May 31, 2026
Let \(\Lambda=\{\lambda_n\}_{n=1}^{\infty}\subset\mathbb{N}\) with \(\lambda_n\) strictly increasing and such that \(\sum_{n=1}^{\infty}\lambda_n^{-1}<\infty\). We show that a Hardy subspace \(H^2 (\mathbb{D}, \Lambda)\) consisting of functions with sparse Fourier spectrum \(\Lambda\) coincides with a Müntz space \(\overline{M^2_{\Lambda}}(\mathbb{D})\) characterized by square-summability of coefficients relative to a biorthogonal family. As consequences, we obtain a new characterization of the Hardy norm in \(H^2 (\mathbb{D}, \Lambda)\) and an integral representation formula for the Fourier coefficients. The proof uses the biorthogonal representation developed in [1].
2020 Mathematics Subject Classifications: 30H10, 41A30
Keywords: Hardy Spaces, Müntz spaces, sparse Fourier spectrum, biorthogonal systems
Let \(\mathbb{D}:=\{z:\, |z|<1\}\) be the unit disk and let \(H^2(\mathbb{D})\) be the Hardy space of functions analytic in \(\mathbb{D}\) such that \[\sup_{0<r<1} \int_{0}^{2\pi} |f(re^{i\theta})|^2\, d\theta<\infty.\] It is well known that an analytic function \(f(z)=\sum_{n=0}^{\infty} a_n z^n\) in \(\mathbb{D}\) belongs to the space \(H^2(\mathbb{D})\) if and only if \(\sum_{n=0}^{\infty} |a_n|^2<\infty\). The standard norm is \[||f||_{H^2} : = \left(\sum_{n=0}^{\infty} |a_n|^2\right)^{1/2}.\] The Taylor coefficients \(a_n\) are equal to the Fourier coefficients \[c_n=\frac{1}{2\pi}\int_{0}^{2\pi} f(e^{i\theta})\cdot e^{-i n\theta}\, d\theta,\qquad n=0, 1, 2, \dots\] of its boundary function \(f(e^{i\theta})\), and the Fourier coefficients \(c_n\) vanish for negative \(n\) ([2]).
For various properties of functions in \(H^p(\mathbb{D})\) spaces one may consult [2]–[5].
Remark 1. Consider a sequence \(\Lambda=\{\lambda_n\}_{n=1}^{\infty}\) which is a subset of the integer set \(\mathbb{N}\), with \(\lambda_n\) strictly increasing, and such that \[\label{convergence} \sum_{n=1}^{\infty} \frac{1}{\lambda_n}<\infty.\qquad{(1)}\] The main result of this paper is Theorem 2. We show that the Hardy subspace \(H^2 (\mathbb{D}, \Lambda)\) with sparse Fourier spectrum \(\Lambda\) is precisely the Müntz space \(\overline{M^2_{\Lambda}}(\mathbb{D})\) whose coefficient sequence \(\{\langle f, r_n\rangle_{L^2 (0,1)}\}\) is square-summable, where \(\{r_n\}_{n=1}^{\infty}\) is a family biorthogonal to the system \(\{t^{\lambda_n}\}_{n=1}^{\infty}\) in \(L^2(0,1)\). This creates a connection between Hardy space theory and the geometry of Müntz spaces, allowing techniques developed for Müntz spaces to be interpreted in terms of sparse Fourier expansions. In particular, the result provides a new characterization of the Hardy norm through biorthogonal coefficients as well as an integral representation formula for the Fourier coefficients.
We let \(H^2 (\mathbb{D}, \Lambda)\) consist of all the functions \(f\in H^2 (\mathbb{D})\) such that their Fourier coefficients vanish whenever \(n\notin \Lambda\). That is,
\[\begin{align} H^2(\mathbb{D}, \Lambda): & = \left\{f\in H^2(\mathbb{D}):\,\, f(z)=\sum_{n=1}^{\infty} c_{\lambda_n} z^{\lambda_n}\right\}\\ & = \left\{\text{f is analytic on \mathbb{D}}:\,\, f(z)=\sum_{n=1}^{\infty} c_{\lambda_n} z^{\lambda_n},\,\, \{c_{\lambda_n}\}\in \ell^2\}\right\}. \end{align}\]
In order to introduce the Müntz space \(\overline{M^2_{\Lambda}}(\mathbb{D})\), we first denote by \(M_{\Lambda}\) the system \(\{e_n(t): = t^{\lambda_n}\}_{n=1}^{\infty}\) and by \(\overline{M_{\Lambda}}\) the closed span of \(M_{\Lambda}\) in \(L^2 (0,1)\). Due to ?? , it follows from the Müntz-Szász theorem that \(\overline{M_{\Lambda}}\) is a proper subspace of \(L^2 (0,1)\). Moreover, as proved by Clarkson and Erdős (see [6] and [7]), any function \(f\in\overline{M_{\Lambda}}\) extends analytically in \(\mathbb{D}\) and it is represented as a power series of the form \[\label{analyticext} f(z)=\sum_{n=1}^{\infty} a_{\lambda_n} z^{\lambda_n}.\tag{1}\]
Remark 2. We denote the space of these analytic extensions by \(\overline{M_{\Lambda}}(\mathbb{D})\).
