Fourier analysis for vector-measures on compact abelian groups

Abstract

We analyze the Fourier transform of vector measures $\nu$ as well as the convolution between scalar and vector-valued regular measures defined on the Borel sets of a compact abelian group. We make special emphasis on the Riemann–Lebesgue lemma and Young’s convolution type results in this setting. Applications to Fourier transform and convolutions between functions in $L^p(\nu )$ and, particularly, for translation invariant type measures $\nu$ are given.

1 Introduction

Let G be a compact abelian group, we write $\mathcal{B}(G)$ for the Borel $\sigma$-algebra of G and $m_G$ for the Haar measure of the group. We denote by $L^0(G)$ the space of Borel measurable functions defined on G and $L^p(G)$ the space of functions in $L^0(G)$ such that $\int \nolimits _G |f|^pdm_G<\infty$.

Given $1<p\le \infty$, a non-negative measure $\lambda$ on $\mathcal{B}(G)$, a Banach space X and a vector measure $\nu : \mathcal{B}(G)\rightarrow X$ we denote by

$$\begin{aligned} \Vert \nu \Vert (A)= \sup _{\Vert x'\Vert =1} |{\langle }\nu ,x'{\rangle }|(A) \end{aligned}$$

the semivariation on a Borel set A and write $\Vert \nu \Vert$ for $\Vert \nu \Vert (G)$, by

$$\begin{aligned} |\nu |(A)=\sup \left\{ \sum _{E\in \pi } \Vert \nu (E)\Vert :\pi \hbox { finite partition of } A\right\} \end{aligned}$$

the variation of $\nu$, by $\Vert \nu \Vert _{p,\lambda }$ the p-semivariation of $\nu$ with respect to $\lambda$ (see [1, p. 246]) defined, for $1<p<\infty$, by

$$\begin{aligned} \Vert \nu \Vert _{p,\lambda }=\sup \left\{ \left\| \sum _{A\in \pi } \alpha _A \nu (A)\right\| _X: \pi \hbox { partition }, {\left\| \sum _{A\in \pi } \alpha _A\chi _A\right\| }_{L^{p'}(\lambda )}\le 1 \right\} , \end{aligned}$$

(1)

and

$$\begin{aligned} \Vert \nu \Vert _{\infty ,\lambda }=\sup _{\lambda (A)>0}\frac{\Vert \nu (A)\Vert }{\lambda (A)}. \end{aligned}$$

We shall use the notation $\mathcal M(G,X)$ for the space of regular vector measures, $\mathcal M_{ac}(G,X)$ for those which are absolutely continuous with respect $m_G$, i.e. $\nu << m_G$,$\mathcal M_{p}(G,X)$ for those with bounded p-semivariation with respect to $m_G$, i.e. $\Vert \nu \Vert _{p,m_G}<\infty$, M(G, X) for those with bounded variation, and finally we write $M_{ac}(G,X)=\mathcal M_{ac}(G,X)\cap M(G,X)$.

As usual, for a given vector measure $\nu$, we write $L_w^1(\nu )$ for the space of functions in $L^0(G)$ such that $\int \nolimits _G |f| d|{\langle }\nu ,x'{\rangle }|<\infty$ for any $x'\in X'$ and we write $L^1(\nu )$ for the subspace of $L_w^1(\nu )$ satisfying that for any $A\in \mathcal{B}(G)$ there exists $x_A\in X$ for which ${\langle }x',x_A{\rangle }=\int \nolimits _A f d{\langle }\nu , x'{\rangle }$. For each $f\in L^1(\nu )$ we denote

$$\begin{aligned} \nu _f(A)=x_A= \int \nolimits _A fd\nu \end{aligned}$$

given as above. Then $\nu _f$ is a vector measure and $\Vert \nu _f\Vert =\Vert f\Vert _{L^1(\nu )}$. We denote $I_\nu$ the integration operator, i.e. $I_\nu :L^1(\nu )\rightarrow X$ is defined by

$$\begin{aligned} I_\nu (f)=\nu _f(G)=\int \nolimits _G fd\nu \end{aligned}$$

and satisfies that $\Vert I_\nu \Vert \le \Vert \nu \Vert$. In the case $f\in L^1_w(\nu )$ we can look of $\nu _f$ as $X''$-valued measure, using ${\langle }\nu _f(A),x'{\rangle }=\int \nolimits _A f d{\langle }\nu ,x'{\rangle }$ for each $A\in \mathcal B(G).$ As usual, for $1<p<\infty$ we denote $L^p(\nu )=\{f\in L^0(G): |f|^p\in L^1(\nu )\}$ and $\Vert f\Vert _{L^p(\nu )}=\Vert |f|^p\Vert ^{1/p}_{L^1(\nu )}$ (see [2]).

The Fourier transform of functions in $L^1(\nu )$ was introduced in [3] as the X-valued function defined on the dual group of $\Gamma$ by

$$\begin{aligned} \hat{f}^\nu (\gamma )=\int \nolimits _G f(t)\overline{\gamma (t)}d\nu (t), \quad \gamma \in \Gamma \end{aligned}$$

(2)

where $f\in L^1(\nu )$ and $\nu$ is any vector measure. The validity of the Riemann–Lebesgue lemma in this setting was considered and it was shown, under the assumption $\nu << m_G$, that the fact $\hat{f}^\nu \in c_0(\Gamma , X)$ for any $f\in L^1(\nu )$ reduces to consider the case $f=\chi _G$ (see [3, Theorem2.5]). The following problems were left open:

• (a) Does it hold that $\hat{f}^\nu \in c_0(\Gamma , X)$ whenever $\nu <<m_G$ and $f\in L^1(\nu )$, for any Banach space X?

• (b) Are there natural subclasses of vector measures for which this version of the Riemann–Lebesgue lemma holds?

• (c) Are there classes of operators that transform vector measures in vector measures satisfying this formulation of the Riemann–Lebesgue lemma?

They also introduce the Fourier transform $\mathcal F_\nu (f)$ of functions $f\in L^1_w(\nu )$ in the case $\nu <<m_G$ as bounded operators in $\mathcal L(X',\ell ^\infty (\Gamma ))$ given by

$$\begin{aligned} \mathcal F_\nu (f)(x')= \widehat{f h_{x'}} \end{aligned}$$

(3)

where $d{\langle }\nu , x'{\rangle }= h_{x'} dm_G$ with $h_{x'}\in L^1(G)$.

We shall understand these cases as particular ones of the Fourier transform of a vector measure $\nu$ defined by

$$\begin{aligned} \hat{\nu }(\gamma )=I_\nu (\bar{\gamma }) \end{aligned}$$

(4)

when dealing with $\nu _f$. The paper is organized into six sections. In Sect. 2 we give some preliminaries on vector measures to be used in the sequel. We shall study some versions of Riemann–Lebesgue lemma in our context and give answers to the above problems in Sect. 3. In Sect. 4 we introduce the convolution of a vector measure $\nu$ and another complex-valued regular measure $\mu \in M(G)$ by means of the formula

$$\begin{aligned} \mu *\nu (A)=\int \nolimits _G \mu (A-t)d\nu (t), \quad A\in \mathcal B(G) \end{aligned}$$

where the map $t\rightarrow \mu (A-t)$ is shown first to be measurable and bounded (and hence in $L^1(\nu )$). This notion is seen to coincide with the symmetric formulation

$$\begin{aligned} \nu *\mu (A)=\int \nolimits _G \nu (A-t)d\mu (t), \quad A\in \mathcal B(G) \end{aligned}$$

for regular measures $\nu \in \mathcal M(G;X)$ and $\mu \in M(G)$.

This concept is when restricted to measures $d\mu _f= f dm_G$ becomes

$$\begin{aligned} f*\nu (A)= I_\nu (\tilde{f}*\chi _A), \quad A\in \mathcal B(G) \end{aligned}$$

(5)

where $\tilde{f}(t)=f(-t)$. Our point of view actually extends the two different convolution maps considered in [3, Definitions4.1and4.5]: If $\nu$ be a vector measure such that $\nu << m_G$, $f\in L^1(G)$ and $g\in L^1_w(\nu )$ the authors introduced $f*_\nu g:X'\rightarrow L^1(G)$ as

$$\begin{aligned} f*_\nu g(x'):= f*(gh_{x'}), \quad x'\in X' \end{aligned}$$

(6)

where $h_{x'}= \frac{d{\langle }\nu ,x'{\rangle }}{dm_G}$.

In the case that $g\in L^1(\nu )$ and $f(t-\cdot )g\in L^1(\nu )$ for $m_G$-almost all $t\in G$ they also defined

$$\begin{aligned} f*^\nu g(t):=\int \nolimits _G f(t-s)g(s) d\nu (s). \end{aligned}$$

(7)

Using the fact that $\nu _g\in \mathcal M(G,X'')$ we actually have

$$\begin{aligned} d{\langle }f*\nu _g,x'{\rangle }= f*_\nu g(x') dm_G, \quad x'\in X' \end{aligned}$$

and also, in the case $g\in L^1(\nu )$ and $f\in C(G)$, we obtain $\nu _g\in \mathcal M(G,X)$ and $f*\nu _g(t)= f*^\nu g(t), \quad t\in G.$

Different formulations of Young’s convolution theorems will be provided which will extend several results in [3] when restricted to measures $\nu _f$. In particular we show that for $1\le p,q \le \infty$ and $1/p+1/q\ge 1$ if $\nu \in \mathcal M_p(G,X)$ and $f\in L^q(G)$ then $\nu *f \in P_r(G,X)$ for $1/p+1/q-1=1/r$ (see definition in Sect. 2).

Section 5 is devoted to analyze the cases where $\nu$ is a translation invariant-type measure. In the paper [4] the notion of “norm integral translation invariant” vector measure was introduced, by the condition

$$\begin{aligned} \Vert I_\nu (\tau _a \phi )\Vert = \Vert I_\nu (\phi )\Vert , \phi \in \mathcal S(G), \quad a\in G \end{aligned}$$

(8)

where $\tau _a(\phi )(s)=\phi (s-a)$. For norm integral translation invariant measures $\nu$ such that $\nu <<m_G$ they showed that $L^1_w(\nu )\subset L^1(G)$ and therefore the convolution and the Fourier transform of functions in $L^1_w(\nu )$ are well defined. One of their main theorems establishes that if $f\in L^1(G)$ and $g\in L^p(\nu )$ then $f*g\in L^p(\nu )$ for $1\le p<\infty$. Later in [3] this notion was generalized and used for more general homeomorphisms $H:G\rightarrow G$, and the particular case of reflection invariant ones (i.e. $H(s)=-s$) played an important role when considering convolution of functions in $L^1(\nu )$. In this paper we shall introduce a weaker but still useful notion to be denoted “semivariation H-invariant” by

$$\begin{aligned} \Vert \nu _f\Vert = \Vert (\nu _{H})_f\Vert , \quad f\in \mathcal S(G) \end{aligned}$$

(9)

where $\nu _H(A)=\nu (H(A))$ for $A\in \mathcal{B}(G)$. This definition will be shown to be different to the “norm integral H-invariant”. However for semivariation translation invariant measures $\nu$ we will still have $L^1(\nu )\subset L^1(G)$. Hence similar results as those in [4] for such a weaker notion will remain valid. We finally include in Sect. 6 several applications of our general theory for vector measures to the study of convolution and Fourier transform of functions in $L^1(\nu )$.

