Abstract

If 𝐹 is a continuous function on the real line and 𝑓=𝐹 is its distributional derivative, then the continuous primitive integral of distribution 𝑓 is 𝑏𝑎𝑓=𝐹(𝑏)𝐹(𝑎). This integral contains the Lebesgue, Henstock-Kurzweil, and wide Denjoy integrals. Under the Alexiewicz norm, the space of integrable distributions is a Banach space. We define the convolution 𝑓𝑔(𝑥)=𝑓(𝑥𝑦)𝑔(𝑦)𝑑𝑦 for 𝑓 an integrable distribution and 𝑔 a function of bounded variation or an 𝐿1 function. Usual properties of convolutions are shown to hold: commutativity, associativity, commutation with translation. For 𝑔 of bounded variation, 𝑓𝑔 is uniformly continuous and we have the estimate 𝑓𝑔𝑓𝑔𝒱, where 𝑓=sup𝐼|𝐼𝑓| is the Alexiewicz norm. This supremum is taken over all intervals 𝐼. When 𝑔𝐿1, the estimate is 𝑓𝑔𝑓𝑔1. There are results on differentiation and integration of convolutions. A type of Fubini theorem is proved for the continuous primitive integral.

1. Introduction and Notation

The convolution of two functions 𝑓 and 𝑔 on the real line is 𝑓𝑔(𝑥)=𝑓(𝑥𝑦)𝑔(𝑦)𝑑𝑦. Convolutions play an important role in pure and applied mathematics in Fourier analysis, approximation theory, differential equations, integral equations, and many other areas. In this paper, we consider convolutions for the continuous primitive integral. This integral extends the Lebesgue, Henstock-Kurzweil, and wide Denjoy integrals on the real line and has a very simple definition in terms of distributional derivatives.

Some of the main results for Lebesgue integral convolutions are that the convolution defines a Banach algebra on 𝐿1 and 𝐿1×𝐿1𝐿1 such that 𝑓𝑔1𝑓1𝑔1. The convolution is commutative, associative, and commutes with translations. If 𝑓𝐿1 and 𝑔𝐶𝑛, then 𝑓𝑔𝐶𝑛 and (𝑓𝑔)(𝑛)(𝑥)=𝑓𝑔(𝑛)(𝑥). Convolutions also have the approximation property that if 𝑓𝐿𝑝 (1𝑝<) and 𝑔𝐿1, then 𝑓𝑔𝑡𝑎𝑓𝑝0 as 𝑡0, where 𝑔𝑡(𝑥)=𝑔(𝑥/𝑡)/𝑡 and 𝑎=𝑔. When 𝑓 is bounded and continuous, there is a similar result for 𝑝=. For these results see, for example, [1]; see [2] for related results with the Henstock-Kurzweil integral. Using the Alexiewicz norm, all of these results have generalizations to continuous primitive integrals that are proven in what follows.

We now define the continuous primitive integral. For this, we need some notation for distributions. The space of test functions is 𝒟=𝐶𝑐()={𝜙𝜙𝐶()andsupp(𝜙)iscompact}. The support of function 𝜙 is the closure of the set on which 𝜙 does not vanish and is denoted supp(𝜙). Under usual pointwise operations, 𝒟 is a linear space over field . In 𝒟, we have a notion of convergence. If {𝜙𝑛}𝒟, then 𝜙𝑛0 as 𝑛 if there is a compact set 𝐾 such that for each 𝑛, supp(𝜙𝑛)𝐾, and for each 𝑚0, we have 𝜙𝑛(𝑚)0 uniformly on 𝐾 as 𝑛. The  distributions  are denoted 𝒟 and are the continuous linear functionals on 𝒟. For 𝑇𝒟 and 𝜙𝒟, we write 𝑇,𝜙. For 𝜙,𝜓𝒟 and 𝑎,𝑏, we have 𝑇,𝑎𝜙+𝑏𝜓=𝑎𝑇,𝜙+𝑏𝑇,𝜓. Moreover, if 𝜙𝑛0 in 𝒟, then 𝑇,𝜙𝑛0 in . Linear operations are defined in 𝒟 by 𝑎𝑆+𝑏𝑇,𝜙=𝑎𝑆,𝜙+𝑏𝑇,𝜙 for 𝑆,𝑇𝒟; 𝑎,𝑏 and 𝜙𝒟. If 𝑓𝐿1loc, then 𝑇𝑓,𝜙=𝑓(𝑥)𝜙(𝑥)𝑑𝑥 defines a distribution 𝑇𝑓𝒟. The integral exists as a Lebesgue integral. All distributions have derivatives of all orders that are themselves distributions. For 𝑇𝒟 and 𝜙𝒟, the  distributional derivative of 𝑇 is 𝑇 where 𝑇,𝜙=𝑇,𝜙. This is also called the  weak derivative. If 𝑝 is a function that is differentiable in the pointwise sense at 𝑥, then we write its derivative as 𝑝(𝑥). If 𝑝 is a 𝐶 bijection such that 𝑝(𝑥)0 for  any 𝑥, then the composition  with  distribution 𝑇 is  defined  by 𝑇𝑝,𝜙=𝑇,(𝜙𝑝1)/(𝑝𝑝1) for all 𝜙𝒟. Translations are a special case. For 𝑥, define the  translation 𝜏𝑥 on  distribution 𝑇𝒟 by 𝜏𝑥𝑇,𝜙=𝑇,𝜏𝑥𝜙 for test function 𝜙𝒟, where 𝜏𝑥𝜙(𝑦)=𝜙(𝑦𝑥). All  of  the  results  on distributions we use can be found in [3].

The following Banach space will be of importance: 𝐶={𝐹𝐹𝐶0(),𝐹()=0,𝐹()}. We use the notation 𝐹()=lim𝑥𝐹(𝑥) and 𝐹()=lim𝑥𝐹(𝑥). The extended real line is denoted by =[,]. The space 𝐶 then consists of functions continuous on with a limit of 0 at . We denote the functions that are continuous on that have real limits at ± by 𝐶0(). Hence, 𝐶 is properly contained in 𝐶0(), which is itself properly contained in the space of uniformly continuous functions on . The space 𝐶 is a Banach space under the uniform norm; 𝐹=sup𝑥|𝐹(𝑥)|=max𝑥|𝐹(𝑥)| for 𝐹𝐶. The continuous primitive integral is defined by taking 𝐶 as the space of primitives. The space of integrable distributions is 𝒜𝐶={𝑓𝒟𝑓=𝐹for𝐹𝐶}. If 𝑓𝒜𝐶, then 𝑏𝑎𝑓=𝐹(𝑏)𝐹(𝑎) for 𝑎,𝑏. The distributional differential equation 𝑇=0 has only constant solutions so the primitive 𝐹𝐶 satisfying 𝐹=𝑓 is unique. Integrable distributions are then tempered and of order one. This integral, including a discussion of extensions to 𝑛, is described in [4]. A more general integral is obtained by taking the primitives to be regulated functions, that is, functions with a left and right limit at each point, see [5].

Examples of distributions in 𝒜𝐶 are 𝑇𝑓 for functions 𝑓 that have a finite Lebesgue, Henstock-Kurzweil, or wide Denjoy integral. We identify function 𝑓 with the distribution 𝑇𝑓. Pointwise function values can be recovered from 𝑇𝑓 at points of continuity of 𝑓 by evaluating the limit 𝑇𝑓,𝜙𝑛 for a delta sequence converging to 𝑥. This is a sequence of test functions {𝜙𝑛}𝒟 such that for each 𝑛, 𝜙𝑛0, 𝜙𝑛=1, and the support of 𝜙𝑛 tends to {𝑥} as 𝑛. Note that if 𝐹𝐶0() is an increasing function with 𝐹(𝑥)=0 for almost all 𝑥, then the Lebesgue integral 𝑏𝑎𝐹(𝑥)𝑑𝑥=0 but 𝐹𝒜𝐶 and 𝑏𝑎𝐹=𝐹(𝑏)𝐹(𝑎). For another example of a distribution in 𝒜𝐶, let 𝐹𝐶0() be continuous and nowhere differentiable in the pointwise sense. Then 𝐹𝒜𝐶 and 𝑏𝑎𝐹=𝐹(𝑏)𝐹(𝑎) for all 𝑎,𝑏.