We point out that \(M_{\Lambda}\) is a minimal system in \(\overline{M_{\Lambda}}\), in other words, every element of \(M_{\Lambda}\) does not belong to the closed span of the remaining elements in \(L^2 (0,1)\) (see [7]). This minimal system has a unique biorthogonal family in \(\overline{M_{\Lambda}}\) , that is there exists a system of functions \(r_{\Lambda}:=\{r_n(t)\}_{n=1}^{\infty}\subset \overline{M_{\Lambda}}\) such that \[\langle e_n, r_m\rangle_{L^2 (0,1)} =\begin{cases} 1, & m=n, \\ 0, & m\not=n,\end{cases}\qquad \text{where}\qquad \langle f, g\rangle_{L^2 (0,1)}:=\int_{0}^{1} f(t)\overline{g(t)}\, dt.\]
In [1] we investigated the properties of the biorthogonal families \(M_{\Lambda}\) and \(r_{\Lambda}\) inside the space \(\overline{M_{\Lambda}}\). Motivated by the Clarkson-Erdős theorem, we proved that for each \(n\in\mathbb{N}\), the \(a_{\lambda_n}\) coefficient appearing in 1 is equal to the inner product \(\langle f, r_{n} \rangle_{L^2 (0,1)}\). Moreover, we showed that the family \(r_{\Lambda}\) is complete in \(\overline{M_{\Lambda}}\), in other words the closed span of \(r_{\Lambda}\) in \(\overline{M_{\Lambda}}\) is equal to \(\overline{M_{\Lambda}}\); hence the families \(r_{\Lambda}\) and \(M_{\Lambda}\) are Markushevich bases for the space \(\overline{M_{\Lambda}}\). We even proved that these families are strong Markushevich bases and applied the results to constructing compact operators on \(\overline{M_{\Lambda}}\) that admit spectral synthesis (see [1]).
A partial result from [1] is the following.
Theorem 1. Let \(\Lambda\subset\mathbb{N}\) such that \(\Lambda\) satisfies \((1.1)\). Then there exists a family of functions \[r_{\Lambda}=\{r_{n}:\,\, n\in\mathbb{N}\}\subset \overline{M_{\Lambda}}\] so that \(r_{\Lambda}\) is the unique biorthogonal sequence to \(M_{\Lambda}\) in the space \(\overline{M_{\Lambda}}\), and such that each function \(f\in \overline{M_{\Lambda}}\) extends as an analytic function in the unit disk \(\mathbb{D}\), hence its analytic extension belongs to \(\overline{M_{\Lambda}}(\mathbb{D})\), so that \[\label{representationf} f(z)=\sum_{n=1}^{\infty}\langle f, r_{n} \rangle_{L^2 (0,1)} z^{\lambda_n},\qquad{(2)}\] converging uniformly on compact subsets of \(\mathbb{D}\).
We now define the particular subspace of \(\overline{M_{\Lambda}}(\mathbb{D})\) which coincides with the space \(H^2(\mathbb{D}, \Lambda)\). Let \[\overline{M^2_{\Lambda}}(\mathbb{D}):=\left\{f\in \overline{M_{\Lambda}}(\mathbb{D}):\,\, \{\langle f, r_n\rangle_{L^2 (0,1)}\}\in \ell^2\right\}.\] Then the following is true. The proof is given in Section 4.