2 Preliminaries on vector measures

Let us start by recalling that a vector measure $\nu$ defined on the Borel $\sigma$-algebra $\mathcal B(G)$ is called regular if for any $\varepsilon >0$ and $A\in \mathcal{B}(G)$ there exists a compact set K and an open set O such that $K\subset A\subset O$ and $\Vert \nu \Vert (O{\setminus }K)<\varepsilon$. It is clear that if $\nu << \lambda$ for some finite regular Borel measure $\lambda$ then $\nu \in \mathcal M(G,X)$. In particular $\nu \in \mathcal M(G,X)$ if and only if any Rybakov control measure $|{\langle }\nu , x_0'{\rangle }|$ is regular.

As usual we denote $\mathcal S(G,X)$ the space of X-valued simple functions and we keep the notation $L^p(G,X)$ for the completion of $\mathcal S(G,X)$ under the norm

$$\begin{aligned} \Vert {\mathbf s}\Vert _{L^p(G,X)}=\left( \int \nolimits _G \Vert {\mathbf s}\Vert ^pdm_G\right) ^{1/p} \end{aligned}$$

in the case $1\le p<\infty$ and write $L^\infty _0(G,X)$ the closure of $\mathcal S(G,X)$ in $L^\infty (G,X)$. It is well known that $L^1(G,X)\subset M_{ac}(G,X)$. Actually, for each ${\mathbf f}\in L^1(G,X)$ we define

$$\begin{aligned} \nu _{\mathbf f}(A)=\int \nolimits _A {\mathbf f} dm_G. \end{aligned}$$

One has that $\nu _{\mathbf f}\in M_{ac}(G,X)$ since $\nu << m_G$ and $|\nu _{\mathbf f}|(A)=\int \nolimits _A \Vert {\mathbf f}\Vert dm_G$ (see [5, p. 46]).

We also have that $\mathcal M(G,X)$ endowed with the norm given by the semivariation becomes a Banach space and that $\mathcal M_{ac}(G,X)$ is a closed subspace of $\mathcal M(G,X).$ It is well known (see [5, p. 159]) that $\mathcal M(G,X)$ is isometric to the space to weakly compact linear operators. To each $\nu \in \mathcal M(G,X)$ corresponds a weakly compact operator $T_\nu :C(G)\rightarrow X$ such that $\Vert T_\nu \Vert = \Vert \nu \Vert$ and we shall write $\int \nolimits _G \phi d\nu =T_\nu (\phi )$ for each $\phi \in C(G)$. We also recall that M(G, X) is isometric to the space of absolutely summing operators $\Pi _1(C(G),X)$(see [5, p. 162]). For $1<p\le \infty$, let us also mention that $\mathcal M_{p}(G,X)$ can be identified with $\mathcal L(L^{p'}(G),X)$ (see [1, p. 259]). In other words, if $\nu \in \mathcal M_{p}(G,X)$ then $T_\nu$ extends to a bounded operator in $\mathcal L(L^{p'}(G),X)$ with $\Vert \nu \Vert _{p,m_G}=\Vert T_\nu \Vert _{\mathcal L(L^{p'}(G),X)}$, and, conversely, for each $T:L^{p'}(G)\rightarrow X$ we associate the vector measure $\nu _{_{T}}:\mathcal B(G) \rightarrow X$ given by $\nu _{_{T}}(A)=T(\chi _{A})$ satisfying that $\nu _{_{T}}\in \mathcal M_p(G, X)$ and $\Vert \nu _{_{T}}\Vert _{p,m_G}=\Vert T\Vert$. This allows to produce easy examples in $\mathcal M_p(G, X)$. For instance, for $X=L^p(G)$ the $L^p(G)$-valued measure

$$\begin{aligned} m_p(A)=\chi _A, \quad A\in \mathcal{B}(G)\end{aligned}$$

belongs to $\mathcal M_{p'}(G, L^{p}(G))$. Another important example is produced using Pettis integrable functions, namely if ${\mathbf f}:G\rightarrow X$ is Pettis integrable and ${\langle }{\mathbf f}, x'{\rangle }\in L^p(G)$ for each $x'\in X'$ then the vector measure

$$\begin{aligned} m_{\mathbf f}(A)=(P)\int \nolimits _{A} \mathbf {f}dm_G, \quad \in \mathcal{B}(G), \end{aligned}$$

(where the integral denotes the Pettis integral of $\mathbf {f}$ over the set A) belongs to $\mathcal M_p(G,X)$ and $\Vert m_{\mathbf {f}}\Vert _{p,m_G}= \sup _{\Vert x'\Vert =1}\Vert {\langle }{\mathbf f}, x'{\rangle }\Vert _{ L^p(G)}$.

For each $1\le p\le \infty$ and $\mathbf s\in \mathcal S(G,X)$ and denote

$$\begin{aligned} \Vert \mathbf s\Vert _{P_p(G,X)}= \Vert \nu _{\mathbf s}\Vert _{p,m_G}=\sup _{\Vert x'\Vert =1} \Vert {\langle }{\mathbf s}, x'{\rangle }\Vert _{ L^p(G)}. \end{aligned}$$

(10)

We define $P_p(G,X)$ the closure of $\mathcal S(G,X)$ in $\mathcal M_p(G,X)$ for $1\le p\le \infty$ where we understand $\mathcal M_1(G,X)=\mathcal M(G,X)$. Since C(G) is dense in $L^p(G)$ for $1\le p<\infty$ and closed for $p=\infty$ we easily see that C(G, X) is dense in $P_p(G,X)$ for $1\le p<\infty$ and C(G, X) is closed in $P_\infty (G,X)$.

It is elementary to see that $L^p(G,X)\subseteq P_p(G,X), 1\le p<\infty$, $L^\infty _0(G,X)\subseteq P_\infty (G,X)$ and $P_{p_2}(G,X)\subseteq P_{p_1}(G,X), p_1\le p_2.$ Finally notice that since the measure $\nu _{\mathbf s}$ defines a finite rank operator on $L^{1}(G)$ into X for each ${\mathbf s}\in \mathcal S(G,X)$, one sees that $P_\infty (G,X)\subseteq L^\infty (G,X)$ (see [5, p.68]).

Let us finish this preliminary section by showing, for the sake of completeness, that C(G) is dense in $L^1(\nu )$ for regular measures $\nu$.

Lemma 2.1

Let $\nu \in \mathcal M(G,X)$ and $1\le p<\infty$. Then C(G) is dense in $L^p(\nu )$. Moreover

$$\begin{aligned} I_\nu (f)=\lim _{n}T_\nu (f_n) \end{aligned}$$

for any $(f_n)\in C(G)$ with $\lim f_n=f\in L^1(\nu )$.

Proof

Assume $\nu$ is regular and let us prove that C(G) is dense in $L^p(\nu )$. Since simple functions are dense in $L^p(\nu )$ it suffices to see that for any $\varepsilon >0$ and $A\in \mathcal{B}(G)$ there exists $\phi \in C(G)$ such that $\Vert \chi _A-\phi \Vert _{L^p(\nu )}<\varepsilon$. Using the regularity of $\nu$ we first select a compact set K and an open set O such that $K\subset A\subset O$ with $\Vert \nu \Vert (O{\setminus }K)<\varepsilon ^p$. Then use Uryshon’s lemma to find $\phi \in C(G)$ such that $0\le \phi \le 1$ and $\phi (t)=1$ for $t\in K$ and $\phi (t)=0$ for $t\notin O$. Finally observe that

$$\begin{aligned} \Vert \chi _A-\phi \Vert _{L^p(\nu )}= & {} \sup _{\Vert x'\Vert =1} \left( \int \nolimits _G |\chi _A-\phi |^pd|{\langle }\nu ,x'{\rangle }|\right) ^{1/p} \\= & {} \sup _{\Vert x'\Vert =1} \left( \int \nolimits _{O{\setminus }K} |\chi _A-\phi |^pd|{\langle }\nu ,x'{\rangle }|\right) ^{1/p}\le \Big (\Vert \nu \Vert (O{\setminus }K)\Big )^{1/p}<\varepsilon . \end{aligned}$$

Let $(f_n)$ be any sequence of continuous functions converging to f in $L^1(\nu )$. Since $I_\nu (f_n)=T_\nu (f_n)$ we have

$$\begin{aligned} \Vert I_\nu (f)-T_\nu (f_n) \Vert \le \Vert f-f_n\Vert _{L^1(\nu )} \end{aligned}$$

and the proof is finished. $\square$

Corollary 2.2

Let $\nu \in \mathcal M(G,X)$ and $A\in \mathcal B(G)$. Then the map

$$\begin{aligned} {\mathbf f}(t)= \nu (A-t) \end{aligned}$$

is (strongly)-measurable and bounded.

Proof

It is obviously bounded by $\Vert \nu \Vert$. We shall show that ${\mathbf f}(t)=\lim _n {\mathbf f}_n(t)$ for some sequence ${\mathbf f}_n\in C(G,X)$. For each $t\in G$ we write $\nu _t$ for the regular measure such that

$$\begin{aligned} \int \nolimits _G \phi (s+t)d\nu (s)= \int \nolimits _G \phi (s)d\nu _t(s), \quad \phi \in C(G). \end{aligned}$$

From Lemma 2.1 select $\phi _n\in C(G)$ such that $\lim _n\Vert \chi _{A} - \phi _n\Vert _{L^1(\nu _t)}=0$. Define ${\mathbf f}_n(t)=\int \nolimits _G \phi _n(s+t)d\nu (s)$ and observe that

$$\begin{aligned} \Vert {\mathbf f}_n(t)- {\mathbf f}_n(t')\Vert \le \sup _{s\in G} |\phi _n(s+t)- \phi _n(s+t')| \Vert \nu \Vert . \end{aligned}$$

Hence ${\mathbf f}_n\in C(G,X)$ and we have

$$\begin{aligned} {\mathbf f}(t)= \int \nolimits _G \chi _{A}(s)d\nu _t(s)=\lim \nolimits _n \int \nolimits _G \phi _n(s)d\nu _t(s)=\lim _n {\mathbf f}_n(t). \end{aligned}$$

This finishes the proof. $\square$

3 Fourier transform and the Riemann–Lebesgue lemma

Definition 3.1

Let $\nu$ be a vector measure. We define the Fourier transform by

$$\begin{aligned} \hat{\nu }(\gamma )=\int \nolimits _G \bar{\gamma }d\nu =I_\nu (\bar{\gamma }), \quad \gamma \in \Gamma . \end{aligned}$$

In the case that $\nu \in \mathcal M(G,X)$ and $T_\nu :C(G)\rightarrow X$ is the corresponding weakly compact operator representing the measure we have $\hat{\nu }(\gamma )=T_\nu (\bar{\gamma }).$

Of course $\hat{f}^\nu (\gamma )=\widehat{\nu _f} (\gamma )$ whenever $f\in L^1(\nu )$ and, in the case $f\in L^1_w(\nu )$ and $\nu <<m_G$, we can consider $\nu _f(A)\in X''$ given by

$$\begin{aligned} {\langle }\nu _f(A),x'{\rangle }= \int \nolimits _A f d{\langle }\nu ,x'{\rangle }\end{aligned}$$

as a $X''$-valued vector measure and then $\mathcal F_\nu (f)(x')(\gamma )= {\langle }\widehat{\nu _f}(\gamma ), x'{\rangle }, \quad \gamma \in \Gamma , x\in X'.$

It is straightforward to see that $\hat{\nu }\in \ell ^\infty (\Gamma , X)$ with $\sup _{\gamma \in \Gamma }\Vert \hat{\nu }(\gamma )\Vert \le \Vert \nu \Vert .$ Due to the Radon–Nikodym theorem in the case $X=\mathbb {C}$ (or even for finite dimensional spaces X) we can say that the Riemman–Lebesgue lemma (see [6]) establishes that $\hat{\nu }\in c_0(\Gamma ,X)$ whenever $\nu << m_G$.

