On the boundedness and norm of certain generalized Hilbert operators in ℓp$\ell ^p$

The Hilbert matrix Hn,m=(n+m+1)−1$\mathcal {H}_{n,m} = (n+m+1)^{-1}$ has been extensively studied in the previous literature. In this paper, we look at generalized Hilbert operators arising from measures on the interval [0,1], such that the Hilbert matrix is obtained by the Lebesgue measure. We provide a necessary and sufficient condition for these operators to be bounded in ℓp$\ell ^p$ and calculate their norm. Interestingly, we obtain complete results in this direction for both p=1,∞$p=1, \infty$ .

It was Hilbert who first proved the above for the special case  =  = 2 in one of his lectures on integral equations.His proof was published by Weyl (1908).The first proof for general ,  > 1 was given by Hardy and Riesz [3].Perhaps, the best reference for the proof is [5].A sharp version of Hilbert's inequality is the following: For (  ) ∞ =0 and (  ) ∞ =0 , both sequences of nonnegative numbers such that (  ) ∈    (1.1) The Hilbert operator and its associated Hilbert matrix are now introduced as follows: Consider the operator  on   given by The fact that this is a bounded operator on   for  > 1 follows from Hilbert's inequality as follows.
We recall that the norm of (  ) ∈    By fixing an arbitrary number  and taking the supremum over all sequences (  ) ∈   such that ‖(  )‖   = , we arrive at the desired result.The associated matrix acting on column vectors whose entries are terms of a sequence in   is another way to look at the operator (we shall not be making a notational distinction between  and its associated matrix).
Further historical notes and properties of the Hilbert matrix can be found in [1].Notice that each entry in  can be written as for ,  integers ranging from 0 to ∞.
The generalized Hilbert matrix can be viewed as an operator on   given by   ∋ (  ) ∞ =0 ↦ (  ) ∞ =0 , with Let us, finally, note that the corresponding question to the one addressed in this paper for the case of generalized Cesàro operators interested G.H. Hardy and was subsequently addressed in [4].The matrix of the standard Cesàro operator is as follows: . In Hardy's work [4], the corresponding entries of the generalized Cesàro operator were where  is such that the transformation is regular † (this, Hardy defines, means that  increases with , lim →0 () = (0) and (1) = 1).In the case  ≡ id [0,1] , the above gives the entries of .
He then goes on to provide a necessary and sufficient criterion for   to be bounded as an operator on   for 1 <  < ∞.The criterion is It is further shown that (), when finite, also equals the norm of the operator   .Throughout this work, our general approach significantly borrows from the ideas presented in [4].At the same time, however, we go beyond Hardy's treatment, in that we introduce the vocabulary required to characterize all finite, positive Borel measures that give rise to bounded operators in   .This, in particular, includes measures with arbitrarily many Dirac masses and forms the content of Section 2.1.† It is interesting to note that, while available at the time, the language and machinery of measure theory was not used in this work.

Preliminaries
In this subsection, we will be introducing a series of lemmata that will become relevant later on.
Proof.That the left-hand side converges (absolutely) is a direct consequence of the ratio test for series.One can use induction on .For  = 0, the result holds.Assume that the result holds for  = .Then we have Since both series above converge absolutely, their product is given by the series which is convergent by Mertens' theorem on the Cauchy product of series.Therefore, all one has to show is that ) , which itself can be shown by a straightforward induction, using the fact that holds for all  nonnegative and  ∈ [0, 1).
Proof.One can consider, for a fixed , the function Differentiating with respect to , we see that the only critical point is at  = 0, whence the result follows.□ Lemma 1.5.The equality holds for any  > 0 and  > 0.
Proof.As long as  is positive, the integral is well defined and convergent.Recall that Setting  =  , we have d = d  and the integral becomes The result follows.□

