2.1 Generalities

In this article, most of our notation is standard, which can be found in [Reference Dahlner6]. We will denote C as a constant, and its value can change from one line to another. For any two positive expressions $f_1$ and $f_2$ , $f_1 \asymp f_2 $ stands for that there are positive constants $C_1$ , $C_2$ such that $C_1 f_1 \leq f_2 \leq C_2 f_1 $ . For $ z \in \mathbb C$ , we use Rez and Imz to denote the real and imaginary parts of z, respectively.

2.2 Lorentz spaces

Let $(X,m) $ be a $ \sigma $ -finite measure space. For $f: X \rightarrow \mathbb C$ a measurable function on $ X $ , the distribution function $ d_f $ defined on $ [0,\infty ) $ is given by $ d_f(\alpha ) = m (\{ x \in X : |f(x)|> \alpha \}) $ . Define for $ p\in [1,\infty ), q \in [1,\infty ]$

(2-1) $$ \begin{align} \|f\|_{p,q} = \begin{cases}\displaystyle \bigg( \dfrac{q}{p} \int_0^\infty [f^{*}(\alpha) \alpha^{{1}/{p}} ]^q \dfrac{d\alpha}{\alpha}\bigg)^{{1}/{q}} &\text{when } q < \infty, \\ \sup_{\alpha> 0} \,\alpha^{{1}/{p}} f^{*}(\alpha) &\text{when } q =\infty, \end{cases} \end{align} $$

where $ f^{*}({s}) =\inf \{ s>0 : d_f(\alpha ) \leq s \} $ is the nonincreasing rearrangement of $ f $ (see [Reference Grafakos13, page 45]). The Lorentz spaces $ L^{p,q}(X) $ consist of all measurable functions $f $ for which $\| f \|_{p,q} $ is finite. For $ p,q \in [1,\infty ),$ the following identity gives an alternative expression of $ \|\cdot \|_{p,q} $ (see [Reference Grafakos13, Proposition 1.4.9]),

$$ \begin{align*} p^{{1}/{q}} \bigg( \int_0^\infty [d_f(\alpha) ^{{1}/{p}}s]^q \frac{ds}{s}\bigg)^{{1}/{q}} = \bigg( \frac{q}{p} \int_0^\infty [f^{*}(\alpha) \alpha^{{1}/{p}} ]^q \frac{d\alpha}{\alpha}\bigg)^{{1}/{q}}. \end{align*} $$

We need the following lemma.

Lemma 2.1. Suppose $1\leq p< q $ and $r \in (p,q) $ . Then for all $ f \in L^{p}(X) \cap L^q(X) $ , there exists a positive constant $ C $ independent of $ f $ such that

(2-2) $$ \begin{align} \|f\|_{r,1} \leq C (\|f\|_p +\|f\|_q). \end{align} $$

Proof. Suppose $ f \in L^{p}(X) \cap L^q(X) $ . From Equation (2-1),

$$ \begin{align*} \|f\|_{r,1}= \frac{1}{r} \int_0^\infty f^{*}(\alpha) \alpha^{{1}/({r}-1)} {\,d\alpha} = \frac{1}{r} \int_0^1 f^{*}(\alpha) \alpha^{{1}/({r}-1)} {\,d\alpha}+ \frac{1}{r} \int_1^\infty f^{*}(\alpha) \alpha^{{1}/({r}-1)} {\,d\alpha}. \end{align*} $$

Next, by Hölder’s inequality and $ 1/q<1/r<1/p $ , the lemma will follow.

2.3 Result from complex analysis

Now we borrow a result from complex analysis, which is a consequence of a log–log-type theorem. For any function F on $\mathbb R$ and $ \alpha>0 $ , we let

$$ \begin{align*} \delta_\infty^{\alpha+}(F)=-\limsup_{t\rightarrow\infty} e^{-{\pi}/{2 \alpha}t}\log|F(t)| \quad\text{and} \quad \delta_\infty^{\alpha-}(F)=-\limsup_{t\rightarrow\infty} e^{-{\pi}/{2 \alpha }t}\log|F(-t)|. \end{align*} $$

Theorem 2.2. Let $M:(0,\infty )\rightarrow (e,\infty )$ be a continuously differentiable decreasing function with

$$ \begin{align*} \lim_{t\rightarrow 0^+}t\log\log M(t)<\infty,\hspace{3mm}\int_0^\infty\log\log M(t)dt<\infty. \end{align*} $$