We would like to study the validity of the Riemann–Lebesgue lemma for measures in $\mathcal M_{ac}(G,X)$. In other words, if we denote

$$\begin{aligned} \mathcal M_{0}(G,X)=\{ \nu \in \mathcal M(G,X): \hat{\nu }\in c_0(\Gamma ,X)\} \end{aligned}$$

we ask ourselves whether or not $\mathcal M_{ac}(G,X)\subset \mathcal M_{0}(G,X)$.

As expected the answer is negative in general as the following easy example shows: Let $G=\mathbb {T}$, $X=\ell ^2(\mathbb {Z})$ and $\nu (A)=(\hat{\chi }_A(n))_{n\in \mathbb {Z}}$. Clearly $T_\nu :C(\mathbb {T})\rightarrow \ell ^2(\mathbb {Z})$ corresponds $T(f)= (\hat{f}(n))_{n\in \mathbb {Z}}$ . Hence $\hat{\nu }(n)= e_n$ where $(e_n)$ is the canonical basis and $\Vert \hat{\nu }(n)\Vert =1$ for each $n\in \mathbb {Z}$.

However, from the classical Riemann–Lebesgue lemma we have the following weak version in the vector-valued setting.

$$\begin{aligned} \nu \in \mathcal M_{ac}(G,X) \Longrightarrow {\langle }\hat{\nu }, x'{\rangle }\in c_o(\Gamma ),\quad x'\in X'. \end{aligned}$$

(11)

Answering question (a) commented in the Introduction we show now that $\mathcal M_{0}(G,X)\subset \mathcal M_{ac}(G,X)$ if and only if X is finite dimensional.

Proposition 3.2

Let X be an infinite dimensional Banach space and $G=\mathbb {T}$. There exists a regular vector measure $\nu : {\mathcal B}(\mathbb {T})\rightarrow X$ such that $\nu << m_\mathbb {T}$ and $\hat{\nu }\notin c_0(\mathbb {Z},X)$.

Proof

Let us first take a sequence $x_n\in X$ such that $1{/}2\le \Vert x_n\Vert \le 1$ satisfying

$$\begin{aligned} \left\| \sum _{n=1}^\infty \alpha _n x_n\right\| \le C \left( \sum _{n=1}^\infty |\alpha _n|^2\right) ^{1/2} \end{aligned}$$

(12)

(see [7, Lemma 1.3]). Define $\nu : \mathcal B(\mathbb {T})\rightarrow X$ by

$$\begin{aligned} \nu (A)= \sum _{n=1}^\infty \hat{\chi }_A(n) x_n. \end{aligned}$$

Since $\sum \nolimits _{n=1}^\infty |\hat{\chi }_A(n)|^2\le \Vert \chi _A\Vert ^2_{L^2(\mathbb {T})}=m_\mathbb {T}(A)$ we have that $\nu$ is well defined. Actually we have $\nu (A)= T(m_2(A))$ where $T:L^2(\mathbb {T})\rightarrow X$ is given by $T(\phi )= \sum \nolimits _{n=1}^\infty \hat{\phi }(n)x_n$ and $m_2: \mathcal B(\mathbb {T}) \rightarrow L^2(\mathbb {T})$ is given by $m_2(A)=\chi _A$ for $A\in \mathcal B(\mathbb {T})$. Hence $\nu \in \mathcal M_{ac}(\mathbb {T},X)$, but $\hat{\nu }(n)= x_n$ does not belong to $c_0(\mathbb {Z},X)$ since $\Vert x_n\Vert \ge 1/2$. $\square$

Next question is to find some natural classes of measures in $\mathcal M_0(G,X)$.

Proposition 3.3

If $\nu \in \mathcal M_{ac}(G,X)$ and $\nu$ has relatively compact range then $\nu \in \mathcal M_0(G,X)$.

Proof

Using that $\nu << m_G$ we conclude that $T^*_\nu :X'\rightarrow L^1(G)$ is given by $T^*_\nu (x')= h_{x'}$ where $d{\langle }\nu ,x'{\rangle }= h_{x'}dm_G$ for each $x'\in X'$. Using now that the unit ball of $L^\infty (G)$ is the closed absolutely convex hull of $\{\chi _A: A\in \mathcal B(G)\}$ and $\nu (A)=T^{**}(\chi _A)$ we obtain that $T^{**}_\nu :L^\infty (G)\rightarrow X''$ is compact (and hence so it is $T_\nu$). This implies that $\{\hat{\nu }(\gamma )=T_\nu (\bar{\gamma }):\gamma \in \Gamma \}$ is relatively compact and, according to (11) also weakly null. Therefore $\hat{\nu }(\gamma )\in c_0(\Gamma , X)$. $\square$

Corollary 3.4

$P_1(G,X)\subset \mathcal M_0(G,X)$.

Let us now study the question of finding classes of bounded operators $T:X\rightarrow Y$ that transform measures in $\mathcal M_{ac}(G,X)$ into measures in $\mathcal M_{0}(G,Y)$. Recall that an operator $T:X\rightarrow Y$ is said to be completely continuous, or Dunford–Pettis, if it maps weakly convergent sequences in X into norm convergent sequences in Y. Hence a simple consequence of (11) and the above definition gives the following result.

Proposition 3.5

Let $T:X\rightarrow Y$ be a completely continuous operator and $\nu \in \mathcal M_{ac}(G,X)$. Then $T(\nu )\in \mathcal M_{0}(G,Y)$.

Let us restrict ourselves to study the version of Riemann–Lebesgue lemma for measures of bounded variation. In general $M_{ac}(G,X)$ is not contained in $\mathcal M_{0}(G,X)$ as it can be seen in the following example: let $G=\mathbb {T}$, $X=L^1(\mathbb {T})$ and $\nu (A)=\chi _A$. Clearly $T_\nu :C(\mathbb {T})\rightarrow L^1(\mathbb {T})$ corresponds to the inclusion map then $\hat{\nu }(n)= \phi _n$ where $\phi _n(t)=e^{int}$ and $\Vert \hat{\nu }(n)\Vert =1$ for each $n\in \mathbb {Z}$.

However there are conditions which allow to have such a version of the Riemann–Lebesgue lemma. For instance, if X has the Radon Nikodym property (see for instance [5] for the definition) then $M_{ac}(G,X)\subset \mathcal M_{0}(G,X)$. Under the RNP we have that $\nu \in M_{ac}(G,X)$ gives $d\nu = {\mathbf f}dm_G$ for some ${\mathbf f}\in L^1(G,X)$ and $\hat{\nu }(n)=\hat{\mathbf f}(n)=\int \nolimits _\mathbb {T}{\mathbf f}(e^{it})e^{-int}dt$ for $n\in \mathbb {Z}$, which belongs to $c_0(\mathbb {Z},X)$.

Definition 3.6

We say that a Banach space satisfies the Riemann–Lebesgue property for measures on G (in short, $X\in (RLP)_G$) if any vector measure $\nu$ satisfying $\nu <<m_G$ and $|\nu |(G)<\infty$ satisfies that $\hat{\nu }\in c_0(\Gamma ,X)$, i.e. $M_{ac}(G,X)\subset \mathcal M_{0}(G,X)$ .

We would like to show that this notion in the case $G=\mathbb {T}$ implies the Riemann–Lebesgue property introduced and considered by Bu and Chill [8] for the case $G=\mathbb {T}$. They worked in the spaces

$$\begin{aligned} L_{1}^{max}(\mathbb {T},X'')=\{{\mathbf f}:\mathbb {T}\rightarrow X'' weak^*-\hbox {meas.}: \sup _{\Vert x'\Vert =1} |{\langle }{\mathbf f},x'{\rangle }|= {\mathbf f}^{max}\in L^1(\mathbb {T})\} \end{aligned}$$

and

$$\begin{aligned} L_{1,X}^{max}(\mathbb {T},X'')=\{{\mathbf f}\in L_{1}^{max}(\mathbb {T},X''): \hat{\mathbf f}(n)\in X\} \end{aligned}$$

where, for a given $weak^*$-measurable function ${\mathbf f}:\mathbb {T}\rightarrow X''$ such that ${\langle }{\mathbf f}, x'{\rangle }\in L^1(\mathbb {T})$ for any $x'\in X'$, the Fourier coefficient $\hat{\mathbf f}(n)\in X''$ is given by

$$\begin{aligned} {\langle }\hat{\mathbf f}(n),x'{\rangle }=\int \nolimits _0^{2\pi } e^{-int}{\langle }{\mathbf f}(e^{it}),x'{\rangle }\frac{dt}{2\pi }. \end{aligned}$$

The Riemann–Lebesgue property of a complex Banach space X was introduced in [8] by the condition $(\hat{\mathbf f}(n))_{n\in \mathbb {Z}}\in c_0(\mathbb {Z},X)$ for any ${\mathbf f}\in L_{1,X}^{max}(\mathbb {T},X'')$.

Proposition 3.7

Let X be a Banach space. If $X\in (RLP)_\mathbb {T}$ then X has the Riemann–Lebesgue property.

Proof

Let ${\mathbf f}\in L_{1,X}^{max}(\mathbb {T},X'')$. We first observe that or any trigonometric polynomial $\psi$

$$\begin{aligned} \left\| \sum _{n=-N}^M \hat{\psi }(n) \hat{\mathbf f}(n)\right\|= & {} \sup _{\Vert x'\Vert =1} \left| \sum _{n=-N}^M \hat{\psi }(n) {\langle }\hat{\mathbf f}(n), x'{\rangle }\right| \\= & {} \sup _{\Vert x'\Vert =1} \left| \int \nolimits _0^{2\pi } {\langle }\hat{\mathbf f}(e^{it}), x'{\rangle }\phi (e^{-it})\frac{dt}{2\pi }\right| \\\le & {} \int \nolimits _0^{2\pi } \sup _{\Vert x'\Vert =1}\left| {\langle }\hat{\mathbf f}(e^{it}), x'{\rangle }| |\phi (e^{-it})\right| \frac{dt}{2\pi }\\\le & {} \Vert \phi \Vert _\infty \Vert \mathbf f^{max}\Vert _1. \end{aligned}$$

Let us define $T_{\mathbf f}(\psi )=\sum \nolimits _{n=-N}^M \hat{\psi }(n) \hat{\mathbf f}(n)\in X$ for any trigonometric polynomial $\psi$. We can use the density of the trigonometric polynomials in $C(\mathbb {T})$, to extend $T_{\mathbf f}: C(\mathbb {T})\rightarrow X$ as a bounded operator. The assumption that $\sup _{\Vert x'\Vert =1} |{\langle }{\mathbf f},x'{\rangle }|\in L^1(\mathbb {T})$ guarantees not only that $T_{\mathbf f}$ is weakly compact (hence there exists a regular measure $\nu$ with $T_\nu =T_{\mathbf f}$) but also that $T_{\mathbf f}$ is absolutely summing (hence $\nu \in M_{ac}(G,X)$). Finally using that $\nu (n)=\hat{\mathbf f}(n)$ for $n\in \mathbb {Z}$ we conclude that $(\hat{\mathbf f}(n))_{n\in \mathbb {Z}}\in c_0(\mathbb {Z})$, from the assumption $X\in (RLP)_\mathbb {T}$. $\square$

Remark 3.1

It was shown (see [8, Proposition 3.4]) that the Riemann–Lebesgue property holds for not only spaces X having RNP but also for spaces satisfying the weak RNP (see [9]) o even the “complete continuity property” (see [10] or [11]).