BOUNDEDNESS OF  𝝁 ON 𝓵 𝒑
In the current section, we shall prove two theorems regarding the boundedness of   as an operator on   spaces.The first concerns the case where there are no Dirac masses at 0 or 1, while the latter allows for the existence of such masses at the endpoints of [0,1].
Theorem 2.1.Fix 1 ⩽  ⩽ ∞.If  has no Dirac masses at 0 or 1, then the matrix   is bounded as an operator from   to itself if and only if Moreover, if that is the case, the operator norm equals    (here we understand that when  = ∞, the integral becomes ∫ d() is finite and there are no Dirac masses at 0 or 1.We will, in the first instance, show that   is bounded from   to   for 1 ⩽  < ∞ and we will calculate its norm.We will then treat the case  = ∞.
We begin by defining, for a given sequence (  ) ∞ =0 ∈   , the function   ∶ [0, 1] → ℝ given by We claim that   is well defined on [0,1] and that there holds, for  ∈ (0, 1), (2.2) To check this, we first look at the case  > 1.Here, Hölder's inequality gives where  =  −1 is the Hölder conjugate of , that is, 1  + 1  = 1.To check that the first term on the right-hand side in (2.3) is convergent, the ratio test for series gives as  → ∞.Hence, for  ∈ [0, 1), we have and therefore, the series is (absolutely) convergent.Finally, for  = 1, the partial sums ∑  =0   = 0, and hence, the series equals zero.This shows that for all  ∈ [0, 1],   () is well defined.
For  = 1, it is enough to show that, for a fixed  ∈ [0, 1] and for any fixed , the sequence   (; ) ∶= is in  ∞ .Consider a biased coin with probability of heads  and probability of tails 1 − .The probability of landing  heads in ( + ) throws for this coin is precisely   (; ).Hence, each term is bounded by 1 and the result follows.This proves that   () is well defined when (  ) ∈  1 .By Hölder's inequality again, we have (2.5) Summing over , we get (for where we have used Lemma 1.3.We now have By Minkowski's inequality, we have or equivalently, To show that    is the optimal constant (and hence the norm of the corresponding operator, when it is finite), we work as follows.Fix any  > 0 small.There exists a  > 0 such that (2.9) That such a  exists follows from the monotone convergence theorem, when applied, for example, to the functions   () = . We choose , ,  (in this order) such that for  ⩾ .• 0 <  < 1  .Moreover, denoting  = 1  + , choose  so that the following two conditions also hold: and such that (2.11) The latter condition can be met for  sufficiently small, since the -series diverges for  = 1 and converges for  > 1.
Finally, we pick the sequence   = ( + 1) − .Using Lemma 1.5, we have (2.12) Using this and (2.1), we can get the following estimate for   (): where the penultimate step was obtained by Lemma 1.4 and the last step from Lemma 1.5.When

𝑐 𝑛
⩽  ⩽ 1 −   , we therefore have (2.14) Using (2.14), we have (2.16)This establishes the "if" direction.For the "only if" direction, assume that   is bounded as an operator on   for 1 ⩽  < ∞.We claim that the integral    is finite.We shall make use of the same sequences as the ones just used.With the preceding notation, for an arbitrarily small but fixed  > 0, we have for the sequences described above: (2.17) and hence, ( ∑ (2.18) From here we observe, given the boundedness of   , that ∫ 1 0 (1 − ) −1  − d() is finite and uniformly bounded above by the operator norm as  → 0 .Moreover, it is not hard to establish that, as  → 0, the way  is chosen means that  → 0 as well and hence the functions pointwise as  → 0. Fix any  > 0 small.An application of the dominated convergence theorem on the interval [, 1 − ] (on this fixed interval, the functions (1 − )(1 − ) −1  − are uniformly bounded above) now implies that the integral is finite and bounded above by the operator norm.As  → 0, an application of the monotone convergence theorem gives the desired result.We finally examine the case  = ∞.Assume that ∫ 1 0 1 1− d() is finite and consider a sequence (  ) ∈  ∞ .We have (2.19) Taking the supremum over , we have (2.20) By taking (  ) to be a constant sequence different than the zero sequence, we see that the constant is optimal.Conversely, if the operator is bounded on  ∞ , picking any constant nonzero sequence (  ), we see that the integral ∫ 1 0 1 1− d() is finite and the result follows.□

2.1
The case where  is allowed to have dirac masses at 0 and/or 1 To allow for the possibility of Dirac masses at 0 or 1, we need to be a bit more careful with semantics and notation.
For a function g defined only on a subset  of [0,1], by a slight abuse of notation, we shall also use   (g) to denote the integral where   is the restriction of the measure on  in the obvious way.For a given measure  and for any function  on [0,1], the following decomposition holds: where ) are the Dirac masses of the measure at 0 and 1, respectively.This, in turn, provides the following decomposition for the matrix   : where  , ∶=  (0,1) (g , ()) and

