1. Main Result
The aim of this paper is to find the general conditions for a complete biorthogonal conjugate system to form a Riesz basis, following the results obtained by Bari [1], Christensen [2], Sarsenbi with coauthors [3–5], San Antolin and Zalik [6], and Guo [7].

Let {un(x)}n=1∞ and {vn(x)}n=1∞ be a complete biorthogonal conjugate system of functions from L2(0,1) space.

By system coefficient space {un(x)}n=1∞ we denote the space X(u) of all the numeric sequences a={an}n=1∞ such that the series ∑n=1∞anun(x) converges in L2(0,1). It is evident that coefficient space X(u) is complete under the norm ∥a∥X(u)=supn∈N∥∑k=1nakuk(x)∥, and a natural basis ɛi={δij}j=1∞, i∈N, where δij is a Kronecker delta, forms a X(u) space basis.

A Banach coordinate space X of numeric sequences a={an}n=1∞ is said to be solid if b∈X follows from a∈X and |bn|≤|an|, n∈N (the inequality ∥b∥X≤∥a∥X, as it is put by the precise definition, is not required here).

It is clear that X(u) space is solid if natural basis is an unconditional basis for X(u). The latter follows from unconditional basicity for a system {un(x)}n=1∞.

Theorem 1.
Let {un(x)}n=1∞ and {vn(x)}n=1∞ be a complete biorthogonal conjugate system of functions that is uniformly bounded:
(1)∫01|un(x)|2dx≤C, ∫01|vn(x)|2dx≤C, n∈N.
Let there be given coefficient spaces X(u) and X(v) which are both solid. Then {un(x)}n=1∞ and {vn(x)}n=1∞ system form a Riesz basis.

Proof.
We consider the series ∑n=1∞anun(x) for a numeric sequence {an}n=1∞∈l2 and show that series converges for almost all choices of signs, that is, series
(2)∑n=1∞anrn(t)un(x),
where {rn(t)}n=1∞ is the Rademacher system and converges for almost all t∈[0,1] in L2 metrics by variable x (e.g., [8, Chapter 2]).

We use the results from [8] claiming that convergence of series ∑n=1∞fn(x) for almost all choices of signs is equivalent to
(3)(∑n=1∞|fn(x)|2)1/2∈L2(0,1).
For the series considered ∑n=1∞anun(x), by Levi’s theorem we have
(4)∫01∑n=1∞|anun(x)|2dx=∑n=1∞|an|2∫01|un(x)|2dx≤C∑n=1∞|an|2<∞,
meaning
(5)(∑n=1∞|anun(x)|2)1/2∈L2(0,1).
Convergence of the series ∑n=1∞anun(x) for almost all choices of signs is shown.

Now take a fixed t0∈[0,1] such that the series ∑n=1∞anrn(t0)un(x) converges in L2(0,1) space. By the solidity condition for coefficient space X(u) in L2(0,1), the series ∑n=1∞anun(x) converges, too.

Thus for any numeric sequence a={an}n=1∞∈l2 the series ∑n=1∞anun(x) converges in L2(0,1). Then the following equivalent inequalities are satisfied:
(6)∥∑n=1∞anun∥2≤B∑n=1∞|an|2, {an}n=1∞∈l2,∑n=1∞|f,un|2≤B∥f∥2, f∈L2(0,1).
This means that {un(x)}n=1∞ is Bessel system.

Besselian property for a system {vn(x)}n=1∞ is proved in the same way. It is clear that Besselian property for both biorthogonal conjugate systems {un(x)}n=1∞ and {vn(x)}n=1∞ implies the Riesz basicity for these systems.

Remark 2.
Note that in Theorem 1 we can replace the coefficient space X(u) with X(u)∩l2 and X(v) with X(v)∩l2.

2. Affine Riesz bases
Let function u:R→R have a support suppu⊂[0,1]. Using the representation n=2k+j, k=0,1,…, j=0,…,2k-1 for n∈N, we assume
(7)un(x)=uk,j(x)=2k/2u(2kx-j).
Besides, we suppose u0(x)=1, x∈[0,1]. System of functions {un(x)}n=0∞ is called an affine system generated by a function u. Here and elsewhere we assume
(8)u∈L2(0,1), ∫01u(x)dx=0.
Note that the classic example of an affine system of functions is the Haar wavelet {hn(x)}n=0∞ generated by the function
(9)h(x)={1,x∈[0,12),-1,x∈[12,1),0,x∉[0,1).
We enumerate the functions of Rademacher system {rk}k=0∞(10)rk=2-k/2∑j=02k-1hk,j, k=0,1,….
We suppose that an affine system {un(x)}n=0∞ generator u can be represented by Rademacher system
(11)u=∑k=0∞akrk, ∑k=0∞|ak|2<∞.
In this case we have the following completeness criterion for a system {un(x)}n=0∞. Let the function
(12)U(z)=∑k=0∞akzk, |z|<1,
be analytic in the unit disk with coefficients ak from (11).

Theorem 3 (see [<xref ref-type="bibr" rid="B9">9</xref>]).
A necessary and sufficient condition for an affine system {un(x)}n=0∞ to be complete in L2(0,1) space is that analytic function U(z) is outer function.

The following results are true for function u in the form (11).

Theorem 4.
System {vn(x)}n=0∞ that is biorthogonal conjugate to the affine system {un(x)}n=0∞ exists and is complete in L2(0,1) space if a0≠0.

Proof.
Suppose
(13)V(z)=1U(z)=∑k=0∞bkzk,
that is,
(14)a0b0=1, ∑ν=0kaνbk-ν=0, k≥1.
Then it follows from the results of [10] that
(15)vn=v(α1,…,αk)=∑ν=0k2-(k-ν)/2bk-νh(α1,…,αν),
where n∈N and n=2k+∑ν=1kαν2k-ν is binary expansion, h(α1,…,αν)=hm is the Haar function for m=2ν+∑μ=1ναμ2ν-μ, and v0(x)=1, x∈[0,1]. The explicit representation (15) shows that vn is a Haar polynomial of degree n. Hence it follows that the system {vn(x)}n=0∞ is complete.

Now we can formulate the Riesz basicity test for affine system {un(x)}n=0∞ with form (11) generator, based on Theorem 1.

Theorem 5.
Let analytic function U(z) have an absolutely convergent Taylor-series expansion
(16)∑k=0∞|ak|<∞
and U(z) does not vanish in the closed unit disk (|z|≤1). Then an affine system of functions {un(x)}n=0∞ forms a Riesz basis.

Proof.
By the conditions of the theorem, U(z) is outer function. By Theorem 3, an affine system {un(x)}n=0∞ is complete in L2(0,1) space. By Theorem 4, biorthogonal conjugate system {vn(x)}n=0∞ is complete, too.

Obviously, ∥un∥≤max{1,∥u∥}. From representation (15) we get
(17)∥vn∥2≤∑k=0∞2-k|bk|2<∞, n∈N.
We need to take into account that by Wiener theorem on absolutely convergent Taylor series we have
(18)∑k=0∞|bk|<∞.
Finally, from results of [11] it follows that X(u)∩l2=l2 and X(v)∩l2=l2, so all the conditions from Theorem 1 including the Remark are satisfied.