© Hindawi Publishing Corp. NEW VERSIONS OF THE NYMAN-BEURLING CRITERION FOR THE RIEMANN HYPOTHESIS

Let ρ(x)=x−[x], χ=χ(0,1), λ(x)=χ(x)logx, and M(x)=ΣK≤x μ(k), where μ is the Mobius function. Norms are in Lp(0,∞), 1 1/p is equivalent to ‖M1‖r<∞ for all r∈(1,p). The space ℬ is the linear space generated by the functions x↦ρ(θ/x) with θ∈(0,1]. Define Gn(x)=∫1/n1M1(θ)ρ(θ/x)θ−1 dθ. For all p∈(1,∞) we prove the following theorems: (I) ‖M1‖p<∞ implies λ∈ℬ¯Lp, and λ∈ℬ¯Lp implies ‖M1‖r<∞ for all r∈(1,p). (II) ‖Gn−λ‖p→0 implies ξ(s)≠0 in ℜs≥1/p, and ξ(s)≠0 in ℜs≥1/p implies ‖Gn−λ‖r→0 for all r∈(1,p).

Recall the usual arithmetical functions M and g given by where µ is the Möbius function.We will denote It is classical number theory that both M(x)x −1 → 0 and g(x) → 0 as x → ∞ are elementarily equivalent to the prime number theorem.A stronger, but still elementary, estimate is M(x) x(log x) −2 .(Heretofore elementary is to be understood in the traditional number theoretical sense, as no analytic function theory, no Fourier analysis.) We also define the less common γ and H p by H p (x) = x 1 M(t)t −2/p dt. (1.10) Summing (1.7) by parts we get and trivially from (1.12) 1.2.The weak Nyman-Beurling theorem.An easy consequence of Wiener's L 2 Tauberian theorem (cf.[19]) is that Ꮽ is dense in L 2 (0, ∞) (see [3]).Nyman [17] for L 2 and Beurling [9] for general L p obtained the much deeper result.
To prove this theorem, Beurling first noted that Ꮿ is dense in L p if and only if −χ ∈ Ꮿ Lp , then showed, quite simply, that −χ ∈ Ꮿ Lp implies ζ(s) ≠ 0 for s > 1/p.
The proof of the converse, which, in his own words, is less trivial, is by contradiction.
If −χ ∈ Ꮿ Lp , then Ꮿ is not dense in the space L p (0, 1).But this, by a highly involved functional analysis argument, implies the existence of a zero with real part greater than 1/p.Later proofs of this fact are illuminating, but just as difficult (see [7,8,11]).
The degree to which the apparent depth of the two sides of the proof is so starkly contrasting has led some authors to doubt the usefulness of the Nyman-Beurling approach (see, e.g., [15]), yet, it has led others to attempt to level off the two sides of the proof.
We say that φ is a generator if φ ∈ L p (0, ∞) for all p ∈ (1, ∞) and These generators are called strong generators in [3], and the term generator is applied when a is allowed to range in (0, ∞) in (1.13).−χ is the simplest example of a generator (the minus sign is immaterial, but more convenient).The function λ defined in (1.1) is also a generator since Clearly, any generator φ may well take the place of −χ in Theorem 1.1.These considerations, together with the fact that for every f ∈ Ꮾ as in (1.3), allow the following minor extension of the Nyman-Beurling theorem (Theorem 1.1), where reference to density of Ꮿ or Ꮾ is dropped.Direct proofs of this theorem for φ = −χ, not depending on deep functional analysis results, were achieved independently by Lee [14], and Balazard and Saias [6].These proofs only make use of standard number theoretical techniques.Thus Lee, perhaps not altogether inappropriately, calls the weak Nyman-Beurling theorem an arithmetical version of Beurling's theorem.The only if part of these proofs depends on identifying natural approximations f n , which we define as sequences in Ꮿ or Ꮾ, such that this weak implication holds for all p ∈ (1, ∞): Balazard and Saias [6] asked the following natural question.
Question 1.4.For a given specific natural approximation {f n }, is it true for some or all p ∈ (1, ∞) that the weak implication (1.16) can be substituted for the strong implication (1.17) We will answer this question, mostly in the negative, in Section 4. The first such natural approximation {B n } ⊂ Ꮾ appeared in [2] defined by .18)This sequence arises rather naturally in more than one way: it is the unique answer to the problem of finding f ∈ Ꮿ as in (1.3) with θ k = 1/k, and Or it can also be seen as a truncation of the fundamental identity referring to earlier results of Nikolski [16], took up the study of the L 2 case in quite some depth for a B n -related sequence {V n } ⊂ Ꮿ defined by (1.21) Vasyunin also conducted numerical studies leading him to state (sic) we can hardly hope that the series converges in the L 2 -norm.That this is indeed the case was first proved in [3].The sequence {S n } ⊂ Ꮾ defined by Furthermore B n is not a series, as defined in (4.24), while V n is the most natural series.Lee [14] proved Theorem 1.3 using V n , 1 < p ≤ 2, and, independently, Balazard and Saias [6] did likewise for B n and S n , 1 < p < ∞.
A further approximating sequence {F n } ⊂ Ꮿ promoted in [2] as the dual approximation, given by is of a different nature, as the θ k are uniformly distributed in (0, 1) as n → ∞.It is proved in [2] that F n + χ 1 → 0, and it can also be shown that F n (x) → −1 for 0 < x ≤ 1.The following question is however open.
Question 1.5.Is F n a natural approximation?