The space 𝒜𝐶 is a Banach space under the Alexiewicz norm; 𝑓=sup𝐼|𝐼𝑓|, where the supremum is taken over all intervals 𝐼. An equivalent norm is 𝑓=sup𝑥|𝑥𝑓|. The continuous primitive integral contains the Lebesgue, Henstock-Kurzweil, and wide Denjoy integrals since their primitives are continuous functions. These three spaces of functions are not complete under the Alexiewicz norm and in fact 𝒜𝐶 is their completion. The lack of a Banach space has hampered application of the Henstock-Kurzweil integral to problems outside of real analysis. As we will see in what follows, the Banach space 𝒜𝐶 is a suitable setting for applications of nonabsolute integration.

We will also need to use functions of bounded variation. Let 𝑔. The variation of 𝑔 is 𝑉𝑔=sup|𝑔(𝑥𝑖)𝑔(𝑦𝑖)| where the supremum is taken over all disjoint intervals {(𝑥𝑖,𝑦𝑖)}. The functions of bounded variation are denoted 𝒱={𝑔𝑉𝑔<}. This is a Banach space under the norm 𝑔𝒱=|𝑔()|+𝑉𝑔. Equivalent norms are 𝑔+𝑉𝑔 and |𝑔(𝑎)|+𝑉𝑔 for each 𝑎. Functions of bounded variation have a left and right limit at each point in and limits at ±, so, as above, we will define 𝑔(±)=lim𝑥±𝑔(𝑥).

If 𝑔𝐿1loc, then the essential variation of 𝑔 is essvar𝑔=sup𝑔𝜙, where the supremum is taken over all 𝜙𝒟 with 𝜙1. Then 𝒱={𝑔𝐿1locessvar𝑔<}. This is a Banach space under the norm 𝑔𝒱=esssup|𝑔|+essvar𝑔. Let 0𝛾1. For 𝑔, define 𝑔𝛾(𝑥)=(1𝛾)𝑔(𝑥)+𝛾𝑔(𝑥+). For left continuity, 𝛾=0 and for right continuity 𝛾=1. The functions  of  normalized bounded variation are 𝒩𝒱𝛾={𝑔𝛾𝑔𝒱}. If 𝑔𝒱, then essvar𝑔=inf𝑉 such that =𝑔 almost everywhere. For each 0𝛾1, there is exactly one function 𝒩𝒱𝛾 such that 𝑔= almost everywhere. In this case, essvar𝑔=𝑉. Changing 𝑔 on a set of measure zero does not affect its essential variation. Each function of essential bounded variation has a distributional derivative that is a signed Radon measure. This will be denoted 𝜇𝑔 where 𝑔,𝜙=𝑔,𝜙=𝑔𝜙=𝜙𝑑𝜇𝑔 for all 𝜙𝒟.

We will see that 𝒜𝐶×𝒱𝐶0() and that 𝑓𝑔𝑓𝑔𝒱. Similarly for 𝑔𝒱. Convolutions for 𝑓𝒜𝐶 and 𝑔𝐿1 will be defined using sequences in 𝒱𝐿1 that converge to 𝑔 in the 𝐿1 norm. It will be shown that 𝒜𝐶×𝐿1𝒜𝐶 and that 𝑓𝑔𝑓𝑔1.

Convolutions can be defined for distributions in several different ways.

Definition 1.1. Let 𝑆,𝑇𝒟 and 𝜙,𝜓𝒟. Define 𝜙(𝑥)=𝜙(𝑥): (i) 𝑇𝜓,𝜙=𝑇,𝜙𝜓, (ii) for each 𝑥, let 𝑇𝜓(𝑥)=𝑇,𝜏𝑥𝜓; (iii) 𝑆𝑇,𝜙=𝑆(𝑥),𝑇(𝑦),𝜙(𝑥+𝑦).

In (i), 𝒟×𝒟𝒟. This definition also applies to other spaces of test functions and their duals, such as the Schwartz space of rapidly decreasing functions or the compactly supported distributions. In (ii), 𝒟×𝒟𝐶. In [1], it is shown that definitions (i) and (ii) are equivalent. In (iii), 𝒟×𝒟𝒟. However, this definition requires restrictions on the supports of 𝑆 and 𝑇. It suffices that one of these distributions has compact support. Other conditions on the supports can be imposed (see [3, 6]). This definition is an instance of the tensor product, 𝑆𝑇,Φ=𝑆(𝑥),𝑇(𝑦),Φ(𝑥,𝑦), where now Φ𝒟(2).

Under (i), 𝑇𝜓 is in 𝐶. It satisfies (𝑇𝜓)𝜙=𝑇(𝜓𝜙), 𝜏𝑥(𝑇𝜓)=(𝜏𝑥𝑇)𝜓=𝑇(𝜏𝑥𝜓), and (𝑇𝜓)(𝑛)=𝑇𝜓(𝑛)=𝑇(𝑛)𝜓. Under (iii), with appropriate support restrictions, 𝑆𝑇 is in 𝒟. It is commutative and associative, commutes with translations, and satisfies (𝑆𝑇)(𝑛)=𝑆(𝑛)𝑇=𝑆𝑇(𝑛). It is weakly continuous in 𝒟, that is, if 𝑇𝑛𝑇 in 𝒟, then 𝑇𝑛𝜓𝑇𝜓 in 𝒟 see [1, 3, 6, 7] for additional properties of convolutions of distributions.

Although elements of 𝒜𝐶 are distributions, we show in this paper that their behavior as convolutions is more like that of integrable functions.

An appendix contains the proof of a type of Fubini theorem.

2. Convolution in 𝒜𝐶×𝒱

In this section, we prove basic results for the convolution when 𝑓𝒜𝐶 and 𝑔𝒱. Under these conditions, 𝑓𝑔 is commutative, continuous on , and commutes with translations. It can be estimated in the uniform norm in terms of the Alexiewicz and 𝒱 norms. There is also an associative property. We first need the result that 𝒱 forms the space of multipliers for 𝒜𝐶, that is, if 𝑓𝒜𝐶, then 𝑓𝑔𝒜𝐶 for all 𝑔𝒱. The integral 𝐼𝑓𝑔 is defined using the integration by parts formula in the appendix. The Hölder inequality (A.5) shows that 𝒱 is the dual space of 𝒜𝐶.

We define the convolution of 𝑓𝒜𝐶 and 𝑔𝒱 as 𝑓𝑔(𝑥)=(𝑓𝑟𝑥)𝑔, where 𝑟𝑥(𝑡)=𝑥𝑡. We write this as 𝑓𝑔(𝑥)=𝑓(𝑥𝑦)𝑔(𝑦)𝑑𝑦.

Theorem 2.1. Let 𝑓𝒜𝐶 and let 𝑔𝒱. Then (a)  𝑓𝑔 exists on .  (b) Let 𝑓𝑔=𝑔𝑓.  (c)  Let  𝑓𝑔|𝑓|inf|𝑔|+𝑓𝑉𝑔𝑓𝑔𝒱.   (d)   Assume 𝑓𝑔𝐶0(), lim𝑥±𝑓𝑔(𝑥)=𝑔(±)𝑓.   (e) If 𝐿1, then 𝑓(𝑔)=(𝑓𝑔)𝐶0().    (f) Let 𝑥,𝑧,   then 𝜏𝑧(𝑓𝑔)(𝑥)=(𝜏𝑧𝑓)𝑔(𝑥)=(𝑓𝜏𝑧𝑔)(𝑥).   (g)   For each  𝑓𝒜𝐶,   define Φ𝑓𝒱𝐶0() by Φ𝑓[𝑔]=𝑓𝑔.   Then Φ𝑓 is a bounded linear operator and Φ𝑓𝑓.   There exists a nonzero distribution  𝑓𝒜𝐶 such that Φ𝑓=𝑓. For each 𝑔𝒱, define Ψ𝑔𝒜𝐶𝐶0() by Ψ𝑔[𝑓]=𝑓𝑔. Then Ψ𝑔 is a bounded linear operator and Ψ𝑔𝑔𝒱. There exists a nonzero function 𝑔𝒱 such that Ψ𝑔=𝑔𝒱.   (h)  supp(𝑓𝑔)supp(𝑓)+supp(𝑔).

