Convolutions with the continuous primitive integral

TL;DR

This paper develops a convolution theory for distributions integrable via the continuous primitive integral, establishing properties, estimates, and a Fubini theorem within this framework.

Contribution

It introduces convolution operations for distributions with the continuous primitive integral, extending classical properties and proving a Fubini theorem in this context.

Findings

Convolution is well-defined for integrable distributions and functions of bounded variation or in L^1.

Key properties like commutativity, associativity, and translation invariance are established.

Provides estimates for the convolution norm and results on differentiation and integration of convolutions.

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 $\int_a^bf=F(b)-F(a)$. 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 $f\ast g(x)=\intinf f(x-y)g(y) dy$ for $f$ an integrable distribution and $g$ a function of bounded variation or an $L^1$ function. Usual properties of convolutions are shown to hold: commutativity, associativity, commutation with translation. For $g$ of bounded variation, $f\ast g$ is uniformly continuous and we have the estimate $\|f\ast g\|_\infty\leq \|f\|\|g\|_\bv$ where $\|f\|=\sup_I|\int_If|$ is the Alexiewicz norm. This supremum is taken over all intervals $I\subset\R$. When $g\in L^1$ the estimate is $\|f\ast g\|\leq \|f\|\|g\|_1$. There are results on differentiation and integration of convolutions. A type of Fubini theorem is proved for the continuous primitive integral.

Equations

$$|f\ast g(t)-f\ast g(x)|$$

$$\sum |g\ast h(s_i)-g\ast h(t_i)|$$

$$f\ast (g\ast h)(x)$$

$$⟨f,\tilde{\psi }\ast \varphi ⟩$$

$$⟨f,{\tau }_x\tilde{\psi }⟩={\int }_{-\infty }^{\infty }f(y)\psi (x-y)dy=f\ast \psi (x).$$

$$⟨f(y),⟨g(x),\varphi (x+y)⟩⟩$$

$$f\ast g(x)={\int }_a^{b}f(x-y)g(y)dy={\int }_{x-b}^{x-a}f(y)g(x-y)dy.$$

$$f\ast g(x)$$

$${\int }_{\alpha }^{\beta }f\ast g_n(x)dx$$

$$=$$

$$\Vert f\ast g_n\Vert \le \Vert f\Vert \Vert g_n{\Vert }_1.$$

$$\Vert f\ast g_m-f\ast g_n\Vert =\Vert f\ast (g_m-g_n)\Vert \le \Vert f\Vert \Vert g_m-g_n{\Vert }_1$$

$${\int }_{\alpha }^{\beta }{f}_n\ast g(x)dx$$

$$\Vert f_n\ast g\Vert \le \Vert f_n\Vert \Vert g{\Vert }_1.$$

$$\Vert f_n\ast g-f\ast g_n\Vert$$

$$\Vert \xi -\eta \Vert$$

$$\Vert (f\ast g)\ast h_n-(f\ast g_n)\ast h_n\Vert$$

$$\Vert (f\ast g_n)\ast h_n-f\ast p_n\Vert$$

$$f\ast g_t(x)={\int }_{-\infty }^{\infty }f(x-y)g\left(\frac{y}{t}\right)\frac{dy}{t}={\int }_{-\infty }^{\infty }f(x-ty)g(y)dy.$$

$$\left|{\int }_{\alpha }^{\beta }\left[f\ast g_t(x)-af(x)\right]dx\right|$$

$$\le$$

$$\Vert f\ast g_t-af\Vert \le \Vert f\ast g_t^{(n)}-a^{(n)}f\Vert +\Vert f\ast g_t^{(n)}-f\ast g_t\Vert +\Vert a^{(n)}f-af\Vert .$$

$$\begin{gathered} \Vert g_t^{(n)}-g_t{\Vert }_1={\int }_{-\infty }^{\infty }\left|g^{(n)}\left(\frac{x}{t}\right)-g\left(\frac{x}{t}\right)\right|\frac{dx}{t} \\ =\Vert g^{(n)}-g{\Vert }_1\to 0\text{ as }n\to \infty . \end{gathered}$$

$${\int }_{\alpha }^{\beta }{\int }_{-\infty }^{\infty }f(x-ty)g(y)dydx$$

$${\int }_{-\infty }^{\infty }{\int }_{\alpha }^{\beta }f(x-ty)g(y)dxdy$$

$$⟨f\ast g,\varphi ⟩$$

$$=$$

$$⟨f\ast g,\varphi ⟩$$

$$f\ast g(x)$$

$$(f\ast g{)}^{\prime }(x)=\underset{h\to 0}{\lim }{\int }_{-\infty }^{\infty }f(y)\left[\frac{g(x+h-y)-g(x-y)}{h}\right]dy.$$

$$V_{y\in \mathbb{R}}\left[\frac{g(x+h-y)-g(x-y)}{h}\right]={\int }_{-\infty }^{\infty }\left|\frac{g^{\prime }(x+h-y)-g^{\prime }(x-y)}{h}\right|dy$$

Full text

Preprint September 18, 2009. To appear in Abstract and Applied Analysis

Convolutions with the continuous primitive integral

Erik Talvila111Supported by the Natural Sciences and Engineering Research Council of Canada.

Department of Mathematics and Statistics

University of the Fraser Valley

Abbotsford, BC Canada V2S 7M8

Erik.Talvila@ufv.ca

Abstract. If $F$ is a continuous function on the real line and $f=F^{\prime}$ is its distributional derivative then the continuous primitive integral of distribution $f$ is $\int_{a}^{b}f=F(b)-F(a)$. 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 $f\ast g(x)=\int^{\infty}_{-\infty}f(x-y)g(y)\,dy$ for $f$ an integrable distribution and $g$ a function of bounded variation or an $L^{1}$ function. Usual properties of convolutions are shown to hold: commutativity, associativity, commutation with translation. For $g$ of bounded variation, $f\ast g$ is uniformly continuous and we have the estimate $\|f\ast g\|_{\infty}\leq\|f\|\|g\|_{\cal BV}$ where $\|f\|=\sup_{I}|\int_{I}f|$ is the Alexiewicz norm. This supremum is taken over all intervals $I\subset{\mathbb{R}}$. When $g\in L^{1}$ the estimate is $\|f\ast g\|\leq\|f\|\|g\|_{1}$. There are results on differentiation and integration of convolutions. A type of Fubini theorem is proved for the continuous primitive integral.

2000 subject classification: 26A39, 42A85, 46E15, 46F10, 46G12

Keywords and phrases: convolution, Schwartz distribution, distributional integral, continuous primitive integral, Henstock–Kurzweil integral

1 Introduction and notation

The convolution of two functions $f$ and $g$ on the real line is $f\ast g(x)=\int^{\infty}_{-\infty}f(x-y)g(y)\,dy$. 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 $L^{1}$ and $\ast\!:\!L^{1}\times L^{1}\to L^{1}$ such that $\|f\ast g\|_{1}\leq\|f\|_{1}\|g\|_{1}$. The convolution is commutative, associative and commutes with translations. If $f\in L^{1}$ and $g\in C^{n}$ then $f\ast g\in C^{n}$ and $(f\ast g)^{(n)}(x)=f\ast g^{(n)}(x)$. Convolutions also have the approximation property that if $f\in L^{p}$ ($1\leq p<\infty$) and $g\in L^{1}$ then $\|f\ast g_{t}-af\|_{p}\to 0$ as $t\to 0$, where $g_{t}(x)=g(x/t)/t$ and $a=\int^{\infty}_{-\infty}g$. When $f$ is bounded and continuous, there is a similar result for $p=\infty$. For these results see, for example, [4]. See [8] for related results with the Henstock–Kurzweil integral. Using the Alexiewicz norm, all of these results have generalisations to continuous primitive integrals that are proven below.