Description of main results.
It is classical knowledge that the Riemann hypothesis, an analytical statement about the location of the zeroes of the zeta function, is equivalent to purely arithmetical statements relative to the distribution of the prime numbers.A classical example is Littlewood's criterion: the Riemann hypothesis is equivalent to M(x) = O(x 1/2+ ) for all > 0. This criterion generalized to the half-planes s > 1/p leads to the following equivalence proved in Section 2.
Proposition 1.6.For any p ∈ (1, ∞) the following are equivalent: It is this condition (ii) on the Möbius function, which in effect, if not always in name, takes the place of the Riemann hypothesis in this paper.This together with their simple and quasi-elementary proofs given in Section 3 has led us to characterize the following two theorems as arithmetical versions of the Nyman-Beurling theorem, in contrast with which, furthermore, the proofs for both the if and the only if implications are of nearly equivalent difficulty.These two theorems stand somewhere between the strong Theorem 1.2 and the weak Theorem 1.3.Theorem 1.7 (arithmetical Nyman-Beurling's theorem I).The following statements are true for all p ∈ (1, ∞): Next we introduce a new "natural approximation" G n which will be seen to belong in every G n arises very naturally from an integrated form of the basic identity for the Möbius numbers which we will prove in Section 3, namely, The above formula can be seen as a formal proof of the Riemann hypothesis since the integral shows the generator λ as a limit of functions of Ꮾ.Then our second main result is the following theorem.

Theorem 1.8 (arithmetical Nyman-Beurling theorem II). The following statements are true for all
Finally, in Section 4, we explore the delicate gap between the weak and strong forms of the Nyman-Beurling theorem.We will show that all the natural approximations B n , V n , S n , F n , and G n diverge in L 2 .We also study the general L p case.The most interesting conclusion is this: if the Riemann hypothesis were not true, and 1/p = sup{ s | ζ(s) = 0}, then S n , V n , and G n would also diverge in L p provided there is a zero of real part 1/p.We have not decided the question for B n and F n .
2. Technical lemma and preliminary propositions.Throughout this section 1 < p < ∞.Some of the results herein may be part of the common folklore and/or stated in less general form than is possible.They are listed here however for the sake of completeness and readability.We thank A. M. Odlyzko for his generous help in these matters.

Technical lemma.
It is assumed that f is a locally bounded complex-valued function defined on [1, ∞), whose Mellin transform f , defined here as has a finite abscissa of convergence α = α f .