Proof. (a) Existence is given via the integration by parts formula (A.1) in the appendix. (b) See [4, Theorem  11] for a change of variables theorem that can be used with 𝑦𝑥𝑦. (c) This inequality follows from the Hölder inequality (A.5). (d) Let 𝑥,𝑡. From (c), we have ||||𝑓𝑔(𝑡)𝑓𝑔(𝑥)𝑓(𝑡)𝑓(𝑥)𝑔𝒱=𝑓(𝑡𝑥)𝑓()𝑔𝒱0as𝑡𝑥.(2.1) The last line follows from continuity in the Alexiewicz norm [4, Theorem  22]. Hence, 𝑓𝑔 is uniformly continuous on . Also, it follows that lim𝑥𝑓(𝑦)𝑔(𝑥𝑦)𝑑𝑦=𝑓(𝑦)lim𝑥𝑔(𝑥𝑦)𝑑𝑦=𝑔()𝑓. The limit 𝑥 can be taken under the integral sign since 𝑔(𝑥𝑦) is of uniform bounded variation, that is, 𝑉𝑦𝑔(𝑥𝑦)=𝑉𝑔. Theorem  22 in [4] then applies. Similarly, as 𝑥. (e) First show 𝑔𝒱. Let {(𝑠𝑖,𝑡𝑖)} be disjoint intervals in . Then ||𝑠𝑔𝑖𝑡𝑔𝑖||||𝑔𝑠𝑖𝑡𝑦𝑔𝑖||||||=𝑦(𝑦)𝑑𝑦||𝑔𝑠𝑖𝑡𝑦𝑔𝑖||||||𝑦(𝑦)𝑑𝑦.(2.2) Hence, 𝑉(𝑔)𝑉𝑔1. The interchange of sum and integral follows from the Fubini-Tonelli theorem. Now (d) shows 𝑓(𝑔)𝐶0(). Write 𝑓(𝑔)(𝑥)=𝑓(𝑦)=𝑔(𝑥𝑦𝑧)(𝑧)𝑑𝑧𝑑𝑦(𝑧)𝑓(𝑦)𝑔(𝑥𝑦𝑧)𝑑𝑦𝑑𝑧=(𝑓𝑔)(𝑥).(2.3) We can interchange orders of integration using Proposition A.3. For (ii) in Proposition A.3, the function 𝑧𝑉𝑦𝑔(𝑥𝑦𝑧)(𝑧)=𝑉𝑔(𝑧) is in 𝐿1 for each fixed 𝑥. Since 𝑔 is of bounded variation, it is bounded so |𝑔(𝑥𝑦𝑧)(𝑧)|𝑔|(𝑧)| and condition (iii) is satisfied. (f) This follows from a linear change of variables as in (a). (g) From (c), we have Φ𝑓=sup𝑔𝒱=1𝑓𝑔sup𝑔𝒱=1𝑓𝑔𝒱=𝑓. Let 𝑓>0 be in 𝐿1. If 𝑔=1, then 𝑔𝒱=1 and 𝑓𝑔(𝑥)=𝑓 so Φ𝑓=𝑓=𝑓1. To prove Ψ𝑔𝑔𝒱, note that Ψ𝑔=sup𝑓=1𝑓𝑔sup𝑓=1𝑓𝑔𝒱=𝑔𝒱. Let 𝑔=𝜒(0,). Then Ψ𝑔=sup𝑓=1𝑓𝑔=sup𝑓=1sup𝑥|𝑥𝑓|=1=𝑔𝒱. (h) Suppose 𝑥supp(𝑓)+supp(𝑔). Note that we can write 𝑓𝑔(𝑥)=𝑔(𝑥𝑦)𝑑𝐹(𝑦) in terms of a Henstock-Stieltjes integral, see [4] for details. This integral is approximated by Riemann sums 𝑁𝑛=1𝑔(𝑥𝑧𝑛)[𝐹(𝑡𝑛)𝐹(𝑡𝑛1)] where 𝑧𝑛[𝑡𝑛1,𝑡𝑛], =𝑡0<𝑡1<<𝑡𝑁= and there is a gauge function 𝛾 mapping to the open intervals in such that [𝑡𝑛1,𝑡𝑛]𝛾(𝑧𝑛). If 𝑧𝑛supp(𝑓), then since supp(𝑓) is open, there is an open interval 𝑧𝑛𝐼supp(𝑓). We can take 𝛾 such that [𝑡𝑛1,𝑡𝑛]𝐼 for all 1𝑛𝑁. Also, 𝐹 is constant on each interval in supp(𝑓). Therefore, 𝑔(𝑥𝑧𝑛)[𝐹(𝑡𝑛)𝐹(𝑡𝑛1)]=0 and only tags 𝑧𝑛supp(𝑓) can contribute to the Riemann sum. However, for all 𝑧𝑛supp(𝑓), we have 𝑥𝑧𝑛supp(𝑔) so 𝑔(𝑥𝑧𝑛)[𝐹(𝑡𝑛)𝐹(𝑡𝑛1)]=0. It follows that 𝑓𝑔(𝑥)=0.

Similar results are proven for 𝑓𝐿𝑝 in [1, Section  8.2].

If we use the equivalent norm 𝑓=sup𝑥|𝑥𝑓|, then Φ𝑓=𝑓. Also, integration by parts gives Φ𝑓𝑓. Now, given 𝑓𝒜𝐶, let 𝑔=𝜒(0,). Then 𝑔𝒱=1,  and 𝑓𝑔(𝑥)=𝑥𝑓. Hence, 𝑓𝑔=𝑓 and Φ𝑓=𝑓. We can have strict inequality in Ψ𝑔𝑔𝒱. For example, let 𝑔=𝜒{0},  then 𝑔𝒱=2 but integration by parts shows 𝑓𝑔=0 for each 𝑓𝒜𝐶.

Remark 2.2. If 𝑓𝒜𝐶 and 𝑔𝒱, one can use Definition A.2 to define 𝑓𝑔(𝑥)=𝑓𝑔𝛾(𝑥) where 𝑔𝛾=𝑔 almost everywhere and 𝑔𝛾𝒩𝒱𝛾. All of the results in Theorem 2.1 and the rest of this paper have analogues. Note that 𝑓𝑔(𝑥)=𝐹()𝑔𝛾()+𝐹𝜇𝑔.

Proposition 2.3. The three definitions of convolution for distributions in Definition 1.1 are compatible with 𝑓𝑔 for 𝑓𝒜𝐶 and 𝑔𝒱.

Proof. Let 𝑓𝒜𝐶, 𝑔𝒱, and 𝜙,𝜓𝒟. Definition 1.1(i) gives 𝑓,𝜓𝜙=𝑓(𝑥)𝜓(𝑦𝑥)𝜙(𝑦)𝑑𝑦𝑑𝑥=𝑓(𝑥)𝜓(𝑦𝑥)𝜙(𝑦)𝑑𝑥𝑑𝑦=𝑓𝜓,𝜙.(2.4) Since 𝜓𝒱 and 𝜙𝐿1, Proposition A.3 justifies the interchange of integrals. Definition 1.1(ii) gives 𝑓,𝜏𝑥𝜓=𝑓(𝑦)𝜓(𝑥𝑦)𝑑𝑦=𝑓𝜓(𝑥).(2.5) Definition 1.1(iii) gives 𝑓(𝑦),𝑔(𝑥),𝜙(𝑥+𝑦)=𝑓(𝑦)=𝑔(𝑥)𝜙(𝑥+𝑦)𝑑𝑥𝑑𝑦𝑓(𝑦)=𝑔(𝑥𝑦)𝜙(𝑥)𝑑𝑥𝑑𝑦𝜙(𝑥)𝑓(𝑦)𝑔(𝑥𝑦)𝑑𝑥𝑑𝑥=𝑓𝑔,𝜙.(2.6) The interchange of integrals is accomplished using Proposition A.3 since 𝑔𝒱 and 𝜙𝐿1.

The locally integrable distributions are defined as 𝒜𝐶(loc)={𝑓𝒟𝑓=𝐹forsome𝐹𝐶0()}. Let 𝑓𝒜𝐶(loc) and let 𝑔𝒱 with support in the compact interval [𝑎,𝑏]. By the Hake theorem [4, Theorem  25], 𝑓𝑔(𝑥) exists if and only if the limits of 𝛽𝛼𝑓(𝑥𝑦)𝑔(𝑦)𝑑𝑦 exist as 𝛼 and 𝛽. This gives

𝑓𝑔(𝑥)=𝑏𝑎𝑓(𝑥𝑦)𝑔(𝑦)𝑑𝑦=𝑥𝑎𝑥𝑏𝑓(𝑦)𝑔(𝑥𝑦)𝑑𝑦.(2.7)

There are analogues of the results in Theorem 2.1. For example, |𝑓𝑔(𝑥)||𝑥𝑎𝑥𝑏𝑓|inf[𝑎,𝑏]|𝑔|+𝑓𝜒[𝑥𝑏,𝑥𝑎]𝑉[𝑎,𝑏]𝑔. There are also versions where the supports are taken to be semi-infinite intervals.