We now define the continuous primitive integral. For this we need some notation for distributions. The space of test functions is ${\cal D}=C^{\infty}_{c}({\mathbb{R}})=\{\phi\!:\!{\mathbb{R}}\to{\mathbb{R}}\mid\phi\in C^{\infty}({\mathbb{R}})\text{ and }{\rm supp}(\phi)\text{ is compact}\}$. The support of function $\phi$ is the closure of the set on which $\phi$ does not vanish and is denoted ${\rm supp}(\phi)$. Under usual pointwise operations ${\cal D}$ is a linear space over field ${\mathbb{R}}$. In ${\cal D}$ we have a notion of convergence. If $\{\phi_{n}\}\subset{\cal D}$ then $\phi_{n}\to 0$ as $n\to\infty$ if there is a compact set $K\subset{\mathbb{R}}$ such that for each $n$, ${\rm supp}(\phi_{n})\subset K$, and for each $m\geq 0$ we have $\phi_{n}^{(m)}\to 0$ uniformly on $K$ as $n\to\infty$. The distributions are denoted ${\cal D}^{\prime}$ and are the continuous linear functionals on ${\cal D}$. For $T\in{\cal D}^{\prime}$ and $\phi\in{\cal D}$ we write $\langle T,\phi\rangle\in{\mathbb{R}}$. For $\phi,\psi\in{\cal D}$ and $a,b\in{\mathbb{R}}$ we have $\langle T,a\phi+b\psi\rangle=a\langle T,\phi\rangle+b\langle T,\psi\rangle$. And, if $\phi_{n}\to 0$ in ${\cal D}$ then $\langle T,\phi_{n}\rangle\to 0$ in ${\mathbb{R}}$. Linear operations are defined in ${\cal D}^{\prime}$ by $\langle aS+bT,\phi\rangle=a\langle S,\phi\rangle+b\langle T,\phi\rangle$ for $S,T\in{\cal D}^{\prime}$; $a,b\in{\mathbb{R}}$ and $\phi\in{\cal D}$. If $f\in L^{1}_{\it loc}$ then $\langle T_{f},\phi\rangle=\int^{\infty}_{-\infty}f(x)\phi(x)\,dx$ defines a distribution $T_{f}\in{\cal D}^{\prime}$. The integral exists as a Lebesgue integral. All distributions have derivatives of all orders that are themselves distributions. For $T\in{\cal D}^{\prime}$ and $\phi\in{\cal D}$ the distributional derivative of $T$ is $T^{\prime}$ where $\langle T^{\prime},\phi\rangle=-\langle T,\phi^{\prime}\rangle$. This is also called the weak derivative. If $p\!:\!{\mathbb{R}}\to{\mathbb{R}}$ is a function that is differentiable in the pointwise sense at $x\in{\mathbb{R}}$ then we write its derivative as $p^{\prime}(x)$. If $p$ is a $C^{\infty}$ bijection such that $p^{\prime}(x)\not=0$ for any $x\in{\mathbb{R}}$ then the composition with distribution $T$ is defined by $\langle T\circ p,\phi\rangle=\langle T,\frac{\phi\circ p^{-1}}{p^{\prime}\circ p^{-1}}\rangle$ for all $\phi\in{\cal D}$. Translations are a special case. For $x\in{\mathbb{R}}$ define the translation $\tau_{x}$ on distribution $T\in{\cal D}^{\prime}$ by $\langle\tau_{x}T,\phi\rangle=\langle T,\tau_{-x}\phi\rangle$ for test function $\phi\in{\cal D}$ where $\tau_{x}\phi(y)=\phi(y-x)$. All of the results on distributions we use can be found in [5].

The following Banach space will be of importance: ${\cal B}_{C}=\{F\!:\!{\mathbb{R}}\to{\mathbb{R}}\mid F\in C^{0}({\mathbb{R}}),F(-\infty)=0,F(\infty)\in{\mathbb{R}}\}$. We use the notation $F(-\infty)=\lim_{x\to-\infty}F(x)$ and $F(\infty)=\lim_{x\to\infty}F(x)$. The extended real line is denoted $\overline{{\mathbb{R}}}=[-\infty,\infty]$. The space ${\cal B}_{C}$ then consists of functions continuous on $\overline{{\mathbb{R}}}$ with a limit of [math] at $-\infty$. We denote the functions that are continuous on ${\mathbb{R}}$ that have real limits at $\pm\infty$ by $C^{0}(\overline{{\mathbb{R}}})$. Hence, ${\cal B}_{C}$ is properly contained in $C^{0}(\overline{{\mathbb{R}}})$, which is itself properly contained in the space of uniformly continuous functions on ${\mathbb{R}}$. The space ${\cal B}_{C}$ is a Banach space under the uniform norm; $\|F\|_{\infty}=\sup_{x\in{\mathbb{R}}}|F(x)|=\max_{x\in\overline{{\mathbb{R}}}}|F(x)|$ for $F\in{\cal B}_{C}$. The continuous primitive integral is defined by taking ${\cal B}_{C}$ as the space of primitives. The space of integrable distributions is ${\cal A}_{C}=\{f\in{\cal D}^{\prime}\mid f=F^{\prime}\text{ for }F\in{\cal B}_{C}\}$. If $f\in{\cal A}_{C}$ then $\int_{a}^{b}f=F(b)-F(a)$ for $a,b\in\overline{{\mathbb{R}}}$. The distributional differential equation $T^{\prime}=0$ has only constant solutions so the primitive $F\in{\cal B}_{C}$ satisfying $F^{\prime}=f$ is unique. Integrable distributions are then tempered and of order one. This integral, including a discussion of extensions to ${\mathbb{R}}^{n}$, is described in [9]. A more general integral is obtained by taking the primitives to be regulated functions, i.e., functions with a left and right limit at each point. See [10].

Examples of distributions in ${\cal A}_{C}$ are $T_{f}$ for functions $f$ that have a finite Lebesgue, Henstock–Kurzweil or wide Denjoy integral. We identify function $f$ with the distribution $T_{f}$. Pointwise function values can be recovered from $T_{f}$ at points of continuity of $f$ by evaluating the limit $\langle T_{f},\phi_{n}\rangle$ for a delta sequence converging to $x\in{\mathbb{R}}$. This is a sequence of test functions $\{\phi_{n}\}\subset{\cal D}$ such that for each $n$, $\phi_{n}\geq 0$, $\int_{-\infty}^{\infty}\phi_{n}=1$, and the support of $\phi_{n}$ tends to $\{x\}$ as $n\to\infty$. Note that if $F\in C^{0}(\overline{{\mathbb{R}}})$ is an increasing function with $F^{\prime}(x)=0$ for almost all $x\in{\mathbb{R}}$ then the Lebesgue integral $\int_{a}^{b}F^{\prime}(x)\,dx=0$ but $F^{\prime}\in{\cal A}_{C}$ and $\int_{a}^{b}F^{\prime}=F(b)-F(a)$. For another example of a distribution in ${\cal A}_{C}$, let $F\in C^{0}(\overline{{\mathbb{R}}})$ be continuous and nowhere differentiable in the pointwise sense. Then $F^{\prime}\in{\cal A}_{C}$ and $\int_{a}^{b}F^{\prime}=F(b)-F(a)$ for all $a,b\in\overline{{\mathbb{R}}}$.

The space ${\cal A}_{C}$ is a Banach space under the Alexiewicz norm; $\|f\|=\sup_{I\subset{\mathbb{R}}}|\int_{I}f|$ where the supremum is taken over all intervals $I\subset{\mathbb{R}}$. An equivalent norm is $\|f\|^{\prime}=\sup_{x\in{\mathbb{R}}}|\int_{-\infty}^{x}f|$. 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 ${\cal A}_{C}$ 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 below, the Banach space ${\cal A}_{C}$ is a suitable setting for applications of nonabsolute integration.

We will also need to use functions of bounded variation. Let $g\!:\!{\mathbb{R}}\to{\mathbb{R}}$. The variation of $g$ is $V\!g=\sup\sum|g(x_{i})-g(y_{i})|$ where the supremum is taken over all disjoint intervals $\{(x_{i},y_{i})\}$. The functions of bounded variation are denoted ${\cal BV}=\{g\!:\!{\mathbb{R}}\to{\mathbb{R}}\mid V\!g<\infty\}$. This is a Banach space under the norm $\|g\|_{\cal BV}=|g(-\infty)|+V\!g$. Equivalent norms are $\|g\|_{\infty}+Vg$ and $|g(a)|+V\!g$ for each $a\in\overline{{\mathbb{R}}}$. Functions of bounded variation have a left and right limit at each point in ${\mathbb{R}}$ and limits at $\pm\infty$ so as above we will define $g(\pm\infty)=\lim_{x\to\pm\infty}g(x)$.