Lemma 2.1 (order lemma). If f (s) has a pole at
for some C ≠ 0. On the other hand the little o condition implies there is an A > 1 such that |f (x)| < (|C|/2)x σ 0 for x > A, so that splitting the right-hand side integral in (2.1) as In particular, f (x) changes sign an infinite number of times as x → ∞.
Proof.It is obviously enough to deal with only one of the above relations.So assume that (2.5) is false.Then for some > 0 there is a C such that is not singular at s = α, but clearly α is also the abscissa of convergence of the lefthand side integral above, which contradicts the theorem that the Laplace transform of a positive measure has a singularity on the real axis at the abscissa of convergence [22,Theorem 5.b].
Proof.It is obviously enough to consider that F ≥ 0. By hypothesis there exists some > 0, and an unbounded set (2.8) Now take an arbitrary x > 1.It is easy to see that there exists y ∈ E such that y > x and so that (2.12)

Some preliminary propositions
Proposition 2.4.The following Mellin transforms are valid at least in the halfplanes indicated: where ω(s) is analytic in s > 0.
Proof.As in Titchmarsh's monograph [20] we write for s > 1 (2.17) This proves (2.13).Proceed likewise with to obtain (2.14).Now using relation (1.12) between M(x), g(x), and γ(x) subtract the preceding two Mellin transforms to get (2.15).Finally, from definition (1.10) and the trivial |M(x)| ≤ x we deduce H p (x) x 2/q .Next note that H p is continuous and piecewise differentiable, which justifies the following integration by parts at least for s > 2/q: Now apply (2.13) to arrive at (2.16).
The following corollary is an immediate consequence of the above Mellin transforms, and the order and oscillation lemmas (Lemmas 2.1 and 2.2). [20]).Sharper results are of course known, for example, that the Mertens hypothesis is false, with M(x) oscillating beyond ± √

Corollary 2.5. Each one of the functions M(x), g(x), γ(x), H p (x), change sign infinitely often as x → ∞. Furthermore, if ζ(s) has some zero on the line
x.This was proven by Odlyzko and te Riele [18].Some further properties of γ(x) needed later are gathered here.
Lemma 2.7.The function γ satisfies: Remark 2.8.In [4] we showed that the existence of lim n→∞ γ(n) is elementarily equivalent to the prime number theorem.
Proof of Lemma 2.7.The prime number theorem and (1.11) yield (i). Decomposing the integral in (ii) in the intervals (k, k + 1) one gets (ii).Letting n → ∞ in (ii) yields (iii).The absolute convergence follows from the elementary estimate M(x) x(log x) −2 .
The result on H p (x) in Corollary 2.5 begets some important consequences for the norms of M 1 .
Proposition 2.9.If ζ(s) has a zero on the line s = 1/p, then (2.20) Remark 2.10.Note therefore that Proof of Proposition 2.9.Take F(x) := M(x)x −2/p .Then F(x) ≠ o(x 1/q ) by Corollary 2.5, so the divergent integral lemma (Lemma 2.3) yields Proof of Proposition 1.6.An extension of Littlewood's well-known criterion for the Riemann hypothesis is that condition (i) is equivalent to (see [10,Proposition IV.21]), so choose p with r < p < p, and then it is obvious how (i) implies (ii).Now we prove that not (i) implies not (ii).So assume there is an s 0 with s 0 = 1/p 1 > 1/p and ζ(s 0 ) = 0. Then by Proposition 2.9 Define the Riemann abscissa β by s. (2.27) Nothing is known beyond 1/2 ≤ β ≤ 1.We do know however that, on the one hand there are no zeroes on the line s = 1 and M 1 1 < ∞, since M(t) t(log t) −2 , and, on the other hand there are zeroes on the line s = 1/2 and M 1 2 = ∞.One could rightly ask the following question.Question 2.12.For β ∈ (1/2, 1), is it true that M 1 1/β < ∞ if and only if ζ(s) ≠ 0 for s = β? 3. Arithmetical versions of the Nyman-Beurling theorem.We define an operator T acting on all L p (0, ∞), p ∈ (1, ∞), by noting that the above integral converges absolutely for f ∈ L p (0, ∞) by Hölder's inequality.Now we show that T is of type (p, p).This does not follow, as could be expected, from the convolution form of the operator, on account of the difference between the measures dx and dx/x in (0, ∞).
Proof.Let f ≥ 0 and x > 0, then splitting the range of integration at x in (3.1) we get The result now emerges from the well-known, elementary Hardy inequalities (see [12,Theorems 327,328]).
The next result establishes the relevance of T for the Nyman-Beurling approach.
(3.3) Remark 3.3.For every f ∈ L p (0, ∞), T f is continuous, so the closure operation on the left-hand side above is necessary.However, for the purpose immediately at hand of proving Theorem 1.7 we only need then T f ∈ Ꮿ Lp .This is the case for f = M 1 by Lemma 2.7(iii).The Riemann sums s n (x) ∈ Ꮽ and s n (x) → T χ [a,b] (x) for each x > 0. Furthermore, it is trivial to see that s n (x) ≤ (b − a)/a for all x > 0, whereas s n (x) = (b − a)/x when x > b, so that

