Fourier Transform for $L^p$-Functions with a Vector Measure on a Homogeneous Space of Compact Groups

Abstract

Let G be a compact group and G/H a homogeneous space where H is a closed subgroup of G. Define an operator $T_H:C(G) \rightarrow C(G/H)$ by $T_Hf(tH)=\int _H f(th) \, dh$ for each $tH \in G/H$. In this paper, we extend $T_H$ to a norm-decreasing operator between $L^p$-spaces with a vector measure for each $1 \le p <\infty$. This extension will be used to derive properties of invariant vector measures on G/H. Moreover, a definition of the Fourier transform for $L^p$-functions with a vector measure is introduced on G/H. We also prove the uniqueness theorem and the Riemann–Lebesgue lemma.

1 Introduction

Let G be a topological group which is compact and Hausdorff. Consider a homogeneous space G/H where H is a closed subgroup of G. If we denote the normalized Haar measures on G and H by m and dh respectively, then there is an induced left invariant Radon measure $\widetilde{m}$ on G/H satisfying Weil’s formula:

$$\begin{aligned} \int _{G/H} \int _H f(th) \, dh \, d\widetilde{m}(tH) = \int _G f \,dm \qquad (f \in C(G)). \end{aligned}$$

In this setting, Farashahi [5] introduced a method to obtain many of the well-known results on G/H from the ones on G. This method relies unavoidably on an extension of the operator $T_H:C(G) \rightarrow C(G/H)$ given by $T_Hf(tH)=\int _H f(th) \, dh$. The extension is in fact a norm-decreasing operator from $L^p(G,m)$ onto $L^p(G/H,\widetilde{m})$ where $1 \le p < \infty$. The crucial property for this method is the surjectivity of the extension as it provides a connection to all $L^p$-functions on G/H to those on G. The extension was used to study abstract Fourier analysis on homogeneous spaces in various aspects such as convolutions, Fourier transform operators, Fourier series and measure algebras, see [6,7,8,9].

A vector measure is a measure taking values in a Banach space. There are many studies about functions in $L^p$-spaces of a compact group associated to a vector measure and invariant properties under the group operations of the vector measure itself. For example, the Fourier transform and the convolution along with invariant properties were studied in [1,2,3] under the condition that G is an abelian compact group. Then they were generalized to a non-abelian case in [13, 14].

Let $\nu$ be a vector measure on G. In this paper, we initiate a study of an extension of the operator $T_H: C(G) \rightarrow C(G/H)$ to an operator with the domain $L^p(G,\nu )$. However, the codomain C(G/H) must be extended as well. For this purpose, we will construct a corresponding vector measure ${\tilde{\nu }}$ on G/H and show that the codomain of the extended operator is $L^p(G/H,{\tilde{\nu }} )$. We investigate whether the extended operator is surjective and whether Weil’s formula is valid. It turns out that these are true for some vector measure $\nu$. Fortunately, it is sufficient for the study of functions in $L^p(G/H,\mu )$ for any vector measure $\mu$ on G/H. We will employ the extension to obtain properties of invariant vector measures on G/H. Moreover, we introduce a new definition of a Fourier transform of functions in $L^1(G,\nu )$ which is a variant definition of [13]. In our definition, $\nu$ is taking values in a Banach space while in [13] $\nu$ is taking values in an operator space. The uniqueness theorem of the Fourier transform and the Riemann–Lebesgue lemma are considered. Finally, we provide an analogous definition for a Fourier transform of functions in $L^1(G/H,\mu )$ and once more employ the extension to obtain relations between the Fourier transforms of functions on G and G/H.

This paper is organized as follows. We give preliminary background in Sect. 2. In Sect. 3, an extension of the operator $T_H$ to the space $L^p(G,\nu )$ is studied along with its properties. Then the obtained properties of the extension will be used to derive properties of invariant vector measures on G/H in Sect. 4. There are three types of invariant vector measures we consider in this paper: translation invariant, norm integral invariant and semivariation invariant measures. Section 5 concerns the Fourier transforms of functions on G and G/H.

2 Preliminaries

2.1 Fourier Analysis with Haar Measures

Let G be a compact group with the normalized Haar measure m. The dual space $\widehat{G}$ of G is the set of all unitary equivalence classes of irreducible unitary representations of G. For each $[\pi ] \in \widehat{G}$, the representation space of $\pi$ is denoted by ${\mathcal {H}}_\pi$ with the dimension $d_\pi = \dim {\mathcal {H}}_\pi$. For $[\pi ] \in \widehat{G}$ and $u,v \in {\mathcal {H}}_\pi$, the function $\pi _{u,v}:G \rightarrow {\mathbb {C}}$ given by $\pi _{u,v}(t)=\langle \pi (t)v,u \rangle$ is called a matrix element of $\pi$. We write $\pi _{ij}$ for $\pi _{e_i,e_j}$. Denote by $\text {Trig}(G)$ the set of all finite linear combinations of matrix elements of irreducible representations. Note that $\text {Trig}(G)$ is dense in C(G) in the uniform norm. For $f \in L^1(G,m)$ and $[\pi ] \in \widehat{G}$, the Fourier transform of f is defined in the weak sense as

$$\begin{aligned} {\mathcal {F}}_G(f)(\pi )=\widehat{f}(\pi )=\int _{G} f(t)\pi (t)^* \, dm(t) \in {\mathcal {B}}({\mathcal {H}}_\pi ). \end{aligned}$$

Given any collection $\{X_i\}_{i \in I}$ of Banach spaces where each $X_i$ is equipped with the norm $\Vert \cdot \Vert _i$. The space $\ell ^\infty (I;X_i) =\{ x \in \prod _{i \in I} X_i: \sup _{i \in I} \Vert x_i\Vert _i < \infty \}$ is a Banach space with the norm $\Vert x\Vert _\infty = \sup _{i \in I} \Vert x_i\Vert _i$. The set $c_0(I;X_i)$ of all $x=(x_i)$ for which $\{i \in I: \Vert x_i\Vert _i > \varepsilon \}$ is finite for any $\varepsilon > 0$ is a closed subspace of $\ell ^\infty (I;X_i)$. By [12, Theorem 28.40], the Fourier transform operator ${\mathcal {F}}_G$ is a bounded linear operator from $L^1(G,m)$ into $c_0(\widehat{G};{\mathcal {B}}({\mathcal {H}}_\pi ))$ with $\Vert \widehat{f}(\pi )\Vert \le \Vert f\Vert _{L^1(G,m)}$. For more details, see [11].

Let H be a closed subgroup of G and G/H the homogeneous space of left cosets equipped with the quotient topology. We denote the quotient map by $q: G \rightarrow G/H$. For $\varphi :G/H \rightarrow {\mathbb {C}}$, we write $\varphi _q:G \rightarrow {\mathbb {C}}$ for a function given by $\varphi _q(t)=\varphi (tH)$. Let dh be the normalized Haar measure on H. It is well-known that there is a unique (up to scalar) invariant Radon measure $\widetilde{m}$ on G/H satisfying Weil’s formula:

$$\begin{aligned} \int _{G/H} \int _H f(th) \, dh \, d\widetilde{m}(tH) = \int _G f \,dm \quad (f \in C(G)). \end{aligned}$$

In fact, $\widetilde{m}$ is the pushforward measure of m by the quotient map q. Define a bounded operator $T_H:C(G) \rightarrow C(G/H)$ by

$$\begin{aligned} T_Hf(tH)=\int _H f(th) \, dh \quad (tH \in G/H, \ f \in C(G)). \end{aligned}$$

According to [5], for any $1 \le p < \infty$, the operator $T_H$ can be extended to a norm-decreasing operator from $L^p(G,m)$ onto $L^p(G/H,\widetilde{m})$ (still denoted by $T_H$) for which the extended Weil’s formula holds:

$$\begin{aligned} \int _{G/H} T_H f \, d\widetilde{m} = \int _G f \,dm \quad \big (f \in L^1(G,m)\big ). \end{aligned}$$

(1)

For more details on Weil’s formula, see [16]. The dual space of G/H is given by $\widehat{G/H}:=\big \{ [\pi ]\in \widehat{G}: T_H^\pi \ne 0 \big \}$ where $T_H^\pi$ is defined in the weak sense as the operator $T_H^\pi :=\int _H \pi (h) \,dh \in {\mathcal {B}}({\mathcal {H}}_\pi )$. For $\varphi \in L^1(G/H,\widetilde{m})$ and $[\pi ] \in \widehat{G/H}$, the Fourier transform of $\varphi$ is defined in the weak sense as

$$\begin{aligned} {\mathcal {F}}_{G/H}(\varphi )(\pi ) = \widehat{\varphi }(\pi ) = \int _{G/H} \varphi (tH)\Gamma _{\pi }(tH)^*\,d\widetilde{m}(tH) \in {\mathcal {B}}({\mathcal {H}}_\pi ), \end{aligned}$$

where $\Gamma _{\pi }(tH)=\pi (t)T^\pi _H$. Then the Fourier transform operator ${\mathcal {F}}_{G/H}$ is a bounded linear operator from $L^1(G/H,\widetilde{m})$ into $c_0(\widehat{G/H};{\mathcal {B}}({\mathcal {H}}_\pi ))$ with $\Vert \widehat{\varphi }(\pi )\Vert \le \Vert \varphi \Vert _{L^1(G/H,\widetilde{m})}$, see [5, Theorem 5.5].

2.2 Vector Measures

Let $(\Omega ,\mathfrak {B}(\Omega ))$ be a Borel measurable space and X a Banach space. The closed unit ball in the dual space $X^*$ is denoted by $B_{X^*}$. A (countably additive) vector measure $\nu$ on $(\Omega ,\mathfrak {B}(\Omega ))$ is an X-valued function $\nu :\mathfrak {B}(\Omega ) \rightarrow X$ such that $\nu (\cup _{n=1}^\infty E_n) = \sum _{n=1}^\infty \nu (E_n)$ in the norm topology for any sequence $(E_n)$ of pairwise disjoint sets in $\mathfrak {B}(\Omega )$. Given $x^* \in X^*$, let ${\langle \nu , x^*\rangle } :\mathfrak {B}(\Omega ) \rightarrow {\mathbb {C}}$ be the complex measure given by ${\langle \nu , x^*\rangle } (E)=\langle \nu (E),x^* \rangle$ for $E \in \mathfrak {B}(\Omega )$. The semivariation $\Vert \nu \Vert$ of $\nu$ is the set function defined by $\Vert \nu \Vert (E)=\sup _{x^* \in B_{X^*}} \vert {\langle \nu , x^*\rangle } \vert (E)$ for $E \in \mathfrak {B}(\Omega )$. A vector measure $\nu$ is said to be regular if for each $E \in \mathfrak {B}(\Omega )$ and $\varepsilon > 0$ there exist a compact set K and an open set O such that $K \subset E \subset O$ and $\Vert \nu \Vert (O\setminus K) < \varepsilon$. We denote by ${\mathcal {M}}(\Omega ,X)$ the set of all regular X-valued measures on $\Omega$.

A measurable function $f:\Omega \rightarrow {\mathbb {C}}$ is said to be $\nu$-integrable if $f \in L^1({\langle \nu , x^*\rangle } )$ for every $x^* \in X^*$ and for each $E \in \mathfrak {B}(\Omega )$ there is an $x_E \in X$ such that $\langle x_E, x^* \rangle = \int _E f \,d{\langle \nu , x^*\rangle }$ for every $x^* \in X^*$. We denote $x_E$ by $\int _E f \, d\nu$. For a measurable function $f:\Omega \rightarrow {\mathbb {C}}$, define

$$\begin{aligned} \Vert f\Vert _{\nu } = \sup _{x^* \in B_{X^*}} \int _\Omega \vert f \vert \, d\vert {\langle \nu , x^*\rangle } \vert \end{aligned}$$

and $\Vert f\Vert _{\nu ,p}:= \Vert \vert f \vert ^p\Vert ^{1/p}_{\nu }$. The space $L^1(\Omega ,\nu )$ of all $\nu$-integrable functions is a Banach space with the norm $\Vert \cdot \Vert _{\nu }$. We say that $f=g$ $\nu$-a.e. if $\Vert f-g\Vert _{\nu }=0$. For each $1 \le p < \infty$, the space $L^p(\Omega ,\nu ):=\{f \in L^1(\Omega ,\nu ): \vert f \vert ^p \in L^1(\Omega ,\nu ) \}$ is a Banach space with the norm $\Vert \cdot \Vert _{L^p(\Omega ,\nu )}:= \Vert \cdot \Vert _{\nu ,p}$. We denote by $S(\Omega )$ the set of all simple functions on $\Omega$. The integral operator $I_\nu :L^1(\Omega ,\nu ) \rightarrow X$ is defined by $I_\nu (f)=\int _\Omega f \, d\nu$ for $f \in L^1(\Omega ,\nu ).$ Then $I_\nu$ is bounded with $\Vert I_\nu (f)\Vert _X \le \Vert f\Vert _{L^1(\Omega ,\nu )}$.