If $g\in L^{1}_{loc}$ then the essential variation of $g$ is ${\rm ess\,var}g=\sup\int^{\infty}_{-\infty}g\phi^{\prime}$ where the supremum is taken over all $\phi\in{\cal D}$ with $\|\phi\|_{\infty}\leq 1$. Then ${\cal EBV}=\{g\in L^{1}_{loc}\mid{\rm ess\,var}g<\infty\}$. This is a Banach space under the norm $\|g\|_{\cal EBV}={\rm ess\,sup}|g|+{\rm ess\,var}g$. Let $0\leq\gamma\leq 1$. For $g\!:\!{\mathbb{R}}\to{\mathbb{R}}$ define $g_{\gamma}(x)=(1-\gamma)g(x-)+\gamma g(x+)$. For left continuity, $\gamma=0$ and for right continuity $\gamma=1$. The functions of normalised bounded variation are ${\cal NBV}_{\gamma}=\{g_{\gamma}\mid g\in{\cal BV}\}$. If $g\in{\cal EBV}$ then ${ess\,var}g=\inf Vh$ such that $h=g$ almost everywhere. For each $0\leq\gamma\leq 1$ there is exactly one function $h\in{\cal NBV}_{\gamma}$ such that $g=h$ almost everywhere. In this case ${ess\,var}g=Vh$. 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 $\mu_{g}$ where $\langle g^{\prime},\phi\rangle=-\langle g,\phi^{\prime}\rangle=-\int^{\infty}_{-\infty}g\phi^{\prime}=\int^{\infty}_{-\infty}\phi\,d\mu_{g}$ for all $\phi\in{\cal D}$.

We will see that $\ast\!:\!{\cal A}_{C}\times{\cal BV}\to C^{0}(\overline{{\mathbb{R}}})$ and that $\|f\ast g\|_{\infty}\leq\|f\|\|g\|_{{\cal BV}}$. Similarly for $g\in{\cal EBV}$. Convolutions for $f\in{\cal A}_{C}$ and $g\in L^{1}$ will be defined using sequences in ${\cal BV}\cap L^{1}$ that converge to $g$ in the $L^{1}$ norm. It will be shown that $\ast\!:\!{\cal A}_{C}\times L^{1}\to{\cal A}_{C}$ and that $\|f\ast g\|\leq\|f\|\|g\|_{1}$.

Convolutions can be defined for distributions in several different ways.

Definition 1

Let $S,T\in{\cal D}^{\prime}$ and $\phi,\psi\in{\cal D}$. Define ${\tilde{\phi}}(x)=\phi(-x)$. (i) $\langle T\ast\psi,\phi\rangle=\langle T,\phi\ast\tilde{\psi}\rangle$. (ii) For each $x\in{\mathbb{R}}$, let $T\ast\psi(x)=\langle T,\tau_{x}\tilde{\psi}\rangle$. (iii) $\langle S\ast T,\phi\rangle=\langle S(x),\langle T(y),\phi(x+y)\rangle\rangle$.

In (i), $\ast\!:\!{\cal D}^{\prime}\times{\cal D}\to{\cal D}^{\prime}$. 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), $\ast\!:\!{\cal D}^{\prime}\times{\cal D}\to C^{\infty}$. In [4] it is shown that definitions (i) and (ii) are equivalent. In (iii), $\ast\!:\!{\cal D}^{\prime}\times{\cal D}^{\prime}\to{\cal D}^{\prime}$. However, this definition requires restrictions on the supports of $S$ and $T$. It suffices that one of these distributions have compact support. Other conditions on the supports can be imposed. See [5] and [11]. This definition is an instance of the tensor product, $\langle S\otimes T,\Phi\rangle=\langle S(x),\langle T(y),\Phi(x,y)\rangle\rangle$ where now $\Phi\in{\cal D}({\mathbb{R}}^{2})$.

Under (i), $T\ast\psi$ is in $C^{\infty}$. It satisfies $(T\ast\psi)\ast\phi=T\ast(\psi\ast\phi)$, $\tau_{x}(T\ast\psi)=(\tau_{x}T)\ast\psi=T\ast(\tau_{x}\psi)$, and $(T\ast\psi)^{(n)}=T\ast\psi^{(n)}=T^{(n)}\ast\psi$. Under (iii), with appropriate support restrictions, $S\ast T$ is in ${\cal D}^{\prime}$. It is commutative and associative, commutes with translations, and satisfies $(S\ast T)^{(n)}=S^{(n)}\ast T=S\ast T^{(n)}$. It is weakly continuous in ${\cal D}^{\prime}$, i.e., if $T_{n}\to T$ in ${\cal D}^{\prime}$ then $T_{n}\ast\psi\to T\ast\psi$ in ${\cal D}^{\prime}$. See [4], [5], [7] and [11] for additional properties of convolutions of distributions.

Although elements of ${\cal A}_{C}$ are distributions, we show in this paper that their behaviour as convolutions is more like that of integrable functions.

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

2 Convolution in ${\cal A}_{C}\times{\cal BV}$

In this section we prove basic results for the convolution when $f\in{\cal A}_{C}$ and $g\in{\cal BV}$. Under these conditions $f\ast g$ is commutative, continuous on $\overline{{\mathbb{R}}}$ and commutes with translations. It can be estimated in the uniform norm in terms of the Alexiewicz and ${\cal BV}$ norms. There is also an associative property. We first need the result that ${\cal BV}$ forms the space of multipliers for ${\cal A}_{C}$, i.e., if $f\in{\cal A}_{C}$ then $fg\in{\cal A}_{C}$ for all $g\in{\cal BV}$. The integral $\int_{I}fg$ is defined using the integration by parts formula in the Appendix. The Hölder inequality (14) shows that ${\cal BV}$ is the dual space of ${\cal A}_{C}$.

We define the convolution of $f\in{\cal A}_{C}$ and $g\in{\cal BV}$ as $f\ast g(x)=\int^{\infty}_{-\infty}(f\circ r_{x})g$ where $r_{x}(t)=x-t$. We write this as $f\ast g(x)=\int^{\infty}_{-\infty}f(x-y)g(y)\,dy$.

Theorem 2

Let $f\in{\cal A}_{C}$ and let $g\in{\cal BV}$. Then (a) $f\ast g$ exists on ${\mathbb{R}}$ (b) $f\ast g=g\ast f$ (c) $\|f\ast g\|_{\infty}\leq|\int_{-\infty}^{\infty}f|\inf_{{\mathbb{R}}}|g|+\|f\|Vg\leq\|f\|\|g\|_{\cal BV}$ (d) $f\ast g\in C^{0}(\overline{{\mathbb{R}}})$, $\lim_{x\to\pm\infty}f\ast g(x)=g(\pm\infty)\int^{\infty}_{-\infty}f$. (e) If $h\in L^{1}$ then $f\ast(g\ast h)=(f\ast g)\ast h\in C^{0}(\overline{{\mathbb{R}}})$. (f) Let $x,z\in{\mathbb{R}}$. Then $\tau_{z}(f\ast g)(x)=(\tau_{z}f)\ast g(x)=(f\ast\tau_{z}g)(x)$. (g) For each $f\in{\cal A}_{C}$ define $\Phi_{f}\!:\!{\cal BV}\to C^{0}(\overline{{\mathbb{R}}})$ by $\Phi_{f}[g]=f\ast g$. Then $\Phi_{f}$ is a bounded linear operator and $\|\Phi_{f}\|\leq\|f\|$. There exists a nonzero distribution $f\in{\cal A}_{C}$ such that $\|\Phi_{f}\|=\|f\|$. For each $g\in{\cal BV}$ define $\Psi_{g}\!:\!{\cal A}_{C}\to C^{0}(\overline{{\mathbb{R}}})$ by $\Psi_{g}[f]=f\ast g$. Then $\Psi_{g}$ is a bounded linear operator and $\|\Psi_{g}\|\leq\|g\|_{\cal BV}$. There exists a nonzero function $g\in{\cal BV}$ such that $\|\Psi_{g}\|=\|g\|_{{\cal BV}}$. (h) ${\rm supp}(f\ast g)\subset{\rm supp}(f)+{\rm supp}(g)$.