Theorem 2. Let \(\Lambda=\{\lambda_n\}_{n=1}^{\infty}\subset\mathbb{N}\) such that \(\Lambda\) satisfies ?? . Let \(r_{\Lambda}\) be the biorthogonal family to \(M_{\Lambda}\) as in Theorem 1. Then \[\label{equivalence} H^2(\mathbb{D}, \Lambda)=\overline{M^2_{\Lambda}}(\mathbb{D}).\qquad{(3)}\] Moreover \[\label{norms} ||f||_{H^2}^2=\sum_{n=1}^{\infty}\left| \langle f, r_n\rangle_{L^2 (0,1)} \right|^2.\qquad{(4)}\] In addition, the following holds for the Fourier \(c_{\lambda_n}\) coefficients of \(f\in H^2 (\mathbb{D}, \Lambda)\): for every fixed \(\theta\in [0,2\pi]\) \[\label{coeff} c_{\lambda_n}=\left(\int_{0}^{1} f(te^{i\theta})\cdot \overline{r_n(t)}\, dt\right)e^{-i\theta\lambda_n}\qquad n=1,2,\dots\qquad{(5)}\]
Now, suppose that the coefficients \(\{a_{\lambda_n}\}\) of \(f\) as in 1 belong to the \(\ell^2\) space: then clearly \(f\in H^2 (\mathbb{D}, \Lambda)\). We examine whether the converse implication holds.
“If \(f\in H^2 (\mathbb{D}, \Lambda)\), is it true that the restriction of \(f\) on the interval \([0,1)\) belongs to \(\overline{M_{\Lambda}}\)?”
The answer is affirmative; in fact it is a special case of the following result.
Theorem 3. Suppose that the Fourier coefficients of a function \(f\in H^2 (\mathbb{D})\) vanish for all \(n\notin A\) where \(A\subset \mathbb{N}\cup\{0\}\). Then the restriction of \(f\) on \([0,1)\) belongs to the closed span of the family \(\{t^n\}_{n\in A}\) in \(L^2 (0,1)\).
Proof. It is a combination of Lemmas 1 and 2. In Lemma 1 we prove that the restriction of \(f\) on \([0,1)\) belongs to the \(L^2 (0,1)\) space, whereas in Lemma 2 we show that if \(f\in L^2 (0,1)\) and \(f(t)=\sum_{n\in A}a_n t^n\) on \([0,1)\) with uniform convergence on compact subsets of \([0,1)\), then \(f\) belongs to the closed span of \(\{t^n\}_{n\in A}\) in \(L^2 (0,1)\). ◻
Next we ask whether the converse of Theorem 3 is true.
“If a function \(f\in L^2(0,1)\) belongs to the closed span of a family \(\{t^n\}_{n\in A}\) in \(L^2 (0,1)\) where \(A\subset
\mathbb{N}\cup\{0\}\), is it true that \(f\) extends analytically in \(\mathbb{D}\) and its extension belongs to the space \(H^2 (\mathbb{D})\), with
\(f\) admitting the representation \(f(z)=\sum_{n\in A} a_n z^{n}\) and \(\{a_n\}\in \ell^2\)?”
Clearly the answer is negative when \(\sum_{n\in A}1/n=\infty\) due to the Müntz-Szász theorem. If such a series diverges then the closed span of \(\{t^n\}_{n\in A}\) is equal to \(L^2(0,1)\).
But what about if \(\sum_{n\in A}1/n<\infty\)? The answer is negative in general, as the following result demonstrates whose proof is given in Section 3.
Theorem 4. Let \(\Lambda=\{\lambda_n\}_{n=1}^{\infty}\) be a lacunary sequence of natural numbers, that is, there exists some \(q>1\) so that \(\lambda_{n+1}/\lambda_n>q\) for all \(n\in\mathbb{N}\). Then \[g(z):=\sum_{n=1}^{\infty} z^{\lambda_n}\] is analytic in \(\mathbb{D}\) but clearly \(g\notin H^2 (\mathbb{D})\). However, \(g(t)\) belongs to \(L^2 (0,1)\) and moreover \(g(t)\) belongs to the closed span of \(\{t^{\lambda_n}\}_{n=1}^{\infty}\) in \(L^2 (0,1)\). Hence, by definition, \(g\) belongs to the space \(\overline{M_{\Lambda}}(\mathbb{D})\).
Therefore, even in the case when \(\Lambda\subset\mathbb{N}\) satisfying \((1.1)\), the condition that a function \(f\) belongs to the space \(\overline{M_{\Lambda}}\) does not guarantee alone that the analytic extension of \(f\) in \(\mathbb{D}\) belongs to \(H^2 (\mathbb{D})\). In other words, the space \(\overline{M_{\Lambda}}(\mathbb{D})\) is not a subspace of \(H^2 (\mathbb{D}, \Lambda)\).
