Basis properties of classical system of exponents

in Lebesgue spaces, are well studied (see e.g. [1-4]). N. K. Bari in her fundamental work [5] raised the issue of the existence of normalized basis in, which is not Riesz basis. The first example of this was given by K. I. Babenko [6]. He proved that the degenerate system of exponents

with forms a basis for, but is not Riesz basis when. This result has been extended by V. F. Gaposhkin [7]. In [8] the condition on the weight was found, which make the system

form a basis for the weight space with a norm

.

Similar problems are considered in [9-13]. Basis properties of a degenerate system of exponents are closely related to the similar properties of an ordinary system of exponents in corresponding weight space. In all the mentioned works the authors consider the cases, when the weight or the degenerate coefficient satisfies the Muckenhoupt condition (see, for example, [14]).

In this paper the basis properties of exponential systems with a degenerate coefficient are studied in the spaces, when the degenerate coefficient does not satisfy the Muckenhoupt condition. A similar problem was considered earlier in [15].

We consider a system of exponents

with a degenerate coefficient

where, are different points.

It is clear that, if and only if. Assume that the function

cancels the system out, that is

where is a complex conjugate. It is clear that, where. Then, it follows directly from (2) that a.e. on and, consequently, a.e. on. Thus, if

then system is complete in

.

Now consider the minimality of system in. If

then it is minimal in and system

is biorthogonal to it. So in this case

. Let. Consider the system

We have

It is clear that in the neighborhood of zero it holds

.

Consequently, the following representation

on is true.

From this representation directly follows that if

then the system (3) belongs to the space. Then from the relation (4) we obtain that the system is minimal in.

Consider the completeness of system

in. Let for some function we have

.

Since, then from this relation follows that

where is some constant. It is clear that

so. Consequently, , hence. As a result we obtain that under the following conditions

system (5) is complete and minimal in. Thus, the system (1) is complete, but it is not minimal in.

Consider the basicity of system (5) in. If the conditions

satisfies, then it is known that (see. e.g. [9-13]) system (1) forms basis for, and in the case it is complete and minimal in. Then it is clear that system (5) is minimal, but is not complete in.

Now, let the condition (6) holds. It is easy to see that the system

is biorthogonal to the system (5) in. Let us show that in this case the system (5) does not form a basis for. Let it forms a basis for.

At first consider the case. Then it is known that (see e.g. [16]) should fulfilled the following conditions

where is an arbitrary norm for. We have

.

Regarding biorthogonal system we get the following condition

So

where is a constant depending only on ( in sequel also). Choose as small as the interval does not contain the points. Then it is absolutely clear that:

.

Consequently

It is clear that for sufficiently great we have. Thus

so. And it contradict the condition (7).

Consider the case. In the absolutely same way as in the previous case, we get

.

Hence it directly follows that

.

Consider the case. We have

.

Let. Take. Consequently

.

In the sequel we should pay attention to the following identities

From these relations and from the fact that the product of cosines expressed in terms of cosines, it directly follows

where , are some constants. It is easy to see that. Taking into account the expressions above we have

.

It is clear that integrals converge. Then from the previous inequality follows that. Thus, the following theorem is true.

Theorem 1. Let the following condition be satisfied

.

Then the system forms a basis for. If the relation (6) holds, then this system is complete, but is not minimal in . In this case system (5) is complete and minimal in but does not form a basis for it.

The following theorem is also true.

Theorem 2. Let the conditions

be satisfied. Then the system is complete and minimal in, but does not form a basis in it. If the conditions

hold, then system (5) is complete and minimal in, but does not form a basis for it.

Proof. If the conditions (8) holds then the system is complete and minimal in. Indeed, it is clear that

.

Consequently, the system forms a basis for, and as a result it is complete in.

is a biorthogonal system to.

It is clear that the system belongs to

, and consequently, it is minimal in

. So the singular operator with the Hilbert kernel is not bounded in, then as a result it follows that this system does not form a basis for. If the conditions (9) hold then in the absolutely same way as in the previous case we establish that the system (5) is complete and minimal in, but does not form a basis for it. Consequently, the system

is complete, but is not minimal inand has a defect equal to 1.

The theorem is proved.

The authors are thankful to the referees for their valuable comments.