Theorem 2.1

[15] Let $f:\Omega \rightarrow {\mathbb {C}}$ be a complex function. Then f is $\nu$-integrable if and only if there is a sequence $(f_n)$ of simple functions which converges pointwise to f and for which $(\int _E f_n \, d\nu )$ is Cauchy for any $E \in \mathfrak {B}(\Omega )$.

Theorem 2.2

[13] Let $\nu \in {\mathcal {M}}(G,X)$. Then C(G) is dense in $L^p(G,\nu )$ for all $1\le p < \infty$.

For Banach spaces X and Y, a linear operator $T:X \rightarrow Y$ is said to be weakly compact if T(B) is a relatively weakly compact subset of Y whenever B is a bounded subset of X. By [4, Corollary VI.2.14], we have that on a compact Hausdorff space there is a one-to-one correspondence between the set of all regular vector measures and the set of all weakly compact operators. To be precise, given a regular vector measure $\nu :\mathfrak {B}(\Omega ) \rightarrow X$, there is a weakly compact operator $T:C(\Omega ) \rightarrow X$ representing $\nu$, that is, $T(f)=\int _\Omega f \, d\nu$ for all $f \in C(\Omega )$, and vice versa.

A vector measure $\nu$ is said to be absolutely continuous with respect to a positive scalar measure $\lambda$, denoted by $\nu \ll \lambda$, if $\nu (E) \rightarrow 0$ in norm as $\lambda (E) \rightarrow 0$ where $E \in \mathfrak {B}(\Omega )$. Note that $\nu \ll \lambda$ if and only if $\nu$ vanishes on all sets of $\lambda$-measure zero, by [4, Theorem I.2.1]. Moreover, $\nu$ vanishes on all sets of $\lambda$-measure zero if and only if $\Vert \nu \Vert$ vanishes on all sets of $\lambda$-measure zero. By Rybakov’s theorem [4], there is a linear functional $x^* \in X^*$ such that $\nu \ll \vert {\langle \nu , x^*\rangle } \vert$. This functional is called a Rybakov functional. For $k \in [0,\infty )$, a vector measure $\nu$ is said to be k-scalarly bounded by m if for any $x^* \in X^*$ and $E \in \mathfrak {B}(\Omega )$, we have $\vert {\langle \nu , x^*\rangle } \vert (E) \le km(E)$.

Let $\tau :G \rightarrow G$ be a homeomorphism. For a measurable function $f:G \rightarrow {\mathbb {C}}$, we denote $f \circ \tau ^{-1}$ by $f_\tau$. For $a \in G$, we define the left translation $L_a$ and the right translation $R_a$ by $L_a(t)=at$ and $R_a(t)=ta^{-1}$ for $t \in G$. In the case that $\tau = L_a$ or $R_a$, we shall write $L_af$ or $R_af$ instead of $f_\tau$. Hence $(L_a f)(t)=f(a^{-1}t)$ and $(R_a f)(t)=f(ta)$ for each $t \in G$.

Definition 1

Let $\tau :G \rightarrow G$ be a homeomorphism and $\nu$ a vector measure on G. We say that $\nu$ is $\tau$-invariant if

$$\begin{aligned} I_\nu (f_\tau ) =I_\nu (f) \quad \text { for all } f \in S(G). \end{aligned}$$

Given a collection ${\mathcal {T}}$ of homeomorphisms on G,  $\nu$ is said to be ${\mathcal {T}}$-invariant if it is $\tau$-invariant for all $\tau \in {\mathcal {T}}$. In particular, if ${\mathcal {T}}=\{L_a: a\in G\}$ (or ${\mathcal {T}}=\{R_a: a\in G\}$), we say that $\nu$ is left (or right) invariant.

Definition 2

Let $\tau :G \rightarrow G$ be a homeomorphism and $\nu$ a vector measure on G. We say that $\nu$ is norm integral $\tau$-invariant if

$$\begin{aligned} \Vert I_\nu (f_\tau ) \Vert =\Vert I_\nu (f) \Vert \quad \text { for all } f \in S(G). \end{aligned}$$

Given a collection ${\mathcal {T}}$ of homeomorphisms on G,  $\nu$ is said to be norm integral ${\mathcal {T}}$-invariant if it is norm integral $\tau$-invariant for all $\tau \in {\mathcal {T}}$. In particular, if ${\mathcal {T}}=\{L_a: a\in G\}$ (or ${\mathcal {T}}=\{R_a: a\in G\}$), we say that $\nu$ is norm integral left (or right) invariant.

Definition 3

Let $\tau :G \rightarrow G$ be a homeomorphism and $\nu$ a vector measure on G. We say that $\nu$ is semivariation $\tau$-invariant if

$$\begin{aligned} \Vert f_\tau \Vert _{L^1(G,\nu )} =\Vert f \Vert _{L^1(G,\nu )} \quad \text { for all } f \in S(G). \end{aligned}$$

Given a collection ${\mathcal {T}}$ of homeomorphisms on G,  $\nu$ is said to be semivariation ${\mathcal {T}}$-invariant if it is semivariation $\tau$-invariant for all $\tau \in {\mathcal {T}}$. In particular, if ${\mathcal {T}}=\{L_a: a\in G\}$ (or ${\mathcal {T}}=\{R_a: a\in G\}$), we say that $\nu$ is semivariation left (or right) invariant.

2.3 Tensor Integration

Let X and Y be any Banach spaces. Recall that the space ${\mathcal {B}}(Y^* \times X^*)$ of bounded bilinear forms on $Y^* \times X^*$ is a Banach space equipped with the norm

$$\begin{aligned} \Vert b\Vert = \sup \{\vert b(y^*,x^*) \vert : y^* \in B_{Y^*}, x^* \in B_{X^*}\}. \end{aligned}$$

Note that we can realize $Y \otimes X$ as a subspace of ${\mathcal {B}}(Y^* \times X^*)$ by considering $u=\sum _{i=1}^n y_i \otimes x_i \in Y \otimes X$ as a bilinear form given by $b_u(y^*,x^*) = \sum y^*(y_i)x^*(x_i)=(y^* \otimes x^*)(u)$ for $y^* \in Y^*$ and $x^* \in X^*$. The injective norm $\Vert \cdot \Vert _\vee$ on $Y \otimes X$ is the norm induced by this embedding, i.e.,

$$\begin{aligned} \Vert u \Vert _\vee = \sup _{y^* \in B_{Y^*}, x^* \in B_{X^*}} \vert (y^* \otimes x^*)(u)\vert . \end{aligned}$$

Moreover, we have alternative formulas for the injective norm

$$\begin{aligned} \Vert u \Vert _\vee = \sup _{y^* \in B_{Y^*}} \left\| \sum y^*(y_i) x_i \right\| _X = \sup _{x^* \in B_{X^*}} \left\| \sum x^*(x_i) y_i \right\| _Y. \end{aligned}$$

The completion of the tensor product space $Y \otimes X$ with the injective norm is called the injective tensor product of Y and X, denoted by $Y \check{\otimes } X$. For more details, see [17].

Now we summarize the concept of tensor integration introduced by [18]. Let $\nu$ be an X-valued vector measure. A function $f:\Omega \rightarrow Y$ is said to be $\nu$-measurable if there is a sequence of Y-valued simple functions $(f_n)$ with $\lim _{n \rightarrow \infty } \Vert f_n - f\Vert _Y = 0$ $\nu$-a.e. We say that a function $f:\Omega \rightarrow Y$ is weakly $\nu$-measurable if for each $y^* \in Y^*$ the function $y^*f$ is $\nu$-measurable. Note that a function $f:\Omega \rightarrow Y$ is $\nu$-measurable if and only if f is $\vert {\langle \nu , x^*\rangle } \vert$-measurable for some Rybakov functional $x^* \in X^*$.

Theorem 2.3

(Pettis’s measurability theorem [4]) Let $\lambda$ be a finite positive measure. A function $f:\Omega \rightarrow Y$ is $\lambda$-measurable if and only if f is weakly $\lambda$-measurable and $\lambda$-essentially separably valued.

Let $E \in \mathfrak {B}(\Omega )$ and $\phi =\sum _{i=1}^n y_i \chi _{A_i}$ be a Y-valued simple function on $\Omega$, where $y_i \in Y$ and $A_i \in \mathfrak {B}(\Omega )$. We define $\int _E \phi \, d\nu = \sum y_i \otimes \nu (E \cap A_i) \in Y \otimes X.$ Then it can be shown that $(y^* \otimes x^*)(\int _E \phi \, d\nu ) = \int _E y^*\phi \,d{\langle \nu , x^*\rangle }$ for $y^* \in Y^*$ and $x^* \in X^*$, hence $\left\| \int _E \phi \, d\nu \right\| _\vee \le \sup _{x^* \in B_{X^*}} \int _E \Vert \phi \Vert \, d\vert \langle \nu ,x^* \rangle \vert .$ For a $\nu$-measurable function $f:\Omega \rightarrow Y$, we let

$$\begin{aligned} \text {N}(f) = \sup _{x^* \in B_{X^*}} \int _\Omega \Vert f\Vert \, d\vert {\langle \nu , x^*\rangle } \vert . \end{aligned}$$

Definition 4

A $\nu$-measurable function $f:\Omega \rightarrow Y$ is $\check{\otimes }$-integrable if there exists a sequence $(f_n)$ of simple functions such that

$$\begin{aligned} \lim _{n \rightarrow \infty } \text {N}(f-\phi _n) =0. \end{aligned}$$

In this case, the sequence $( \int _E \phi _n \,d\nu )$ is a Cauchy sequence in $Y \check{\otimes } X$ for each $E \in \mathfrak {B}(\Omega )$. By the completeness of $Y \check{\otimes } X$, the limit of $( \int _E \phi _n \,d\nu )$ is denoted by $\int _E f \, d\nu$ and is called the $\check{\otimes }$-integral of f over E with respect to $\nu$.

Note that if f is $\check{\otimes }$-integrable, then $(y^* \otimes x^*)(\int _E f \, d\nu ) = \int _E y^*f \,d{\langle \nu , x^*\rangle }$ for $E \in \mathfrak {B}(\Omega )$, $y^* \in Y^*$ and $x^* \in X^*$ and $\Vert \int _\Omega f \,d\nu \Vert _\vee \le N(f)$.

Theorem 2.4

[18] A $\nu$-measurable function f is $\check{\otimes }$-integrable if and only if $\Vert f\Vert$ is $\nu$-integrable.