Remark 3. This clearly changes if \(f\in \overline{M^2_{\Lambda}}(\mathbb{D})\).
Let \[f(z)=\sum_{n=0}^{\infty} c_n z^n\] be a function in the Hardy space \(H^2(\mathbb{D})\). We study the radial integrals \[\int_0^1 |f(te^{i\theta})|^2\, dt\] for a fixed \(\theta\in[0,2\pi]\).
Although radial limits of \(H^2\) functions need not exist for every \(\theta\), the result below shows that the radial function \[t \mapsto f(te^{i\theta})\] belongs to \(L^2(0,1)\) for each \(\theta\in [0, 2\pi]\) and its norm is uniformly controlled by the \(H^2\) norm of \(f\).
Lemma 1. If \(f\in H^2(\mathbb{D})\), then for every \(\theta\in[0,2\pi]\), \[\label{upperbound} \int_{0}^{1} |f(te^{i\theta})|^2\, dt \le \pi ||f||_{H^2}^2.\qquad{(6)}\]
Proof. We have \(f(z)=\sum_{n=0}^{\infty}c_n z^n\) with \(\{c_n\}\in\ell^2\). Fix \(\theta\) and write \[f(te^{i\theta}) = \sum_{n=0}^{\infty} c_ne^{i\theta n} t^n.\]
Hence for any \(\rho\in (0,1)\) we have \[\begin{align} \int_{0}^{\rho} |f(te^{i\theta})|^2\, dt & = \int_{0}^{\rho} \left|\sum_{n=0}^{\infty} c_ne^{i\theta n} t^n\right|^2 dt\\ & = \int_{0}^{\rho} \sum_{n=0}^{\infty} \sum_{k=0}^{\infty}c_n\overline{c_k}e^{i\theta (n-k)}t^{n+k}\, dt\\ & \le \sum_{n=0}^{\infty} \sum_{k=0}^{\infty}\frac{|c_n||c_k|}{n+k+1}. \end{align}\]
Since \(\{c_n\}\in\ell^2\) then the double series \[\sum_{n=0}^{\infty} \sum_{k=0}^{\infty}\frac{|c_n||c_k|}{n+k+1}\] converges as a special case of Hilbert’s Inequality [8] and in fact we have \[\sum_{n=0}^{\infty} \sum_{k=0}^{\infty}\frac{|c_n||c_k|}{n+k+1}\le \pi \sum_{n=0}^{\infty}|c_n|^2=\pi ||f||^2_{H^2}.\]
Combining the above shows that for any \(\rho\in (0,1)\) we have \[\int_{0}^{\rho} |f(te^{i\theta})|^2\, dt \le \pi ||f||_{H^2}^2.\] This uniform upper bound implies that ?? is true. ◻
The following result is known in the case when \(\Lambda=\{\lambda_n\}_{n=1}^{\infty}\) is a sequence of positive real numbers in an increasing order such that ?? holds (see [7] and [1]). We reprove the result as given by us in [1] \(\boldsymbol{w}ithout\) taking into assumption the convergence of the series in ?? . In other words, the result holds even when the series \(\sum_{n=1}^{\infty} 1/\lambda_n=\infty\). This \(extension\) is needed for proving Theorem 3.
Lemma 2. Let \(\Lambda=\{\lambda_n\}_{n=1}^{\infty}\subset\mathbb{N}\) with the \(\lambda_n\) in a strictly increasing order. Suppose that \(f\in L^2 (0,1)\) and \(f(t)=\sum_{n=1}^{\infty} c_n t^{\lambda_n}\) for \(t\in [0,1)\) with the series converging uniformly on compact subsets of \([0, 1)\). Then \(f\) belongs to the closed span of the family \(\{t^{\lambda_n}\}_{n=1}^{\infty}\) in \(L^2 (0,1)\).