Proof: (a) Existence is given via the integration by parts formula (13) in the Appendix. (b) See [9, Theorem 11] for a change of variables theorem that can be used with $y\mapsto x-y$. (c) This inequality follows from the Hölder inequality (14). (d) Let $x,t\in{\mathbb{R}}$. From (c) we have

[TABLE]

The last line follows from continuity in the Alexiewicz norm [9, Theorem 22]. Hence, $f\ast g$ is uniformly continuous on ${\mathbb{R}}$. And, $\lim_{x\to\infty}\int^{\infty}_{-\infty}f(y)g(x-y)\,dy=\int^{\infty}_{-\infty}f(y)\lim_{x\to\infty}g(x-y)\,dy=g(\infty)\int^{\infty}_{-\infty}f$. The limit $x\to\infty$ can be taken under the integral sign since $g(x-y)$ is of uniform bounded variation, i.e., $V_{y\in{\mathbb{R}}}g(x-y)=V\!g$. Theorem 22 in [9] then applies. Similarly as $x\to-\infty$. (e) First show $g\ast h\in{\cal BV}$. Let $\{(s_{i},t_{i})\}$ be disjoint intervals in ${\mathbb{R}}$. Then

[TABLE]

Hence, $V(g\ast h)\leq V\!g\,\|h\|_{1}$. The interchange of sum and integral follows from the Fubini–Tonelli theorem. Now (d) shows $f\ast(g\ast h)\in C^{0}(\overline{{\mathbb{R}}})$. Write

[TABLE]

We can interchange orders of integration using Proposition 17. For (ii) in Proposition 17, the function $z\mapsto V_{y\in{\mathbb{R}}}g(x-y-z)h(z)=V\!g\,h(z)$ is in $L^{1}$ for each fixed $x\in{\mathbb{R}}$. Since $g$ is of bounded variation it is bounded so $|g(x-y-z)h(z)|\leq\|g\|_{\infty}|h(z)|$ and condition (iii) is satisfied. (f) This follows from a linear change of variables as in (a). (g) From (c) we have $\|\Phi_{f}\|=\sup_{\|g\|_{\cal BV}=1}\|f\ast g\|_{\infty}\leq\sup_{\|g\|_{\cal BV}=1}\|f\|\|g\|_{\cal BV}=\|f\|$. Let $f>0$ be in $L^{1}$. If $g=1$ then $\|g\|_{\cal BV}=1$ and $f\ast g(x)=\int^{\infty}_{-\infty}f$ so $\|\Phi_{f}\|=\|f\|=\|f\|_{1}$. To prove $\|\Psi_{g}\|\leq\|g\|_{\cal BV}$, note that $\|\Psi_{g}\|=\sup_{\|f\|=1}\|f\ast g\|_{\infty}\leq\sup_{\|f\|=1}\|f\|\|g\|_{\cal BV}=\|g\|_{\cal BV}$. Let $g=\chi_{(0,\infty)}$. Then $\|\Psi_{g}\|=\sup_{\|f\|=1}\|f\ast g\|_{\infty}=\sup_{\|f\|=1}\sup_{x\in{\mathbb{R}}}|\int_{-\infty}^{x}f|=1=\|g\|_{\cal BV}$. (h) Suppose $x\notin{\rm supp}(f)+{\rm supp}(g)$. Note that we can write $f\ast g(x)=\int^{\infty}_{-\infty}g(x-y)\,dF(y)$, in terms of a Henstock–Stieltjes integral. See [9] for details. This integral is approximated by Riemann sums $\sum_{n=1}^{N}g(x-z_{n})[F(t_{n})-F(t_{n-1})]$ where $z_{n}\in[t_{n-1},t_{n}]$, $-\infty=t_{0}<t_{1}<\ldots<t_{N}=\infty$ and there is a gauge function $\gamma$ mapping $\overline{{\mathbb{R}}}$ to the open intervals in $\overline{{\mathbb{R}}}$ such that $[t_{n-1},t_{n}]\subset\gamma(z_{n})$. If $z_{n}\notin{\rm supp}(f)$ then since ${\mathbb{R}}\setminus{\rm supp}(f)$ is open there is an open interval $z_{n}\subset I\subset{\mathbb{R}}\setminus{\rm supp}(f)$. We can take $\gamma$ such that $[t_{n-1},t_{n}]\subset I$ for all $1\leq n\leq N$. And, $F$ is constant on each interval in ${\mathbb{R}}\setminus{\rm supp}(f)$. Therefore, $g(x-z_{n})[F(t_{n})-F(t_{n-1})]=0$ and only tags $z_{n}\in{\rm supp}(f)$ can contribute to the Riemann sum. But for all $z_{n}\in{\rm supp}(f)$ we have $x-z_{n}\notin{\rm supp}(g)$ so $g(x-z_{n})[F(t_{n})-F(t_{n-1})]=0$. It follows that $f\ast g(x)=0$. $\quad\blacksquare$

Similar results are proven for $f\in L^{p}$ in [4, § 8.2].

If we use the equivalent norm $\|f\|^{\prime}=\sup_{x\in{\mathbb{R}}}|\int_{-\infty}^{x}f|$ then $\|\Phi_{f}\|=\|f\|^{\prime}$. For, integration by parts gives $\|\Phi_{f}\|\leq\|f\|^{\prime}$. Now given $f\in{\cal A}_{C}$ let $g=\chi_{(0,\infty)}$. Then $\|g\|_{\cal BV}=1$. And, $f\ast g(x)=\int_{-\infty}^{x}f$. Hence, $\|f\ast g\|_{\infty}=\|f\|^{\prime}$ and $\|\Phi_{f}\|=\|f\|^{\prime}$. We can have strict inequality in $\|\Psi_{g}\|\leq\|g\|_{\cal BV}$. For example, let $g=\chi_{\{0\}}$. Then $\|g\|_{\cal BV}=2$ but integration by parts shows $f\ast g=0$ for each $f\in{\cal A}_{C}$.

Remark 3

If $f\in{\cal A}_{C}$ and $g\in{\cal EBV}$ we can use Definition 16 to define $f\ast g(x)=f\ast g_{\gamma}(x)$ where $g_{\gamma}=g$ almost everywhere and $g_{\gamma}\in{\cal NBV}_{\gamma}$. All of the results in Theorem 2 and the rest of the paper have analogues. Note that $f\ast g(x)=F(\infty)g_{\gamma}(-\infty)+F\ast\mu_{g}$. **

Proposition 4

The three definitions of convolution for distributions in Definition 1 are compatible with $f\ast g$ for $f\in{\cal A}_{C}$ and $g\in{\cal BV}$.

Proof: Let $f\in{\cal A}_{C}$, $g\in{\cal BV}$ and $\phi,\psi\in{\cal D}$. Definition 1(i) gives

[TABLE]

Since $\psi\in{\cal BV}$ and $\phi\in L^{1}$, Proposition 17 justifies the interchange of integrals. Definition 1(ii) gives

[TABLE]

Definition 1(iii) gives

[TABLE]

The interchange of integrals is accomplished using Proposition 17 since $g\in{\cal BV}$ and $\phi\in L^{1}$. $\quad\blacksquare$

The locally integrable distributions are defined as ${\cal A}_{C}(loc)=\{f\in{\cal D}^{\prime}\mid f=F^{\prime}\text{ for some }F\in C^{0}({\mathbb{R}})\}$. Let $f\in{\cal A}_{C}(loc)$ and let $g\in{\cal BV}$ with support in the compact interval $[a,b]$. By the Hake theorem [9, Theorem 25], $f\ast g(x)$ exists if and only if the limits of $\int_{\alpha}^{\beta}f(x-y)g(y)\,dy$ exist as $\alpha\to-\infty$ and $\beta\to\infty$. This gives

[TABLE]

There are analogues of the results in Theorem 2. For example, $|f\ast g(x)|\leq|\int_{x-b}^{x-a}f|\inf_{[a,b]}|g|+\|f\chi_{[x-b,x-a]}\|V_{[a,b]}g$. There are also versions where the supports are taken to be semi-infinite intervals.

We can also define the distributions with bounded primitive as ${\cal A}_{C}(bd)=\{f\in{\cal D}^{\prime}\mid f=F^{\prime}\text{ for some bounded }F\in C^{0}({\mathbb{R}})\text{ with }F(0)=0\}$. Let $f\in{\cal A}_{C}(bd)$ and let $F$ be its unique primitive. If $g\in{\cal BV}$ such that $g(\pm\infty)=0$ then

[TABLE]

It follows that $\|f\ast g\|_{\infty}\leq\|F\|_{\infty}Vg$.

