Vert p_{kjn}\Vert _{2,{\mathbb {D}}}^2 =\sum _{n\in I_{kj}} \int _{{\partial {\mathbb {D}}}}\int _0^1 |a_{kjn}z^nt^n|^2 dt d\sigma (z)=\sum _{n\in I_{kj}} \frac{|a_{kjn}|^2}{2n+1}. \end{aligned}$$

We conclude with the following estimation

$$\begin{aligned} \displaystyle \sum _{k,j,n}\left( \frac{|a_{kjn}|^2}{2n+1}\right) ^s\ge & {} \displaystyle \sum _{k,j}\left( \sum _{n \in I_{kj}}\frac{|a_{kjn}|^2}{2n+1}\right) ^s=\displaystyle \sum _{k=1}^{\infty } \sum _{j=0}^{N_k} \Vert P_{kj}\Vert _{2,{\mathbb {D}}}^{2s}\\&{\mathop {>}\limits ^{(9)}}&\displaystyle \sum _{k=1}^{\infty } \left\| \varPhi - \mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right\| _{1,{\partial {\mathbb {D}}}}^{k(s-1)}\sum _{j=0} ^{N_k} \Vert P_{kj}\Vert _{2,{\mathbb {D}}}^{2}\\= & {} \displaystyle \sum _{k=1}^{\infty } \left\| \varPhi - \mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right\| _{1,{\partial {\mathbb {D}}}}^{k(s-1)}\Vert Q_k\Vert _{2,{\mathbb {D}}}^{2} \\&{\mathop {>}\limits ^{(p5)}}&\displaystyle \sum _{k=1}^{\infty }\frac{1}{2} \left\| \varPhi - \mathcal {R}^2\left( \sum _{m=0}^{k-1} Q_m\right) \right\| _{1,{\partial {\mathbb {D}}}}^{k(s-1)+1} \\\ge & {} \displaystyle \sum _{k>1/(1-s)}\frac{1}{2}=\infty \end{aligned}$$

for $s \in [0,1)$. This completes the proof. $\square$

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.

4 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 $\phi$ such that $G=e^\phi$. Solving a standard Dirichlet problem for $\phi$ we obtain a harmonic function $u_\phi \in \mathcal {C}(\overline{{\mathbb {D}}})$ such that $u_\phi = \phi$ on ${\partial {\mathbb {D}}}$. Then, the Schwarz integral for $u_\phi$ produces a function $\mathcal {S}(u_\phi ) \in \mathcal {O}({\mathbb {D}})$ such that ${\text {Re}}(\mathcal {S}(u_\phi ))\in \mathcal {C}(\overline{{\mathbb {D}}})$ and ${\text {Re}}(\mathcal {S}(u_\phi )) = \phi ~~\text {on} ~~{\partial {\mathbb {D}}}$ (see e.g. [3, 7]). Now it is enough to take $g:=e^{\mathcal {S}(u_\phi )}$. Here we observe that if $f,g \in \mathcal {O}({\mathbb {D}})$ are solutions of (1) and (2) respectively, then both functions |g| and $\mathcal {R}^p(f)$ are continuous up to the boundary ${\partial {\mathbb {D}}}$ 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 ${\partial {\mathbb {D}}}$ 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 $f(z):=\sum _{k=1}^{\infty } (n_kp+1)^{1/p}Q_k(z)z^{n_k}$, 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 $m \in {\mathbb {N}}$ there exist $R_m \in (0,1)$ and $n_m \in {\mathbb {N}}$ sufficiently large such that

$$\begin{aligned} \left| \sum _{k=1}^{m-1} (n_kp+1)^{1/p}Q_k(z)z^{n_k}\right| <(n_mp+1)^{1/p}R_m^{n_m}|Q_m(z)|\quad \text {for }z \in \partial {\mathbb {D}}_{R_m}. \end{aligned}$$

Then by Rouché’s theorem, polynomial $\sum _{k=1}^{m} (n_kp+1)^{1/p}Q_k(z)z^{n_k}$ has the same number of zeros as $(n_mp+1)^{1/p}Q_m(z)z^{n_m}$ in the disc ${\mathbb {D}}_{R_m}$. Choosing suitably increasing sequences $\{R_{m}\}_{m\in {\mathbb {N}}}\subset (0,1)$ and $\{n_{m}\}_{m\in {\mathbb {N}}} \subset {\mathbb {N}}$ we may guarantee that f has infinitely many zeros in ${\mathbb {D}}$.