Proof. Fatou’s Lemma and changing variables gives
\[\begin{align} \int_{0}^{1}|f(t)|^2\,dt\le \liminf_{\rho\to 1^-}\int_{0}^{1}|f(\rho t)|^2\,dt & \le \limsup_{\rho\to 1^-}\int_{0}^{1}|f(\rho t)|^2\,dt \\ & = \limsup_{\rho\to 1^-}\int_{0}^{\rho}|f(u)|^2\, \frac{du}{\rho}\\ & \le \limsup_{\rho\to 1^-}\frac{1}{\rho}\cdot \int_{0}^{1}|f(u)|^2\, du \\ & = \int_{0}^{1}|f(t)|^2\,dt. \end{align}\]
The above implies that \(\lim_{\rho\to 1^-}\int_{0}^{1}|f(\rho t)|^2\,dt\) exists and
\[\label{limit} \lim_{\rho\to 1^-}\int_{0}^{1}|f(\rho t)|^2\, dt = \int_{0}^{1}|f(t)|^2\, dt.\tag{2}\] Obviously one has \[|f(\rho t)-f(t)|^2\to 0\,\,\text{as}\,\, \rho\to 1^-\] and \[|f(\rho t)-f(t)|^2\le 2(|f(\rho t)|^2 + |f(t)|^2).\] Clearly \[2(|f(\rho t)|^2 + |f(t)|^2)\to 4|f(t)|^2\,\,\text{as}\,\, \rho\to 1^-,\] and from 2 we get \[\int_{0}^{1} 2(|f(\rho t)|^2 + |f(t)|^2)\, dt \to \int_{0}^{1} 4|f(t)|^2\, dt \,\,\text{as}\,\, \rho\to 1^-.\] It then follows from the \[\text{\it{Generalized Lebesgue Convergence Theorem}}\] that \[\lim_{\rho\to 1^-}\int_{0}^{1}|f(\rho t)-f(t)|^2\, dt = 0.\] Therefore, for every \(\epsilon>0\) there is some \(0<\delta_{\epsilon}<1\) so that for all \(\rho\in (\delta_{\epsilon}, 1)\) one has \[\label{epsilon} \int_{0}^{1}|f(\rho t)-f(t)|^2\, dt < \epsilon.\tag{3}\]
Next, for fixed \(\epsilon\), \(\delta_{\epsilon}\), as well as \(\rho \in (\delta_{\epsilon}, 1)\), the series
\[f(\rho t) = \sum_{n=1}^{\infty} c_{n} \cdot \rho^{\lambda_n}\cdot t^{\lambda_n}\] converges uniformly on the interval \([0,1]\). Hence, there is some positive integer \(N\), depending on \(\epsilon\) and \(\rho\), such that \[\int_{0}^{1}\left| f(\rho t) - \sum_{n=1}^{N} c_{n} \cdot \rho^{\lambda_n}\cdot t^{\lambda_n}\right|^2\, dt<\epsilon.\] Combining this with 3 and applying Minkowski’s inequality shows that \[\int_{0}^{1}\left| f(t) - \sum_{n=1}^{N} c_{n} \cdot \rho^{\lambda_n}\cdot t^{\lambda_n}\right|^2\, dt<2\epsilon.\] Clearly now we conclude that \(f\) belongs to the space \(\overline{M_{\Lambda}}\). ◻
We have \[|g(t)|^2=\left(\sum_{n=1}^{\infty} t^{\lambda_n}\right)\left(\sum_{k=1}^{\infty} t^{\lambda_k}\right)= \sum_{n=1}^{\infty}\sum_{k=1}^{\infty} t^{\lambda_n + \lambda_k}.\] Since \(\int_{0}^{1} t^{\lambda_n + \lambda_k}\, dt<1/(\lambda_n + \lambda_k)\) we need to evaluate the double series \[S=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{\lambda_n + \lambda_k}.\] If \(S\) is a positive finite number then for any \(\rho\in (0,1)\) we have \[\int_{0}^{\rho}|g(t)|^2=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\int_{0}^{\rho} t^{\lambda_n + \lambda_k} \, dt \le \sum_{n=1}^{\infty}\sum_{k=1}^{\infty} \frac{1}{\lambda_n + \lambda_k}=S<\infty\] and this shows that \(g\in L^2 (0,1)\).