4 Convolution for vector measures

From Corollary 2.2 we have that $t\rightarrow \mu (A-t)$ is measurable and bounded for each $\mu \in M(G)$ (and hence in $L^1(\nu )$ for any vector measure $\nu$). This allows us to give the following definition.

Definition 4.1

Let $\nu$ be a vector valued measure and $\mu \in M(G)$ we define the vector valued set function $\mu *\nu (A)$ given by

$$\begin{aligned} \mu *\nu (A)=\int \nolimits _G \mu (A-t)d\nu (t), \quad A\in \mathcal B(G). \end{aligned}$$

Let us see that $\mu *\nu$ is always a vector measure.

Proposition 4.2

If $\nu$ is a vector measure and $\mu \in M(G)$ then $\mu *\nu$ is a vector measure. Moreover

$$\begin{aligned} \Vert \mu *\nu \Vert \le |\mu |(G)\Vert \nu \Vert . \end{aligned}$$

(13)

Proof

Let $(A_n)$ pairwise disjoint sets in $\mathcal B(G)$ with $A=\cup _n A_n$ and $t\in G$. To show that $\mu *\nu (A)=\sum \nolimits _n \mu *\nu (A_n)$, due to the Orlicz–Pettis theorem (see [5, p. 7]), we simply need to see that $\sum \nolimits _n \mu *\nu (A_n)$ is weakly unconditionally convergent to $\mu *\nu (A)$. Let $x'\in X'$ and note that for $A\in \mathcal B(G)$,

$$\begin{aligned} {\langle }\mu *\nu , x'{\rangle }(A)=\int \nolimits _G \mu (A-t)d{\langle }\nu ,x'{\rangle }(t)= \mu *{\langle }\nu ,x'{\rangle }(A). \end{aligned}$$

On the one hand

$$\begin{aligned} \sum _n |{\langle }\mu *\nu (A_n),x'{\rangle }|\le & {} \sum _n |\mu |* |{\langle }\nu ,x'{\rangle }|(A_n)\\= & {} |\mu |* |{\langle }\nu ,x'{\rangle }|(A)\\\le & {} |\mu |(G)|{\langle }\nu ,x'{\rangle }|(G). \end{aligned}$$

On the other hand

$$\begin{aligned} \left| \sum _{n=1}^m {\langle }\mu *\nu (A_n),x'{\rangle }-{\langle }\mu *\nu (A), x'{\rangle }\right|= & {} \left| \int \nolimits _G \mu (\cup _{n=m+1}^\infty A_n-t)d{\langle }\nu ,x'{\rangle }(t)\right| \\\le & {} \int \nolimits _G |\mu |(\cup _{n=m+1}^\infty A_n-t)d|{\langle }\nu ,x'{\rangle }|(t). \end{aligned}$$

Let $\phi _m(t)=|\mu |(\cup _{n=m+1}^\infty A_n-t)$. We have that $\lim _{m\rightarrow \infty }\phi _m(t)= 0$ for each $t\in G$ and $\phi _m(t)\le |\mu |(G)$ for each $m\in \mathbb {N}$ and $t\in G$. Hence the Lebesgue dominated convergence theorem shows that $\sum \nolimits _n {\langle }\mu *\nu (A_n),x'{\rangle }= {\langle }\mu *\nu (A),x'{\rangle }$.

To show (13) we use that

$$\begin{aligned} |{\langle }\mu *\nu , x'{\rangle }| (G)\le & {} |\mu |*|{\langle }\nu ,x'{\rangle }|(G)\\\le & {} |\mu |(G)|{\langle }\nu , x'{\rangle }|(G). \end{aligned}$$

Now taking supremum over the unit ball of $X'$ we get the desired estimate. $\square$

Lemma 4.3

If $\nu \in \mathcal M(G,X)$ and $\mu \in M(G)$ then $\mu *\nu \in \mathcal M(G,X)$. Moreover $T_{\mu *\nu }=T_\nu \circ C_\mu$, where $C_\mu (g)(t)=\int \nolimits _G g(t+s)d\mu (s)$ for $g\in C(G)$ and

$$\begin{aligned} \Vert T_{\mu *\nu }\Vert \le \Vert T_{\nu }\Vert |\mu |(G). \end{aligned}$$

Proof

It is immediate to see that $C_\mu$ is continuous from C(G) into itself and the composition $T_\nu \circ C_\mu$ defines a weakly compact operator from C(G) into X whose representing measure is given by

$$\begin{aligned} \eta (A)=(T_\nu \circ C_\mu )^{**}(\chi _A)= T^{**}_\nu \circ C^{**}_\mu (\chi _A). \end{aligned}$$

Let us show that $\eta =\mu *\nu .$ Recall that for each $\lambda \in M(G)$ we have that $\mu *\lambda \in M(G)$ which is defined by

$$\begin{aligned} \int \nolimits _G g(u)d(\mu *\lambda )(u)=\int \nolimits _G\int \nolimits _G g(t+s) d\mu (s)d\lambda (t)= \int \nolimits _G C_\mu (g)(t)d\lambda (t). \end{aligned}$$

Therefore $C_\mu ^*(\lambda )= \lambda *\mu$. We also have that $C^{**}_\mu (\chi _A)\in (M(G))'$ with

$$\begin{aligned} C^{**}_\mu (\chi _A)(\lambda )= & {} \lambda *\mu (A)\\= & {} \int \nolimits _G \mu (A-t)d\lambda (t). \end{aligned}$$

We conclude that the element $C^{**}_\mu (\chi _A)$ is represented by the measurable function $t\rightarrow \mu (A-t)$, and taking into account that $M(G)'\subset L^1(\nu )$ we obtain $\eta (A)= T^{**}_\nu (\mu (A-\cdot ))=I_\nu (\mu (A-\cdot ))=\mu *\nu (A)$. Finally using that $\Vert C_\mu \Vert \le |\mu |(G)$ the proof is completed. $\square$

Making use again of Corollary 2.2 we can also define the convolution as follows.

Definition 4.4

Let $\nu \in \mathcal M(G,X)$ and $\mu \in M(G)$ we define the vector valued set function $\nu *\mu (A)$ given by

$$\begin{aligned} \nu *\mu (A)=\int \nolimits _G \nu (A-t)d\mu (t), \quad A\in \mathcal B(G) \end{aligned}$$

where the map $t\rightarrow \nu (A-t)$ is (strongly)-measurable and bounded (and hence in $L^1(\mu )$).

Proposition 4.5

If $\nu \in \mathcal M(G,X)$ and $\mu \in M(G)$ then $\nu *\mu =\mu *\nu$.

Proof

It suffices to show that $\langle \nu *\mu (A), x'\rangle =\langle \mu *\nu (A), x'\rangle$ for any $A\in \mathcal B(G)$ and $x'\in X'$. This now follows from the scalar-valued case, using that $\langle \nu *\mu (A), x'\rangle = \langle \nu , x'\rangle * \mu (A)$ and $\langle \mu *\nu (A), x'\rangle = \mu *\langle \nu , x'\rangle (A)$ for each $A\in \mathcal B(G)$ and the easy observation that if $\mu _1, \mu _2\in M(G)$ then, for each $g\in C(G)$,

$$\begin{aligned} \int \nolimits _G \Big (\int \nolimits _G g(t+s) d\mu _1(t)\Big ) d\mu _2(s) =\int \nolimits _G \Big (\int \nolimits _Gg(t+s) d\mu _2(s)\Big ) d\mu _1(t), \end{aligned}$$

which shows that $\mu _1*\mu _2=\mu _2*\mu _1$. $\square$

Following the classical argument we obtain the following easy fact.

Proposition 4.6

Let $\mu \in M(G)$ and $\nu \in \mathcal M(G,X)$ . Then

$$\begin{aligned} \widehat{\mu *\nu }(\gamma )= \hat{ \mu }(\gamma )\hat{\nu }(\gamma ), \quad \gamma \in \Gamma . \end{aligned}$$

(14)

Proof

Let $\gamma \in \Gamma$. Then

$$\begin{aligned} \widehat{\mu *\nu }(\gamma )= & {} T_{\nu }(C_\mu (\bar{\gamma }))=T_\nu \left( \int \nolimits _G \bar{\gamma }(\cdot +s)d\mu (s)\right) \\= & {} T_\nu \left( \bar{\gamma }\int \nolimits _G \bar{\gamma }(s)d\mu (s)\right) =\hat{\mu }(\gamma )\hat{\nu }(\gamma ). \end{aligned}$$

This gives the result. $\square$

Let us now restrict to some classes of measures in M(G) and $\mathcal M(G,X)$.

Remark 4.1

For $f\in L^1(G)$ and ${\mathbf g}\in L^1(G,X)$ we write

$$\begin{aligned} f*{\mathbf g}(s)=\int \nolimits _G f(s-t){\mathbf g}(t)dm_G(t),\quad m_G-a.e \end{aligned}$$

It is well known that $f*{\mathbf g}\in L^1(G,X)$. If we write $d\mu _f=fdm_G$ and $d\nu _{\mathbf g}= {\mathbf g}dm_G$ then $d(\mu _f*\nu _{\mathbf g})= d\nu _{f*\tilde{\mathbf g}}=(f*\tilde{\mathbf g})dm_G$, where we use the notation $\tilde{\mathbf g}(u)={\mathbf g}(-u)$. Indeed

$$\begin{aligned} \mu _f*d\nu _{\mathbf g}(A)= & {} \int \nolimits _G \mu _f(A-t)d\nu _{\mathbf g}(t)\\= & {} \int \nolimits _G \Big (\int \nolimits _{A-t}f(s)dm_G(s)\Big ){\mathbf g}(t)dm_G(t)\\= & {} \int \nolimits _G \Big (\int \nolimits _{A}f(s+t)dm_G(s)\Big ){\mathbf g}(t)dm_G(t)\\= & {} \int \nolimits _A \Big (\int \nolimits _{G}f(s+t){\mathbf g}(t)dm_G(t)\Big ) dm_G(s)\\= & {} \int \nolimits _A \Big (\int \nolimits _{G}f(s-u)\tilde{\mathbf g}(u)dm_G(u)\Big ) dm_G(s). \end{aligned}$$

Remark 4.2

For $f\in L^1(G)$ and a vector measure $\nu$ we have

$$\begin{aligned} \mu _f*\nu (A)= & {} \int \nolimits _G \Big (\int \nolimits _{A-t}f(s)dm_G(s)\Big ) d\nu (t)\\= & {} \int \nolimits _G \Big (\int \nolimits _{G}\chi _A(s+t)f(s)dm_G(s)\Big ) d\nu (t)\\= & {} \int \nolimits _G \Big (\int \nolimits _{G}\chi _A(t-s)\tilde{f}(s)dm_G(s)\Big ) d\nu (t)\\= & {} I_\nu (\chi _A* \tilde{f}). \end{aligned}$$

Hence $\mu _f*\nu (A)= I_\nu (\chi _A* \tilde{f})$ for all $A\in \mathcal B(G).$

If $\nu$ is a vector measure and $f\in L^1(G)$ we denote $\mu _f*\nu =f*\nu$ and we say that $f*\nu \in C(G,X)$ whenever there exists ${\mathbf f_\nu }\in C(G,X)$ such that $d(f*\nu )={\mathbf f_\nu }dm_G$.