Proof of Proposition 3.2. Fix p ∈ (1, ∞). For any bounded interval
, and the above inequalities show it converges in L pnorm to the function ρ(α/x).If α = a the modification to the above proof is obvious.
We next introduce the essential, elementary identity.

.11)
Proof.Here we denote χ(S) = 1 if the statement S is true, otherwise χ(S) = 0. We start from the well-known elementary identity which we multiply by 1/t and integrate, thus (3.13) Proposition 3.6.For every x > 0, the following identity holds true as an absolutely convergent integral, without any assumptions on the L p -norms of M 1 : Proof.The upper limit of integration in (3.11) can trivially be substituted by ∞, so we get from Lemma 2.7(iii), and the (absolute) convergence of the last integral, again due to M(t) t(log t) −2 .Now make the change of variables t = 1/θ, and in the formula obtained substitute x 1/x.Remark 3.7.In [4] we show that the existence of is elementarily equivalent to the prime number theorem.

The proofs of the main theorems.
We now prove our first main result, Theorem 1.7.
Remark 3.8.The proof of (a) is elementary, and, interestingly, it corresponds to the hard necessity part of the Nyman-Beurling theorem (Theorem 1.2).Note however that the strong form of (a) is connected with the fact that the hypothesis implies by (2.21) that there are no zeroes of ζ(s) in s ≥ 1/p.On the other hand (b), a weak statement, corresponding to the easy sufficiency part of the Nyman-Beurling theorem (Theorem 1.2), is proved essentially by the traditional argument.
We now prove our second main result, Theorem 1.8.

Now let us prove statement (c). If ζ(s)
≠ 0 for s > 1/p, then, by Proposition 1.6, Lr → λ by the L p -continuity of T (Lemma 3.1).To prove (d) we proceed by contradiction.Assume there is s 0 such that ζ(s 0 ) = 0 and s 0 = 1/p 1 ≥ 1/p.Therefore γ(n) ≠ o(n −1/q 1 ) by Corollary 2.5.Now by definition (1.24) of G n and (ii) in Lemma 2.7 we have (3.17) so that Remark 3.9.The proof of (c), a weak statement corresponding to the hard necessity part of the Nyman-Beurling theorem (Theorem 1.2), is easy and quasi elementary.On the other hand the proof of (d), a strong statement, corresponding to the easy sufficiency part of the Nyman-Beurling theorem, is rather easy, but not elementary.Remark 3.10.At least, formally we can apply the operators to (c), and let h ↓ 0 to obtain the corresponding Balazard-Saias result for S n in [6].The difficulty in formalizing this argument stems from the fact that, for Ᏸ h as an operator from L p to itself, Ᏸ h → ∞ except when p = 1.A rigorous proof would be desirable.

Behavior of
G n also behaves nicely pointwise and in L 1 , as the original natural approximations.To see this we first need a lemma.
Proof of the lemma.For θ > n, we have (3.21) Proposition 3.12.G n satisfies the properties Proof.The first statement follows easily from the fact that the integral in (3.14) is absolutely convergent.Changing variables in the first iterated integral below, we get Now we split the outer integral on the right-hand side as The first one easily evaluates to |γ(n)| log n taking into account Lemma 2.7(ii) and (iii).This term converges to zero in view of (1.11) and an elementary error term in the prime number theorem.The second one is bounded by where we have applied in succession Fubini's theorem, Lemma 3.11, and an elementary error term for the prime number theorem of the form M(x) x(log x) −3 .