We can also define the distributions with bounded primitive as 𝒜𝐶(𝑏𝑑)={𝑓𝒟𝑓=𝐹forsomebounded𝐹𝐶0()with𝐹(0)=0}. Let 𝑓𝒜𝐶(𝑏𝑑) and let 𝐹 be its unique primitive. If 𝑔𝒱 such that 𝑔(±)=0, then

𝑓𝑔(𝑥)=lim𝛼𝛽𝛽𝛼𝑓(𝑥𝑦)𝑔(𝑦)𝑑𝑦=lim𝛼𝛽𝐹(𝑥𝛼)𝑔(𝛼)𝐹(𝑥𝛽)𝑔(𝛽)+𝛽𝛼=𝐹(𝑥𝑦)𝑑𝑔(𝑦)𝐹(𝑥𝑦)𝑑𝑔(𝑦)=𝐹(𝑦)𝑑𝑔(𝑥𝑦).(2.8)

It follows that 𝑓𝑔𝐹𝑉𝑔.

It is possible to formulate other existence criteria. For example, if 𝑓(𝑥)=log|𝑥|sin(𝑥) and 𝑔(𝑥)=|𝑥|𝛼 for some 0<𝛼<1, then 𝑓 and 𝑔 are not in 𝒜𝐶, 𝒱 or 𝐿𝑝 for any 1𝑝 but 𝑓𝑔 exists on because 𝑓,𝑔𝐿1loc and if 𝐹(𝑥)=𝑥0𝑓, then lim|𝑥|𝐹(𝑥)𝑔(𝑥)=0.

The following example shows that 𝑓𝑔 needs not to be of bounded variation and hence not absolutely continuous. Let 𝑔=𝜒(0,). For 𝑓𝒜𝐶, we have 𝑓𝑔(𝑥)=𝑥𝑓=𝐹(𝑥), where 𝐹𝐶 is the primitive of 𝑓. However, 𝐹 needs not to be of bounded variation or even of local bounded variation. For example, let 𝑓(𝑥)=sin(𝑥2)2𝑥2cos(𝑥2) and let 𝐹 be its primitive in 𝐶. Finally, although 𝑓𝑔 is continuous, it needs not to be integrable over . For example, let 𝑔=1, then 𝑓𝑔(𝑥)=𝑓 and 𝑓𝑔 only exists if 𝑓=0.

3. Convolution in 𝒜𝐶×𝐿1

We now extend the convolution 𝑓𝑔 to 𝑓𝒜𝐶 and 𝑔𝐿1. Since there are functions in 𝐿1 that are not of bounded variation, there are distributions 𝑓𝒜𝐶 and functions 𝑔𝐿1 such that the integral 𝑓(𝑥𝑦)𝑔(𝑦)𝑑𝑦 does not exist. The convolution is then defined as the limit in of a sequence 𝑓𝑔𝑛 for 𝑔𝑛𝒱𝐿1 such that 𝑔𝑛𝑔 in the 𝐿1 norm. This is possible since 𝒱𝐿1 is dense in 𝐿1. We also give an equivalent definition using the fact that 𝐿1 is dense in 𝒜𝐶. Take a sequence {𝑓𝑛}𝐿1 such that 𝑓𝑛𝑓0. Then 𝑓𝑔 is the limit in of 𝑓𝑛𝑔. In this more general setting of convolution defined in 𝒜𝐶×𝐿1, we now have an Alexiewicz norm estimate for 𝑓𝑔 in terms of estimates of 𝑓 in the Alexiewicz norm and 𝑔 in the 𝐿1 norm. There is associativity with 𝐿1 functions and commutativity with translations.

Definition 3.1. Let 𝑓𝒜𝐶 and let 𝑔𝐿1. Let {𝑔𝑛}𝒱𝐿1 such that 𝑔𝑛𝑔10. Define 𝑓𝑔 as the unique element in 𝒜𝐶 such that 𝑓𝑔𝑛𝑓𝑔0.

To see that the definition makes sense, first note that 𝒱𝐿1 is dense in 𝐿1 since step functions are dense in 𝐿1. Hence, the required sequence {𝑔𝑛} exists. Let [𝛼,𝛽] be a compact interval. Let 𝐹𝐶 be the primitive of 𝑓. Then

𝛽𝛼𝑓𝑔𝑛(𝑥)𝑑𝑥=𝛽𝛼𝑓(𝑦)𝑔𝑛=(𝑥𝑦)𝑑𝑦𝑑𝑥𝑓(𝑦)𝛽𝛼𝑔𝑛(=𝑥𝑦)𝑑𝑥𝑑𝑦(3.1)𝑓(𝑦)𝛽𝑦𝛼𝑦𝑔𝑛(𝑥)𝑑𝑥𝑑𝑦=𝐹(𝑦)𝑑𝛽𝑦𝛼𝑦𝑔𝑛=(3.2)𝑔𝐹(𝑦)𝑛(𝛽𝑦)𝑔𝑛=(𝛼𝑦)𝑑𝑦𝛽𝑦𝛼𝑦𝑓𝑔𝑛(𝑦)𝑑𝑦.(3.3) The interchange of orders of integration in (3.1) is accomplished with Proposition A.3 using 𝑔(𝑥,𝑦)=𝑔𝑛(𝑥𝑦)𝜒[𝛼,𝛽](𝑥). Integration by parts gives (3.2) since lim𝑦𝛽𝑦𝛼𝑦𝑔𝑛=0. As 𝐹 is continuous and the function 𝑦𝛽𝑦𝛼𝑦𝑔𝑛 is absolutely continuous, we get (3.3). Taking the supremum over 𝛼,𝛽 gives

𝑓𝑔𝑛𝑔𝑓𝑛1.(3.4) We now have

𝑓𝑔𝑚𝑓𝑔𝑛=𝑔𝑓𝑚𝑔𝑛𝑔𝑓𝑚𝑔𝑛1(3.5)

and {𝑓𝑔𝑛} is a Cauchy sequence in 𝒜𝐶. Since 𝒜𝐶 is complete, this sequence has a limit in 𝒜𝐶 which we denote 𝑓𝑔. The definition does not depend on the choice of sequence {𝑔𝑛}, thus if {𝑛}𝒱𝐿1 such that 𝑛𝑔10, then 𝑓𝑔𝑛𝑓𝑛𝑓(𝑔𝑛𝑔1+𝑛𝑔1)0 as 𝑛. The previous calculation also shows that if 𝑔𝒱𝐿1, then the integral definition 𝑓𝑔(𝑥)=𝑓(𝑥𝑦)𝑔(𝑦)𝑑𝑦 and the limit definition agree.

Definition 3.2. Let 𝑓𝒜𝐶 and let 𝑔𝐿1. Let {𝑓𝑛}𝐿1 such that 𝑓𝑛𝑓0. Define 𝑓𝑔 as the unique element in 𝒜𝐶 such that 𝑓𝑛𝑔𝑓𝑔0.

To show this definition makes sense, first show 𝐿1 is dense in 𝒜𝐶.

Proposition 3.3. 𝐿1 is dense in 𝒜𝐶.

Proof. Let 𝐴𝐶() be the functions that are absolutely continuous on each compact interval and which are of bounded variation on the real line. Then, 𝑓𝐿1 if and only if there exists 𝐹𝐴𝐶() such that 𝐹(𝑥)=𝑓(𝑥) for almost all 𝑥. Let 𝑓𝒜𝐶 be given. Let 𝐹𝐶 be its primitive. For 𝜖>0, take 𝑀>0 such that |F(𝑥)|<𝜖 for 𝑥<𝑀 and |𝐹(𝑥)𝐹()|<𝜖 for 𝑥>𝑀. Due to the Weierstrass approximation theorem, there is a continuous function 𝑃 such that 𝑃(𝑥)=𝐹(𝑀) for 𝑥𝑀, 𝑃(𝑥)=𝐹(𝑀) for 𝑥𝑀, |𝑃(𝑥)𝐹(𝑥)|<𝜖 for |𝑥|𝑀 and 𝑃 is a polynomial on [𝑀,𝑀]. Hence, 𝑃𝐴𝐶() and 𝑃𝑓<3𝜖.