It is possible to formulate other existence criteria. For example, if $f(x)=\log|x|\sin(x)$ and $g(x)=|x|^{-\alpha}$ for some $0<\alpha<1$ then $f$ and $g$ are not in ${\cal A}_{C}$, ${\cal BV}$ or $L^{p}$ for any $1\leq p\leq\infty$ but $f\ast g$ exists on ${\mathbb{R}}$ because $f,g\in L^{1}_{loc}$ and if $F(x)=\int_{0}^{x}f$ then $\lim_{|x|\to\infty}F(x)g(x)=0$.

The following example shows that $f\ast g$ need not be of bounded variation and hence not absolutely continuous. Let $g=\chi_{(0,\infty)}$. For $f\in{\cal A}_{C}$ we have $f\ast g(x)=\int_{-\infty}^{x}f=F(x)$ where $F\in{\cal B}_{C}$ is the primitive of $f$. But $F$ need not be of bounded variation or even of local bounded variation. For example, let $f(x)=\sin(x^{-2})-2x^{-2}\cos(x^{-2})$ and let $F$ be its primitive in ${\cal B}_{C}$. Finally, although $f\ast g$ is continuous it need not be integrable over ${\mathbb{R}}$. For example, let $g=1$ then $f\ast g(x)=\int^{\infty}_{-\infty}f$ and $\int^{\infty}_{-\infty}f\ast g$ only exists if $\int^{\infty}_{-\infty}f=0$.

3 Convolution in ${\cal A}_{C}\times L^{1}$

We now extend the convolution $f\ast g$ to $f\in{\cal A}_{C}$ and $g\in L^{1}$. Since there are functions in $L^{1}$ that are not of bounded variation, there are distributions $f\in{\cal A}_{C}$ and functions $g\in L^{1}$ such that the integral $\int^{\infty}_{-\infty}f(x-y)g(y)\,dy$ does not exist. The convolution is then defined as the limit in $\|\cdot\|$ of a sequence $f\ast g_{n}$ for $g_{n}\in{\cal BV}\cap L^{1}$ such that $g_{n}\to g$ in the $L^{1}$ norm. This is possible since ${\cal BV}\cap L^{1}$ is dense in $L^{1}$. We also give an equivalent definition using the fact that $L^{1}$ is dense in ${\cal A}_{C}$. Take a sequence $\{f_{n}\}\subset L^{1}$ such that $\|f_{n}-f\|\to 0$. Then $f\ast g$ is the limit in $\|\cdot\|$ of $f_{n}\ast g$. In this more general setting of convolution defined in ${\cal A}_{C}\times L^{1}$ we now have an Alexiewicz norm estimate for $f\ast g$ in terms of estimates of $f$ in the Alexiewicz norm and $g$ in the $L^{1}$ norm. There is associativity with $L^{1}$ functions and commutativity with translations.

Definition 5

Let $f\in{\cal A}_{C}$ and let $g\in L^{1}$. Let $\{g_{n}\}\subset{\cal BV}\cap L^{1}$ such that $\|g_{n}-g\|_{1}\to 0$. Define $f\ast g$ as the unique element in ${\cal A}_{C}$ such that $\|f\ast g_{n}-f\ast g\|\to 0$.

To see that the definition makes sense, first note that ${\cal BV}\cap L^{1}$ is dense in $L^{1}$ since step functions are dense in $L^{1}$. Hence, the required sequence $\{g_{n}\}$ exists. Let $[\alpha,\beta]\subset{\mathbb{R}}$ be a compact interval. Let $F\in{\cal B}_{C}$ be the primitive of $f$. Then

[TABLE]

The interchange of orders of integration in (3) is accomplished with Proposition 17 using $g(x,y)=g_{n}(x-y)\chi_{[\alpha,\beta]}(x)$. Integration by parts gives (2) since $\lim_{y\to\infty}\int_{\alpha-y}^{\beta-y}g_{n}=0$. As $F$ is continuous and the function $y\mapsto\int_{\alpha-y}^{\beta-y}g_{n}$ is absolutely continuous we get (3). Taking the supremum over $\alpha,\beta\in{\mathbb{R}}$ gives

[TABLE]

We now have

[TABLE]

and $\{f\ast g_{n}\}$ is a Cauchy sequence in ${\cal A}_{C}$. Since ${\cal A}_{C}$ is complete this sequence has a limit in ${\cal A}_{C}$ which we denote $f\ast g$. The definition does not depend on the choice of sequence $\{g_{n}\}$, for if $\{h_{n}\}\subset{\cal BV}\cap L^{1}$ such that $\|h_{n}-g\|_{1}\to 0$ then $\|f\ast g_{n}-f\ast h_{n}\|\leq\|f\|(\|g_{n}-g\|_{1}+\|h_{n}-g\|_{1})\to 0$ as $n\to\infty$. The above calculation also shows that if $g\in{\cal BV}\cap L^{1}$ then the integral definition $f\ast g(x)=\int^{\infty}_{-\infty}f(x-y)g(y)\,dy$ and the limit definition agree.

Definition 6

Let $f\in{\cal A}_{C}$ and let $g\in L^{1}$. Let $\{f_{n}\}\subset L^{1}$ such that $\|f_{n}-f\|\to 0$. Define $f\ast g$ as the unique element in ${\cal A}_{C}$ such that $\|f_{n}\ast g-f\ast g\|\to 0$.

To show this definition makes sense, first show $L^{1}$ is dense in ${\cal A}_{C}$.

Proposition 7

$L^{1}$* is dense in ${\cal A}_{C}$.*

Proof: Let $AC(\overline{{\mathbb{R}}})$ be the functions that are absolutely continuous on each compact interval and which are of bounded variation on the real line. Then $f\in L^{1}$ if and only if there exists $F\in AC(\overline{{\mathbb{R}}})$ such that $F^{\prime}(x)=f(x)$ for almost all $x\in{\mathbb{R}}$. Let $f\in{\cal A}_{C}$ be given. Let $F\in{\cal B}_{C}$ be its primitive. For $\epsilon>0$, take $M>0$ such that $|F(x)|<\epsilon$ for $x<-M$ and $|F(x)-F(\infty)|<\epsilon$ for $x>M$. Due to the Weierstrass approximation theorem there is a continuous function $P\!:\!{\mathbb{R}}\to{\mathbb{R}}$ such that $P(x)=F(-M)$ for $x\leq-M$, $P(x)=F(M)$ for $x\geq M$, $|P(x)-F(x)|<\epsilon$ for $|x|\leq M$ and $P$ is a polynomial on $[-M,M]$. Hence, $P\in AC(\overline{{\mathbb{R}}})$ and $\|P^{\prime}-f\|<3\epsilon$. $\quad\blacksquare$

In Definition 6, the required sequence $\{f_{n}\}\subset L^{1}$ exists. Let $[\alpha,\beta]\subset{\mathbb{R}}$ be a compact interval. Then, by the usual Fubini–Tonelli theorem in $L^{1}$,

[TABLE]

Take the supremum over $\alpha,\beta\in{\mathbb{R}}$ and use the $L^{1}-L^{\infty}$ Hölder inequality to get

[TABLE]

It now follows that $\{f_{n}\ast g\}$ is a Cauchy sequence. It then converges to an element of ${\cal A}_{C}$. Inequality (5) also shows this limit is independent of the choice of $\{f_{n}\}$. To see that Definition 5 and Definition 6 agree, take $\{f_{n}\}\subset L^{1}$ with $\|f_{n}-f\|\to 0$ and $\{g_{n}\}\subset{\cal BV}\cap L^{1}$ with $\|g_{n}-g\|_{1}\to 0$. Then

[TABLE]

Letting $n\to\infty$ shows the limits of $f_{n}\ast g$ in Definition 6 and $f\ast g_{n}$ in Definition 5 are the same.