To evaluate \(S\), split the double sum into the diagonal terms \(m=n\) and the off–diagonal terms: due to symmetry, we have
\[S=\sum_{n=1}^{\infty}\frac{1}{\lambda_n + \lambda_n} +2\sum_{n=1}^{\infty}\sum_{k=n+1}^{\infty}\frac{1}{\lambda_n + \lambda_k}.\]
We easily get
\[\sum_{n=1}^{\infty}\frac{1}{\lambda_n + \lambda_n} =\sum_{n=1}^{\infty}\frac{1}{2\lambda_n}<\infty.\]
Regarding \(\sum_{n=1}^{\infty}\sum_{k=n+1}^{\infty}\frac{1}{\lambda_n + \lambda_k}\), for each fixed \(n\) and \(k\ge n+1\) we write \(k=n+j\) for \(j=1,2,\dots\). Since \(\{\lambda_{n}\}\) is lacunary with \(\lambda_{n+1}/\lambda_n>q>1\) then \(\lambda_{n+j}>q^j\lambda_n\). Thus \(\lambda_n+\lambda_{n+j}>q^j\lambda_n\). Hence
\[\sum_{n=1}^{\infty}\sum_{k=n+1}^{\infty}\frac{1}{\lambda_n + \lambda_k} < \sum_{n=1}^{\infty}\sum_{j=1}^{\infty} \frac{1}{\lambda_n}\cdot q^{-j}=\left(\sum_{n=1}^{\infty}\frac{1}{\lambda_n}\right)\left(\sum_{j=1}^{\infty}q^{-j}\right)<\infty.\]
Combining the above shows that \(S\) is a positive finite number thus \(g\in L^2 (0,1)\). It then follows from Lemma 2 that \(g\) belongs to the closed span of \(\{t^{\lambda_n}\}_{n=1}^{\infty}\) in \(L^2 (0,1)\).
Suppose first that \(f(z)\in \overline{M^2_{\Lambda}}(\mathbb{D})\) thus by definition \[f(z)=\sum_{n=1}^{\infty} \langle f, r_n\rangle_{L^2 (0,1)} z^{\lambda_n}\] and \[\{\langle f, r_n\rangle_{L^2 (0,1)}\} \in\ell^2.\] Then obviously \(f\in H^2 (\mathbb{D}, \Lambda)\).
Suppose now that \(f\in H^2 (\mathbb{D}, \Lambda)\) thus \(f(z)=\sum_{n=1}^{\infty} c_{\lambda_n} z^{\lambda_n}\) and \(\{c_{\lambda_n}\}\in\ell^2\). From Lemma 1 we know that \(f(t)\) belongs to the space \(L^2 (0,1)\). It then follows from Lemma 2 that \(f(t)\in\overline{M_{\Lambda}}\), thus from ?? we have the series representation \(f(z)=\sum_{n=1}^{\infty}\langle f, r_n\rangle_{L^2 (0,1)} z^{\lambda_n}\). But the coefficients of a power series are unique, thus for each \(n\in\mathbb{N}\) we have \(c_{\lambda_n}=\langle f, r_n\rangle_{L^2 (0,1)}\). Since \(\{c_{\lambda_n}\}\in\ell^2\) then \(\{\langle f, r_n\rangle_{L^2 (0,1)}\}\in\ell^2\). Therefore we conclude that \(f(z)\in \overline{M^2_{\Lambda}}(\mathbb{D})\).
Hence ?? holds, we have \(H^2(\mathbb{D}, \Lambda)=\overline{M^2_{\Lambda}}(\mathbb{D})\).
Next, if \(f\in H^2 (\mathbb{D}, \Lambda)\) then \(||f||_{H^2}^2=\sum_{n=1}^{\infty}|c_{\lambda_n}|^2\). Since \(c_{\lambda_n}=\langle f, r_n\rangle_{L^2
(0,1)}\) then relation ?? is true.
Finally we deal with relation ?? . Let \(f\) belong to \(H^2 (\mathbb{D}, \Lambda)\) thus \(f(z)=\sum_{n=1}^{\infty} c_{\lambda_n} z^{\lambda_n}\). Fix some \(\theta\in [0,2\pi)\) so \(f(te^{i\theta})=\sum_{n=1}^{\infty} c_{\lambda_n}\cdot e^{i\lambda_n\theta}\cdot t^{\lambda_n}\). By Lemma 1 we know that \(f(te^{i\theta})\) belongs to the space \(L^2 (0,1)\). It follows from Lemma 2 that \(f(te^{i\theta})\) belongs to \(\overline{M_{\Lambda}}\), thus from ?? we have \[f(te^{i\theta})=\sum_{n=1}^{\infty}\left(\int_{0}^{1} f(te^{i\theta})\cdot \overline{r_n(t)}\, dt\right)\cdot t^{\lambda_n},\] with the series converging uniformly on compact subsets of \([0,1)\). By the uniqueness of coefficients of power series we conclude that \[c_{\lambda_n}\cdot e^{i\lambda_n\theta}=\int_{0}^{1} f(te^{i\theta})\cdot \overline{r_n(t)}\, dt,\qquad n=1,2,\dots.\] This proves ?? .
The proof of Theorem 2 is now complete.