3 Extensions of the Operator $T_H$

In this section, we study extensions of the operator $T_H: C(G) \rightarrow C(G/H)$. Given a vector measure $\nu \in {\mathcal {M}}(G,X)$, we can naturally construct a vector measure on G/H as follows. Let $T_\nu :C(G) \rightarrow X$ be the corresponding weakly compact operator for $\nu$, i.e.,

$$\begin{aligned} T_\nu (f)=\int _{G} f \, d\nu \quad (f \in C(G)). \end{aligned}$$

Define $T_{\tilde{\nu }} :C(G/H) \rightarrow X$ by

$$\begin{aligned} T_{\tilde{\nu }} (\varphi )=T_\nu (\varphi _q)=\int _G \varphi _q \,d\nu \quad (\varphi \in C(G/H)). \end{aligned}$$

Then $T_{\tilde{\nu }}$ is weakly compact since $\Vert \varphi \Vert _{\sup }=\Vert \varphi _q\Vert _{\sup }$ for all $\varphi \in C(G/H)$. Hence there is a representing vector measure ${\tilde{\nu }} \in {\mathcal {M}}(G/H,X)$. Moreover, we immediately have that

$$\begin{aligned} \int _{G/H} \varphi \, d{\tilde{\nu }} = \int _G \varphi _q \, d\nu \quad (\varphi \in C(G/H)). \end{aligned}$$

(2)

We shall begin with some basic properties of ${\tilde{\nu }}$.

Proposition 3.1

Let $\nu \in {\mathcal {M}}(G,X)$.

• 1. The vector measure ${\tilde{\nu }}$ is the pushforward (vector) measure of $\nu$ by the quotient map q, i.e., ${\tilde{\nu }} (E)=\nu (q^{-1}(E))$ for all $E \in \mathfrak {B}(G/H)$. Moreover, the Eq. (2) holds for all $\varphi \in L^1(G/H,{\tilde{\nu }} )$ provided that $\varphi _q \in L^1(G,\nu )$.

• 2. For any $x^* \in X^*$ and $E \in \mathfrak {B}(G/H)$, $\vert {\langle {\tilde{\nu }} , x^*\rangle } \vert (E) \le \vert {\langle \nu , x^*\rangle } \vert (q^{-1}(E))$. Then $\Vert \varphi \Vert _{L^p(G/H,{\tilde{\nu }} )} \le \Vert \varphi _q\Vert _{\nu ,p}$ for any $1\le p < \infty$ and $\varphi \in L^p(G/H,{\tilde{\nu }} )$.

Proof

• 1. Let $\lambda$ be the pushforward measure of $\nu$ by the quotient map q. It follows from Eq. (2) that

  $$\begin{aligned} \int _{G/H} \varphi \, d\lambda =\int _G \varphi _q \, d\nu =\int _{G/H} \varphi \, d{\tilde{\nu }} \end{aligned}$$

  for all $\varphi \in C(G/H)$. Hence ${\tilde{\nu }} =\lambda$. Next observe that Eq. (2) holds for all $\varphi \in S(G/H)$. Let $0 \le \varphi \in L^1(G/H,{\tilde{\nu }} )$. Then there is a sequence of positive simple functions $\varphi _n \uparrow \varphi$ pointwise. By the monotone convergence theorem, for each $x^* \in X^*$

  $$\begin{aligned} \int _{G/H} \varphi \, d{\langle {\tilde{\nu }} , x^*\rangle }&= \lim _{n\rightarrow \infty } \int _{G/H} \varphi _n \, d{\langle {\tilde{\nu }} , x^*\rangle } \\&= \lim _{n\rightarrow \infty } \int _G (\varphi _n)_q \, d{\langle \nu , x^*\rangle } \\&= \int _G \varphi _q \, d{\langle \nu , x^*\rangle } . \end{aligned}$$

  This identity easily extends to $\varphi \in L^1(G/H,{\tilde{\nu }} )$. If we assume that $\varphi _q \in L^1(G,\nu )$, then

  $$\begin{aligned} \left\langle \int _{G/H} \varphi \, d{\tilde{\nu }} , x^* \right\rangle = \int _{G/H} \varphi \, d{\langle {\tilde{\nu }} , x^*\rangle } = \int _G \varphi _q \, d{\langle \nu , x^*\rangle } = \left\langle \int _{G} \varphi _q \, d\nu , x^* \right\rangle \end{aligned}$$

  for all $x^* \in X^*$, which proves the Eq. (2).

• 2. Let $E \in \mathfrak {B}(G/H)$. Consider any disjoint partition $\{E_n\}_{n=1}^{k}$ of E where $E_n \in \mathfrak {B}(G/H)$. Since $\{q^{-1}(E_n)\}_{n=1}^{k}$ forms a disjoint partition of $q^{-1}(E)$,

  $$\begin{aligned} \sum _{n=1}^k \vert {\langle {\tilde{\nu }} , x^*\rangle } (E_n)\vert = \sum _{n=1}^k \vert {\langle \nu , x^*\rangle } (q^{-1}(E_n))\vert \le \vert {\langle \nu , x^*\rangle } \vert (q^{-1}(E)). \end{aligned}$$

  Hence $\vert {\langle {\tilde{\nu }} , x^*\rangle } \vert (E) \le \vert {\langle \nu , x^*\rangle } \vert (q^{-1}(E))$. Consequently,

  $$\begin{aligned} \int _{G/H} \varphi \, d\vert {\langle {\tilde{\nu }} , x^*\rangle } \vert \le \int _{G} \varphi _q \, d\vert {\langle \nu , x^*\rangle } \vert \end{aligned}$$

  holds for any simple function $\varphi \ge 0$ on G/H. Then for any $\varphi \in L^1(G/H,\mu )$, the monotone convergence theorem implies that

  $$\begin{aligned} \int _{G/H} \vert \varphi \vert \, d\vert {\langle {\tilde{\nu }} , x^*\rangle } \vert \le \int _{G} \vert \varphi _q\vert \, d\vert {\langle \nu , x^*\rangle } \vert . \end{aligned}$$

  Therefore, $\Vert \varphi \Vert _{L^p(G/H,{\tilde{\nu }} )} \le \Vert \varphi _q\Vert _{\nu ,p}$ for any $1\le p < \infty$ and $\varphi \in L^p(G/H,{\tilde{\nu }} )$.

$\square$

Example 1

Let $1 \le p < \infty$ and $S:L^p(G,m) \rightarrow X$ be any bounded linear map, where m is the normalized Haar measure on G. Define a vector measure $\nu :\mathfrak {B}(G) \rightarrow X$ corresponding to S by $\nu (E) = S(\chi _E)$ for $E \in \mathfrak {B}(G)$. Then by Proposition 3.1.1 the vector measure ${\tilde{\nu }}$ is given by ${\tilde{\nu }} (F)=S(\chi _{q^{-1}(F)})$ for $F \in {\mathcal {B}}(G/H)$.

• 1. Let $X = {\mathbb {C}}$ and $S:L^1(G,m) \rightarrow {\mathbb {C}}$ be given by $S(f)=\int _G f \,dm$ for any $f \in L^1(G,m)$. In this case, $\nu = m$. Moreover, ${\tilde{\nu }} = \widetilde{m}$ since ${\tilde{\nu }} (F) = \int _G \chi _{q^{-1}(F)} \, dm = \int _{G/H} \chi _F \, d\widetilde{m} = \widetilde{m}(F)$ for all $F \in \mathfrak {B}(G/H)$, where $\widetilde{m}$ is the pushforward measure of m.

• 2. Let $X = L^1(G,m)$ and $S = \text {Id}_{L^1(G,m)}$. Then ${\tilde{\nu }} (F)=\chi _{q^{-1}(F)}$ for $F \in \mathfrak {B}(G/H)$.

• 3. Let $\lambda$ be a complex regular measure on G and $1 \le p < \infty$. We define $S:L^p(G,m) \rightarrow L^p(G,m)$ by $S(f) = f * \lambda$ where $(f * \lambda )(t)=\int _G f(ts^{-1}) \,d\lambda (s)$ for $t \in G$. Then ${\tilde{\nu }} (F)=\chi _{q^{-1}(F)}* \lambda$ for $F \in \mathfrak {B}(G/H)$.

• 4. Let $1 \le p \le 2$ and $S:L^p(G,m) \rightarrow \ell ^{p'}(\widehat{G};{\mathcal {B}}({\mathcal {H}}_\pi ))$ be defined by $S(f)={\mathcal {F}}_G(f)$. Then ${\tilde{\nu }} (F)={\mathcal {F}}_G(\chi _{q^{-1}(F)})={\mathcal {F}}_{G/H}(\chi _F)$ for $F \in \mathfrak {B}(G/H)$.

Let $\tau :G/H \rightarrow G/H$ be a homeomorphism. For example, one can consider a left translation $L_a:G/H \rightarrow G/H$ by $a \in G$ given by $L_a(tH)=atH$ for each $tH \in G/H$. For a measurable function $\varphi :G/H \rightarrow {\mathbb {C}}$, we denote $\varphi \circ \tau ^{-1}$ by $\varphi _\tau$. In the case that $\tau =L_a$ where $a \in G$, we shall denote $\varphi \circ (L_a)^{-1}$ by $L_a \varphi$ and by definition we have $(L_a \varphi )(tH)=\varphi (a^{-1}tH)$ for all $tH \in G/H$.

Definition 5

Let $\tau :G/H \rightarrow G/H$ be a homeomorphism. For any vector measure $\mu$ on G/H, we say that $\mu$ is norm integral $\tau$-invariant if

$$\begin{aligned} \Vert I_\mu (\varphi _\tau ) \Vert =\Vert I_\mu (\varphi ) \Vert \quad \text { for all } \varphi \in S(G/H). \end{aligned}$$

Given a collection ${\mathcal {T}}$ of homeomorphisms on G/H,  $\mu$ is said to be norm integral ${\mathcal {T}}$-invariant if it is norm integral $\tau$-invariant for all $\tau \in {\mathcal {T}}$. In particular, if ${\mathcal {T}}=\{L_a: a\in G\}$, we say that $\mu$ is norm integral left invariant.

This proposition is merely a consequence of Proposition 3.1.

Proposition 3.2

Let $\nu \in {\mathcal {M}}(G,X)$.

• 1. For $a \in G$, if $\nu$ is norm integral $L_a$-invariant, then so is ${\tilde{\nu }}$.

• 2. If $\nu \ll m$, then ${\tilde{\nu }} \ll \widetilde{m}$.

• 3. If $\nu$ is k-scalarly bounded by m, then ${\tilde{\nu }}$ is k-scalarly bounded by $\widetilde{m}$.

Proof

• 1. For $\varphi \in S(G/H)$, by Proposition 3.1.1, Eq. (2) holds for simple functions, we have

  $$\begin{aligned} \Vert I_{\tilde{\nu }} (L_a\varphi ) \Vert = \Vert I_\nu ((L_a\varphi )_q)\Vert = \Vert I_\nu (L_a(\varphi _q))\Vert = \Vert I_\nu (\varphi _q)\Vert = \Vert I_{\tilde{\nu }} (\varphi )\Vert . \end{aligned}$$

• 2. For any $F \in \mathfrak {B}(G/H)$, $\widetilde{m}(F) = m(q^{-1}(F))$ and ${\tilde{\nu }} (F)=\nu (q^{-1}(F))$. If $\widetilde{m}(F) \rightarrow 0$, then also $m(q^{-1}(F)) \rightarrow 0$, and hence ${\tilde{\nu }} (F)=\nu (q^{-1}(F)) \rightarrow 0$ since $\nu \ll m$.

• 3. It follows immediately from the fact that

  $$\begin{aligned} \vert {\langle {\tilde{\nu }} , x^*\rangle } \vert (E)\le \vert {\langle \nu , x^*\rangle } \vert (q^{-1}(E)) \le k m(q^{-1}(E)) = k\widetilde{m}(E) \end{aligned}$$

  for any $E \in \mathfrak {B}(G/H)$.

$\square$

Now we prove an existence of an extension of $T_H$ to an operator from $L^p(G,\nu )$ into $L^p(G/H,{\tilde{\nu }} )$ for each $1\le p < \infty$. This is a generalization of Theorem 3.2 in [5].

Theorem 3.3

Let $1\le p < \infty$ and ${\mathcal {R}} = \{R_h: h \in H\}$. Suppose that $\nu$ is semivariation ${\mathcal {R}}$-invariant. Then the operator $T_H:C(G) \rightarrow C(G/H)$ satisfies

$$\begin{aligned} \Vert T_H f\Vert _{L^p(G/H,{\tilde{\nu }} )} \le \Vert f\Vert _{L^p(G,\nu )} \quad \text { for all } f \in C(G), \end{aligned}$$

hence it has a unique extension to a norm-decreasing operator $T_{H,\nu }:L^p(G,\nu ) \rightarrow L^p(G/H,{\tilde{\nu }} )$.