Theorem 8

Let $f\in{\cal A}_{C}$ and $g\in L^{1}$. Define $f\ast g$ as in Definition 5. Then (a) $\|f\ast g\|\leq\|f\|\|g\|_{1}$. (b) Let $h\in L^{1}$. Then $(f\ast g)\ast h=f\ast(g\ast h)\in{\cal A}_{C}$. (c) For each $z\in{\mathbb{R}}$, $\tau_{z}(f\ast g)=(\tau_{z}f)\ast g=(f\ast\tau_{z}g)$. (d) For each $f\in{\cal A}_{C}$ define $\Phi_{f}\!:\!L^{1}\to{\cal A}_{C}$ by $\Phi_{f}[g]=f\ast g$. Then $\Phi_{f}$ is a bounded linear operator and $\|\Phi_{f}\|\leq\|f\|$. There exists a nonzero distribution $f\in{\cal A}_{C}$ such that $\|\Phi_{f}\|=\|f\|$. For each $g\in L^{1}$ define $\Psi_{g}\!:\!{\cal A}_{C}\to{\cal A}_{C}$ by $\Psi_{g}[f]=f\ast g$. Then $\Psi_{g}$ is a bounded linear operator and $\|\Psi_{g}\|\leq\|g\|_{1}$. There exists a nonzero function $g\in L^{1}$ such that $\|\Psi_{g}\|=\|g\|_{{\cal BV}}$. (e) Define $g_{t}(x)=g(x/t)/t$ for $t>0$. Let $a=\int^{\infty}_{-\infty}g_{t}(x)\,dx=\int^{\infty}_{-\infty}g$. Then $\|f\ast g_{t}-af\|\to 0$ as $t\to 0$. (f) ${\rm supp}(f\ast g)\subset{\rm supp}(f)+{\rm supp}(g)$.

Proof: Let $\{g_{n}\}$ be as in Definition 5. (a) Since $\|f\ast g_{n}\|\to\|f\ast g\|$, equation (4) shows $\|f\ast g\|\leq\|f\|\|g\|_{1}$. (b) Let $\{h_{n}\}\subset{\cal BV}\cap L^{1}$ such that $\|h_{n}-h\|_{1}\to 0$. Then $(f\ast g)\ast h:=\xi\in{\cal A}_{C}$ such that $\|(f\ast g)\ast h_{n}-\xi\|\to 0$. Since $g\ast h\in L^{1}$ there is $\{p_{n}\}\subset{\cal BV}\cap L^{1}$ such that $\|p_{n}-g\ast h\|_{1}\to 0$. Then $f\ast(g\ast h):=\eta\in{\cal A}_{C}$ such that $\|f\ast p_{n}-\eta\|\to 0$. Now,

[TABLE]

Using (4),

[TABLE]

Finally, use Theorem 2(e) and (4) to write

[TABLE]

(c) The Alexiewicz norm is invariant under translation [9, Theorem 28] so $\tau_{z}(f\ast g)\in{\cal A}_{C}$. Use Theorem 2(f) to write $\|\tau_{z}(f\ast g)-\tau_{z}(f\ast g_{n})\|=\|f\ast g-f\ast g_{n}\|=\|\tau_{z}(f\ast g)-(\tau_{z}f)\ast g_{n})\|=\|\tau_{z}(f\ast g)-f\ast(\tau_{z}g_{n})\|$. Translation invariance of the $L^{1}$ norm completes the proof. (d) From (a) we have $\|\Phi_{f}\|=\sup_{\|g\|_{1}=1}\|f\ast g\|\leq\sup_{\|g\|_{1}=1}\|f\|\|g\|_{1}=\|f\|$. We get equality by considering $f$ and $g$ to be positive functions in $L^{1}$. To prove $\|\Psi_{g}\|\leq\|g\|_{1}$, note that $\|\Psi_{g}\|=\sup_{\|f\|=1}\|f\ast g\|\leq\sup_{\|f\|=1}\|f\|\|g\|_{1}=\|g\|_{1}$. We get equality by considering $f$ and $g$ to be positive functions in $L^{1}$. (e) First consider $g\in{\cal BV}\cap L^{1}$. We have

[TABLE]

For $-\infty<\alpha<\beta<\infty$,

[TABLE]

By dominated convergence we can take the limit $t\to 0$ inside the integral (3). Continuity of $f$ in the Alexiewicz norm then shows $\|f\ast g_{t}-af\|\to 0$ as $t\to 0$.

Now take a sequence $\{g^{(n)}\}\subset{\cal BV}\cap L^{1}$ such that $\|g^{(n)}-g\|_{1}\to 0$. Define $g_{t}^{(n)}(x)=g^{(n)}(x/t)/t$ and $a^{(n)}=\int^{\infty}_{-\infty}g^{(n)}(x)\,dx$. We have

[TABLE]

By the inequality in (a), $\|f\ast g_{t}^{(n)}-f\ast g_{t}\|\leq\|f\|\|g_{t}^{(n)}-g_{t}\|_{1}$. Whereas,

[TABLE]

And, $\|a^{(n)}f-af\|=|a^{(n)}-a|\|f\|=\|g^{(n)}-g\|_{1}\|f\|$. Given $\epsilon>0$ fix $n$ large enough so that $\|f\ast g_{t}^{(n)}-f\ast g_{t}\|+\|a^{(n)}f-af\|<\epsilon$. Now let $t\to 0$ in (8).

The interchange of order of integration in (6) is justified as follows. A change of variables and Proposition 17 give

[TABLE]

And,

[TABLE]

Note that $\int_{\alpha}^{\beta}\int^{\infty}_{-\infty}f(x)g(y)\,dy\,dx=\int^{\infty}_{-\infty}\int_{\alpha}^{\beta}f(x)g(y)\,dx\,dy$ by Corollary 18. (f) This follows from the equivalence of Definition 1 and Definition 5, proved in Proposition 9. See Theorem 5.4-2 and Theorem 5.3-1 in [11].$\quad\blacksquare$

Young’s inequality states that $\|f\ast g\|_{p}\leq\|f\|_{p}\|g\|_{1}$ when $f\in L^{p}$ for some $1\leq p\leq\infty$ and $g\in L^{1}$. Part (a) of Theorem 8 extends this to $f\in{\cal A}_{C}$. See [4] for other results when $f\in L^{p}$.

The fact that convolution is linear in both arguments, together with (b), shows that ${\cal A}_{C}$ is an $L^{1}$-module over the $L^{1}$ convolution algebra. See [3] for the definition. It does not appear that ${\cal A}_{C}$ is a Banach algebra under convolution.

We now show that Definition 1(iii) and the above definitions agree.

Proposition 9

Let $f\in{\cal A}_{C}$, $g\in L^{1}$ and $\phi\in{\cal D}$. Define $F(y)=\int_{-\infty}^{y}f$ and $G(x)=\int_{-\infty}^{x}g$. Definition 1 and Definition 5 both give

[TABLE]

Proof: Let $\Phi(y)=\int^{\infty}_{-\infty}g(x)\phi(x+y)\,dx$. Then $\Phi\in C^{\infty}({\mathbb{R}})$ and $\Phi^{\prime}(y)=\int^{\infty}_{-\infty}g(x)\phi^{\prime}(x+y)\,dx$. And, $\int^{\infty}_{-\infty}|\Phi^{\prime}(y)|\,dy\leq\int^{\infty}_{-\infty}|g(x)|\int^{\infty}_{-\infty}|\phi^{\prime}(x+y)|\,dy\,dx\leq\|g\|_{1}\|\phi^{\prime}\|_{1}$ so $\Phi\in AC(\overline{{\mathbb{R}}})$. Dominated convergence then shows $\lim_{|y|\to\infty}\Phi(y)=0$. Integration by parts now gives (9) and (10).