Main result
In this section, we will show the following: Theorem 2.2.Assume that  contains Dirac masses  0 at 0 and  1 at 1, with | 0 | + | 1 | > 0. For 1 ⩽  < ∞, the only case when   can be a bounded operator on   is when  1 ≠ 0 =  0 and  = 1.
In that case, the operator is bounded if and only if is finite.Moreover, this expression, when finite, equals the norm of the operator.Finally, when  = ∞, the operator is bounded if and only if  0 ≠ 0 =  1 and the integral is finite.Moreover, the above, when finite, equals the norm of the operator on  ∞ .
Proof.We begin by establishing the following two lemmata.
Lemma 2.3.If the measure  contains a Dirac mass at 0, then the matrix is unbounded on   for 1 ⩽  < ∞.
This holds for the following reason.The matrix  for  0 ≠ 0 does not even map   to itself, since and clearly, as the sequence on the right-hand side eventually consists of the same nonzero (as long as  0 ≠ 0) constant, it cannot be in   .Finally, since the matrix  comprises only positive entries, the sum  +  also fails to map   to   .
Lemma 2.4.If the measure  contains a Dirac mass at 1 and no Dirac mass at 0, then the matrix   is unbounded as an operator on   for 1 <  < ∞.
To establish this, notice that the first term of the sequence on the right-hand side of (2.22) is  1 ( 0 +  1 + … ).In general, the   spaces are nested, in the sense that   ⊂   for  <  and hence for each  > 1, there exists a sequence in   that is not in  1 .Since  consists of only positive entries, it again follows that   does not map   to itself when  1 ≠ 0 =  0 and  > 1.
The only case left to examine, when the measure contains a Dirac mass, is when  1 ≠ 0 =  0 and  is viewed as an operator on  1 .In this case, we claim that the operator is bounded if and only if is finite and that, when finite, the expression above equals the operator norm.
Indeed, assume first that the expression above is finite.Proceeding as in (2.6), we get, for 0 <  < 1: To show that this is the optimal constant, we again use the sequences described in the previous section.With the same notation as before, with   = ( + 1) − , we have (2.27) Letting  → 0, we obtain the desired result.Finally, let us suppose that the operator   is bounded on  1 .The same argument as in the previous case shows the finiteness of ∫ (0,1] 1  d().We finally examine the case  = ∞.The proof is similar in spirit to the previous results, although a bit simpler.Indeed, if  1 ≠ 0, then it is clear from (2.21) that   does not map  ∞ to itself.
Assume that ∫ [0,1) 1 1− d() is finite and consider a sequence (  ) ∈  ∞ .We have ] ‖(  )‖  ∞ +  0   . (2.28) Taking the supremum of (2.28) over , we have (2.29) By taking (  ) to be a constant sequence different from the zero sequence, we see that the constant is optimal.Conversely, assuming that the operator is bounded from  ∞ to itself, again picking a nonzero constant sequence, we see that (2.30) From this, the finiteness of the operator norm implies the finiteness of and the result follows.□

Future problems
It is an interesting question to study such generalized Hilbert operators on different spaces.The natural next step, in the author's opinion, is to study   as operators on Hardy and/or Bergman spaces of analytic functions on the unit disk.

A C K N O W L E D G M E N T S
I would like to thank Aristomenis Siskakis for introducing me to this problem and for various beneficial suggestions regarding the format of the paper.I would also like to thank him, as well as Vasilis Daskalogiannis, Petros Galanopoulos, and Giorgos Stylogiannis for several fruitful discussions on the topic.Finally, I thank Vasilis for useful comments and suggestions on the introduction.

J O U R N A L I N F O R M AT I O N
The Bulletin of the London Mathematical Society is wholly owned and managed by the London Mathematical Society, a not-for-profit Charity registered with the UK Charity Commission.All surplus income from its publishing programme is used to support mathematicians and mathematics research in the form of research grants, conference grants, prizes, initiatives for early career researchers and the promotion of mathematics.

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)      d() +  0   = Let  be a finite, positive Borel measure on [0,1].We define the generalized Hilbert matrix   by d, where g , () ∶= ( +  ) (1 − )    .Motivated by this integral representation, in the current paper, we shall be looking at a class of generalized Hilbert operators, arising from replacing d with finite, positive Borel measures.† The definition is as follows.Definition 1.2.