Divergence of approximations to λ.
The main result needs no proof as it is just the counterpositive of statement (d) in Theorem 1.8, namely: Remark 4.2.This proposition shows that in general the weak implication, (c) ζ(s) ≠ 0, s > 1/p implies G n − λ r → 0 for all r ∈ (1,p) in Theorem 1.8 cannot be made stronger to include r = p.The hypothesis of (c) can hold only if 1 ≤ p ≤ 2. Although we resolved at the outset to keep 1 < p < ∞, our resolve is weak, so we note that for p = 1 the strong version is true because of Proposition 3.12.For p = 2 the strong statement is definitely false for there are zeroes on s = 1/2.In the case 1 < p < 2 a simple logical analysis shows that the only interesting case is β = 1/p.Now, either there are roots on the line s = β, then the strong statement is false; or else, there are no roots on that line, then we can say nothing at present.This is related to Question 2.12.
Remark 4.3.By Corollary 2.5 there is a subsequence of zero crossings of γ(n) where clearly |γ(n)| < 1/n.For this subsequence the contradiction (3.18) would not hold.Thus the possibility remains open that there is a subsequence of G n that does converge.This peculiarity is common to all natural approximations discussed here.But there are reasons to believe this is may be a mirage.
To probe a little into the possible mirage we now bring to bear the existence of an isometry (it is actually a unitary operator, but that is not relevant here) of L 2 (0, ∞) denoted by U in [3] satisfying the following conditions, where we let ρ 1 (x) = ρ(1/x): UK a = K a U , (a > 0), (4.1) If we apply this to the Riemann sums of T f , when f is continuous of compact support, and make the obvious modifications to the reasoning in Lemma 3.1 and Proposition 3.2, we obtain the following lemma.
Moreover, the right-hand side defines a continuous extension to all L p (0, ∞).
Remark 4.5.At present we will use this lemma only in L 2 .It is however interesting to see how UT extends to all L p 's given the fact that U cannot be extended continuously to any L p other than for p = 2 (see [5]).When restricted to f ∈ L 2 (0, 1) the integral of the right-hand side of (4.5) is the Hilbert-Schmidt operator studied by Alcántara-Bode in [1] where it is shown at the outset that the Riemann Hypothesis is equivalent to the injectivity of this operator.Lemma 4.4 leads to the simple calculation which spells further trouble for the L 2 convergence of subsequences of G n .
Proposition 4.6.The divergence of G n in L 2 can be sharpened as follows: Remark 4.7.Since there are roots of ζ(s) on s = 1/2, neither n 1/2 γ(n) nor n 1/2 H 2 (n) converge to zero by Corollary 2.5, and most likely they are unbounded as n → ∞.However, optimism about almost periodicity of these functions may induce the idea that their zero crossings implied also by Corollary 2.5 will be close together an infinite number of times.
Proof of Proposition 4.6.That λ−G n 2 n 1/2 γ(n) is simply (3.17) for p = 2.For the second part we use (4.6), (4.8) A finer analysis of selected intervals in (1/n, ∞) seems likely to produce an infinite number of barriers increasing the lower bound in (4.8), so that one may be inclined to think that all subsequences of G n diverge in L 2 .
An even more natural looking approximation of λ is obtained by writing the simplest Riemann sum of the integral (3.14), namely Yet another approximation could be defined by truncation, say

.10)
We will not pursue this matter here either, but it seems to deserve some attention.