Let $\{g_{n}\}\subset{\cal BV}\cap L^{1}$ such that $\|g_{n}-g\|_{1}\to 0$. Since convergence in $\|\cdot\|$ implies convergence in ${\cal D}^{\prime}$, we have

[TABLE]

Proposition 17 allows interchange of the iterated integrals. Define $\Phi_{n}(y)=\int^{\infty}_{-\infty}g_{n}(x)\phi(x+y)\,dx$. As above, $V\Phi_{n}\leq\|g_{n}\|_{1}\|\phi^{\prime}\|_{1}\leq(\|g\|_{1}+1)\|\phi^{\prime}\|_{1}$ for large enough $n$. Hence, $\Phi_{n}$ is of uniform bounded variation. Theorem 22 in [9] then gives $\langle f\ast g,\phi\rangle=\int^{\infty}_{-\infty}f(y)\lim_{n\to\infty}\Phi_{n}(y)\,dy=\int^{\infty}_{-\infty}f(y)\int^{\infty}_{-\infty}g(x-y)\phi(x)\,dx\,dy$. The last step follows since $\|g_{n}-g\|_{1}\to 0$.$\quad\blacksquare$

If $g\in L^{1}\setminus{\cal BV}$ then $f\ast g$ need not be continuous or bounded. For example, take $1/2\leq\alpha<1$ and let $f(x)=g(x)=x^{-\alpha}\chi_{(0,1)}(x)$. Then $f\in L^{1}\subset{\cal A}_{C}$ and $g\in L^{1}\setminus{\cal BV}$. We have $f\ast g(x)=0$ for $x\leq 0$. For $0<x\leq 1$ we have $f\ast g(x)=\int_{0}^{x}y^{-\alpha}(x-y)^{-\alpha}\,dy=x^{1-2\alpha}\int_{0}^{1}y^{-\alpha}(1-y)^{-\alpha}\,dy=x^{1-2\alpha}\Gamma^{2}(1-\alpha)/\Gamma(2-2\alpha)$. Hence, $f\ast g$ is not continuous at [math]. If $1/2<\alpha<1$ then $f\ast g$ is unbounded at [math].

As another example, consider $f(x)=\sin(\pi x)/\log|x|$ and $g(x)=\chi_{(0,1)}(x)$. Then $f\in{\cal A}_{C}$ and for each $1\leq p\leq\infty$ we have $g\in{\cal BV}\cap L^{p}$. And,

[TABLE]

Therefore, by Theorem 2(d), $f\ast g\in C^{0}(\overline{{\mathbb{R}}})$ and $\lim_{|x|\to\infty}f\ast g(x)=0$ but for each $1\leq p<\infty$ we have $f\ast g\not\in L^{p}$.

4 Differentiation and integration

If $g$ is sufficiently smooth then the pointwise derivative is $(f\ast g)^{\prime}(x)=f\ast g^{\prime}(x)$. Recall the definition $AC(\overline{{\mathbb{R}}})$ of primitives of $L^{1}$ functions given in the proof of Proposition 7. In the following theorem we require pointwise derivatives of $g$ to exist at each point in ${\mathbb{R}}$.

Theorem 10

Let $f\in{\cal A}_{C}$, $n\in{\mathbb{N}}$ and $g^{(k)}\in AC(\overline{{\mathbb{R}}})$ for each $0\leq k\leq n$. Then $f\ast g\in C^{n}({\mathbb{R}})$ and $(f\ast g)^{(n)}(x)=f\ast g^{(n)}(x)$ for each $x\in{\mathbb{R}}$.

Proof: First consider $n=1$. Let $x\in{\mathbb{R}}$. Then

[TABLE]

To take the limit inside the integral we can show that the bracketed term in the integrand is of uniform bounded variation for $0<|h|\leq 1$. Let $h\not=0$. Since $g\in AC(\overline{{\mathbb{R}}})$ it follows that the variation is given by the Lebesgue integrals

[TABLE]

Since $g^{\prime}\in AC(\overline{{\mathbb{R}}})$ we have $g^{\prime\prime}\in L^{1}$. The second integral on the right of (12) gives the $L^{1}$ derivative of $g^{\prime}$ in the limit $h\to 0$. See [4, p. 246]. Hence, in (11) we can use Theorem 22 in [9] to take the limit under the integral sign. This then gives $(f\ast g)^{\prime}(x)=f\ast g^{\prime}(x)$. Theorem 2(d) now shows $(f\ast g)^{\prime}\in C^{0}(\overline{{\mathbb{R}}})$. Induction on $n$ completes the proof.$\quad\blacksquare$

For similar results when $f\in L^{1}$, see [4, Proposition 8.10].

Note that $g^{\prime}\in AC(\overline{{\mathbb{R}}})$ does not imply $g\in AC(\overline{{\mathbb{R}}})$. For example, $g(x)=x$. The conditions $g^{(k)}\in{\cal BV}$ for $0\leq k\leq n+1$ imply those in Theorem 10. To see this it suffices to consider $n=1$. If $g^{\prime},g^{\prime\prime}\in{\cal BV}$ then $g^{\prime\prime}$ exists at each point and is bounded. Hence, the Lebesgue integral $g^{\prime}(x)=g^{\prime}(0)+\int_{0}^{x}g^{\prime\prime}(y)\,dy$ exists for each $x\in{\mathbb{R}}$ and $g^{\prime}$ is absolutely continuous. Since $g^{\prime}\in{\cal BV}$ we then have $g^{\prime}\in AC(\overline{{\mathbb{R}}})$. Similarly for $n>1$. The example $g(x)=|x|^{1.5}\sin(1/[1+x^{2}])$ shows the $AC(\overline{{\mathbb{R}}})$ condition in the theorem is weaker that the above ${\cal BV}$ condition since $g,g^{\prime}\in AC(\overline{{\mathbb{R}}})$ but $g^{\prime\prime}(0)$ does not exist so $g^{\prime\prime}\not\in{\cal BV}$.

We found that when $g\in{\cal BV}\cap L^{1}$ then $f\ast g\in{\cal A}_{C}$. We can compute the the distributional derivative $(F\ast g)^{\prime}=f\ast g$ where $F$ is a primitive of $f$.

Proposition 11

Let $F\in C^{0}(\overline{{\mathbb{R}}})$ and write $f=F^{\prime}\in{\cal A}_{C}$. Let $g\in{\cal BV}\cap L^{1}$. Then $F\ast g\in C^{0}(\overline{{\mathbb{R}}})$ and $(F\ast g)^{\prime}=f\ast g\in{\cal A}_{C}$.

Proof: Let $x,t\in{\mathbb{R}}$. Then by the usual Hölder inequality,

[TABLE]

Hence, $F\ast g$ is continuous on ${\mathbb{R}}$. Dominated convergence shows that $\lim_{x\to\pm\infty}F\ast g(x)=F(\pm\infty)\int^{\infty}_{-\infty}g$. Therefore, $F\ast g\in C^{0}(\overline{{\mathbb{R}}})$.

Let $\phi\in{\cal D}$. Then

[TABLE]

Integrate by parts and use the change of variables $x\mapsto x+y$ to get

[TABLE]

This gives an alternate definition of $f\ast g$ for $f\in{\cal A}_{C}$ and $g\in L^{1}$.

Theorem 12

Let $f\in{\cal A}_{C}$, let $F\in{\cal B}_{C}$ be the primitive of $f$ and let $g\in L^{1}$. Define $f\ast g$ as in Definition 5. Then $(F\ast g)^{\prime}=f\ast g\in{\cal A}_{C}$.

Proof: Let $-\infty<\alpha<\beta<\infty$. Let $\{g_{n}\}\subset{\cal BV}\cap L^{1}$ such that $\|g_{n}-g\|_{1}\to 0$. By Proposition 11 we have

[TABLE]

As in (3), $\int_{\alpha}^{\beta}f\ast g_{n}=\int^{\infty}_{-\infty}F(y)\left[g_{n}(\beta-y)-g_{n}(\alpha-y)\right]\,dy$. Hence,

[TABLE]

Therefore, $\|(F\ast g)^{\prime}-f\ast g_{n}\|\leq 2\|f\|\|g_{n}-g\|_{1}\to 0$ as $n\to\infty$. $\quad\blacksquare$

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

Theorem 13

Let $f\in{\cal A}_{C}$ and let $g\in L^{1}$. Define $F(x)=\int_{-\infty}^{x}f$ and $G(x)=\int_{-\infty}^{x}g$. Then $f\ast G\in C^{0}(\overline{{\mathbb{R}}})$ and $f\ast G(x)=F\ast g(x)$ for all $x\in{\mathbb{R}}$.

Proof: Since $G\in AC(\overline{{\mathbb{R}}})$, Theorem 2(d) shows $f\ast G\in C^{0}(\overline{{\mathbb{R}}})$. We have

[TABLE]

Proposition 17 justifies the interchange of orders of integration.$\quad\blacksquare$

Corollary 14

(a) $f\ast g=(F\ast g)^{\prime}=(f\ast G)^{\prime}$ (b) For all $-\infty\leq\alpha<\beta\leq\infty$ we have $\int_{\alpha}^{\beta}f\ast g=F\ast g(\beta)-F\ast g(\alpha)=f\ast G(\beta)-f\ast G(\beta)$.