Proposition 4.7

Let $\nu$ be a vector measure.

• (a) If $f\in C(G)$ then $f*\nu \in C(G,X)$ and

  $$\begin{aligned} \Vert f*\nu \Vert _{C(G,X)}\le \Vert f\Vert _{C(G)} \Vert \nu \Vert . \end{aligned}$$

  (15)

• (b) If $f\in L^1(G)$ then $f*\nu \in P_1(G,X)$ and

  $$\begin{aligned} \Vert f*\nu \Vert _{P_1(G,X)}\le \Vert f\Vert _{L^{1}(G)} \Vert \nu \Vert . \end{aligned}$$

  (16)

Proof

We define ${\mathbf f_\nu }(t)\in X$ by

$$\begin{aligned} {\mathbf f_\nu }(t)=\int \nolimits _G f(t+s)d\nu (s)= I_\nu (\tau _{-t}f), \quad t\in G. \end{aligned}$$

(17)

Let us see first that ${\mathbf f_\nu }\in C(G,X)$. For each $t,t'\in G$ we have

$$\begin{aligned} \Vert {\mathbf f_\nu }(t)- {\mathbf f_\nu }(t')\Vert = \Vert I_\nu (\tau _{-t}f-\tau _{-t'}f)\Vert \le \Vert \nu \Vert \Vert \tau _{-t}f-\tau _{-t'}f\Vert _{C(G)}. \end{aligned}$$

Now the result follows using that the map $G\rightarrow C(G)$ given by $t\rightarrow \tau _{-t}f$ is uniformly continuous.

Let us now show that $d(f*\nu )={\mathbf f_\nu }dm_G$. Let $A\in \mathcal B(G)$ and $x'\in X'$ and note that

$$\begin{aligned} \left\langle \int \nolimits _A{\mathbf f_\nu }dm_G, x'\right\rangle= & {} \left\langle \int \nolimits _A\Big (\int \nolimits _G f(t+s)d\nu (s)\Big ) dm_G(t),x'\right\rangle \\= & {} \int \nolimits _A\Big (\int \nolimits _G f(t+s)d{\langle }\nu ,x'{\rangle }(s)\Big ) dm_G(t)\\= & {} \int \nolimits _G\Big (\int \nolimits _A f(t+s)dm_G(t)\Big ) d{\langle }\nu ,x'{\rangle }(s)\\= & {} \left\langle \int \nolimits _G \Big (\int \nolimits _{A-s} f(t)dm_G(t)\Big )d{\langle }\nu , x'{\rangle }\right\rangle (s) \\= & {} \left\langle \mu _f*\nu (A), x'\right\rangle \end{aligned}$$

Finally (15) follows trivially since

$$\begin{aligned} \sup _{t\in G} \Vert \mathbf f_\nu (t)\Vert = \sup _{t\in G, \Vert x'\Vert =1} \left| \int \nolimits _G f(t+s)d{\langle }\nu , x'{\rangle }(s)\right| \le \Vert f\Vert _{C(G)}\Vert \nu \Vert . \end{aligned}$$

(b) Assume now that $f\in L^1(G)$. We first find $f_n\in C(G)$ such that $\Vert f-f_n\Vert _{L^1(G)}\rightarrow 0$. Using the previous case, the estimate (13) and the fact $|\mu _f|(G)=\Vert f\Vert _{L^1(G)}$ we conclude that

$$\begin{aligned} \Vert \mu _{f_n}*\nu -\mu _f*\nu \Vert \le \Vert f_n-f\Vert _{L^1(G)}\Vert \nu \Vert . \end{aligned}$$

Then $f*\nu \in P_1(G,X)$ and $\Vert f*\nu \Vert _{P_1(G,X)}= \Vert \mu _f*\nu \Vert \le \Vert f\Vert _{L^1(G)}\Vert \nu \Vert$. $\square$

Let us now look for some Young’s convolution result when assuming that either $\nu \in \mathcal M_p(G,X)$ or $f\in L^p(G)$. Let us mention first that any measure $\nu$ with bounded p-semivariation with respect to $m_G$ for some $1<p\le \infty$ necessarily belongs to $\mathcal M_{ac}(G,X)$. This is due to the fact that it satisfies $\Vert \nu (A)\Vert \le m_G(A)^{1/p'}$ for each $A\in \mathcal B(G)$ which implies $\nu << m_G$ and, in particular $\nu$ is automatically regular.

To work with measures in $\mathcal M_p(G,X)$ we shall use some lemmas whose proofs we include for the sake of completeness.

Lemma 4.8

Let $1<p\le \infty$ and let $\nu$ be a vector measure. Then $\nu \in \mathcal M_p(G,X)$ if and only if $\nu \in \mathcal M_{ac}(G;X)$ and $T^*_\nu (X')\subset L^{p}(G)$.

Proof

Assume $\nu \in \mathcal M_p(G,X)$. Now (1) gives that $T_\nu$ extends to a bounded operator from $L^{p'}(G)$ into X. Hence $T_\nu ^*$ is bounded from $X'$ into $L^{p}(G)$ and therefore $T^*_\nu (X')\subset L^{p}(G)$.

Assume now that $\nu \in \mathcal M_{ac}(G,X)$ and $T^*_\nu (X')\subset L^{p}(G)$. Use now that ${\langle }\nu , x'{\rangle }=T^*_\nu (x')$ and then $d|{\langle }\nu , x'{\rangle }|= |h_{x'}| dm_G$ for some $h_{x'}\in L^p(G)$. This implies that for $\Vert \sum \nolimits _{A\in \pi } \alpha _A\chi _A\Vert _{L^{p'}(m_G)}\le 1$ we have

$$\begin{aligned} \left\| \sum _{A\in \pi } \alpha _A \nu (A)\right\| _X\le \sup _{\Vert x'\Vert =1} \int \nolimits _G\left( \sum _{A\in \pi } |\alpha _A|\chi _A \right) |h_{x'}|dm_G = \sup _{\Vert x'\Vert =1} \Vert h_{x'}\Vert _{L^p(G)}. \end{aligned}$$

Hence we have $\nu$ is of bounded p-semivariation. $\square$

Next result extends [3, Theorem 3.9] to the case $1<p<\infty$.

Proposition 4.9

Let $1<p\le \infty$ and let $\nu \in \mathcal M_{ac}(G,X)$. The following statements are equivalent:

• (i) $\nu \in \mathcal M_{p}(G,X)$.

• (ii) $L^{p'}(G)\subset L^1(\nu )$.

• (iii) $L^{p'}(G)\subset L^1_w(\nu )$. Moreover $\Vert \nu \Vert _{p.m_G}= \Vert Id\Vert _{L^{p'}(G)\rightarrow L^1(\nu )}.$

Proof

(i) $\Longrightarrow$ (ii) Assume that $\nu$ has bounded p-semivariation with respect to $m_G$. Hence

$$\begin{aligned} \left\| \int \nolimits _G \phi d\nu \Vert \le \Vert \nu \Vert _{p,m_G}\Vert \phi \right\| _{L^{p'}(G)}, \quad \phi \hbox { simple }. \end{aligned}$$

This gives, due to the density of simple functions in $L^{p'}(G)$ and $L^1(\nu )$, that $L^{p'}(G)\subset L^1(\nu )$.

(ii) $\Longrightarrow$ (iii) It is obvious.

(iii) $\Longrightarrow$ (i) Assume $L^{p'}(G)\subset L^1_w(\nu )$. Since $\nu <<m_G$, using Radon–Nikodym theorem we have, for each $x'\in X'$ the existence of $h_{x'} \in L^1(G)$ for which

$$\begin{aligned} d{\langle }\nu , x'{\rangle }= h_{x'}dm_G. \end{aligned}$$

(18)

Therefore for each $f\in L^{p'}(G)$ we have

$$\begin{aligned} \Vert f\Vert _{L^1_w(\nu )}= \sup _{\Vert x'\Vert =1}\int \nolimits _G |f| |h_{x'}|dm_G\le K\Vert f\Vert _{L^{p'}(G)}. \end{aligned}$$

This implies that $h_{x'} \in L^p(G)$ for all $x'\in X'$ and

$$\begin{aligned} \sup _{ \Vert x'\Vert =1} \Vert h_{x'}\Vert _{L^p(G)}\le K. \end{aligned}$$

(19)

This gives $\Vert \nu \Vert _{p,m_G}\le K$. $\square$

Theorem 4.10

Let $1<p<\infty$ and let $\nu$ be a vector measure.

• (a) If $\nu \in \mathcal M_p(G,X)$ and $f\in L^{p'}(G)$ then $f*\nu \in C(G,X)$ and

  $$\begin{aligned} \Vert f*\nu \Vert _{C(G,X)}\le \Vert f\Vert _{L^{p'}(G)} \Vert \nu \Vert _{p,m_G}. \end{aligned}$$

  (20)

• (b) If $\nu \in \mathcal M_{ac}(G,X)$ and $f\in L^{\infty }(G)$ then $f*\nu \in C(G,X)$ and

  $$\begin{aligned} \Vert f*\nu \Vert _{C(G,X)}\le \Vert f\Vert _{L^{\infty }(G)} \Vert \nu \Vert . \end{aligned}$$

  (21)

• (c) If $f\in L^p(G)$ then $f*\nu \in P_p(G,X)$. Moreover

  $$\begin{aligned} \Vert f*\nu \Vert _{P_p(G,X)}\le \Vert f\Vert _{L^p(G)}\Vert \nu \Vert . \end{aligned}$$

  (22)

• (d) If $f\in L^q(G)$ and $\nu \in \mathcal M_p(G,X)$ with $q'> p$ then $f*\nu \in P_r(G,X)$ for $1/r=1/p-1/q'$. Moreover

  $$\begin{aligned} \Vert f*\nu \Vert _{P_r(G,X)}\le \Vert f\Vert _{L^q(G)}\Vert \nu \Vert _{p,m_G}. \end{aligned}$$

  (23)

Proof

• (a) Using Proposition 4.9 we have $L^{p'}(G)\subset L^1(\nu )$. Hence that $I_\nu$ is well defined on $L^{p'}(G)$ and, denoting $f_t(s)=f(t+s)=\tau _{-t}f(s)$, we observe that ${\mathbf f_\nu }(t)=I_\nu (f_t)$ makes sense for each value of $t\in G$. Repeating the argument in Proposition 4.7 part (a) and using now that $G\rightarrow L^{p'}(G)$ given by $t\rightarrow f_t$ is uniformly continuous we obtain that ${\mathbf f_\nu }$ is continuous. And also we have

  $$\begin{aligned} \Vert f*\nu (t)\Vert \le \Vert I_\nu \Vert _{L^{p'}\rightarrow X} \Vert f_t\Vert _{L^{p'}(G)}= \Vert \nu \Vert _{p,m_G} \Vert f\Vert _{L^{p'}(G)}. \end{aligned}$$

• (b) If $f\in L^\infty (G)$ then value $I_\nu (f_t)$ makes sense for any $t\in G$. Observe that if $\nu \in \mathcal M_{ac}(G,X)$ and $f=\chi _A$ we have

  $$\begin{aligned} \int \nolimits _G \chi _A(t+s)d\nu (s)= \nu (A-t) \end{aligned}$$

  is continuous. Hence $f*\nu \in C(G,X)$ for any simple function f and it satisfies $\Vert I_\nu (f)\Vert \le \Vert f\Vert _{L^\infty (G)}\Vert \nu \Vert$. Finally using the density of simple functions in $L^\infty (G)$ we have the desired result.