Proof

Let $f \in C(G)$. By Proposition 3.1.2 and $\nu$ being semivariation ${\mathcal {R}}$-invariant,

$$\begin{aligned} \Vert T_H f\Vert _{L^p(G/H,{\tilde{\nu }} )}^p&\le \Vert (T_H f)_q\Vert _{L^p(G,\nu )}^p \\&= \sup _{x^* \in B_{X^*}} \int _{G} \vert (T_H f)(tH)\vert ^p \, d\vert {\langle \nu , x^*\rangle } \vert (t) \\&\le \sup _{x^* \in B_{X^*}} \int _G \int _H \vert f(th)\vert ^p \, dh \, d\vert {\langle \nu , x^*\rangle } \vert (t) \\&= \sup _{x^* \in B_{X^*}} \int _H \int _G \vert f(th)\vert ^p \, d\vert {\langle \nu , x^*\rangle } \vert (t) \, dh \\&\le \int _H \bigg (\sup _{x^* \in B_{X^*}} \int _G \vert f(th)\vert ^p \, d\vert {\langle \nu , x^*\rangle } \vert (t)\bigg ) \, dh \\&= \int _H \Vert R_h f\Vert _{L^p(G,\nu )}^p \, dh \\&= \int _H \Vert f\Vert _{L^p(G,\nu )}^p \, dh \\&= \Vert f\Vert _{L^p(G,\nu )}^p. \end{aligned}$$

By the density of C(G) in $L^p(G,\nu )$, the operator $T_H$ can be extended uniquely to a bounded linear map from $L^p(G,\nu )$ to $L^p(G/H,{\tilde{\nu }} )$. To verify that $T_{H,\nu }$ is norm-decreasing, let $f \in L^p(G,\nu )$ with $f_n \rightarrow f$ in $L^p(G,\nu )$ where $f_n \in C(G)$. Then

$$\begin{aligned} \Vert T_{H,\nu } f\Vert _{L^p(G/H,{\tilde{\nu }} )} = \lim _{n\rightarrow \infty } \Vert T_H f_n\Vert _{L^p(G/H,{\tilde{\nu }} )} \le \lim _{n\rightarrow \infty } \Vert f_n\Vert _{L^p(G,\nu )} = \Vert f\Vert _{L^p(G,\nu )} \end{aligned}$$

as desired. $\square$

Remark 1

If there is no ambiguity, we shall denote $T_{H,\nu }$ by $T_H$. Secondly, it is worth noting that even though the extensions of $T_H:C(G) \rightarrow C(G/H)$ to $L^p(G,\nu )$ and $L^q(G,\nu )$ might be different operators if $p \ne q$, they coincide on the intersection of the domains. Suppose that we denote the extension of $T_H$ to $L^p(G,\nu )$ by $T_{H,p}$ for $1 \le p < \infty$. Consider $1 \le p< q < \infty$. Note that it follows from [15, Proposition 3.31(ii)] that for any vector measure $\mu$ on $\Omega$, $L^q(\Omega ,\mu ) \subset L^p(\Omega ,\mu )$ with $\Vert f\Vert _{L^p(\Omega ,\mu )} \le K \Vert f\Vert _{L^q(\Omega ,\mu )}$ for some constant $K>0$. Now we show that the extensions $T_{H,p}$ and $T_{H,q}$ coincide on $L^q(G,\nu ) \subset L^p(G,\nu )$. Let $f_n \rightarrow f$ in $L^q(G,\nu )$ where $f_n \in C(G)$. Then $T_{H,p} f_n \rightarrow T_{H,p} f$ in $L^p(G/H,{\tilde{\nu }} )$ and also $T_{H,q} f_n \rightarrow T_{H,q} f$ in $L^p(G/H,{\tilde{\nu }} )$. Since $T_{H,p}$ and $T_{H,q}$ agree on C(G), we have that $T_{H,p} f = T_{H,q} f$ in $L^p(G/H,{\tilde{\nu }} )$ which implies $T_{H,p} f = T_{H,q} f$ ${\tilde{\nu }}$-a.e. Thus there is no ambiguity to denote any extension $T_{H,p}$ for any $1\le p < \infty$ by $T_H$.

Now we prove that the extension $T_H$ is norm-decreasing in the sense of the norm in X.

Theorem 3.4

Let $\nu$ be norm integral ${\mathcal {R}}$-invariant where ${\mathcal {R}} = \{R_h: h \in H\}$. Then

$$\begin{aligned} \left\| \int _{G/H} T_H f \, d{\tilde{\nu }} \right\| _X \le \left\| \int _G f \, d\nu \right\| _X \quad \big (f \in L^1(G,\nu )\big ). \end{aligned}$$

Proof

Let $f \in C(G)$. For $x^* \in B_{X^*}$, by Eq. (2)

$$\begin{aligned} \int _{G/H} T_H f \, d{\langle {\tilde{\nu }} , x^*\rangle }&= \int _G( T_H f)_q \, d{\langle \nu , x^*\rangle } \quad \\&= \int _G \int _H f(th) \,dh \,d{\langle \nu , x^*\rangle } (t) \\&= \int _H \int _G (R_h f)(t) \,d{\langle \nu , x^*\rangle } (t) \,dh. \end{aligned}$$

Hence

$$\begin{aligned} \left\| \int _{G/H} T_H f \, d{\tilde{\nu }} \right\| _X&=\sup _{x^* \in B_{X^*}} \left| \int _{G/H} T_H f \, d{\langle {\tilde{\nu }} , x^*\rangle } \right| \\&=\sup _{x^* \in B_{X^*}} \left| \int _H \int _G (R_h f)(t) \,d{\langle \nu , x^*\rangle } (t) \,dh \right| \\&\le \int _H \bigg ( \sup _{x^* \in B_{X^*}} \left| \int _G (R_h f)(t) \,d{\langle \nu , x^*\rangle } (t) \right| \bigg ) \,dh \\&=\int _H \left\| \int _G R_h f \, d\nu \right\| _X dh \\&=\left\| \int _G f \, d\nu \right\| _X. \end{aligned}$$

For any $f \in L^1(G,\nu )$, let $f_n$ be a sequence of continuous functions converging to f in $L^1(G,\nu )$. Then

$$\begin{aligned} \Vert I_{\tilde{\nu }} (T_H f) \Vert _X = \lim _{n \rightarrow \infty } \Vert I_{\tilde{\nu }} (T_H f_n) \Vert _X \le \lim _{n \rightarrow \infty } \Vert I_\nu (f_n) \Vert _X = \Vert I_\nu (f) \Vert _X \end{aligned}$$

that is $\Vert \int _{G/H} T_H f \, d{\tilde{\nu }} \Vert _X \le \Vert \int _G f \, d\nu \Vert _X$ as desired. $\square$

We have investigated the properties of the extension $T_H:L^p(G,\nu ) \rightarrow L^p(G/H,{\tilde{\nu }} )$ for a given vector measure $\nu \in {\mathcal {M}}(G,X)$. However, in general, to study Fourier analysis on homogeneous spaces, it is essential to consider the space $L^p(G/H,\mu )$ for a given vector measure $\mu$ on G/H instead of the space $L^p(G/H,{\tilde{\nu }} )$. To deal with this situation, we will define a corresponding measure ${\breve{\mu }}$ on G and study the extension $T_H:L^p(G,{\breve{\mu }} ) \rightarrow L^p(G/H,\mu )$.

Let $\mu \in {\mathcal {M}}(G/H,X)$ and $T_\mu :C(G/H) \rightarrow X$ be the corresponding weakly compact operator given by

$$\begin{aligned} T_\mu (\varphi ) = \int _{G/H} \varphi \, d\mu \quad (\varphi \in C(G/H)). \end{aligned}$$

Observe that the operator $T_\mu \circ T_H:C(G) \rightarrow X$ is weakly compact since $T_H$ is bounded and $T_\mu$ is weakly compact. Then there is a representing regular vector measure on G. Denote the representing vector measure by ${\breve{\mu }} \in {\mathcal {M}}(G,X)$ and $T_\mu \circ T_H$ by $T_{\breve{\mu }}$. Hence we immediately have that

$$\begin{aligned} \int _G f \,d{\breve{\mu }} = \int _{G/H} T_Hf \, d\mu \quad (f \in C(G)). \end{aligned}$$

(3)

Remark 2

Let $\Phi :{\mathcal {M}}(G,X) \rightarrow {\mathcal {M}}(G/H,X)$ be defined by $\Phi (\nu )={\tilde{\nu }}$ for $\nu \in {\mathcal {M}}(G,X)$ and $\Psi :{\mathcal {M}}(G/H,X) \rightarrow {\mathcal {M}}(G,X)$ by $\Psi (\mu )={\breve{\mu }}$ for $\mu \in {\mathcal {M}}(G/H,X)$. Then the following diagram commutes:

In other words, $\Phi \circ \Psi = \text {Id}_{{\mathcal {M}}(G/H,X)}$ or equivalently $\tilde{{\breve{\mu }} }=\mu$ for any $\mu \in {\mathcal {M}}(G/H,X)$. This can be proved by observing that

$$\begin{aligned} \int _{G/H} \varphi \, d\tilde{{\breve{\mu }} } = \int _G \varphi _q \, d{\breve{\mu }} = \int _{G/H} T_H(\varphi _q) \, d\mu = \int _{G/H} \varphi \, d\mu \end{aligned}$$

for all $\varphi \in C(G/H)$. Note that the commutativity of the diagram also implies that $\Phi$ is surjective and $\Psi$ is injective.

Proposition 3.5

Let ${\mathcal {R}} = \{R_h: h \in H\}$ and $x^* \in B_{X^*}$. Then ${\breve{\mu }}$ and $\vert {\langle {\breve{\mu }} , x^*\rangle } \vert$ are ${\mathcal {R}}$-invariant.

Proof

To show that ${\breve{\mu }}$ is ${\mathcal {R}}$-invariant, let $h \in H$ and $f\in C(G)$. Observe that

$$\begin{aligned} T_H(R_h f)(tH) = \int _H (R_hf)(th') \, dh' = \int _H f(th') \, dh' = T_H(f)(tH). \end{aligned}$$

Hence

$$\begin{aligned} T_{{\breve{\mu }} _{R_h}}( f) = T_{\breve{\mu }} (R_h f) = T_\mu (T_H(R_h f)) = T_\mu (T_H f) = T_{\breve{\mu }} (f). \end{aligned}$$

Hence ${\breve{\mu }} _{R_h}={\breve{\mu }}$, that is, ${\breve{\mu }}$ is $R_h$-invariant.

Now let $x^* \in B_{X^*}$, $E \in \mathfrak {B}(G)$ and $h \in H$. For any disjoint partition $\{E_n\}_{n=1}^{k}$ of E where $E_n \in \mathfrak {B}(G)$, note that $\{R_hE_n\}_{n=1}^{k}$ forms a disjoint partition of $R_hE$ and

$$\begin{aligned} \sum _{n=1}^{k}\vert {\langle {\breve{\mu }} , x^*\rangle } (E_n)\vert = \sum _{n=1}^{k}\vert {\langle {\breve{\mu }} , x^*\rangle } (R_hE_n)\vert \le \vert {\langle {\breve{\mu }} , x^*\rangle } \vert (R_hE). \end{aligned}$$

Hence $\vert {\langle {\breve{\mu }} , x^*\rangle } \vert (E) \le \vert {\langle {\breve{\mu }} , x^*\rangle } \vert (R_hE)$. Taking E as $R_hE$ and h as $h^{-1}$, we also get $\vert {\langle {\breve{\mu }} , x^*\rangle } \vert (R_h E) \le \vert {\langle {\breve{\mu }} , x^*\rangle } \vert (E)$. $\square$

This proposition particularly implies that ${\breve{\mu }}$ is semivariation ${\mathcal {R}}$-invariant. Hence we can apply Theorem 3.3 to get that the operator $T_H$ has an extension to a norm-decreasing operator from $L^p(G,{\breve{\mu }} )$ to $L^p(G/H,\mu )$ for any $1 \le p < \infty$. Moreover, Eq. (3) extends to $L^1(G,{\breve{\mu }} )$

$$\begin{aligned} \int _G f \,d{\breve{\mu }} = \int _{G/H} T_H f \, d\mu \quad \big (f \in L^1(G,{\breve{\mu }} )\big ). \end{aligned}$$

(4)

Indeed, if $f_n \rightarrow f$ in $L^1(G,{\breve{\mu }} )$ where $f_n \in C(G)$, then $I_\mu (T_H f) = \lim _{n\rightarrow \infty } T_\mu (T_H(f_n)) =\lim _{n\rightarrow \infty } T_{\breve{\mu }} (f_n)=I_{\breve{\mu }} (f)$. Now we prove that the Eq. (4) is also true for the total variation of the associated complex measures.

