ON THE DEFICIENCY OF CERTAIN SETS OF EXPONENTIALS
For cer ta in functions £ € L([-1,1]) we show that the system iz t Ce ) Vidiere (ζ ) are the zeros of the Fourier transform of f nëN η η ® or a natura l ly re la ted function has f i n i t e deficiency in L 2 ( [ -1 ,1] ) . The deficiency is calculated. 42 A 65, 42 C 15, 42 C 30 1. A system of osc i l l a to r s can be analysed by decomposing a l l inputs and outpucs according to i t s e igenstates . If these are not known or i f no sa t i s fy ing mathematical model of the process in question does e x i s t , 348 Brüning Petzold as for example in the case of electroencephalography, one tries to find characteristic features of the output by using one of the standard transforms. One may ask, however, for procedures which are more naturally adapted to the output and do not impose artificial frequencies on the system. It is the purpose of this note to propose one such possibility. More precisely, if f ε L2([-1,1]) we try to decompose f according iznt to the system (e ^ e r e (^rPnâN a r e the zeros of the Fourier transform of f or some other function related to f. Obviously such a system cannot be complete in L2([-1,1]). Our observation is that under simple assumptions on f the deficiency can be calculated explicitly and turns out to be finite. This means that nonharmonic Fourier analysis (which is, of course, a well established theory, see e. g. [1],[5]) could be a useful tool in many questions of the above type. We wish to thank the referee whose valuable remarks greatly improved the presentation of this work. Cki 2. Throughout Sections 25 f denotes a function of class C ([-1,1]), k ä o, such that f(l^(-1) Φ 0 for some index i < k and f ^ d ) Φ 0 for some index j < k, where these derivatives are interpreted in the onesided sense. We denote by a the .smallest index i suchthat ί^(-1)Φθ and by b the smallest index j such that f ^ ( 1 ) Φ 0, and we set, by definition, ρ := minta,b}. Thus, ρ is the smallest index i such that f ^ M ) Φ 0 or f U )(1) Φ 0. According to a theorem of Pôlya [2] the Fourier transform 1 -izt F(z) := J e -at)dt 1 has infinitely many distinct zeros zn, η ε Ν. Let ζ have multiplicity n^ and put Deficiency of Certain Sets of Exponentials 349 q := Σ (mn-1) n=1 iz t which might be infinite. The system of exponentials (e η spans a proper subspace of L([-1,1]) the codimension of which is called the deficiency of the system. Our main result reads as follows: THEOREM Unden. the above hypothueA on f we have q < «> and the de.fiiz t ciency of (e n ) n g N ¿6 p+q+1. An important aspect of the Theorem is the fact that the ζeJWi, ate. not counted according to thein. rmttiplicity. In all previous results of this general type known to us Cef. [3]) the zeros are repeated, i. e. one studies the deficiency of the system . iz t Ct e n ϊ o^jSn^-l Our Theorem makes use of the pure exponential terms only. It is because of this feature that the theory of nonharmonic Fourier analysis is available to study the convergence, as explained in Section 6. 3. For the proof of the Theorem it suffices to establish the following two lemmas. LENMA 1 The ¿yitem (1) hai deficiency p+1. LENMA 2 The function F(z) in the Theotiyn hai only finitely many multiple zexoi. In passing from Lemna 1 to the Theorem we omit n^-1 terms in (1) for each zero of multiplicity n^ > 1. Hence the set of omitted terms spans a subspace of dimension q, vii ich is finite by Lemma 2, and the Theorem follows. 350 Brüning Petzold Before turning to the proof of the lemmas we make, two preliminary re-izt marks. First, since f(-t) is orthogonal to e whenever f(t) is orthoizt gonal to e , the completeness properties of the sets (e ) and -iz t Ce n ) are exactly the same. It is more convenient to deal with -iz t (e n ) in the sequel, and we shall do so. Second, if h e L([-1,1]), an elementary theorem of Titchmarsh([4], Lemma 2.2) gives 1 ; 1 (2) I e"h(t)dt I = 0(e ) ΐθ uniformly in 6 as r -» where ζ = re .We use this repeatedly with various choices of h. 4. Proof of Lemma 1 By partial integration (3) F(z) = eA(z) + e"BCz) + — } e~ f (t)dt (iz) -1 where A(z) and B(z) are polynomials in 1/z of the form (4) A(z) = + ..., B(z) = -¿L+ ..., αβ * 0, ζ ζ where the terms not written involve only higher powers of 1/z. . -iz t If g is orthogonal to (t· e n then o<j <11̂ -1 1 . -iz t J t e g(t)dt = 0, η e N, o<j<m-1 , 1 n that is, the Fourier transform G of g vanishes at all zeros of F and at least of the same order. By the Hadamard factorization theorem G(z) = F(z)H(z) where H if of order 1 and finite type. For fixed Θ, 0 < θ < π, we see from (2) and (3) that Deficiency of Certain Sets of Exponentials 351 |F( r e i e)| f o r r ^ J with some constant C > o, and by (2) again n r ίθ-, , r sin Θ-. G (re ) = o(e ) , r -» <» . These inequalities combine to give HCre^) = o(r^+^).An entirely similar argument with the roles of A and Β interchanged gives ίθ 3.+1 H(re ) = o(r ) as r <» and -π < θ < 0. The two inequalities together with the Phragmên-Lindelôf theorem show f i r s t that H is a polynomial, and second that the degree of this polynomial is at most p. Hence the deficiency is at most p+1. On the other hand by partial integration ( iz) jF(z) = } e " l z t f (J) ( t)dt, o í j <p . 1 Since the functions ( f ^ i t ) ) ^ ^ are obviously linearly independent this shows that the deficiency is at least p+1. Tliis proves Lemma 1. Q Let use not«uiat this result is consistent with Theorem 22 of [3] a f ' îg to which the deficiency is |m| i f m is the largest integer -ach that F(x)/(1+x' Z)m/2 does not belong to L2 OR). In the present case this largest integer is m = (1+p), and therefore the deficiency is 1+P. 5. Proof of Lemma 2 Since ? ( t ) := f ( t ) also satisf ies the assumptions of the Theorem, i t suffices to show that F(z) has only finitely many multiple zeros in the upper half plane, Im ζ > 0. Suppose then that Im ζ £ 0. I f the third term on the right in (3) is denoted by I(z) we have (5) F(z) = e lzA(z) + e" l zE(z) where E(z) = B(z) + e l z I ( z ) . For sufficiently large |z| clearly 352 Brüning Petzold