• (c) From Proposition 4.7 and $f\in C(G)$ we know that $d(f*\nu )={\mathbf f_\nu }dm_G$ with ${\mathbf f_\nu }\in C(G,X)$. Moreover, for each $x'\in X'$ ,

  $$\begin{aligned} \int \nolimits _G |{\langle }{\mathbf f_\nu }(t), x'{\rangle }|^p dm_G(t)\le & {} \int \nolimits _G \Big (\int \nolimits _G |f(t+s)|d|{\langle }\nu , x'{\rangle }|(s)\Big )^pdm_G(t) \\\le & {} \int \nolimits _G (|{\langle }\nu , x'{\rangle }|(G))^{p-1}\Big (\int \nolimits _G |f(t+s)|^pd|{\langle }\nu , x'{\rangle }|(s)\Big )dm_G(t) \\= & {} (|{\langle }\nu , x'{\rangle }|(G))^{p-1} \int \nolimits _G \Big (\int \nolimits _G |f(t+s)|^p dm_G(t)\Big )d|{\langle }\nu , x'{\rangle }|(s)\\= & {} \Vert f\Vert ^p_{L^p(G)}(|{\langle }\nu , x'{\rangle }|(G))^p. \end{aligned}$$

  This shows (22) for continuous functions. Let $f\in L^p(G)$. We first find $f_n\in C(G)$ such that $\Vert f-f_n\Vert _{L^p(G)}\rightarrow 0$ and denote $\nu _n=f_n*\nu$. From the previous case conclude that

  $$\begin{aligned} \Vert \nu _n -\nu _m\Vert _{p,m_G} = \Vert {\mathbf f_{\nu _n}}-{\mathbf f_{\nu _m}}\Vert _{P_p(G,X)}\le \Vert f_n-f_m\Vert _{L^p(G)}\Vert \nu \Vert . \end{aligned}$$

  Therefore $\nu _n$ is a Cauchy sequence in $\mathcal M_p(G,X)$. Since it converges to $f*\nu$ in $\mathcal M(G,X)$ we conclude that $f*\nu \in {P_p(G,X)}$ and

  $$\begin{aligned} \Vert f*\nu \Vert _{P_p(G,X)}= \lim _n \Vert f_n*\nu \Vert _{P_p(G,X)}\le \lim _n \Vert f_n\Vert _{L^p(G)}\Vert \nu \Vert = \Vert f\Vert _{L^p(G)}\Vert \nu \Vert . \end{aligned}$$

• (d) Since $\nu \in \mathcal M_p(G,X)$, using Lemma 4.8 we have $d{\langle }\nu , x'{\rangle }= h_{x'}dm_G$ with $h_{x'}\in L^p(G)$. Assume again first that $f\in C(G)$ and $\Vert x'\Vert =1$. Hence

  $$\begin{aligned} {\langle }f*\nu (t), x'{\rangle }= \int \nolimits _G f(t+s) h_{x'}(s)dm_G(s)=f* \tilde{h}_{x'}(t). \end{aligned}$$

  Therefore $|{\langle }f*\nu (t), x'{\rangle }|\le |f|* |\tilde{h}_{x'}|(t)$ and we can apply the classical Young’s inequality

  $$\begin{aligned} \left( \int \nolimits _G |{\langle }f*\nu (t), x'{\rangle }|^rdm_G(t)\right) ^{1/r}\le & {} \Vert |f|* |\tilde{h}_{x'}|\Vert _{L^r(G)}\\\le & {} \Vert f\Vert _{L^q(G)}\Vert h_{x'}\Vert _{L^p(G)}\\\le & {} \Vert f\Vert _{L^q(G)}\Vert \nu \Vert _{p,m_G} \end{aligned}$$

  This gives (23) for continuous functions. The argument is finished using density as above.

$\square$

5 Invariance under homomorphisms

Throughout this section $H:G\rightarrow G$ denotes a homeomorphism and we write $R(s)=-s$ for the reflection and $\tau _a(s)=s+a$ for the translation.

If $f\in L^0(G)$ we shall use the notation $f_H(s)=f(H^{-1}s)$ and, in particular, $\tau _af(s)=f(s-a)$ and $\tilde{f}(s)=f(-s)$.

If $\nu$ is a vector measure we denote $\nu _H(A)=\nu (H(A))$ for $A\in \mathcal B(G)$, in particular, $\tau _a\nu (A)=\nu (A+a)$ and $\tilde{\nu }(A)=\nu (-A)$. It is elementary to see that $\nu _H$ is also a vector measure with $\Vert \nu _H\Vert =\Vert \nu \Vert$. If $\nu \in \mathcal M(G,X)$ then $\nu _H\in \mathcal M(G,X)$ and $T_{\nu _H}= T_\nu \circ \Phi _H$ where $\Phi _H:C(G)\rightarrow C(G)$ is the induced operator $g\rightarrow g_H= g\circ H^{-1}$.

Let us point out some useful formulae to be used later on. From the fact $(\chi _A)_H=\chi _{H(A)}$ and $I_{\nu _H}(\chi _A)=\nu _H(A)=\nu (H(A))=I_\nu ((\chi _A)_H)$ we conclude

$$\begin{aligned} I_{\nu _H}(f)=I_\nu (f_H), \quad f\in \mathcal S (G). \end{aligned}$$

(24)

Also we have

$$\begin{aligned} (\nu _H)_f=(\nu _{f_H})_H \hbox { and } (\nu _f)_H= (\nu _H)_{f_{H^{-1}}} \hbox { for any } f\in \mathcal S(G). \end{aligned}$$

(25)

This follows by linearity and the obvious case

$$\begin{aligned} (\nu _H)_f(A)=\int \nolimits _{A}fd\nu _H= \int \nolimits _{G}(f\chi _A)_{H}d\nu =\int \nolimits _{H(A)}f_{H}d\nu = (\nu _{f_H})_H(A). \end{aligned}$$

In particular

$$\begin{aligned} \widetilde{\nu _f}=\tilde{\nu }_{\tilde{f}}, \quad \tau _a(\nu _f)= (\tau _a\nu )_{\tau _{-a} f}, \quad a\in G. \end{aligned}$$

(26)

Taking into account (25) we conclude that $f\in L^1(\nu _H)$ if and only if $f_H\in L^1(\nu )$. Moreover with

$$\begin{aligned} \Vert f\Vert _{L^1(\nu _H)}=\Vert (\nu _H)_f\Vert = \Vert \nu _{f_H}\Vert =\Vert f_H\Vert _{L^1(\nu )}, \quad f\in \mathcal S(G). \end{aligned}$$

(27)

Definition 5.1

Let $\nu$ be a vector measure and $\mathcal H$ a family of homeomorphisms $H:G\rightarrow G$. We say that $\nu$ is $\mathcal H$- invariant whenever $\nu _H=\nu$ for any $H\in \mathcal H$. In particular, we say that $\nu$ is translation invariant (respect. reflection invariant) whenever $\tau _a\nu =\nu$ for any $a\in G$ (respect. $\tilde{\nu }=\nu$.)

Given a vector measure $\nu \in \mathcal M(G,X)$ we can define

$$\begin{aligned} \nu _{inv}(A)=\int \nolimits _G \tau _t\nu (A)dm_G(t). \end{aligned}$$

It is not difficult to show that $\nu _{inv} \in \mathcal M(G,X)$. Clearly $\nu _{inv}$ is translation invariant. Actually case of translation invariant measures, as in the scalar valued case, reduces to the $xm_G$ for some $x\in X$.

Proposition 5.2

Let $\nu \in \mathcal M(G,X)$ with $\nu (G)\ne 0$. Then $\nu$ is translation invariant if and only if $\nu =\nu (G) m_G$.

Proof

Only the direct implication needs a proof. Assume that $\nu$ is translation invariant, that is $\nu =\nu _{inv}$. We shall show that

$$\begin{aligned} \nu _{inv}(A)=\nu (G)m_G(A), \quad \forall A\in \mathcal B(G). \end{aligned}$$

It suffices to see that $T_{\nu _{inv}}(g)= (\int \nolimits _G g dm_G)\nu (G)$ for all $g\in C(G).$ This follows by noticing that

$$\begin{aligned} T_{\nu _{inv}}(g)= & {} \int \nolimits _G T_{\tau _t\nu }(g)dm_G(t)=\int \nolimits _G T_{\nu }(\tau _t g)dm_G(t)\\= & {} T_{\nu }\left( \int \nolimits _G (\tau _t g)dm_G(t)\right) = T_{\nu }\left( \left( \int \nolimits _G g dm_G\right) \chi _G\right) \\= & {} \nu (G) \left( \int \nolimits _G g dm_G\right) . \end{aligned}$$

This gives the desired implication. $\square$

Using (24) we easily formulate the H-invariance as follows:

Remark 5.1

Let $\nu$ be a vector measure and $H:G\rightarrow G$ an homomorphism. The following statements are equivalent.

• (i) $\nu$ is H-invariant.

• (ii) $I_{\nu _H}(f)=I_\nu (f)$ for any $f\in \mathcal S(G).$

• (iii) $L^1(\nu )=L^1(\nu _H)$ and $I_{\nu }(f_H)= I_{\nu }(f)$ for any $f\in L^1(\nu ).$

Definition 5.3

(see [3, 4]) Given an homeomorphism $H:G\rightarrow G$ and $f\in L^0(G)$ a vector measure $\nu$ is said to be a “norm integral H-invariant” whenever

$$\begin{aligned} \Vert I_{\nu _H}(f)\Vert = \Vert I_\nu (f)\Vert , \forall f\in \mathcal S(G). \end{aligned}$$

(28)

Given a family of homeomorphisms on G, say $\mathcal H$, we shall say that $\nu$ is “norm integral $\mathcal H$-invariant” whenever it is norm integral H-invariant for any $H\in \mathcal H$. We say “norm integral reflection invariant” and “norm integral translation invariant” in the cases of $\mathcal H=\{R\}$ and $\mathcal H=\{\tau _a:a\in G\}$ respectively.

Proposition 5.4

Let $H:G\rightarrow G$ be an homeomorphism and $\nu \in \mathcal M(G,X)$. The following are equivalent.

• (i) $\nu$ is norm integral H-invariant.

• (ii) $\Vert I_\nu (f_H)\Vert =\Vert I_\nu (f)\Vert , \forall f\in L^1(\nu ).$

• (iii) $\Vert T_{\nu _H}(f)\Vert =\Vert T_\nu (f)\Vert , \forall f\in C(G).$

• (iv) $L^1(\nu )=L^1(\nu _H)$ and $\Vert I_{\nu _H}(f)\Vert = \Vert I_{\nu }(f)\Vert$ for any $f\in L^1(\nu )$.

Proof

• (i) $\Longrightarrow$ (ii). It was shown in [3, Thm 3.3], due to the fact $I_\nu (f_H)=I_{\nu _H}(f)$ for all $f\in \mathcal S(G)$.