Lemma 3.6

For $x^* \in B_{X^*}$ and $f \in L^1(G,{\breve{\mu }} )$,

$$\begin{aligned} \int _G f \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert = \int _{G/H} T_H f \, d\vert {\langle \mu , x^*\rangle } \vert . \end{aligned}$$

In particular, $\Vert T_H \vert f \vert \Vert _{L^1(G/H,\mu )}=\Vert f\Vert _{L^1(G,{\breve{\mu }} )}$ for any $f \in L^1(G,{\breve{\mu }} )$.

Proof

It suffices to prove that for any $f \in L^1(G,{\breve{\mu }} )$ and $x^* \in B_{X^*}$

$$\begin{aligned} \int _G \vert f \vert \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert = \int _{G/H} T_H \vert f\vert \, d\vert {\langle \mu , x^*\rangle } \vert . \end{aligned}$$

We first claim that for each $E \in \mathfrak {B}(G)$, $T_H(\chi _E) \ge 0$ $\vert {\langle \mu , x^*\rangle } \vert$-a.e that is the set $F=\{tH \in G/H: T_H(\chi _E) < 0\}$ is $\vert {\langle \mu , x^*\rangle } \vert$-null. Let $f_n \rightarrow \chi _E$ in $L^1(G,{\breve{\mu }} )$ where $f_n \in C(G)$ is positive (which exists by using Urysohn’s lemma together with the regularity of ${\breve{\mu }}$). Since $T_H(f_n) \ge 0$ for all $n \in {\mathbb {N}}$,

$$\begin{aligned} \int _F \vert T_H(\chi _E)\vert \, d\vert {\langle \mu , x^*\rangle } \vert&\le \int _F \vert T_H(\chi _E)-T_H(f_n)\vert \, d\vert {\langle \mu , x^*\rangle } \vert \\&\le \Vert T_H(\chi _E - f_n)\Vert _{L^1(G/H,\mu )} \end{aligned}$$

which implies that $\vert {\langle \mu , x^*\rangle } \vert (F) = 0$ as desired. Now fix $E \in \mathfrak {B}(G)$ and consider any disjoint partition $\{E_n\}_{n=1}^{k}$ of E where $E_n \in \mathfrak {B}(G)$. By Eq. (4) and the claim,

$$\begin{aligned} \sum _{n=1}^k \vert {\langle {\breve{\mu }} , x^*\rangle } (E_n)\vert&= \sum _{n=1}^k \left| \int _{G/H} T_H(\chi _{E_n}) \, d{\langle \mu , x^*\rangle } \right| \\&\le \sum _{n=1}^k \int _{G/H} T_H(\chi _{E_n}) \, d\vert {\langle \mu , x^*\rangle } \vert \\&= \int _{G/H} T_H(\chi _E) \, d\vert {\langle \mu , x^*\rangle } \vert . \end{aligned}$$

Hence $\int _G \chi _E \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert \le \int _{G/H} T_H(\chi _E) \, d\vert {\langle \mu , x^*\rangle } \vert .$ It follows immediately that for any $f \in S(G)$,

$$\begin{aligned} \int _G \vert f\vert \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert \le \int _{G/H} T_H\vert f\vert \, d\vert {\langle \mu , x^*\rangle } \vert \end{aligned}$$

which can be extended to any $f\in L^1(G,{\breve{\mu }} )$ by using the density of S(G) in $L^1(G,{\breve{\mu }} )$.

Conversely, by Propositions 3.1.2 and 3.5, for $f \in C(G)$

$$\begin{aligned} \int _{G/H} T_H\vert f\vert \, d\vert {\langle \mu , x^*\rangle } \vert&\le \int _G (T_H\vert f\vert )_q \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert \\&= \int _G \int _H \vert f(th)\vert \,dh \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert (t) \\&= \int _H \int _G \vert f(th)\vert \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert (t) \,dh \\&= \int _H \int _G \vert f(t)\vert \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert (t) \,dh \\&= \int _G \vert f(t)\vert \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert (t). \end{aligned}$$

Hence by the density of C(G) in $L^1(G,{\breve{\mu }} )$, for any $f \in L^1(G,{\breve{\mu }} )$

$$\begin{aligned} \int _{G/H} T_H\vert f\vert \, d\vert {\langle \mu , x^*\rangle } \vert \le \int _G \vert f\vert \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert . \end{aligned}$$

$\square$

For any $\nu \in {\mathcal {M}}(G,X)$, we cannot find an example of an operator $T_{H,\nu }$ constructed in the manner of Theorem 3.3 which is not surjective. However, we know that if $\nu$ is in the form of ${\breve{\mu }}$, where $\mu \in {\mathcal {M}}(G/H,X)$, then the operator $T_{H,{\breve{\mu }} }$ is certainly surjective as shown in the following theorem.

Theorem 3.7

Let $1 \le p < \infty$. The extension $T_H:L^p(G,{\breve{\mu }} ) \rightarrow L^p(G/H,\mu )$ satisfies the formula $T_H f(tH)=\int _H f(th) \, dh$ $\mu$-a.e. for all $f \in L^p(G,{\breve{\mu }} )$. Moreover, the extension $T_H:L^p(G,{\breve{\mu }} ) \rightarrow L^p(G/H,\mu )$ is surjective.

Proof

Claim that for a lower semicontinuous function $\phi \ge 0$ and $x^* \in B_{X^*}$,

$$\begin{aligned} \int _G \phi \,d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert =\int _{G/H}\int _H \phi (th) \, dh \, d\vert {\langle \mu , x^*\rangle } \vert (tH). \end{aligned}$$

Let $\Phi = \{g \in C(G): 0 \le g \le \phi \}$. By [10, Proposition 7.12] and Lemma 3.6,

$$\begin{aligned} \int _G \phi \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert&= \sup _{g \in \Phi } \int _G g \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert \\&= \sup _{g \in \Phi } \int _{G/H} T_H g \, d\vert {\langle \mu , x^*\rangle } \vert \ \\&= \int _{G/H} \left( \sup _{g \in \Phi } \, T_H g \right) d\vert {\langle \mu , x^*\rangle } \vert \\&= \int _{G/H} \left( \sup _{g \in \Phi } \int _H g(th) \, dh \right) d\vert {\langle \mu , x^*\rangle } \vert (tH) \\&= \int _{G/H} \left( \sup _{\tilde{g} \in \Phi (tH)} \int _H \tilde{g}(h) \, dh \right) d\vert {\langle \mu , x^*\rangle } \vert (tH) \\&= \int _{G/H} \int _H \phi (th) \, dh \, d\vert {\langle \mu , x^*\rangle } \vert (tH) \end{aligned}$$

where $\Phi (tH):= \{\tilde{g} \in C(H): 0 \le \tilde{g}(h) \le \phi (th) \text { for } h \in H\}$. Hence for any measurable function F and any lower semicontinuous function $\phi \ge \vert F\vert$,

$$\begin{aligned} \int _{G/H} \int _H \vert F(th)\vert \, dh \, d\vert {\langle \mu , x^*\rangle } \vert (tH)&\le \int _{G/H} \int _H \phi (th) \, dh \, d\vert {\langle \mu , x^*\rangle } \vert (tH) \\&= \int _G \phi \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert . \end{aligned}$$

Hence by [10, Proposition 7.14]

$$\begin{aligned} \int _{G/H} \int _H \vert F(th)\vert \, dh \, d\vert {\langle \mu , x^*\rangle } \vert (tH) \le \int _G \vert F\vert \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert . \end{aligned}$$

(5)

Let $f \in L^p(G,{\breve{\mu }} )$ and $f_n \rightarrow f$ in $L^p(G,{\breve{\mu }} )$ where $f_n \in C(G)$. Define a function $\tilde{f}:G/H \rightarrow {\mathbb {C}}$ by $\tilde{f}(tH) =\int _H f(th) \, dh$ for $tH \in G/H$. By taking $F=\vert f-f_n\vert ^p$ in (5), we have

$$\begin{aligned} \Vert \tilde{f} - T_H f_n \Vert ^p_{L^p(G/H,\mu )}&= \sup _{x^* \in B_{X^*}} \int _{G/H} \big \vert \tilde{f} - T_H f_n\big \vert ^p \,d\vert {\langle \mu , x^*\rangle } \vert \\&\le \sup _{x^* \in B_{X^*}} \int _{G/H} \int _H \vert f - f_n\vert ^p(th)\,dh \,d\vert {\langle \mu , x^*\rangle } \vert (tH) \\&\le \sup _{x^* \in B_{X^*}} \int _G \vert f-f_n\vert ^p \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert \\&= \Vert f-f_n\Vert ^p_{L^p(G,{\breve{\mu }} )} \end{aligned}$$

which shows that $\tilde{f}$ is well-defined and $\tilde{f}=T_Hf$ $\mu$-a.e.

To show that $T_H$ is surjective, we first claim that $\Vert \phi _q\Vert _{L^p(G,{\breve{\mu }} )} = \Vert \phi \Vert _{L^p(G/H,\mu )}$ for $\phi \in L^p(G/H,\mu )$. Let $\phi _n \uparrow \vert \phi \vert$ pointwise where $\phi _n \in S(G/H)$. Then $(\phi _n)_q \uparrow \vert \phi \vert _q$ pointwise. Applying the monotone convergence theorem and Lemma 3.6 to each $x^* \in B_{X^*}$, we get

$$\begin{aligned} \int _G \vert \phi _q\vert ^p \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert&= \lim _{n\rightarrow \infty } \int _G \vert (\phi _n)_q\vert ^p \, d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert \\&= \lim _{n\rightarrow \infty } \int _{G/H} \vert \phi _n\vert ^p \, d\vert {\langle \mu , x^*\rangle } \vert \\&= \int _G \vert \phi \vert ^p \, d\vert {\langle \mu , x^*\rangle } \vert \end{aligned}$$

which proves the claim. Now let $\varphi \in L^p(G/H,\mu )$. If we can show that $\varphi _q \in L^p(G,{\breve{\mu }} )$, then by the formula of $T_H$ we have $T_H(\varphi _q)=\varphi$. Let $\varphi _n \rightarrow \varphi$ in $L^p(G/H,\mu )$ where $\varphi _n \in S(G/H)$. Then it follows by the claim that $\Vert \varphi _q - (\varphi _n)_q\Vert _{L^p(G,{\breve{\mu }} )} = \Vert \varphi -\varphi _n\Vert _{L^p(G/H,\mu )} \rightarrow 0$. Hence $\varphi _q \in L^p(G,{\breve{\mu }} )$ by the completeness of $L^p(G,{\breve{\mu }} )$. $\square$

Corollary 3.8

• 1. Weil’s formula holds for all $f \in L^1(G,{\breve{\mu }} )$

  $$\begin{aligned} \int _{G/H}\int _H f(th) \, dh \, d\mu (tH) = \int _G f \,d{\breve{\mu }} . \end{aligned}$$

  Moreover, for all $x^* \in X^*$ and $f \in L^1(G,{\breve{\mu }} )$

  $$\begin{aligned} \int _{G/H}\int _H f(th) \, dh \, d\vert {\langle \mu , x^*\rangle } \vert (tH) = \int _G f \,d\vert {\langle {\breve{\mu }} , x^*\rangle } \vert . \end{aligned}$$

• 2. For $1 \le p < \infty$, if $\varphi \in L^p(G/H,\mu )$, then $\varphi _q \in L^p(G,{\breve{\mu }} )$ with $\Vert \varphi _q\Vert _{L^p(G,{\breve{\mu }} )} = \Vert \varphi \Vert _{L^p(G/H,\mu )}$.