Hence, the convolution $f\ast g$ can be evaluated by taking the distributional derivative of the Lebesgue integral $F\ast g$. Since $f\ast G\in C^{0}(\overline{{\mathbb{R}}})$ when $f\in{\cal A}_{C}$ and $G\in{\cal BV}$ we can use the equation $f\ast g=(f\ast G)^{\prime}$ to define $f\ast g$ for $f\in{\cal A}_{C}$ and $g=G^{\prime}$ for $G\in{\cal BV}$. In this case, $g$ will be a signed Radon measure. As $G(x)=\int_{-\infty}^{x}g$ and this integral is a regulated primitive integral [10], we will save this case for discussion elsewhere.

5 Appendix

The integration by parts formula is as follows. If $f\in{\cal A}_{C}$ and $g\in{\cal BV}$ it gives the integral of $fg$ in terms of a Henstock–Stieltjes integral:

[TABLE]

See [9] and [6, p. 199].

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

Corollary 15

Let $F\in C^{0}(\overline{{\mathbb{R}}})$. Let $g\in{\cal EBV}$. Fix $0\leq\gamma\leq 1$. Take $g_{\gamma}\in{\cal NBV}_{\gamma}$ such that $g_{\gamma}=g$ almost everywhere. Let $\mu_{g}$ be the signed Radon measure given by $g^{\prime}$. Then $\int^{\infty}_{-\infty}F\,dg_{\gamma}=\int^{\infty}_{-\infty}F\,d\mu_{g}$.

Proof: The distributional derivative of $g$ is $\langle g^{\prime},\phi\rangle=-\langle g,\phi^{\prime}\rangle=-\int^{\infty}_{-\infty}g\phi^{\prime}=\int^{\infty}_{-\infty}\phi\,d\mu_{g}$ for all $\phi\in{\cal D}$. Note that $g_{\gamma}$ is unique and $\mu_{g}=\mu_{g_{\gamma}}$. Suppose $\phi\in{\cal D}$ with ${\rm supp}(\phi)\subset[A,B]\subset{\mathbb{R}}$. Then, using integration by parts for the Henstock–Stieltjes integral,

[TABLE]

Let $F\in C^{0}(\overline{{\mathbb{R}}})$. There is a uniformly bounded sequence $\{\phi_{n}\}\subset{\cal D}$ such that $\phi_{n}\to F$ pointwise on ${\mathbb{R}}$. By dominated convergence,

[TABLE]

Corollary 15 now justifies the following definition.

Definition 16

Let $f\in{\cal A}_{C}$ and let $F\in{\cal B}_{C}$ be its primitive. Let $g\in{\cal EBV}$. Fix $0\leq\gamma\leq 1$ and take $g_{\gamma}\in{\cal NBV}_{\gamma}$ such that $g_{\gamma}=g$ almost everywhere. Define

[TABLE]

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

The Hölder inequality is

[TABLE]

and is valid for all $f\in{\cal A}_{C}$ and $g\in{\cal BV}$. For $g\in{\cal EBV}$ we replace $g$ with $g_{\gamma}$. This gives

[TABLE]

See [8, Lemma 24] for a proof using the Henstock–Kurzweil integral. The same proof works for the continuous primitive integral.

A Fubini theorem has been established in [1] 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 on page $58$ in [2] for the wide Denjoy integral on compact intervals.

Proposition 17

Let $f\in{\cal A}_{C}$. Let $g\!:\!{\mathbb{R}}\times{\mathbb{R}}\to{\mathbb{R}}$ be measurable. Assume (i) for each $x\in{\mathbb{R}}$ the function $y\mapsto g(x,y)$ is in ${\cal BV}$; (ii) the function $x\mapsto V_{y\in{\mathbb{R}}}g(x,y)$ is in $L^{1}$; (iii) there is $M\in L^{1}$ such that for each $y\in{\mathbb{R}}$ we have $|g(x,y)|\leq M(x)$. Then the iterated integrals exist and are equal, $\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}f(y)g(x,y)\,dy\,dx=\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}f(y)g(x,y)\,dx\,dy$.

Proof: Let $F\in{\cal B}_{C}$ be the primitive of $f$. For each $x\in{\mathbb{R}}$ we have

[TABLE]

where $d_{2}(x,y)$ indicates a Henstock–Stieltjes integral with respect to $y$. Then,

[TABLE]

The integral $\int^{\infty}_{-\infty}g(x,\infty)\,dx$ exists due to condition (iii). The iterated integral in (16) converges absolutely since

[TABLE]

Now, show the function $y\mapsto\int^{\infty}_{-\infty}g(x,y)\,dx$ is in ${\cal BV}$. Let $\{(s_{i},t_{i})\}_{i=1}^{n}$ be disjoint intervals in ${\mathbb{R}}$. Then

[TABLE]

The interchange of summation and integration follows from condition (ii) and the usual Fubini–Tonelli theorem. Hence, the function $y\mapsto\int^{\infty}_{-\infty}g(x,y)\,dx$ is in ${\cal BV}$ and the iterated integral $\int^{\infty}_{-\infty}f(y)\int^{\infty}_{-\infty}g(x,y)\,dx\,dy$ exists.

Integrate by parts,

[TABLE]

In (18), we have $\lim_{y\to\infty}\int^{\infty}_{-\infty}g(x,y)\,dx=\int^{\infty}_{-\infty}g(x,\infty)\,dx$ due to dominated convergence and condition (iii). To complete the proof we need to show the integrals in (16) and (18) are equal. First consider the case when $F=\chi_{(a,b)}$ for an interval $(a,b)\subset{\mathbb{R}}$. Then (16) becomes $\int^{\infty}_{-\infty}\int_{a}^{b}d_{2}g(x,y)\,dx=\int^{\infty}_{-\infty}\left[g(x,b)-g(x,a)\right]dx$. And now (18) becomes $\int_{a}^{b}d\left[\int^{\infty}_{-\infty}g(x,y)\,dx\right]=\int^{\infty}_{-\infty}g(x,b)\,dx-\int^{\infty}_{-\infty}g(x,a)\,dx$. Hence, when $F$ is a step function, $F(y)=\sum_{i=1}^{n}c_{i}\chi_{I_{i}}(y)$ for some $n\in{\mathbb{N}}$, disjoint intervals $\{I_{i}\}_{i=1}^{n}$ and real numbers $\{c_{i}\}_{i=1}^{n}$, we have the desired equality of (16) and (18). But $F\in{\cal B}_{C}$ is uniformly continuous on $\overline{{\mathbb{R}}}$, i.e., for each $\epsilon>0$ there is $\delta>0$ such that for all $0\leq|x-y|<\delta$ we have $|F(x)-F(y)|<\epsilon$, for all $x<-1/\delta$ we have $|F(x)|<\epsilon$ and for all $x>1/\delta$ we have $|F(x)-F(\infty)|<\epsilon$. It then follows from the compactness of $\overline{{\mathbb{R}}}$ that the step functions are dense in ${\cal B}_{C}$. Hence, there is a sequence of step functions $\{\sigma_{N}\}$ such that $\|F-\sigma_{N}\|_{\infty}\to 0$. In (16) we have

[TABLE]

The $N$ limit can be brought inside the $x$ integral using dominated convergence and (ii) since $|\int^{\infty}_{-\infty}\sigma_{N}(y)\,d_{2}g(x,y)|\leq(\|F\|_{\infty}+1)V_{y\in{\mathbb{R}}}g(x,y)$ for large enough $N$. The $N$ limit can be brought inside the $y$ integral using dominated convergence since $|\sigma_{N}(y)|\leq(\|F\|_{\infty}+1)$ for large enough $N$. In (18) we have

[TABLE]

The $N$ limit can be brought inside the $y$ integral since $\{\sigma_{N}\}$ converges to $F$ uniformly on $\overline{{\mathbb{R}}}$ and $d\left[\int^{\infty}_{-\infty}g(x,y)\,dx\right]$ is a finite signed measure. $\quad\blacksquare$

Corollary 18

If $f$ has compact support we can replace (iii) with: (iv) for each $y\in{\rm supp}(f)$ the function $x\mapsto g(x,y)$ is in $L^{1}$.