• (ii) $\Longrightarrow$ (iii) It follows using that $C(G)\subset L^1(\nu )$ and $I_\nu (f)=T_\nu (f)$ for $f\in C(G)$.

• (iii) $\Longrightarrow$ (iv) Let us show that $\Vert f\Vert _{L^1(\nu )}=\Vert f\Vert _{L^1(\nu )}$ for any $f\in C(G)$. Indeed, for each $f\in C(G)$ we can write

  $$\begin{aligned} \Vert f\Vert _{L^1(\nu )}= & {} \Vert \nu _f\Vert =\Vert T_{\nu _f}\Vert \\= & {} \sup \{\Vert T_{\nu _f}(g)\Vert :\Vert g\Vert _{C(G)}=1\}=\sup \{\Vert T_{\nu }(fg)\Vert :\Vert g\Vert _{C(G)}=1\} \\= & {} \sup \{\Vert T_{\nu _H}(fg)\Vert :\Vert g\Vert _{C(G)}=1\}=\sup \{\Vert T_{(\nu _H)_f}(g)\Vert :\Vert g\Vert _{C(G)}=1\} \\= & {} \Vert (\nu _H)_f\Vert = \Vert f\Vert _{L^1(\nu _H)}. \end{aligned}$$

  Finally use Lemma 2.1 to obtain $\Vert I_{\nu _H}(f)\Vert =\Vert I_\nu (f)\Vert$ to all $f\in L^1(\nu )$.

• (iv) $\Longrightarrow$ (i) It is immediate. $\square$

Let us also consider some weaker notions still good enough for our purposes.

Definition 5.5

Let $\nu$ be a vector measure and let $\mathcal H$ be a family of homeomorphisms $H:G\rightarrow G$. Then $\nu$ is said to be “semivariation $\mathcal H$-invariant” whenever

$$\begin{aligned} \Vert \nu _f\Vert = \Vert (\nu _{H})_f\Vert , \quad f\in \mathcal S(G), H\in \mathcal H. \end{aligned}$$

(29)

In particular $\nu$ is said to be “semivariation translation invariant” and “semivariation reflection invariant” in the cases of $\mathcal H=\{\tau _a: a\in G\}$ and $\mathcal H=\{R\}$ respectively.

Remark 5.2

If $\nu$ is norm integral translation invariant (respect. semivariation translation invariant) then $\tilde{\nu }$ is also norm integral translation invariant (respect. semivariation translation invariant).

Assume first that $\nu$ is norm integral translation invariant, $f\in \mathcal S(G)$ and $a\in G$. Then

$$\begin{aligned} \Vert I_{\tilde{\nu }}(\tau _a f)\Vert = \Vert I_\nu (\tau _{-a}\tilde{f})\Vert =\Vert I_\nu (\tilde{f})\Vert =\Vert I_{\tilde{\nu }}(f)\Vert . \end{aligned}$$

Assume now that $\nu$ is semivariation translation invariant, $f\in \mathcal S(G)$ and $a\in G$. Then applying (26) we obtain $(\tau _a\tilde{\nu })_f=(\widetilde{\tau _{-a}\nu })_f=\widetilde{(\tau _{-a}\nu )_{\tilde{f}}}$ and therefore

$$\begin{aligned} \Vert (\tau _a\tilde{\nu })_f\Vert =\Vert (\tau _{-a}\nu )_{\tilde{f}}\Vert =\Vert \nu _{\tilde{f}}\Vert = \Vert \widetilde{\nu _{\tilde{f}}}\Vert =\Vert (\tilde{\nu })_f\Vert . \end{aligned}$$

Proposition 5.6

Let $\nu$ be a vector measure and let $H:G\rightarrow G$ be an homeomorphism. The following statements are equivalent:

• (i) $\nu$ is semivariation H-invariant.

• (ii) $L^1(\nu )=L^1(\nu _H)$ isometrically.

• (iii) $\Vert T_{\nu _H}\circ M_f\Vert =\Vert T_\nu \circ M_f\Vert , \forall f\in C(G)$ where $M_f:C(G)\rightarrow C(G)$ stands for the multiplication operator $g\rightarrow fg$.

Proof

(i) $\Longleftrightarrow$ (ii) follows using that $\Vert f \Vert _{L^1(\nu )}=\Vert \nu _f\Vert =\Vert (\nu _H)_f\Vert =\Vert f \Vert _{L^1(\nu _H)}$ and the density of simple functions in $L^1(\nu )$ and $L^1(\nu _H)$.

(i) $\Longleftrightarrow$ (iii) follows observing that $\Vert T_\nu \circ M_f\Vert =\Vert \nu _f\Vert$ whenever $f\in C(G)$ and Lemma 2.1. $\square$

In [3, Theorem 3.3] and [4, Proposition 3.5] it was shown that if $\nu$ is norm integral H-invariant then it is semivariation H-invariant. We shall see now that the converse is not true in general.

Proposition 5.7

Let $G=\mathbb {T}$ and $X=L^1(\mathbb {T})$. For each $n\in \mathbb {Z}{\setminus }\{0\}$ denote $\phi _n(t)=t^n$ for $t\in \mathbb {T}$ and define

$$\begin{aligned} \nu _{(n)}(A)= \hat{\chi }_A(n)\phi _n, A\in \mathcal B(\mathbb {T}). \end{aligned}$$

Then $\nu _{(n)}$ is semivariation reflection invariant but not norm integral reflection-invariant.

Proof

To see that $\nu _{(n)}$ is not norm integral reflection-invariant we use an argument similar to that of example 3.6 (c) in [3]. Note that $T_{\nu _{(n)}}(g)= \hat{g}(n) \phi _n$ for any $g\in C(\mathbb {T})$ and $T_{\widetilde{\nu _{(n)}}}(g)= \hat{g}(-n) \phi _n$ for any $g\in C(\mathbb {T})$. Hence $T_{\nu _{(n)}}(\phi _n)=\phi _n$ and $T_{\nu _{(n)}}(\tilde{\phi }_n)=0.$ Hence $\Vert T_{\nu _{(n)}}(\tilde{\phi }_n)\Vert \ne \Vert T_{\nu _ {(n)}}(\phi _n)\Vert$.

We shall see that $\Vert {(\nu _{(n)})}_f\Vert = \Vert f\Vert _1$ for any $f\in C(\mathbb {T})$. Due to (26) this gives $\Vert {(\nu _{(n)})}_f\Vert =\Vert {(\widetilde{\nu _{(n)}})}_{ f}\Vert$ for any $f\in C(\mathbb {T})$.

First we notice that if $f,g\in C(\mathbb {T})$ we have

$$\begin{aligned} T_{{(\nu _{(n)})}_f}(g)=\widehat{fg}(n) \phi _n. \end{aligned}$$

Hence for each $\psi \in X'=L^\infty (\mathbb {T})$ we obtain

$$\begin{aligned} {\langle }T_{{(\nu _{(n)})}_f}(g),\psi {\rangle }= \widehat{fg}(n)\hat{\psi }(n)=\int \nolimits _\mathbb {T}g(t)\Big (\hat{\psi }(n)f(t)\bar{\phi }_n(t)\Big ) dm_\mathbb {T}(t). \end{aligned}$$

This shows that $d{\langle }{(\nu _{(n)})}_f, \psi {\rangle }= \hat{\psi }(n)f\bar{\phi }_{n}dm_\mathbb {T}$ and

$$\begin{aligned} \Vert {(\nu _{(n)})}_f\Vert = \sup _{\Vert \psi \Vert _{L^\infty (\mathbb {T})}=1} |{\langle }{(\nu _{(n)})}_f, \psi {\rangle }|(\mathbb {T})=\sup _{\Vert \psi \Vert _{L^\infty (\mathbb {T})}=1} \Vert f\Vert _1|\hat{\psi }(n)|=\Vert f\Vert _1. \end{aligned}$$

The proof is now complete. $\square$

Proposition 5.8

Let $\nu _{\mathbf f}(A)=\int \nolimits _A {\mathbf f}(s)dm_G(s)$ with ${\mathbf f}\in L^\infty (G,X)$ non constant function satisfying that

$$\begin{aligned} \Vert {\mathbf f}(t)\Vert =1, \quad t\in G \end{aligned}$$

and there exists $A\in \mathcal B(G)$ and $a\in G$ for which

$$\begin{aligned} \nu _{\mathbf f}(A)=0, \quad \nu _{\mathbf f}(A+a)\ne 0. \end{aligned}$$

Then $\nu _{\mathbf f}$ is semivariation translation invariant but not norm integral translation invariant.

Proof

Note that $\tau _t \nu _{\mathbf f}= \nu _{\tau _t {\mathbf f} }$ and $\tau _t {\mathbf f}\in L^\infty (G,X)$ for each $t\in G$. In particular $\tau _t \nu _{\mathbf f}$ is of bounded variation and $d|\tau _t \nu _{\mathbf f}|= \tau _t \Vert {\mathbf f}\Vert dm_G= dm_G$. Hence $L^1(\nu )=L^1(\tau _t \nu )=L^1(m_G)$ for any $t\in G$. Invoking now Proposition 5.6 we obtain that $\nu$ is semivariation translation invariant.

On the other hand $I_\nu (g)= \int \nolimits _G g{\mathbf f} dm_G$ and we have $\Vert I_{\nu }(\tau _a \chi _A)\Vert \ne 0$ while $\Vert I_{\nu }(\chi _A)\Vert =0$, showing that $\nu _{\mathbf f}$ is not norm integral translation invariant. $\square$

Remark 5.3

Select $X=\mathbb {C}$, $G=\mathbb {T}$, ${\mathbf f}(s)=\chi _{[0,1/2)}(e^{2\pi is})- \chi _{[1/2,1)}(e^{2\pi is})$, $A=\{e^{2\pi is}: 1/4\le s<1/2\}$ and $a=e^{i\pi /2}$) to have an example satisfying conditions of Proposition 5.8.

One of the basic properties of semivariation translation invariant measures is the following fact.

Lemma 5.9

Let $1\le p<\infty$, let $\nu \in \mathcal M(G,X)$ be a semivariation translation invariant measure and $f\in L^p(\nu )$. Then $a\rightarrow \tau _a f$ is uniformly continuous from G into $L^p(\nu )$.