Proof

The first two equations follow by applying the formula of $T_H$ to Eq. (4) and Lemma 3.6 while the last assertion is in the proof of the theorem. $\square$

Corollary 3.9

For $1 \le p < \infty$, C(G/H) is dense in $L^p(G/H,\mu )$.

Proof

Let $\varphi \in L^p(G/H,\mu )$. Then $\varphi _q \in L^p(G,{\breve{\mu }} )$. By the density of C(G) in $L^p(G,{\breve{\mu }} )$, there is a sequence $f_n \rightarrow \varphi _q$ in $L^p(G,{\breve{\mu }} )$ with $f_n \in C(G)$. Hence $T_H f_n \rightarrow T_H(\varphi _q)=\varphi$ in $L^p(G/H,\mu )$. $\square$

It is straightforward to see that $\breve{\widetilde{m}}=m$. Hence $T_{H,m} = T_{H,\breve{\widetilde{m}}}$ is the same operator $T_H$ given by Farashahi in [5]. Now we provide a relation between the extensions $T_{H,m}$ and $T_{H,{\breve{\mu }} }$.

Proposition 3.10

If $\mu \ll \widetilde{m}$, then $T_{H,m} f = T_{H,{\breve{\mu }} } f$ $\mu$-a.e. for all $f \in L^1(G,m) \cap L^1(G,{\breve{\mu }} )$, and hence ${\breve{\mu }} \ll m$.

Proof

Let $f \in L^1(G,m) \cap L^1(G,{\breve{\mu }} )$. Then $T_{H,m} f(tH) = \int _H f(th) \, dh$ $\widetilde{m}$-a.e.; in particular, $T_{H,m} f(tH) = \int _H f(th) \, dh$ $\mu$-a.e. since $\mu \ll \widetilde{m}$. By Theorem 3.7, we also have $T_{H,{\breve{\mu }} } f(tH) = \int _H f(th) \, dh$ $\mu$-a.e., so $T_{H,m} f = T_{H,{\breve{\mu }} } f$ $\mu$-a.e.

Given $E \in \mathfrak {B}(G)$ with $m(E)=0$. Then $T_{H,m} \chi _E = 0$ $\widetilde{m}$-a.e. and hence $T_{H,{\breve{\mu }} } \chi _E = 0$ $\mu$-a.e. By Lemma 3.6, we get $\Vert {\breve{\mu }} \Vert (E)=0$. We conclude that ${\breve{\mu }} \ll m$. $\square$

Example 2

Let $1 \le p < \infty$ and $S:L^p(G/H,\widetilde{m}) \rightarrow X$ be any bounded linear map. Define a vector measure $\mu :\mathfrak {B}(G/H) \rightarrow X$ corresponding to S by $\mu (E) = S(\chi _E)$ for $E \in \mathfrak {B}(G/H)$. Then the vector measure ${\breve{\mu }}$ is given by ${\breve{\mu }} (F) = \int _{G/H} T_{H,{\breve{\mu }} }\chi _F \, d\mu$ for $F \in {\mathcal {B}}(G)$. Note that for $\varphi \in L^p(G/H,\widetilde{m})$, $\varphi$ is $\mu$-integrable and $\int _{G/H} \varphi \,d\mu = S(\varphi )$, see [15, Proposition 4.4]. Hence it follows from Proposition 3.10 that ${\breve{\mu }} (F) = \int _{G/H} T_{H,m}\chi _F \, d\mu = S(T_{H,m}\chi _F)$.

• 1. Let $X = {\mathbb {C}}$ and $S:L^1(G/H,\widetilde{m}) \rightarrow {\mathbb {C}}$ be given by $S(\varphi )=\int _{G/H} \varphi \,d\widetilde{m}$ for any $\varphi \in L^1(G/H,\widetilde{m})$. In this case, $\mu = \widetilde{m}$. Moreover, ${\breve{\mu }} = m$ since ${\breve{\mu }} (F) = \int _{G/H} T_{H,m}\chi _F \, d\widetilde{m} = \int _G \chi _F \,dm = m(F)$ for all $F \in \mathfrak {B}(G)$.

• 2. If $X = L^1(G/H,\widetilde{m})$ and $S = Id_{L^1(G/H,\widetilde{m})}$, then ${\breve{\mu }} (F)=T_{H,m}\chi _F$ for $F \in \mathfrak {B}(G)$.

• 3. If we let $1 \le p \le 2$ and define $S:L^p(G/H,\widetilde{m}) \rightarrow \ell ^{p'}(\widehat{G/H};{\mathcal {B}}({\mathcal {H}}_\pi ))$ by $S(\varphi )={\mathcal {F}}_{G/H}(\varphi )$, then ${\breve{\mu }} (F)(\pi )={\mathcal {F}}_{G/H}(T_{H,m}\chi _F)(\pi )=T^\pi _H\widehat{\chi _F}(\pi )$ for $F \in \mathfrak {B}(G)$ and $[\pi ] \in \widehat{G/H}$, by [5, Proposition 5.3].

Finally, we give relations between $\mu$ and ${\breve{\mu }}$ in terms of invariant properties.

Definition 6

Let $\mu$ be a vector measure on G/H. For each $a \in G$, $\mu$ is said to be $L_a$-invariant if $\mu (aE)=\mu (E)$ for all $E \in \mathfrak {B}(G/H)$. We say that $\mu$ is left invariant if it is $L_a$-invariant for all $a \in G$.

Definition 7

Let $\tau :G/H \rightarrow G/H$ be a homeomorphism. For any vector measure $\mu$ on G/H, we say that $\mu$ is semivariation $\tau$-invariant if

$$\begin{aligned} \Vert \varphi _\tau \Vert _{L^1(G/H,\mu )} =\Vert \varphi \Vert _{L^1(G/H,\mu )} \quad \text { for all } \varphi \in S(G/H). \end{aligned}$$

Given a collection ${\mathcal {T}}$ of homeomorphisms on G/H,  $\mu$ is said to be semivariation ${\mathcal {T}}$-invariant if it is semivariation $\tau$-invariant for all $\tau \in {\mathcal {T}}$. In particular, if ${\mathcal {T}}=\{L_a: a\in G\}$, we say that $\mu$ is semivariation left invariant.

Proposition 3.11

Let $a \in G$.

• 1. $\mu$ is $L_a$-invariant if and only if ${\breve{\mu }}$ is $L_a$-invariant.

• 2. $\mu$ is norm integral $L_a$-invariant if and only if ${\breve{\mu }}$ is norm integral $L_a$-invariant.

• 3. $\mu$ is semivariation $L_a$-invariant if and only if ${\breve{\mu }}$ is semivariation $L_a$-invariant.

Proof

• 1. Suppose that $\mu$ is $L_a$-invariant. Then by the Weil formula (4), for any $f \in C(G)$,

  $$\begin{aligned} \int _G L_a f \,d{\breve{\mu }}&= \int _{G/H} T_H (L_a f) \, d\mu = \int _{G/H} L_a (T_H f) \, d\mu \\&= \int _{G/H} T_H f \, d\mu = \int _G f \,d{\breve{\mu }} . \end{aligned}$$

  Hence ${\breve{\mu }}$ is $L_a$-invariant. Conversely, suppose that ${\breve{\mu }}$ is $L_a$-invariant. Then for any $\varphi \in S(G/H)$

  $$\begin{aligned} \int _{G/H} L_a \varphi \, d\mu = \int _G L_a \varphi _q \, d{\breve{\mu }} = \int _G \varphi _q \, d{\breve{\mu }} = \int _{G/H} \varphi \, d\mu \end{aligned}$$

  Hence $\mu$ is $L_a$-invariant.

• 2. Suppose that $\mu$ is norm integral $L_a$-invariant. Then by [2, Theorem 3.3], we have $\Vert I_\mu (L_a \varphi )\Vert = \Vert I_\mu (\varphi )\Vert$ for all $\varphi \in L^1(G/H,\mu )$. Hence by the Weil formula (4)

  $$\begin{aligned} \Vert I_{\breve{\mu }} (L_a f) \Vert = \Vert I_\mu (T_H(L_a f))\Vert = \Vert I_\mu (L_a T_Hf)\Vert = \Vert I_\mu (T_H f)\Vert = \Vert I_{\breve{\mu }} f \Vert \end{aligned}$$

  for any $f \in S(G)$. Hence ${\breve{\mu }}$ is norm integral left invariant. The converse is proved in Proposition 3.2.

• 3. Suppose that $\mu$ is semivariation $L_a$-invariant. It is routine to check that $\Vert L_a \varphi \Vert _{L^1(G/H,\mu )} = \Vert \varphi \Vert _{L^1(G/H,\mu )}$ for all $\varphi \in L^1(G/H,\mu )$. So

  $$\begin{aligned} \Vert L_a f\Vert _{L^1(G,{\breve{\mu }} )} = \Vert T_H\vert L_a f \vert \Vert _{L^1(G/H,\mu )} = \Vert T_H\vert f \vert \Vert _{L^1(G/H,\mu )} = \Vert f\Vert _{L^1(G,{\breve{\mu }} )} \end{aligned}$$

  for any $f \in S(G)$. Conversely, if ${\breve{\mu }}$ is semivariation $L_a$-invariant then

  $$\begin{aligned} \Vert L_a \varphi \Vert _{L^1(G/H,\mu )} = \Vert L_a \varphi _q\Vert _{L^1(G,{\breve{\mu }} )} = \Vert \varphi _q\Vert _{L^1(G,{\breve{\mu }} )} = \Vert \varphi \Vert _{L^1(G/H,\mu )} \end{aligned}$$

  for any $\varphi \in S(G/H)$.

$\square$

4 Invariant Measures

In this section, we provide properties of invariant measures on G and their analogies on G/H. The following proposition generalizes Proposition 5.2 in [1].

Proposition 4.1

Let $\nu \in {\mathcal {M}}(G,X)$. The following are equivalent:

• 1. $\nu$ is left (or right) invariant

• 2. ${\langle \nu , x^*\rangle }$ is left (or right) invariant for all $x^* \in X^*$

• 3. $\nu = \nu (G)m$.

Proof

We only show that 2 implies 3; the other directions are trivial. Assume that ${\langle \nu , x^*\rangle }$ is left invariant for all $x^* \in X^*$. Then the real part ${\langle \nu , x^*\rangle } _r$ is left invariant. Let $G = P \cup N$ be a Hahn decomposition for ${\langle \nu , x^*\rangle } _r$ where P is positive and N is negative. Note that $G = aP \cup aN$ is also a Hahn decomposition for ${\langle \nu , x^*\rangle } _r$ for any $a \in G$. Hence ${\langle \nu , x^*\rangle } _r^+(aE) = {\langle \nu , x^*\rangle } _r(aE \cap aP) = {\langle \nu , x^*\rangle } _r(E \cap P) = {\langle \nu , x^*\rangle } _r^+(E)$ for any $a \in G$ and $E \in \mathfrak {B}(G)$. This shows that ${\langle \nu , x^*\rangle } _r^+$ is left invariant. By the uniqueness of the left Haar measure, ${\langle \nu , x^*\rangle } _r^+ =\alpha _r^+(x^*) m$ for some $\alpha _r^+(x^*) \ge 0$. Applying the same argument to all parts of ${\langle \nu , x^*\rangle }$, we obtain that ${\langle \nu , x^*\rangle } =\alpha (x^*) m$ for some $\alpha (x^*) \in {\mathbb {C}}$. Hence $\langle \nu (E), x^* \rangle = \alpha (x^*) m(E)=\langle \nu (G), x^* \rangle m(E) =\langle \nu (G)m(E), x^* \rangle$ for any $E \in \mathfrak {B}(G)$. Since this equation holds for all $x^* \in X^*$, we have that $\nu = \nu (G)m$. A similar argument can be applied to the case of right invariance. $\square$

Proposition 4.2

Let $\mu \in {\mathcal {M}}(G/H,X)$. The following are equivalent:

• 1. $\mu$ is left invariant

• 2. ${\langle \mu , x^*\rangle }$ is left invariant for all $x^* \in X^*$

• 3. $\mu = \mu (G/H)\widetilde{m}$.