In Definition 3.2, the required sequence {𝑓𝑛}𝐿1 exists. Let [𝛼,𝛽] be a compact interval. Then, by the usual Fubini-Tonelli theorem in 𝐿1,

𝛽𝛼𝑓𝑛𝑔(𝑥)𝑑𝑥=𝛽𝛼𝑓𝑛=(𝑥𝑦)𝑔(𝑦)𝑑𝑦𝑑𝑥𝑔(𝑦)𝛽𝛼𝑓𝑛(𝑥𝑦)𝑑𝑥𝑑𝑦.(3.6)

Take the supremum over 𝛼,𝛽 and use the 𝐿1𝐿 Hölder inequality to get

𝑓𝑛𝑓𝑔𝑛𝑔1.(3.7) It now follows that {𝑓𝑛𝑔} is a Cauchy sequence. It then converges to an element of 𝒜𝐶. However, (3.7) also shows that this limit is independent of the choice of {𝑓𝑛}. To see that Definitions 3.1 and 3.2 agree, take {𝑓𝑛}𝐿1 with 𝑓𝑛𝑓0 and {𝑔𝑛}𝒱𝐿1 with 𝑔𝑛𝑔10. Then

𝑓𝑛𝑔𝑓𝑔𝑛=𝑓𝑛𝑔𝑓𝑔𝑓𝑛𝑓𝑔𝑛+𝑔𝑓𝑔𝑓𝑛𝑓𝑔𝑛𝑓𝑔1𝑔+𝑓𝑛𝑔1(3.8)

Letting 𝑛 shows that the limits of 𝑓𝑛𝑔 in Definition 3.2 and 𝑓𝑔𝑛 in Definition 3.1 are the same.

Theorem 3.4. Let 𝑓𝒜𝐶 and 𝑔𝐿1. Define 𝑓𝑔 as in Definition 3.1. Then (a)𝑓𝑔𝑓𝑔1.  (b)  Let 𝐿1.  Then (𝑓𝑔)=𝑓(𝑔)𝒜𝐶.  (c) For each 𝑧, 𝜏𝑧(𝑓𝑔)=(𝜏𝑧𝑓)𝑔=(𝑓𝜏𝑧𝑔).  (d) For each 𝑓𝒜𝐶, define Φ𝑓𝐿1𝒜𝐶 by Φ𝑓[𝑔]=𝑓𝑔. Then Φ𝑓 is a bounded linear operator and Φ𝑓𝑓. There exists a nonzero distribution 𝑓𝒜𝐶 such that Φ𝑓=𝑓. For each 𝑔𝐿1, define Ψ𝑔𝒜𝐶𝒜𝐶 by Ψ𝑔[𝑓]=𝑓𝑔. Then Ψ𝑔 is a bounded linear operator and Ψ𝑔𝑔1. There exists a nonzero function 𝑔𝐿1 such that Ψ𝑔=𝑔𝒱.  (e) Define 𝑔𝑡(𝑥)=𝑔(𝑥/𝑡)/𝑡 for 𝑡>0. We have 𝑎=𝑔𝑡(𝑥)𝑑𝑥=𝑔. Then 𝑓𝑔𝑡𝑎𝑓0 as 𝑡0.  (f)  Let supp(𝑓𝑔)supp(𝑓)+supp(𝑔).

Proof. Let {𝑔𝑛} be as in Definition 3.1. (a) Since 𝑓𝑔𝑛𝑓𝑔, (3.4) shows 𝑓𝑔𝑓𝑔1. (b) Let {𝑛}𝒱𝐿1 such that 𝑛10. Then (𝑓𝑔)=𝜉𝒜𝐶 such that (𝑓𝑔)𝑛𝜉0. Since 𝑔𝐿1, there is {𝑝𝑛}𝒱𝐿1 such that 𝑝𝑛𝑔10. Then 𝑓(𝑔)=𝜂𝒜𝐶 such that 𝑓𝑝𝑛𝜂0. Now, 𝜉𝜂(𝑓𝑔)𝑛+𝜉𝑓𝑝𝑛+𝜂(𝑓𝑔)𝑛𝑓𝑔𝑛𝑛+𝑓𝑔𝑛𝑛𝑓𝑝𝑛.(3.9) Using (3.4), (𝑓𝑔)𝑛𝑓𝑔𝑛𝑛=𝑓𝑔𝑔𝑛𝑛𝑔𝑓𝑛𝑔1𝑛10as𝑛.(3.10) Finally, use Theorem 2.1(e) and (3.4) to write 𝑓𝑔𝑛𝑛𝑓𝑝𝑛=𝑔𝑓𝑛𝑛𝑝𝑛𝑔𝑓𝑛𝑔1𝑛1+𝑔1𝑛1+𝑝𝑛𝑔10as𝑛.(3.11)(c) The Alexiewicz norm is invariant under translation [4, Theorem  28] so 𝜏𝑧(𝑓𝑔)𝒜𝐶. Use Theorem 2.1(f) to write 𝜏𝑧(𝑓𝑔)𝜏𝑧(𝑓𝑔𝑛)=𝑓𝑔𝑓𝑔𝑛=𝜏𝑧(𝑓𝑔)(𝜏𝑧𝑓)𝑔𝑛)=𝜏𝑧(𝑓𝑔)𝑓(𝜏𝑧𝑔𝑛). Translation invariance of the 𝐿1 norm completes the proof. (d) From (a), we have Φ𝑓=sup𝑔1=1𝑓𝑔sup𝑔1=1𝑓𝑔1=𝑓. We get equality by considering 𝑓 and 𝑔 to be positive functions in 𝐿1. To prove Ψ𝑔𝑔1, note that Ψ𝑔=sup𝑓=1𝑓𝑔sup𝑓=1𝑓𝑔1=𝑔1. We get equality by considering 𝑓 and 𝑔 to be positive functions in 𝐿1. (e) First consider 𝑔𝒱𝐿1. We have 𝑓𝑔𝑡(𝑥)=𝑦𝑓(𝑥𝑦)𝑔𝑡𝑑𝑦𝑡=𝑓(𝑥𝑡𝑦)𝑔(𝑦)𝑑𝑦.(3.12) For <𝛼<𝛽<, ||||𝛽𝛼𝑓𝑔𝑡||||=||||(𝑥)𝑎𝑓(𝑥)𝑑𝑥𝛽𝛼[𝑓]𝑔||||=||||(𝑥𝑡𝑦)𝑓(𝑥)(𝑦)𝑑𝑦𝑑𝑥𝛽𝛼[]||||𝑓(𝑥𝑡𝑦)𝑓(𝑥)𝑔(𝑦)𝑑𝑥𝑑𝑦(3.13)𝜏𝑡𝑦||||𝑓𝑓𝑔(𝑦)𝑑𝑦2𝑓𝑔1.(3.14) By dominated convergence, we can take the limit 𝑡0 inside the integral (3.14). Continuity of 𝑓 in the Alexiewicz norm then shows 𝑓𝑔𝑡𝑎𝑓0 as 𝑡0.
Now take a sequence {𝑔(𝑛)}𝒱𝐿1 such that 𝑔(𝑛)𝑔10. Define 𝑔𝑡(𝑛)(𝑥)=𝑔(𝑛)(𝑥/𝑡)/𝑡 and 𝑎(𝑛)=𝑔(𝑛)(𝑥)𝑑𝑥. We have
𝑓𝑔𝑡𝑎𝑓𝑓𝑔𝑡(𝑛)𝑎(𝑛)𝑓+𝑓𝑔𝑡(𝑛)𝑓𝑔𝑡+𝑎(𝑛)𝑓𝑎𝑓.(3.15) By the inequality in (a), 𝑓𝑔𝑡(𝑛)𝑓𝑔𝑡𝑓𝑔𝑡(𝑛)𝑔𝑡1. Whereas, 𝑔𝑡(𝑛)𝑔𝑡1=|||𝑔(𝑛)𝑥𝑡𝑥𝑔𝑡|||𝑑𝑥𝑡=𝑔(𝑛)𝑔10as𝑛,(3.16) and 𝑎(𝑛)𝑓𝑎𝑓=|𝑎(𝑛)𝑎|𝑓=𝑔(𝑛)𝑔1𝑓. Given 𝜖>0 fix 𝑛 large enough so that 𝑓𝑔𝑡(𝑛)𝑓𝑔𝑡+𝑎(𝑛)𝑓𝑎𝑓<𝜖. Now let 𝑡0 in (3.15).
The interchange of order of integration in (3.13) is justified as follows. A change of variables and Proposition A.3 give
𝛽𝛼𝑓(𝑥𝑡𝑦)𝑔(𝑦)𝑑𝑦𝑑𝑥=𝛽𝛼𝑓(𝑦)𝑔𝑥𝑦𝑡𝑑𝑦𝑡=𝑑𝑥𝛽𝛼𝑓(𝑦)𝑔𝑥𝑦𝑡𝑑𝑥𝑑𝑦𝑡,𝛽𝛼𝑓(𝑥𝑡𝑦)𝑔(𝑦)𝑑𝑥𝑑𝑦=𝑓𝑦(𝑥)𝑔𝑡𝜒(𝛼𝑦,𝛽𝑦)(𝑥)𝑑𝑥𝑑𝑦𝑡=𝑦𝑓(𝑥)𝑔𝑡𝜒(𝛼𝑦,𝛽𝑦)(𝑥)𝑑𝑦𝑡=𝑑𝑥𝛽𝛼𝑓(𝑥)𝑔𝑦𝑥𝑡𝑑𝑦𝑡𝑑𝑥.(3.17) Note that 𝛽𝛼𝑓(𝑥)𝑔(𝑦)𝑑𝑦𝑑𝑥=𝛽𝛼𝑓(𝑥)𝑔(𝑦)𝑑𝑥𝑑𝑦 by Corollary A.4. (f) This follows from the equivalence of Definitions 1.1 and 3.1, proved in Proposition 3.5, see [6, Theorems  5.4-2 and 5.3-1].