Let $\Lambda $ be a collection of bounded holomorphic functions on $ \mathbf {S}^\circ _\alpha $ such that

$$ \begin{align*} \inf_{F\in \Lambda}\delta^{\alpha+}_\infty(F)=\inf_{F\in \Lambda}\delta^{\alpha-}_\infty(F)=0. \end{align*} $$

Suppose H is a function that satisfies the following estimates for some nonnegative integer N:

$$ \begin{align*} |H(z)|\leq &(1+|z|)^NM(d(z,\partial \mathbf{S}_\alpha)),\quad z\in\mathbb{C}\backslash \mathbf{S}_\alpha,\\ |F(z)H(z)|\leq & (1+|z|)^NM(d(z,\partial \mathbf{S}_\alpha)), \quad z\in \mathbf{S}_\alpha^\circ \quad\text{for all } F\in \Lambda. \end{align*} $$

1. (1) If, in addition, $ H $ is a holomorphic function on $\mathbf {S}_\alpha \backslash \{ \pm \alpha \}$ , then $ H $ is dominated by a polynomial outside a bounded neighbourhood of $\{ \pm \alpha \}.$

2. (2) If $ H $ is an entire function, then it is a polynomial.

2.4 Real variable theory on semisimple Lie groups of rank one

Let $ G $ be a noncompact connected semisimple real-rank-one Lie group with finite centre, with Lie algebra $\mathfrak g$ . Let $ \theta $ be a Cartan involution of $ \mathfrak {g} $ and $ \mathfrak {g} =\mathfrak {k+p} $ be the associated Cartan decomposition. Let $ K=\exp \mathfrak {k} $ be a maximal compact subgroup of $ G $ . Let $ \mathfrak {a} $ be a maximal abelian subspace of $ \mathfrak {p} $ . Since the group $ G $ is of real rank one, dim $\mathfrak {a} =1 $ . Let $ \Sigma $ be the set of nonzero roots of the pair $ (\mathfrak {g,a}) $ , and let $ W $ be the associated Weyl group. For the rank-one case, it is well known that either $ \Sigma =\{-\alpha ,\alpha \} $ or $ \{-2\alpha ,-\alpha ,\alpha ,2\alpha \} $ , where $ \alpha $ is a positive root and the Weyl group $ W $ associated to $ \Sigma $ is {-Id, Id}, where Id is the identity operator. Let $ \mathfrak {a}^+ =\{ H \in \mathfrak {a} : \alpha (H)>0 \} $ be a positive Weyl chamber, and let $ \Sigma ^+ $ be the corresponding set of positive roots. In our case, $ \Sigma ^+ = \{\alpha \}$ or $ \{ \alpha , 2\alpha \} $ . For any root $ \beta \in \Sigma $ , let $ \mathfrak {g}_\beta $ be the root space associated to $ \beta $ . Let $ \mathfrak {n} =\sum _{\beta \in \Sigma ^+} \mathfrak {g}_\beta $ and $ {N =\exp \mathfrak {n}}$ . Then the group $ G $ has an Iwasawa decomposition $ G= K (\exp \mathfrak {a})N $ and a Cartan decomposition $ G=K(\exp \overline {\mathfrak {a}^+}) K.$ These decompositions are unique. For each $ g \in G $ , we denote $ H(g) \in \mathfrak {a} $ and $ g^+ \in \overline {\mathfrak {a}^+} $ as the unique elements such that

$$ \begin{align*} g=k \exp H(g) n, \quad k\in K, n\in N, \end{align*} $$

and

(2-3) $$ \begin{align} g=k_1 \exp (g^+) k_2, \quad k_1, k_2\in K. \end{align} $$

Let $H_0$ be the unique element in $\mathfrak a$ such that $\alpha (H_0)=1$ , and through this, we identify $\mathfrak a$ with $\mathbb R$ as $t\leftrightarrow tH_0$ and $\mathfrak a_+= \{H\in \mathfrak a: \alpha (H)>0\}$ is identified with the set of positive real numbers. We also identify $\mathfrak a^{*}$ and its complexification $\mathfrak {a}^{*}_{\mathbb {C}}$ with $\mathbb R$ and $\mathbb C$ , respectively, by $t\leftrightarrow t\alpha $ and $z\leftrightarrow z\alpha $ , $t\in \mathbb R$ , $z\in \mathbb C$ . Let $A=\exp \mathfrak {a}=\{a_t:=\exp (t H_0): t\in \mathbb R\}$ and ${A^+=\{a_t: t>0\}}$ . Let $m_1=\dim \mathfrak {g}_\alpha $ and $m_{2}=\dim \mathfrak {g}_{2\alpha }$ , where $\mathfrak g_\alpha $ and $\mathfrak g_{2\alpha }$ are the root spaces corresponding to $\alpha $ and $2\alpha $ . Let $\rho =\frac 12(m_1+2m_2)\alpha $ denote the half sum of the positive roots. By abuse of notation, we denote $\rho (H_0)=\frac 12(m_1+2m_2)$ by $\rho $ .