Proof

The first two assertions follow from the fact that ${\langle {\breve{\mu }} , x^*\rangle } = {\langle \mu , x^*\rangle } ^\smallsmile$ for all $x^* \in X^*$. Next, assume that $\mu$ is left invariant. Then ${\breve{\mu }}$ is left invariant. By Proposition 4.1, ${\breve{\mu }} ={\breve{\mu }} (G) m$. Since $\mu$ and $\widetilde{m}$ are the pushforward measures of ${\breve{\mu }}$ and m, we have $\mu ={\breve{\mu }} (G) \widetilde{m}=\mu (G/H) \widetilde{m}$. This finishes the proof. $\square$

The following proposition improves Lemma 3.4 in [3].

Proposition 4.3

Let $\nu$ be a vector measure on G. The following are equivalent.

• 1. $\nu$ is norm integral left (or right) invariant.

• 2. For each $x^* \in X^*$ and $a \in G$, there exists $x^*_a \in X^*$ such that $\Vert x^*_a\Vert \le \Vert x^*\Vert$ and ${\langle \nu , x^*\rangle } (aE)=\langle \nu , x^*_{a}\rangle (E)$ (or ${\langle \nu , x^*\rangle } (Ea)=\langle \nu , x^*_{a}\rangle (E)$) for all $E \in {\mathcal {B}}(G)$.

Moreover, $x^*_a \in X^*$ is unique in the sense that if there is another such functional then they must agree on $I_\nu (S(G))$.

Proof

We shall prove only for the case of norm integral left invariance as the other case is similar. The proof of 1 implies 2 follows by the same argument of [3, Lemma 3.4]. For the converse, let $f \in S(G)$ and $a \in G$. Then

$$\begin{aligned} \Vert I_\nu (L_a f)\Vert = \sup _{x^* \in B_{X^*}} \vert x^*I_\nu (L_a f)\vert = \sup _{x^* \in B_{X^*}} \vert x^*_a I_\nu (f)\vert \le \Vert I_\nu (f)\Vert . \end{aligned}$$

This also implies $\Vert I_\nu (f)\Vert = \Vert I_\nu (L_{a^{-1}} (L_a f))\Vert \le \Vert I_\nu (L_a f)\Vert$.

For the uniqueness, suppose there is another functional $y^* \in X^*$ such that $\Vert y^*\Vert \le \Vert x^*\Vert$ and ${\langle \nu , x^*\rangle } (aE)=\langle \nu , y^*\rangle (E)$ for all $E \in {\mathcal {B}}(G).$ Then $x^*_a(\nu (E))=\langle \nu , x^*_{a}\rangle (E)=\langle \nu , y^*\rangle (E)=y^*(\nu (E))$ for all $E \in {\mathcal {B}}(G).$ By the linearity of $x^*_a$ and $y^*$, we have that $x^*_a = y^*$ on $I_\nu (S(G))$. $\square$

Proposition 4.4

Let $\mu$ be a vector measure on G/H. The following are equivalent.

• 1. $\mu$ is norm integral left invariant.

• 2. For each $x^* \in X^*$ and $a \in G$, there exists $x^*_a \in X^*$ such that $\Vert x^*_a\Vert \le \Vert x^*\Vert$ and ${\langle \mu , x^*\rangle } (aE)=\langle \mu , x^*_{a}\rangle (E)$ for all $E \in {\mathcal {B}}(G/H)$.

Moreover, $x^*_a \in X^*$ is unique in the sense that if there is another such functional then they must agree on $I_\mu (S(G/H))$.

Proof

It can be proven by the same argument as in Proposition 4.3. However, if $\mu$ is also assumed to be regular, we can employ Proposition 4.3 with ${\breve{\mu }}$ and obtain the result immediately. $\square$

The following result can be proved by the same argument as in [13, Theorem 5.6] and [1, Theorem 5.10]. Hence the proof is omitted.

Proposition 4.5

Let $1 \le p < \infty$. Suppose that $\nu \in {\mathcal {M}}(G,X)$ is semivariation left (or right) invariant with $\nu (G) \ne 0$. Then $L^p(G,\nu ) \subset L^p(G,m)$ with $\Vert f\Vert _{L^p(G,m)} \le \Vert \nu (G)\Vert ^{-1/p} \Vert f\Vert _{L^p(G,\nu )}$ for $f \in L^p(G,\nu )$.

Proposition 4.6

Let $1 \le p < \infty$. Suppose that $\mu \in {\mathcal {M}}(G/H,X)$ is semivariation left invariant with $\mu (G/H) \ne 0$. Then $L^p(G/H,\mu ) \subset L^p(G/H,\widetilde{m})$ with $\Vert \varphi \Vert _{L^p(G/H,\widetilde{m})} \le \Vert \mu (G/H)\Vert ^{-1/p} \Vert \varphi \Vert _{L^p(G/H,\mu )}$ for $\varphi \in L^p(G/H,\mu )$.

Proof

Since ${\breve{\mu }}$ is semivariation left invariant, by Proposition 4.5, $L^p(G,{\breve{\mu }} ) \subset L^p(G,m)$ with $\Vert f \Vert ^p_{L^p(G,m)} \le \Vert {\breve{\mu }} (G)\Vert ^{-1} \Vert f\Vert ^p_{L^p(G,{\breve{\mu }} )}$ for $f \in L^p(G,{\breve{\mu }} ).$ Hence

$$\begin{aligned} \Vert \varphi \Vert ^p_{L^p(G/H,\widetilde{m})} = \Vert \varphi _q \Vert ^p_{L^p(G,m)}&\le \Vert {\breve{\mu }} (G)\Vert ^{-1} \Vert \varphi _q\Vert ^p_{L^p(G,{\breve{\mu }} )} \\&= \Vert \mu (G/H)\Vert ^{-1} \Vert \varphi \Vert ^p_{L^p(G/H,\mu )} \end{aligned}$$

for $\varphi \in L^p(G/H,\mu )$. $\square$

5 Fourier Transforms

In this section, we define a Fourier transform of functions in $L^1(G,\nu )$ and $L^1(G/H,\mu )$. Our definition is motivated by Definition 4.1 in [13]; however, X is not considered as an operator space. Let $\nu$ be a vector measure on G.

Definition 8

For $f \in L^1(G,\nu )$ and $[\pi ] \in \widehat{G}$, we define the Fourier transform of f as

$$\begin{aligned} \widehat{f}^\nu (\pi ) = \int _{G} f(t)\pi (t)^* \, d\nu \in {\mathcal {B}}({\mathcal {H}}_\pi )\check{\otimes } X. \end{aligned}$$

To see that the definition is well-defined, we have to show that the function $g:G \rightarrow {\mathcal {B}}({\mathcal {H}}_\pi )$ given by $g(t) = f(t)\pi (t)^*$ is $\nu$-measurable and $\check{\otimes }$-integrable. Let $x^* \in X^*$ be a Rybakov functional. Clearly, g is weakly $\vert {\langle \nu , x^*\rangle } \vert$-measurable since $y^*g(\cdot ) = f(\cdot )y^*\pi (\cdot )^*$ is a product of $\vert {\langle \nu , x^*\rangle } \vert$-measurable functions for all $y^* \in {\mathcal {B}}({\mathcal {H}}_\pi )^*$. Moreover, ${\mathcal {B}}({\mathcal {H}}_\pi )$ is separable. Thus, by Pettis’s measurability theorem, g is $\vert {\langle \nu , x^*\rangle } \vert$-measurable and hence is $\nu$-measurable. Since the function $\Vert g\Vert =\vert f \vert$ is $\nu$-integrable, g is $\check{\otimes }$-integrable. This immediately implies the following proposition.

Proposition 5.1

Define the operator ${\mathcal {F}}^\nu _G: L^1(G,\nu ) \rightarrow \ell ^\infty (\widehat{G};{\mathcal {B}}({\mathcal {H}}_\pi )\check{\otimes } X )$ by ${\mathcal {F}}^\nu _G(f)(\pi )=\widehat{f}^\nu (\pi )$ for $f \in L^1(G,\nu )$ and $[\pi ] \in \widehat{G}$. Then the Fourier transform operator ${\mathcal {F}}^\nu _G$ is bounded with $\Vert \widehat{f}^\nu (\pi )\Vert _\vee \le \Vert f\Vert _{L^1(G,\nu )}$.

Remark 3

If we take $\nu$ to be ${\langle \nu , x^*\rangle }$, then

$$\begin{aligned} \widehat{f}^{\, {\langle \nu , x^*\rangle } }(\pi ) = \int _{G} f(t)\pi (t)^* \, d{\langle \nu , x^*\rangle } = (Id_{{\mathcal {B}}({\mathcal {H}}_\pi )} \otimes x^*)\big (\widehat{f}^\nu (\pi )\big ). \end{aligned}$$

This can be considered as a generalization of Definition 4.6 in [13].

Remark 4

If G is abelian, then ${\mathcal {B}}({\mathcal {H}}_\pi ) \cong {\mathbb {C}}$ for any $[\pi ] \in \widehat{G}$. In this case, note that ${\mathbb {C}} \check{\otimes } X \cong X$ isometrically via the map $\alpha \otimes x \mapsto \alpha x$ and $\text {N}(\cdot )=\Vert \cdot \Vert _{L^1(G,\nu )}$. Hence our definition generalizes Definition 2.1 in [2].

Definition 9

We say that the Fourier transform ${\mathcal {F}}^\nu _G$ satisfies the Riemann–Lebesgue lemma if ${\mathcal {F}}^\nu _G(f) \in c_{0}(\widehat{G};{\mathcal {B}}({\mathcal {H}}_\pi )\check{\otimes } X )$ for all $f \in L^1(G,\nu )$.

The Fourier transform ${\mathcal {F}}^\nu _G$ need not satisfy the Riemann–Lebesgue lemma even if G is abelian as shown in [2, Example 2.4]. Now we give a necessary condition for ${\mathcal {F}}^\nu _G$ to satisfy the Riemann–Lebesgue lemma and also a stronger condition for the sufficiency.

Theorem 5.2

Let ${\mathcal {M}}= \{ \pi _{ij}: [\pi ] \in \widehat{G}, 1 \le i,j \le d_\pi \}$.

• 1. If ${\mathcal {F}}^\nu _G$ satisfies the Riemann–Lebesgue lemma, then the set $\{ \psi \in {\mathcal {M}}: \Vert \int _G \phi (t) \overline{\psi (t)} \, d\nu \Vert _X > \varepsilon \}$ is finite for any $\varepsilon > 0$ and $\phi \in {\mathcal {M}}$.

• 2. Moreover, if $\nu$ is regular and $\{ \psi =\pi _{ij}\in {\mathcal {M}}: d_\pi ^2\Vert \int _G \phi (t) \overline{\psi (t)} \, d\nu \Vert _X > \varepsilon \}$ is finite for any $\varepsilon > 0$ and $\phi \in {\mathcal {M}}$, then ${\mathcal {F}}^\nu _G$ satisfies the Riemann–Lebesgue lemma.