Young's inequality states that 𝑓𝑔𝑝𝑓𝑝𝑔1 when 𝑓𝐿𝑝 for some 1𝑝 and 𝑔𝐿1. Part (a) of Theorem 3.4 extends this to 𝑓𝒜𝐶. see [1] for other results when 𝑓𝐿𝑝.

The fact that convolution is linear in both arguments, together with (b), shows that 𝒜𝐶 is an 𝐿1-module over the 𝐿1 convolution algebra, see [8] for the definition. It does not appear that 𝒜𝐶 is a Banach algebra under convolution.

We now show that Definition 1.1(iii) and the aforementioned definitions agree.

Proposition 3.5. Let 𝑓𝒜𝐶, 𝑔𝐿1, and 𝜙𝒟. Define 𝐹(𝑦)=𝑦𝑓 and 𝐺(𝑥)=𝑥𝑔. Definitions 1.1 and 3.1 both give 𝑓𝑔,𝜙=𝑓(𝑦)=𝑔(𝑥)𝜙(𝑥+𝑦)𝑑𝑥𝑑𝑦𝐹(𝑦)𝐺(𝑥)𝜙(𝑥+𝑦)𝑑𝑥𝑑𝑦.(3.18)

Proof. Let Φ(𝑦)=𝑔(𝑥)𝜙(𝑥+𝑦)𝑑𝑥. Then Φ𝐶() and Φ(𝑦)=𝑔(𝑥)𝜙(𝑥+𝑦)𝑑𝑥. Also, |Φ(𝑦)|𝑑𝑦|𝑔(𝑥)||𝜙(𝑥+𝑦)|𝑑𝑦𝑑𝑥𝑔1𝜙1, so Φ𝐴𝐶(). Dominated convergence then shows lim|𝑦|Φ(𝑦)=0. Integration by parts now gives (3.18).
Let {𝑔𝑛}𝒱𝐿1 such that 𝑔𝑛𝑔10. Since convergence in implies convergence in 𝒟, we have
𝑓𝑔,𝜙=lim𝑛𝑓𝑔𝑛,𝜙=lim𝑛𝑓(𝑦)𝑔𝑛(𝑥𝑦)𝜙(𝑥)𝑑𝑦𝑑𝑥=lim𝑛𝑓(𝑦)𝑔𝑛(𝑥𝑦)𝜙(𝑥)𝑑𝑥𝑑𝑦.(3.19) Proposition A.3 allows interchange of the iterated integrals. Define Φ𝑛(𝑦)=𝑔𝑛(𝑥)𝜙(𝑥+𝑦)𝑑𝑥. Then, 𝑉Φ𝑛𝑔𝑛1𝜙1(𝑔1+1)𝜙1 for large enough 𝑛. Hence, Φ𝑛 is of uniform bounded variation. Theorem  22 in [4], then gives 𝑓𝑔,𝜙=𝑓(𝑦)lim𝑛Φ𝑛(𝑦)𝑑𝑦=𝑓(𝑦)𝑔(𝑥𝑦)𝜙(𝑥)𝑑𝑥𝑑𝑦. The last step follows since 𝑔𝑛𝑔10.

If 𝑔𝐿1𝒱, then 𝑓𝑔 needs not to be continuous or bounded. For example, take 1/2𝛼<1 and let 𝑓(𝑥)=𝑔(𝑥)=𝑥𝛼𝜒(0,1)(𝑥). Then, 𝑓𝐿1𝒜𝐶 and 𝑔𝐿1𝒱. We have 𝑓𝑔(𝑥)=0 for 𝑥0. For 0<𝑥1, we have 𝑓𝑔(𝑥)=𝑥0𝑦𝛼(𝑥𝑦)𝛼𝑑𝑦=𝑥12𝛼10𝑦𝛼(1𝑦)𝛼𝑑𝑦=𝑥12𝛼Γ2(1𝛼)/Γ(22𝛼). Hence, 𝑓𝑔 is not continuous at 0. If 1/2<𝛼<1, then 𝑓𝑔 is unbounded at 0.

As another example, consider 𝑓(𝑥)=sin(𝜋𝑥)/log|𝑥| and 𝑔(𝑥)=𝜒(0,1)(𝑥). Then 𝑓𝒜𝐶 and for each 1𝑝, we have 𝑔𝒱𝐿𝑝. And,

𝑓𝑔(𝑥)=𝑥𝑥1sin(𝜋𝑦)=log(𝑦)𝑑𝑦for𝑥2cos(𝜋(𝑥1))𝜋log(𝑥1)cos(𝜋𝑥)1𝜋log(𝑥)𝜋𝑥𝑥1cos(𝜋𝑦)𝑦log2(𝑦)𝑑𝑦2cos(𝜋𝑥)𝜋log(𝑥)as𝑥.(3.20)

Therefore, by Theorem 2.1(d), 𝑓𝑔𝐶0(), and lim|𝑥|𝑓𝑔(𝑥)=0 but for each 1𝑝<, we have 𝑓𝑔𝐿𝑝.

4. Differentiation and Integration

If 𝑔 is sufficiently smooth, then the pointwise derivative is (𝑓𝑔)(𝑥)=𝑓𝑔(𝑥). Recall the definition 𝐴𝐶() of primitives of 𝐿1 functions given in the proof of Proposition 3.3. In the following theorem, we require pointwise derivatives of 𝑔 to exist at each point in .

Theorem 4.1. Let 𝑓𝒜𝐶, 𝑛, and 𝑔(𝑘)𝐴𝐶() for each 0𝑘𝑛. Then 𝑓𝑔𝐶𝑛() and (𝑓𝑔)(𝑛)(𝑥)=𝑓𝑔(𝑛)(𝑥) for each 𝑥.

Proof. First consider 𝑛=1. Let 𝑥. Then (𝑓𝑔)(𝑥)=lim0𝑓(𝑦)𝑔(𝑥+𝑦)𝑔(𝑥𝑦)𝑑𝑦.(4.1) To take the limit inside the integral we can show that the bracketed term in the integrand is of uniform bounded variation for 0<||1. Let 0. Since 𝑔𝐴𝐶() it follows that the variation is given by the Lebesgue integrals 𝑉𝑦𝑔(𝑥+𝑦)𝑔(𝑥𝑦)=||||𝑔(𝑥+𝑦)𝑔(𝑥𝑦)||||𝑑𝑦||𝑔||(𝑦)𝑑𝑦+||||𝑔(𝑥+𝑦)𝑔(𝑥𝑦)𝑔||||(𝑥𝑦)𝑑𝑦.(4.2) Since 𝑔𝐴𝐶(), we have 𝑔𝐿1. The second integral on the right of (4.2) gives the 𝐿1 derivative of 𝑔 in the limit 0; see [1, page 246]. Hence, in (4.1), we can use [4, Theorem  22] to take the limit under the integral sign. This then gives (𝑓𝑔)(𝑥)=𝑓𝑔(𝑥). Theorem 2.1(d) now shows (𝑓𝑔)𝐶0(). Induction on 𝑛 completes the proof.

For similar results when 𝑓𝐿1, see [1, Proposition  8.10].