Divergence of approximations to −χ.
We may treat S n and V n together, defined in (1.22) and (1.21), since S n − V n 2 → 0. The following proposition is then the corresponding divergence result for S n .Proposition 4.9.If there is some zero of ζ(s) with real part 1/p then S n and V n diverge in L p .In particular S n and V n diverge in L 2 .
Proof.The hypothesis on the zero of ζ(s) implies, by Corollary 2.5, that Now assume by contradiction that S n converges in L p , so it must converge to −χ.On the other hand, noting that kx > 1 when x > 1/m and k > m, we get (4.12) Then letting n → ∞ we obtain Since the left-hand side goes to zero when m → ∞ this contradicts (4.11).
Remark 4.10.Proposition 4.9 implies that in general the weak implication of Balazard-Saias ((i) implies (vii) in [6], see also [14]) cannot be made stronger to include r = p.An analysis analogous to that carried out for G n in Remark 4.2 is also possible here.Mutatis mutandis the conclusions are the same.But a cautionary note is in order.We have not been able to treat the L p case for B n , other than for p = 1 or 2.
Remark 4.11.Again, the existence of a subsequence of zero crossings of g(n) given by Corollary 2.5 indicates that this subsequence is still a candidate in the running to converge in L p -norm to −χ.However as with G n we now prove a stricter failure for S n in the L 2 case.Proposition 4.12.There exists a constant C > 0 such that Proof.For p = 2, inequality (4.13) translates into On the other hand, if we apply U to S n we get and for some positive constant C.
Odlyzko and te Riele [18] have conjectured that in which case S n would not even be bounded in L 2 , endangering also the possibility of a strong version of condition (vi) in Balazard-Saias's work [6].On the other hand, by Corollary 2.5 there is a subsequence where M(n) = −2, and we know there is a subsequence where g(n) crosses zero, with g(n) ≤ 1/n.Nevertheless, as for G n , one may suspect that there is no L 2 -convergent subsequence of S n .The initial natural approximation B n is more resilient.We already remarked that it is not equivalent to S n , neither it is a series as defined in (4.24).The fact that B n (x) = −1 in (1/n, 1) destroys the possibility of using the same argument of Proposition 4.9.However with the help of the operator U we can dispose of the L 2 -case both for B n and F n .Analogous considerations as for S n apply in relation to the possibility of a subsequence of B n or of F n converging in L 2 .
To round off the presumption of divergence of the natural approximations in L 2 , we prove Proposition 4.14, a result, suggested by M. Balazard (personal communication), stating that no series of a certain kind in Ꮿ can converge to −χ in L 2 .
Denote by Ꮿ nat the subspace generated by the linearly independent functions {e k | k ≥ 2}, where We can now state the following proposition.
Proposition 4.14.No series in Ꮿ nat converges in L p (0, 1) to −χ if there is a zero of ζ(s) with real part 1/p.In particular, no series in Ꮿ nat converges to −χ in L 2 (0, 1).
To achieve the proof of this theorem we need a lemma.Assume by contradiction that χ + f n p → 0. For the step functions involved, this clearly implies pointwise convergence, then, from Lemma 4.15, we get c k = µ(k) for each k ≥ 2, which, by the way, forces (4.26) to hold for k = 1 too.But this immediately implies that f n = V n .However, V n diverges in L p by Proposition 4.9, so we have obtained a contradiction.

A
series in Ꮿ nat is defined as any sequence of type

Lemma 4 . 15 .Remark 4 . 16 .
Let f n be a sequence in Ꮿ nat converging pointwise to −χ.Assume f n is written asf n (x) = n k=1 a n,k ρ 1 kx , (4.25) then a n,j → µ(j), (n → ∞), (4.26)for every j ≥ 1.Proof.Each f n ∈ Ꮿ, so condition (1.4) implies that it is the right-continuous, step functionf n (x) = − j = 1, 2,.... (4.29)Now we proceed by induction.For j = 1 it is clear that (4.29) gives a n,1 → 1 = µ(1).Next assume for j > 1 that a n,k → µ(k) for 1 ≤ k ≤ j −1, then the limit in (4.29) yields desired a n,j → µ(j) as n → ∞.In some sense Lemma 4.15 shows the inevitability of the natural approximation S n .Proof of Proposition 4.14.We have triviallyf n (x) [21]a nontrivial zero-free half-plane if and only if χ + B n p → 0 for some p > 1, which conferred some legitimacy to the question.In related work Vasyunin[21], which led us to ask whether the strong or the weak implications, (1.16), and (1.17), were true for f n = B n , 1 < p ≤ 2. A mild positive answer was [2, Proposition 2.4] that ζ(s) .9) which happens to be a Beurling function in Ꮿ with an uncanny resemblance to the dual approximation F n defined by (1.23).But bear in mind that the integral (3.14) is not a proper Riemann integral, and we have not yet been able to show that R n is a natural approximation, in the sense that it satisfies a weak Beurling theorem such as Theorem 1.8, so we state the following true theorem without proof.