Proof

Observe that for $F:G \rightarrow Y={\mathcal {B}}({\mathcal {H}}_\pi )$

$$\begin{aligned} \left\| \int _G F \,d\nu \right\| _\vee = \sup _{y^* \in B_{Y^*}} \left\| \int _G y^*F \,d\nu \right\| _X. \end{aligned}$$

Hence if we write $y^* \in B_{Y^*}$ as $y^* = \sum _{1 \le i,j \le d_\pi } \alpha _{ij} e^*_{ij}$, we have

$$\begin{aligned} \max _{i,j}\left\| \int _G e^*_{ij}F \,d\nu \right\| _X \le \left\| \int _G F \,d\nu \right\| _\vee \le d_\pi ^2 \max _{i,j}\left\| \int _G e^*_{ij}F \,d\nu \right\| _X. \end{aligned}$$

• 1. Let $\varepsilon > 0$ and $\phi \in {\mathcal {M}} \subset L^1(G,\nu )$. Suppose that ${\mathcal {F}}^\nu _G$ satisfies the Riemann–Lebesgue lemma. If $\pi _{ij} \in {\mathcal {M}}$ satisfies $\Vert \int _G \phi (t) \overline{\pi _{ij}(t)} \, d\nu \Vert _X > \varepsilon$, by the observation above with $F(t) = \phi (t) \pi (t)^*$, we have $\Vert \widehat{\phi }^\nu (\pi )\Vert _\vee > \varepsilon$. Hence if $\{ \psi \in {\mathcal {M}}: \Vert \int _G \phi (t) \overline{\psi (t)} \, d\nu \Vert _X > \varepsilon \}$ is infinite, then so does the set $\{[\pi ] \in \widehat{G}: \Vert \widehat{\phi }^\nu (\pi ) \Vert _\vee > \varepsilon \}$, which is a contradiction.

• 2. Let $\phi \in {\mathcal {M}}$ and $\varepsilon > 0$. Suppose that $\{ \psi \in {\mathcal {M}}: d_\pi ^2\Vert \int _G \phi (t) \overline{\psi (t)} \, d\nu \Vert _X > \varepsilon \}$ is finite. If $[\pi ] \in \widehat{G}$ satisfies $\Vert \widehat{\phi }^\nu (\pi )\Vert _\vee > \varepsilon$, then $d_\pi ^2 \Vert \int _G \phi (t) \overline{\pi _{ji}(t)} \, d\nu \Vert _X > \varepsilon$ for some i, j. Hence we must have that $\widehat{\phi }^\nu \in c_{0}(\widehat{G};{\mathcal {B}}({\mathcal {H}}_\pi )\check{\otimes } X )$. Note that the linear span of ${\mathcal {M}}$ is $\text {Trig}(G)$ and $\text {Trig}(G)$ is dense in $L^1(G,\nu )$. By the continuity, ${\mathcal {F}}^\nu _G$ satisfies the Riemann–Lebesgue lemma.

$\square$

Remark 5

If G is abelian and $\nu$ is regular, then ${\mathcal {F}}^\nu _G$ satisfies the Riemann–Lebesgue lemma if and only if the set $\{ \psi \in {\mathcal {M}}: \Vert \int _G \phi (t) \overline{\psi (t)} \, d\nu \Vert _X > \varepsilon \}$ is finite for any $\varepsilon > 0$ and $\phi \in {\mathcal {M}}$.

We now prove the uniqueness theorem for the Fourier transform ${\mathcal {F}}^\nu _G$.

Theorem 5.3

Let $\nu \in {\mathcal {M}}(G,X)$ and $f \in L^1(G,\nu )$. If $\widehat{f}^\nu (\pi )=0$ for all $[\pi ] \in \widehat{G}$, then $f=0$ $\nu$-a.e.

Proof

Suppose that $\widehat{f}^\nu (\pi )=0$ for all $[\pi ] \in \widehat{G}$. Fix a Rybakov functional $x^* \in X^*$ and write $d{\langle \nu , x^*\rangle } = g \, d\vert {\langle \nu , x^*\rangle } \vert$ where $g \in L^1(G,\vert {\langle \nu , x^*\rangle } \vert )$. Then $\int _G f(t) y^*\pi (t)^* \, d{\langle \nu , x^*\rangle } = 0$ for any $y^* \in {\mathcal {B}}({\mathcal {H}}_\pi )^*$ and $[\pi ] \in \widehat{G}$. In particular, $\int _G \overline{\pi _{ij}(t)} (fg)(t) \, d\vert {\langle \nu , x^*\rangle } \vert = 0$ for any $[\pi ] \in \widehat{G}$ and $1 \le i,j \le d_\pi$. Since $\overline{\pi _{ij}}$ is a matrix element of the contragradient representation of $\pi$, $\int _G \phi (f g) \, d\vert {\langle \nu , x^*\rangle } \vert =0$ for any $\phi \in \text {Trig}(G)$. By the density of $\text {Trig}(G)$ in C(G) in the uniform norm, $fg \, d\vert {\langle \nu , x^*\rangle } \vert =0$ as a measure. Hence $fg=0$ $\vert {\langle \nu , x^*\rangle } \vert$-a.e. However $\vert g\vert = 1$ $\vert {\langle \nu , x^*\rangle } \vert$-a.e. Then it must be the case that $f = 0$ $\vert {\langle \nu , x^*\rangle } \vert$-a.e. Therefore $f=0$ $\nu$-a.e. since $\nu \ll \vert {\langle \nu , x^*\rangle } \vert$. $\square$

Now we give a definition of a Fourier transform of functions on G/H with a vector measure. This definition is motivated by [5]. Let $\mu$ be a vector measure on G/H.

Definition 10

For $\varphi \in L^1(G/H,\mu )$ and $[\pi ] \in \widehat{G/H}$, we define the Fourier transform of $\varphi$ at $[\pi ]$ as

$$\begin{aligned} \widehat{\varphi }^\mu (\pi ) = \int _{G/H} \varphi (tH)\Gamma _{\pi }(tH)^* \, d\mu (tH) \in {\mathcal {B}}({\mathcal {H}}_\pi )\check{\otimes } X, \end{aligned}$$

where $\Gamma _{\pi }(tH)=\pi (t)T^\pi _H$.

Let $g:G/H \rightarrow {\mathcal {B}}({\mathcal {H}}_\pi )$ be defined by $g(tH)=\varphi (tH)\Gamma _{\pi }(tH)^*$ for $tH \in G/H$. Then the $\mu$-measurability of g can be verified similarly to case of compact groups. Moreover, $\Vert \Gamma _{\pi }(tH)\Vert ^2 = \Vert \Gamma _{\pi }(tH)^* \Gamma _{\pi }(tH)\Vert = \Vert (T^\pi _H)^*T^\pi _H\Vert = \Vert T^\pi _H\Vert ^2 = 1,$ so $\Vert g\Vert$ is $\mu$-integrable. Hence the definition is well-defined.

Proposition 5.4

Define the operator ${\mathcal {F}}^\mu _{G/H}: L^1(G/H,\mu ) \rightarrow \ell ^\infty (\widehat{G/H};{\mathcal {B}}({\mathcal {H}}_\pi )\check{\otimes } X )$ by ${\mathcal {F}}^\mu _{G/H}(\varphi )(\pi )=\widehat{\varphi }^\nu (\pi )$ for $\varphi \in L^1(G/H,\mu )$ and $[\pi ] \in \widehat{G/H}$. Then the Fourier transform operator ${\mathcal {F}}^\mu _{G/H}$ is bounded with $\Vert {\mathcal {F}}^\mu _{G/H}(\varphi )(\pi )\Vert _\vee \le \Vert \varphi \Vert _{L^1(G/H,\mu )}$.

Proposition 5.5

Let $\mu \in {\mathcal {M}}(G/H,X)$, $\varphi \in L^1(G/H,\mu )$. Then $\widehat{\varphi }^\mu (\pi ) = \widehat{\varphi _q}^{\breve{\mu }} (\pi )$ for each $[\pi ] \in \widehat{G/H}$.

Proof

Recall that $T^\pi _H =\int _H \pi (h) \,dh$ is a bounded linear operator on ${\mathcal {H}}_\pi$ defined in the weak sense that is $\langle T^\pi _H u,v \rangle = \int _H \langle \pi (h) u,v \rangle \,dh$ for $u,v \in {\mathcal {H}}_\pi$. For $y^* \in {\mathcal {B}}({\mathcal {H}}_\pi )^*$, write $y^* = \sum _{i,j} \alpha _{ij} e_{ij}^*$. Since $\int _H e_{ij}^*\pi (th)^* \,dh = e_{ij}^* \int _H \pi (th)^* \,dh = e_{ij}^* (T^\pi _H \pi (t)^*)$ for any i, j, we have

$$\begin{aligned} T_H(y^*\pi (t)^*)&= \int _H y^*\pi (th)^* \,dh \\&= \sum _{i,j} \alpha _{ij} \int _H e_{ij}^*\pi (th)^* \,dh \\&= \sum _{i,j} \alpha _{ij} e_{ij}^* (T^\pi _H \pi (t)^* ) \\&= y^* (T^\pi _H \pi (t)^*) \end{aligned}$$

for any $t \in G$. Hence $T_H(y^*\pi (\cdot )^*) = y^* (T^\pi _H \pi (\cdot )^*)$. Consider for $x^* \in X^*$ and $y^* \in {\mathcal {B}}({\mathcal {H}}_\pi )^*$,

$$\begin{aligned} (y^* \otimes x^*)(\widehat{\varphi }^\mu (\pi ))&= \int _{G/H} \varphi (tH) y^*(T^\pi _H \pi (t)^*) \, d{\langle \mu , x^*\rangle } (tH) \\&= \int _{G/H} T_H(\varphi _q (\cdot ) y^*\pi (\cdot )^* )\, d{\langle \mu , x^*\rangle } (tH) \\&= \int _{G} \varphi _q(t) y^*\pi (t)^* \, d{\langle {\breve{\mu }} , x^*\rangle } (t) \\&= (y^* \otimes x^*)(\widehat{\varphi _q}^{\breve{\mu }} (\pi )). \end{aligned}$$

Hence $\widehat{\varphi }^\mu (\pi ) = \widehat{\varphi _q}^{\breve{\mu }} (\pi )$. $\square$

Definition 11

We say that the Fourier transform ${\mathcal {F}}^\mu _{G/H}$ satisfies the Riemann–Lebesgue lemma if ${\mathcal {F}}^\mu _{G/H}(\varphi ) \in c_{0}(\widehat{G/H};{\mathcal {B}}({\mathcal {H}}_\pi )\check{\otimes } X )$ for all $\varphi \in L^1(G/H,\mu )$.

Corollary 5.6

If ${\mathcal {F}}^{\breve{\mu }} _G$ satisfies the Riemann–Lebesgue lemma, then so does ${\mathcal {F}}^\mu _{G/H}$.

The Fourier transform ${\mathcal {F}}^\mu _{G/H}$ also satisfies the uniqueness theorem.

Theorem 5.7

Let $\mu \in {\mathcal {M}}(G/H,X)$ and $\varphi \in L^1(G/H,\mu )$. If $\widehat{\varphi }^\mu (\pi )=0$ for all $[\pi ] \in \widehat{G/H}$, then $\varphi =0$ $\mu$-a.e.

Proof

Suppose that $\widehat{\varphi }^\mu (\pi )=0$ for all $[\pi ] \in \widehat{G/H}$. Then $\widehat{\varphi _q}^{\breve{\mu }} (\pi )=0$ for all $[\pi ] \in \widehat{G/H}$. Moreover, if $[\pi ] \in \widehat{G}$ but $[\pi ] \notin \widehat{G/H}$, then $\widehat{\varphi _q}^{\breve{\mu }} (\pi )=0$. Indeed, for any $x^* \in X^*$ and $y^* \in {\mathcal {B}}({\mathcal {H}}_\pi )^*$,

$$\begin{aligned} (y^* \otimes x^*)(\widehat{\varphi _q}^{\breve{\mu }} (\pi ))&= \int _{G} \varphi _q(t) y^*\pi (t)^* \, d{\langle {\breve{\mu }} , x^*\rangle } (t) \\&= \int _{G/H} \varphi (tH) y^*(T^\pi _H \pi (t)^*) \, d{\langle \mu , x^*\rangle } (tH) = 0 \end{aligned}$$

since $T_H(y^*\pi (\cdot )^*) = y^* (T^\pi _H \pi (\cdot )^*)=0$. Then one can apply Theorem 5.3 and obtains that $\varphi _q = 0$ ${\breve{\mu }}$-a.e. Hence $\varphi = T_H(\varphi _q) = 0$ $\mu$-a.e. $\square$