Note that 𝑔𝐴𝐶() does not imply 𝑔𝐴𝐶(). For example, 𝑔(𝑥)=𝑥. The conditions 𝑔(𝑘)𝒱 for 0𝑘𝑛+1 imply those in Theorem 4.1. To see this, it suffices to consider 𝑛=1. If 𝑔,𝑔𝒱, then 𝑔 exists at each point and is bounded. Hence, the Lebesgue integral 𝑔(𝑥)=𝑔(0)+𝑥0𝑔(𝑦)𝑑𝑦 exists for each 𝑥 and 𝑔 is absolutely continuous. Since 𝑔𝒱, we then have 𝑔𝐴𝐶(). Similarly, for 𝑛>1. The example 𝑔(𝑥)=|𝑥|1.5sin(1/[1+x2]) shows that the 𝐴𝐶() condition in the theorem is weaker than the aforementioned 𝒱 condition since 𝑔,𝑔𝐴𝐶() but 𝑔(0) does not exist so 𝑔𝒱.

We found that when 𝑔𝒱𝐿1, then 𝑓𝑔𝒜𝐶. We can compute the distributional derivative (𝐹𝑔)=𝑓𝑔, where 𝐹 is a primitive of 𝑓.

Proposition 4.2. Let 𝐹𝐶0() and write 𝑓=𝐹𝒜𝐶. Let 𝑔𝒱𝐿1. Then 𝐹𝑔𝐶0() and (𝐹𝑔)=𝑓𝑔𝒜𝐶.

Proof. Let 𝑥,𝑡. Then by the usual Hölder inequality, ||||=||||𝐹𝑔(𝑥)𝐹𝑔(𝑡)[]||||𝐹(𝑥𝑦)𝐹(𝑡𝑦)𝑔(𝑦)𝑑𝑦𝐹(𝑥)𝐹(𝑡)𝑔10as𝑡𝑥since𝐹isuniformlycontinuouson.(4.3) Hence, 𝐹𝑔 is continuous on . Dominated convergence shows that lim𝑥±𝐹𝑔(𝑥)=𝐹(±)𝑔. Therefore, 𝐹𝑔𝐶0().
Let 𝜙𝒟. Then
(𝐹𝑔),𝜙=𝐹𝑔,𝜙=𝐹(𝑥𝑦)𝑔(𝑦)𝜙(𝑥)𝑑𝑦𝑑𝑥=𝑔(𝑦)𝐹(𝑥𝑦)𝜙(𝑥)𝑑𝑥𝑑𝑦(Fubini-Tonellitheorem).(4.4) Integrate by parts and use the change of variables 𝑥𝑥+𝑦 to get (𝐹𝑔)=,𝜙𝑔(𝑦)=𝑓(𝑥)𝜙(𝑥+𝑦)𝑑𝑥𝑑𝑦𝑓(𝑥)=𝑔(𝑦)𝜙(𝑥+𝑦)𝑑𝑦𝑑𝑥(byPropositionA.3)𝑓(𝑥)=𝑔(𝑦𝑥)𝜙(𝑦)𝑑𝑦𝑑𝑥𝜙(𝑦)𝑓(𝑥)𝑔(𝑦𝑥)𝑑𝑥𝑑𝑦(byPropositionA.3)=𝑓𝑔,𝜙.(4.5)

This gives an alternate definition of 𝑓𝑔 for 𝑓𝒜𝐶 and 𝑔𝐿1.

Theorem 4.3. Let 𝑓𝒜𝐶, let 𝐹𝐶 be the primitive of 𝑓 and let 𝑔𝐿1. Define 𝑓𝑔 as in Definition 3.1. Then (𝐹𝑔)=𝑓𝑔𝒜𝐶.

Proof. Let <𝛼<𝛽<. Let {𝑔𝑛}𝒱𝐿1 such that 𝑔𝑛𝑔10. By Proposition 4.2, we have 𝛽𝛼(𝐹𝑔)=𝐹𝑔(𝛽)𝐹𝑔(𝛼)=[]𝐹(𝑦)𝑔(𝛽𝑦)𝑔(𝛼𝑦)𝑑𝑦.(4.6) As in (3.3), 𝛽𝛼𝑓𝑔𝑛=𝐹(𝑦)[𝑔𝑛(𝛽𝑦)𝑔𝑛(𝛼𝑦)]𝑑𝑦. Hence, ||||𝛽𝛼(𝐹𝑔)𝑓𝑔𝑛||||=||||𝐹(𝑦)𝑔(𝛽𝑦)𝑔𝑛(𝛽𝑦)𝑔(𝛼𝑦)𝑔𝑛(||||𝛼𝑦)𝑑𝑦𝐹𝑔(𝛽)𝑔𝑛(𝛽)1+𝑔(𝛼)𝑔𝑛(𝛼)1𝑔=2𝑓𝑛𝑔1.(4.7) Therefore, (𝐹𝑔)𝑓𝑔𝑛2𝑓𝑔𝑛𝑔10 as 𝑛.

The following theorem and its corollary give results on integrating convolutions.

Theorem 4.4. Let 𝑓𝒜𝐶 and let 𝑔𝐿1. Define 𝐹(𝑥)=𝑥𝑓 and 𝐺(𝑥)=𝑥g. Then, 𝑓𝐺𝐶0() and 𝑓𝐺(𝑥)=𝐹𝑔(𝑥) for all 𝑥.

Proof. Since 𝐺𝐴𝐶(), Theorem 2.1(d) shows 𝑓𝐺𝐶0(). We have 𝑓𝐺(𝑥)=𝑓(𝑦)𝑥𝑦=𝑔(𝑧)𝑑𝑧𝑑𝑦𝑓(𝑦)𝜒(,𝑥𝑦)=(𝑧)𝑔(𝑧)𝑑𝑧𝑑𝑦𝑓(𝑦)𝜒(,𝑥𝑦)=(𝑧)𝑔(𝑧)𝑑𝑦𝑑𝑧𝑔(𝑧)𝑥𝑧𝑓(𝑦)𝑑𝑦𝑑𝑧=𝐹𝑔(𝑥).(4.8) Proposition A.3 justifies the interchange of orders of integration.

Corollary 4.5. Note that the following hold: (a)𝑓𝑔=(𝐹𝑔)=(𝑓𝐺),(b) for all 𝛼<𝛽, one has 𝛽𝛼𝑓𝑔=𝐹𝑔(𝛽)𝐹𝑔(𝛼)=𝑓𝐺(𝛽)𝑓𝐺(𝛽).

Hence, the convolution 𝑓𝑔 can be evaluated by taking the distributional derivative of the Lebesgue integral 𝐹𝑔. Since 𝑓𝐺𝐶0(), when 𝑓𝒜𝐶 and 𝐺𝒱, we can use the equation 𝑓𝑔=(𝑓𝐺) to define 𝑓𝑔 for 𝑓𝒜𝐶 and 𝑔=𝐺 for 𝐺𝒱. In this case, 𝑔 will be a signed Radon measure. As 𝐺(𝑥)=𝑥𝑔 and this integral is a regulated primitive integral [5], we will save this case for discussion elsewhere.

Appendix

The integration by parts formula is as follows. If 𝑓𝒜𝐶 and 𝑔𝒱, it gives the integral of 𝑓𝑔 in terms of a Henstock-Stieltjes integral:

𝑓𝑔=𝐹()𝑔()𝐹𝑑𝑔,(A.1) see [4] and [9, page 199].

We have the following corollary for functions of essential bounded variation.

Corollary A.1. Let 𝐹𝐶0(). Let 𝑔𝒱. Fix 0𝛾1. Take 𝑔𝛾𝒩𝒱𝛾 such that 𝑔𝛾=𝑔 almost everywhere. Let 𝜇𝑔 be the signed Radon measure given by 𝑔. Then 𝐹𝑑𝑔𝛾=𝐹𝑑𝜇𝑔.

Proof. The distributional derivative of 𝑔 is 𝑔,𝜙=𝑔,𝜙=𝑔𝜙=𝜙𝑑𝜇𝑔 for all 𝜙𝒟. Note that 𝑔𝛾 is unique and 𝜇𝑔=𝜇𝑔𝛾. Suppose 𝜙𝒟 with supp(𝜙)[𝐴,𝐵]. Then, using integration by parts for the Henstock-Stieltjes integral: 𝑔𝛾,𝜙=𝐵𝐴𝑔𝛾𝜙=𝑔𝛾(𝐵)𝜙(𝐵)𝑔𝛾(𝐴)𝜙(𝐴)𝐵𝐴𝜙𝑑𝑔𝛾=𝜙𝑑𝑔𝛾=𝑔𝛾,𝜙=𝜙𝑑𝜇𝑔𝛾=𝜙𝑑𝜇𝑔.(A.2)
Let 𝐹𝐶0(). There is a uniformly bounded sequence {𝜙𝑛}𝒟 such that 𝜙𝑛𝐹 pointwise on . By dominated convergence,
lim𝑛𝜙𝑛𝑑𝑔𝛾=𝐹𝑑𝑔𝛾=lim𝑛𝜙𝑛𝑑𝜇𝑔=𝐹𝑑𝜇𝑔.(A.3)

Corollary A.1 now justifies the following definition.