Proof

Invoking Lemma 2.1, for each $\varepsilon >0$ we find $g\in C(G)$ such that $\Vert f-g\Vert _{L^p(\nu )}<\varepsilon$. Now use the standard argument

$$\begin{aligned} \Vert \tau _a f- \tau _b f\Vert _{L^p(\nu )}\le & {} \Vert \tau _a (f- g)\Vert _{L^p(\nu )}+\Vert \tau _b (f- g)\Vert _{L^p(\nu )}+\Vert \tau _a g- \tau _b g\Vert _{L^p(\nu )}\\= & {} \Vert f- g\Vert _{L^p(\nu _a)}+\Vert f- g\Vert _{L^p(\nu _b)}+\Vert \tau _a g- \tau _b g\Vert _{L^p(\nu )}\\ \end{aligned}$$

Hence, using Proposition 5.6, we conclude

$$\begin{aligned} \Vert \tau _a f- \tau _b f\Vert _{L^p(\nu )}\le 2\varepsilon + \Vert \tau _a g- \tau _bg\Vert _{C(G)}\Vert \nu \Vert ^{1/p} \end{aligned}$$

and the proof finishes using that g is uniformly continuous. $\square$

Theorem 5.10

Let $1\le p<\infty$ and let $\nu \in \mathcal M(G,X)$ be semivariation translation invariant with $\nu (G)\ne 0$. Then $L^p(\nu )\subset L^p(G)$ and

$$\begin{aligned} \Vert f\Vert _{L^p(G)}\le \Vert f\Vert _{L^p(\nu )} \Vert \nu (G)\Vert ^{-1/p}. \end{aligned}$$

Proof

Using that $\nu _{inv}= \nu (G) m_G$, in particular we know that $L^p(\nu _{inv})=L^p(G)$ and $\Vert f\Vert _{L^p(\nu _{inv})}= \Vert \nu (G)\Vert ^{1/p}\Vert f\Vert _{L^p(G)}$. Therefore it suffices to show that $L^p(\nu )\subset L^p(\nu _{inv})$ and $\Vert f\Vert _{L^p(\nu _{inv})}\le \Vert f\Vert _{L^p(\nu )}.$ We first point out the following trivial estimate between positive measures

$$\begin{aligned} |{\langle }\nu _{inv},x'{\rangle }|\le \int \nolimits _G |{\langle }\tau _t \nu ,x'{\rangle }| dm_G(t) \end{aligned}$$

for any $x'\in X'$. Hence, for each $f\in L^p(\nu )=L^p(\tau _t\nu )$ for all $t\in G$, we have

$$\begin{aligned} \Vert f\Vert _{L^p(\nu _{inv})}= & {} \sup _{\Vert x'\Vert =1} \left( \int \nolimits _G |f|^pd |{\langle }\nu _{inv},x'{\rangle }|\right) ^{1/p}\\\le & {} \sup _{\Vert x'\Vert =1}\left( \int \nolimits _G \left( \int \nolimits _G |f|^p d|{\langle }\tau _t \nu ,x'{\rangle }|\right) dm_G(t)\right) ^{1/p}\\\le & {} \left( \int \nolimits _G \left( \sup _{\Vert x'\Vert =1}\int \nolimits _G |f|^p d|{\langle }\tau _t \nu ,x'{\rangle }|\right) dm_G(t)\right) ^{1/p}\\= & {} \left( \int \nolimits _G \Vert f\Vert ^p_{L^p(\tau _t \nu )} dm_G(t)\right) ^{1/p}\\= & {} \Vert f\Vert _{L^p(\nu )}. \end{aligned}$$

This finishes the proof. $\square$

Proposition 5.11

Let $\nu \in \mathcal M(G,X)$ a semivariation translation invariant measure and $f\in L^1(\nu )$. Then $f*\nu \in C(G, X)$ and

$$\begin{aligned} \Vert f*\nu \Vert _{ C(G, X)}\le \Vert f\Vert _{L^1(\nu )}\Vert \nu \Vert . \end{aligned}$$

(30)

Proof

Recall that

$$\begin{aligned} \mathbf f_\nu (t)=\int \nolimits _G f(t+s)d\nu (s)=I_\nu (\tau _{-t}f)=I_{\tau _{-t}\nu }(f) \end{aligned}$$

which it is well defined for each $t\in G$ using that $L^1(\nu )= L^1(\tau _{-t}\nu )$. Moreover

$$\begin{aligned} \Vert f*\nu (t)\Vert \le \Vert \tau _{-t} f\Vert _{L^1(\nu )}\Vert \tilde{\nu }\Vert = \Vert f\Vert _{L^1(\nu )}\Vert \nu \Vert . \end{aligned}$$

Using now Lemma 5.9 we have that $t\rightarrow \tau _{-t}f$ is continuous from G into $L^1(\nu )$ which shows that $f*\nu \in C(G, X)$. $\square$

Corollary 5.12

Let $\nu \in \mathcal M(G,X)$. Then $\nu$ is norm integral translation invariant if and only if for each $f\in L^1(\nu )$ one has that $f*\nu \in C(G,X)$ with

$$\begin{aligned} \Vert f*\nu (t)\Vert =\left\| \int \nolimits _G f d\nu \right\| \quad \forall t\in G. \end{aligned}$$

Proof

Assume that $\nu$ is norm integral translation invariant. From Proposition 5.11 we have $f*\nu \in C(G,X)$. Moreover in this case

$$\begin{aligned} \Vert f*\nu (t)\Vert = \Vert I_\nu (\tau _{-t} f)\Vert = \Vert I_\nu ( f)\Vert =\left\| \int \nolimits _G f d\nu \right\| \quad \forall t\in G. \end{aligned}$$

The converse is immediate because any simple function f belongs to $L^1(\nu )$ and by assumption

$$\begin{aligned} \Vert I_{\nu } (\tau _{t} f)\Vert = \Vert f* \nu (-t)\Vert =\Vert \int \nolimits _G f d\nu \Vert =\Vert I_\nu (f)\Vert . \end{aligned}$$

$\square$

6 Applications to Fourier analysis on $L^1(\nu )$

In this section we analyze the properties of $\nu _f$, where $\nu _f(A)=\int \nolimits _A f d\nu$ for each $A\in \mathcal{B}(G)$, in terms of those of $\nu$ and $f\in L^1(\nu )$ and apply them to recover and improve the results in [3, 4].

Proposition 6.1

Let $\nu$ be vector measure.

• (i) If $\nu \in \mathcal M(G,X)$ and $f\in L^1(\nu )$ then $\nu _f\in \mathcal M(G,X)$.

• (ii) If $1<p\le \infty$, $\nu \in \mathcal M_p(G,X)$ and $f\in L^q(G)$ for some $p'\le q\le \infty$ then $\nu _f\in \mathcal M_r(G,X)$ for $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}.$

Proof

• (i) Note from Lemma 2.1 we have that $T_{\nu _f}=\lim _n T_{\nu _{f_n}}$ where $f_n\in C(G)$ for each n. It suffices to see that $T_{\nu _g}:C(G)\rightarrow X$ given by $\varphi \rightarrow \int \varphi g d\nu$ is weakly compact for any $g\in C(G)$. On the other hand $T_{\nu _g}= T_\nu \circ M_g$ where $M_g$ stands for the multiplication operator on C(G) given by $M_g(\phi )=g \phi$. Therefore, since $\nu$ is regular, hence $T_\nu$ is weakly compact and then we obtain that $\nu _f$ is regular.

• (ii) We use Proposition 4.9 to have that $L^q(G)\subset L^{p'}(G)\subset L^1(\nu )$. Hence $\nu _f$ is well defined. Now use the fact that $T^*_{\nu _f}= M_f\circ T^*_\nu$ where $M_f:L^{p}(G)\rightarrow L^{r}(G)$ is the multiplication operator $M_f(g)=fg$ and the fact that $\nu _f\in \mathcal M_r(G,X)$ is equivalent to $T^*_{\nu _f}\in \mathcal L(X',L^r(G))$ to finish the proof.

$\square$

Corollary 6.2

Let $1\le p<\infty$ and $\nu \in \mathcal M(G,X)$.

• (i) If $f\in L^p(G)$ and $g\in L^1(\nu )$ then $f*^\nu g\in P_p(G,X)$. Moreover

  $$\begin{aligned} \Vert f*^\nu g\Vert _{P_p(G,X)}\le \Vert f\Vert _{L^p(G)} \Vert g\Vert _{L^1(\nu )}(see [3]). \end{aligned}$$

• (ii) If $\nu \in \mathcal M_{p_1}(G,X)$, $g\in L^{p_2}(G)$ and $f\in L^{p_3}(G)$ with $\frac{1}{p_1}+\frac{1}{p_2}\le 1$ and $\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}\ge 1$ then $f*^\nu g\in P_r(G,X)$ for $\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}=\frac{1}{r'}$. Moreover

  $$\begin{aligned} \Vert f*^\nu g\Vert _{P_r(G,X)}\le \Vert \nu \Vert _{p_1,m_G}\Vert f\Vert _{L^{p_2}(G)} \Vert g\Vert _{L^{p_3}(G)}. \end{aligned}$$

Proof

As mentioned in the introduction $f*^\nu g= f*\nu _g$. Now both cases follow combining Proposition 6.1 and Theorem 4.10. $\square$

From Theorem 5.10 when assuming that $\nu \in \mathcal M(G,X)$ is semivariation translation invariant we have $L^p(\nu )\subset L^p(G)$ for any $p\ge 1$. Hence, as pointed out in [4], we can consider the classical convolution

$$\begin{aligned} f*_Gg(t)=\int \nolimits _G f(t-s)g(s)dm_G(s) \end{aligned}$$

between $f\in L^1(G)$ and $g\in L^p(\nu )$ and between $f\in L^p(\nu )$ and $g\in L^q(\nu )$.

Theorem 6.3

Let $1\le p<\infty$ and let $\nu \in \mathcal M(G,X)$ semivariation translation invariant.

• (i) If $f\in L^1(G)$ and $g\in L^p(\nu )$ then $f*_Gg\in L^p(\nu )$. Moreover

  $$\begin{aligned} \Vert f*_Gg\Vert _{L^p(\nu )} \le \Vert f\Vert _{L^1(G)}\Vert g\Vert _{L^p(\nu )}. \end{aligned}$$

  (31)

• (ii) If $f\in L^p(G)$ and $g\in L^1(\nu )$ then $f*_Gg\in L^p(\nu )$. Moreover

  $$\begin{aligned} \Vert f*_Gg\Vert _{L^p(\nu )} \le \Vert f\Vert _{L^p(G)}\Vert g\Vert _{L^1(\nu )}. \end{aligned}$$

  (32)

Proof

We first analyze the case $p=1$. Note that for $f,g\in C(G)$ we have $s\rightarrow \tau _s g$ is continuous function with values in $L^1(\nu )$. We write the $L^1(\nu )$-valued Riemann integral

$$\begin{aligned} f*_Gg=\int \nolimits _G \tau _s g f(s)dm_G(s). \end{aligned}$$

Using Minkowsky’s inequality and Proposition 5.6 we get

$$\begin{aligned} \Vert f*_Gg\Vert _{L^1(\nu )}\le \int \nolimits _G \Vert \tau _s g\Vert _{L^1(\nu )} |f(s)|dm_G(s)= \Vert f\Vert _{L^1(G)}\Vert g\Vert _{L^1(\nu )}. \end{aligned}$$

To extend to general functions, we use that C(G) is dense in $L^1(\nu )$ and in $L^1(G)$.

Assume now $p>1$. As above we start with $g\in C(G)$ and use Hölder’s inequality together with $L^1(\nu )\subset L^1(G)$ to have

$$\begin{aligned} |f*_Gg(t)|^p\le \min \{\Vert f\Vert _{L^1(G)}^{p-1} (|f|*_G |g|^p)(t), \Vert g\Vert _{L^1(\nu )}^{p-1}(|f|^p*_G |g|)(t)\}. \end{aligned}$$

Therefore (i) follow from the case $p=1$. Indeed,

$$\begin{aligned} \Vert f*_Gg\Vert ^p_{L^p(\nu )}\le \Vert f\Vert _{L^1(G)}^{p-1}\Vert |f|*_G |g|^p\Vert ^p_{L^1(\nu )}\le \Vert f\Vert ^p_{L^1(G)} \Vert |g|^p\Vert _{L^1(\nu )}=\Vert f\Vert ^p_{L^1(G)}\Vert g\Vert ^p_{L^p(\nu )}. \end{aligned}$$

The case (ii) is analogue. $\square$