On spectral properties of the modified convolution operator
© Jumabayeva et al.; licensee Springer 2012
Received: 16 December 2011
Accepted: 6 June 2012
Published: 25 June 2012
We investigated the s-number of the modified convolution operator and obtained the following results
where , , G is a set of all segments Q from , F is a set of all compacts from , is the measure of a set Q.
Recall that the sequence (s-numbers of operator A) are numerated eigenvalues of the operator .
where depends only on , , , .
Given that the eigenvalues of the operator coincide with the Fourier coefficients of the kernel K with respect to the trigonometric system, in the case this problem reduces to the well-known problem of Fourier series multipliers. Let and be the sequence of its Fourier coefficients with respect to the trigonometric system . It is assumed that K is such that , . Let . The problem is to determine conditions on the function f ensuring the boundedness of the operator .
We obtain sufficient conditions on a multiplier f ensuring that it belongs to the space . These conditions are expressed in terms of Lorentz and Besov spaces. We also construct examples showing the sharpness of the obtained constants for corresponding embedding theorems.
2 Main results
is a nonincreasing rearrangement of f.
In the following theorem the cases , are considered. The upper and the lower estimates of the norm () are obtained.
whereis the measure of a set Q.
We shall define the class of generalized monotone functions for which the upper and the lower estimates coincide.
holds. The class of such functions is denoted by .
In case parameters , are both either less or greater than 2, we use the space of smooth functions.
for . Here are the Fourier coefficients of the function f by trigonometric system , , and is the integer part of .
This class is called the Nikol’skii-Besov space.
Now consider the case .
Theorem 4 Let, , , , , .
3 Properties of class
where nonincreasing rearrangement of the sequence .
Lemma 1 (See )
Let and be the spaces obtained by the method of real interpolation of Banach pairs of spaces , respectively.
Lemma 2 (See )
Here, , , , .
and we have to describe the class of those functions f with Fourier coefficients , for which Inequality (1) holds.
- (1)Let, , , . Then
- (2)Let, , , then
Proof The proof of the first statement follows from Remark and from the fact that , where is a complex conjugate of the function f. Now we prove (2).
and the condition implies the condition , where .
The proof is complete. ?
By (2), in particular, the following proposition follows.
where and are any.
4 Proof of main results
In order to prove our main result, we need two lemmas in .
Lemma 3 (Bilinear interpolation, the complex method, see )
where, , are the spaces obtained by the method of complex interpolation of Banach pairs of spaces, , respectively.
Lemma 4 (Bilinear interpolation, the real method, see )
where are Fourier coefficients of the function f.
where , i.e., .
where , , , i.e., , .
Since the parameters , , are arbitrary in Inequality (5), it guarantees the arbitrary of the corresponding parameters in Inequality (4).
where are Fourier coefficients of the function f and , . According to Remark, this inequality is equivalent to the statement of Theorem 1. ?
where constants c and depend only on parameters p and q.
The proof of obtaining an upper estimate follows from Theorem 1 and the embedding , for .
Proof of Corollary 1 Let Q be an arbitrary compact from F.
Proof of Theorem 3 Let . For a sequence of numbers and a function we consider the mapping T of the form , where is the sequence of Fourier coefficients on the trigonometric system of functions f. This map is bilinear and from Karadzhov’s theorem  and Remark it follows that it is bounded from to .
i.e., T is a bounded mapping from to , where , , . The arbitrary choice of parameters guarantees the arbitrary of the parameters available in the theorem. ?
The case follows from the statements proved above and the property .
where , , .
where , , .
To complete the proof we fix the function f and the parameters r, s, a and we choose the parameters , , satisfying (11). We use Lemma 2 to get . ?
The case , as in the proof of Theorem 3, will follow from .
5 Examples demonstrating the sharpness of the results
Proposition 1 Let, , . If, then for anythere existssuch that, ifthere existssuch that.
, and therefore . Since Fourier coefficients of are the sequence it follows that .
Hence , and therefore . At the same time taking into account the monotonicity of the sequence and Hardy-Littlewood theorem, we have that . The statement is proved. ?
Theorem 6 Let, , . Then for anythere existandsuch that, .
It is obvious that , and belongs to .
and consequently, . Therefore .
. The proof that , is similar to the proof of the first part. ?
The authors thank the referees for their careful reading and valuable comments on how to improve the article. This paper was partly supported by the grant MTM 2011-27637.
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