Letter to the Editor Comment on (On the Frame Properties of Degenerate System of Sines)

The proof of Theorem 3.1 of the paper “On the Frame Properties of Degenerate System of Sines” (see (Bilalov and Guliyeva, 2012)) published earlier in this journal contains a gap; the reasoning given there to prove this theorem is not enough to state the validity of the mentioned theorem. To overcome this shortage we state the most general fact on the completeness of sine system which implies in particular the validity of this fact. It is shown in this note that the system {ω(t)φn(t)}, where {φn(t)} is an exponential or trigonometric (cosine or sine) systems, becomes complete in the corresponding Lebesgue space Lp(−π, π) or Lp(0, π), respectively, whenever {ω(t)φn(t)} belongs to the corresponding Lebesgue space for all indices n (under the evident natural condition mes{t : ω(t) = 0} = 0). It is also shown that the same conclusion does not remain valid for, in general, any complete or complete orthonormal system {φn(t)}. Besides it, the largest class of functions ω(t) for which the system {ω(t) sin nt}n∈N is complete in Lp(0, π) space is determined.

The aim of this note is the determination of the largest class of functions () for which the system {()  ()}, where {  ()} is an exponential or trigonometric (cosine or sine) system, becomes complete in the corresponding Lebesgue space   (−, ) or   (0, ), respectively.
Note that the observation similar to one that is given before the proposition is not valid for the sine system; more precisely, () sin  ∈   (0, ),  ∈  does not always imply () ∈   (0, ).Therefore, the scheme used to prove the above proposition does not work in the case of the system {() sin } ∈ .This feature of sine system was overlooked in the proof of one of the results of the paper [1]; but the results of this note show that the statement of the mentioned result from [1] is true.
It should be mentioned that the proposition given above and the analogous result for sine system which will be proven in this note show that the system {()  ()}, where {  ()} is an exponential or trigonometric (cosine or sine) system, becomes complete in the corresponding Lebesgue space   (−, ) or   (0, ), respectively, whenever mes{ : () = 0} = 0 and {()  ()} belongs to the corresponding Lebesgue space for all indices .This conclusion may give rise to the impression that if the function () is such that mes{ : () = 0} = 0 and ()  () ∈   (, ) for all indices , where {  ()} is any complete system in   (, ), then the system {()  ()} is also complete in the space   (, ).The arguments given below show that, in general, this is not true.

Proof.
Necessity.Let the relation () sin  ∈   (0, ),  ∈  hold.Write the function () sin  in the following form: Consider an auxiliary function It is evident that the function Φ() is continuous and never vanishes at the segment [0, ].Therefore, there is a positive number  such that |Φ()| >  for all  ∈ [0, ].Using these inequalities in (1), we obtain the following estimation: This estimation implies ( − )() ∈   (0, ) since () sin  ∈   (0, ) by the condition of the Lemma.The necessity part of the lemma is proven.
Taking into account that the proof of the necessity part of this lemma relies only on the relation () sin  ∈   (0, ) and using the sufficiency part of the same lemma we obtain the validity of the following.Lemma 3. Let () be any measurable function on (0, ).The relation () sin  ∈   (0, ), ∀ ∈  is possible if and only if () sin  ∈   (0, ).