Definition A.2. Let 𝑓𝒜𝐶 and let 𝐹𝐶 be its primitive. Let 𝑔𝒱. Fix 0𝛾1 and take 𝑔𝛾𝒩𝒱𝛾 such that 𝑔𝛾=𝑔 almost everywhere. Define 𝑓𝑔=𝑔𝛾()𝐹()𝐹𝑑𝜇𝑔=𝑓𝑔𝛾.(A.4)

Since limits at infinity are not affected by the choice of 𝛾, the definition is independent of 𝛾.

The Hölder inequality is

||||||||||||𝑓𝑔𝑓||||inf||𝑔||+𝑓𝑉𝑔𝑓𝑔𝒱,(A.5) and is valid for all 𝑓𝒜𝐶 and 𝑔𝒱. For 𝑔𝒱, we replace 𝑔 with 𝑔𝛾. This gives

||||||||||||𝑓𝑔𝑓||||inf||𝑔𝛾||+𝑓𝑉𝑔𝛾𝑓𝑔𝒱,(A.6) see [2, Lemma  24] for a proof using the Henstock-Kurzweil integral. The same proof works for the continuous primitive integral.

Fubini theorem has been established in [10] for the continuous primitive integral on compact intervals. This says that if a double integral exists in the plane, then the two iterated integrals exist and are equal. Of more utility for the case at hand is to show directly that iterated integrals are equal without resorting to the double integral. The following theorem extends a type of Fubini theorem proved in [11, page  58] for the wide Denjoy integral on compact intervals.

Proposition A.3. Let 𝑓𝒜𝐶. Let 𝑔× be measurable. Assume (i) for each 𝑥 the function 𝑦𝑔(𝑥,𝑦) is in 𝒱; (ii) the function 𝑥𝑉𝑦𝑔(𝑥,𝑦) is in 𝐿1; (iii) there is 𝑀𝐿1 such that for each 𝑦, one has |𝑔(𝑥,𝑦)|𝑀(𝑥). Then the iterated integrals exist and are equal, 𝑓(𝑦)𝑔(𝑥,𝑦)𝑑𝑦𝑑𝑥=𝑓(𝑦)𝑔(𝑥,𝑦)𝑑𝑥𝑑𝑦.

Proof. Let 𝐹𝐶 be the primitive of 𝑓. For each 𝑥, we have 𝑓(𝑦)𝑔(𝑥,𝑦)𝑑𝑦=𝐹()𝑔(𝑥,)𝐹(𝑦)𝑑2𝑔(𝑥,𝑦),(A.7) where 𝑑2(𝑥,𝑦) indicates a Henstock-Stieltjes integral with respect to 𝑦. Then, 𝑓(𝑦)𝑔(𝑥,𝑦)𝑑𝑦𝑑𝑥=𝐹()𝑔(𝑥,)𝑑𝑥𝐹(𝑦)𝑑2𝑔(𝑥,𝑦)𝑑𝑥.(A.8) The integral 𝑔(𝑥,)𝑑𝑥 exists due to condition (iii). The iterated integral in (A.8) converges absolutely since ||||𝐹(𝑦)𝑑2||||𝑔(𝑥,𝑦)𝑑𝑥||||𝐹(𝑦)𝑑2||||𝑔(𝑥,𝑦)𝑑𝑥𝐹𝑉𝑦𝑔(𝑥,𝑦)𝑑𝑥.(A.9)
Now, show the function 𝑦𝑔(𝑥,𝑦)𝑑𝑥 is in 𝒱. Let {(𝑠𝑖,𝑡𝑖)}𝑛𝑖=1 be disjoint intervals in . Then
𝑛𝑖=1||||𝑔𝑥,𝑠𝑖𝑑𝑥𝑔𝑥,𝑡𝑖||||𝑑𝑥𝑛𝑖=1||𝑔𝑥,𝑠𝑖𝑔𝑥,𝑡𝑖||=𝑑𝑥𝑛𝑖=1||𝑔𝑥,𝑠𝑖𝑔𝑥,𝑡𝑖||𝑑𝑥𝑉𝑦𝑔(𝑥,𝑦)𝑑𝑥.(A.10) The interchange of summation and integration follows from condition (ii) and the usual Fubini-Tonelli theorem. Hence, the function 𝑦𝑔(𝑥,𝑦)𝑑𝑥 is in 𝒱 and the iterated integral 𝑓(𝑦)𝑔(𝑥,𝑦)𝑑𝑥𝑑𝑦 exists.
Integrate by parts
𝑓(𝑦)𝑔(𝑥,𝑦)𝑑𝑥𝑑𝑦(A.11)=𝐹()𝑔(𝑥,)𝑑𝑥𝐹(𝑦)𝑑𝑔(𝑥,𝑦)𝑑𝑥.(A.12) In (A.11), we have lim𝑦𝑔(𝑥,𝑦)𝑑𝑥=𝑔(𝑥,)𝑑𝑥 due to dominated convergence and condition (iii). To complete the proof, we need to show that the integrals in (A.8) and (A.12) are equal. First, consider the case when 𝐹=𝜒(𝑎,𝑏) for an interval (𝑎,𝑏). Then (A.8) becomes 𝑏𝑎𝑑2𝑔(𝑥,𝑦)𝑑𝑥=[𝑔(𝑥,𝑏)𝑔(𝑥,𝑎)]𝑑𝑥. Moreover, now (A.12) becomes 𝑏𝑎𝑑[𝑔(𝑥,𝑦)𝑑𝑥]=𝑔(𝑥,𝑏)𝑑𝑥𝑔(𝑥,𝑎)𝑑𝑥. Hence, when 𝐹 is a step function, 𝐹(𝑦)=𝑛𝑖=1𝑐𝑖𝜒𝐼𝑖(𝑦) for some 𝑛, disjoint intervals {𝐼𝑖}𝑛𝑖=1 and real numbers {𝑐𝑖}𝑛𝑖=1, we have the desired equality of (A.8) and (A.12). However, 𝐹𝐶 is uniformly continuous on , that is, for each 𝜖>0, there is 𝛿>0 such that for all 0|𝑥𝑦|<𝛿, we have |𝐹(𝑥)𝐹(𝑦)|<𝜖, for all 𝑥<1/𝛿, we have |𝐹(𝑥)|<𝜖 and for all 𝑥>1/𝛿, we have |𝐹(𝑥)𝐹()|<𝜖. It then follows from the compactness of that the step functions are dense in 𝐶. Hence, there is a sequence of step functions {𝜎𝑁} such that 𝐹𝜎𝑁0. In (A.8), we have lim𝑁𝜎𝑁(𝑦)𝑑2𝑔(𝑥,𝑦)𝑑𝑥=𝐹(𝑦)𝑑2𝑔(𝑥,𝑦)𝑑𝑥.(A.13) The 𝑁 limit can be brought inside the 𝑥 integral using dominated convergence and (ii) since |𝜎𝑁(𝑦)𝑑2𝑔(𝑥,𝑦)|(𝐹+1)𝑉𝑦𝑔(𝑥,𝑦) for large enough 𝑁. The 𝑁 limit can be brought inside the 𝑦 integral using dominated convergence since |𝜎𝑁(𝑦)|(𝐹+1) for large enough 𝑁. In (A.12), we have lim𝑁𝜎𝑁(𝑦)𝑑=𝑔(𝑥,𝑦)𝑑𝑥𝐹(𝑦)𝑑𝑔(𝑥,𝑦)𝑑𝑥.(A.14) The 𝑁 limit can be brought inside the 𝑦 integral since {𝜎𝑁} converges to 𝐹 uniformly on and 𝑑[𝑔(𝑥,𝑦)𝑑𝑥] is a finite-signed measure.

Corollary A.4. If 𝑓 has compact support, one can replace (iii) with (iv): for each 𝑦supp(𝑓) the function 𝑥𝑔(𝑥,𝑦) is in 𝐿1.

Acknowledgment

Supported by the Natural Sciences and Engineering Research Council of Canada.