Let $ dg $ , $ dn $ , $ dk $ and $ dm $ be the Haar measures of $ G, N, K $ and $ M $ , respectively, where $ \int _K dk =1 $ and $ \int _m dm =1 $ . We have the following integral formula corresponding to the Cartan decomposition, which holds for any integrable function f:

$$ \begin{align*} \int_Gf(g)dg=\int_K\int_{\mathbb R^+}\int_K f(k_1a_tk_2) \Delta(t)\,dk_1\,dt\,dk_2, \end{align*} $$

where $\Delta (t)=(2\sinh t)^{m_1 + m_2}(2\cosh t)^{m_2}$ . A function f is called K-biinvariant if

(2-4) $$ \begin{align} f(k_1xk_2)=f(x) \quad\text{for all } x\in G, k_1, k_2\in K.\end{align} $$

For a class of functions $\mathcal {F} $ on G, we denote the corresponding subclass of K-biinvariant functions by $\mathcal F (G//K)$ .

2.5 Spherical function

Let $\mathbb D(G/K)$ be the algebra of G-invariant differential operators on $G/K$ . The elementary spherical functions $\phi $ are $C^\infty $ functions and are joint eigenfunctions of all $D\in \mathbb D(G/K)$ for some complex eigenvalue $\lambda (D)$ . That is,

$$ \begin{align*}D\phi=\lambda(D)\phi, \quad D\in\mathbb D(G/K).\end{align*} $$

They are parametrized by $\lambda \in \mathbb C$ . The algebra $\mathbb D(G/K)$ is generated by the Laplace–Beltrami operator $ L$ . Then we have, for all $\lambda \in \mathbb C $ , that $\phi _\lambda $ is a $C^\infty $ solution of

(2-5) $$ \begin{align} L\phi=-(\lambda^2 + \rho^2)\phi. \end{align} $$

The A-radial part of the Laplace–Beltrami operator is given by

$$ \begin{align*} L_Af(a_t):=\frac{d^2}{dt^2}f (a_t)+((m_1+ m_2)\coth t + m_2\tanh t)\frac{d}{dt}f(a_t),\quad t>0. \end{align*} $$

For $\lambda \not =-i, -2i, \ldots ,$ we have another solution $\Phi _\lambda $ of Equation (2-5) on $(0,\infty )$ given by (see [Reference Pusti and Samanta21, Equation (2.7)]),

$$ \begin{align*} \Phi_\lambda(a_t)&= (2\cosh t)^{i\lambda-\rho} \,_2F_1\bigg(\frac{\rho-i\lambda}{2}, \frac{m_1 +2}{4} -\frac{i\lambda}{2}; 1-i\lambda; \cosh^{-2} t\bigg), \end{align*} $$

where $ _2F_1 $ is the Gauss hypergeometric function.

The function $\Phi _\lambda $ has a series representation, called the Harish-Chandra series for $t>0$ , and has a singularity at $t=0$ . Using the Cartan decomposition, we extend $\Phi _\lambda $ as a K-biinvariant function on $G/K \setminus \{eK\}$ . Therefore, $\Phi _\lambda $ is a solution of Equation (2-5) on $G/K \setminus \{eK\}$ and we also have for $t\rightarrow \infty $ ,

(2-6) $$ \begin{align} \Phi_\lambda(a_t)=e^{(i\lambda-\rho)t}(1 + O(1)). \end{align} $$

For $\lambda \in \mathbb C\setminus i\mathbb Z$ , $\Phi _\lambda $ and $\Phi _{-\lambda }$ are two linearly independent solutions. Therefore, for $ {\lambda \in \mathbb C\setminus i\mathbb Z }$ , $\phi _\lambda $ is a linear combination of both $\Phi _\lambda $ and $\Phi _{-\lambda }$ ; that is,

$$ \begin{align*}\phi_\lambda=c(\lambda)\Phi_\lambda + c(-\lambda)\Phi_{-\lambda},\end{align*} $$

where $c(\lambda )$ is the Harish-Chandra c-function given by

$$ \begin{align*} c(\lambda)=\frac{2^{\rho-i\lambda} \Gamma\big(\frac{m_1 + m_2 +1}{2}\big)\Gamma(i\lambda)}{\Gamma\big(\frac{\rho+ i\lambda}{2}\big) \Gamma\big(\frac{m_1+ 2}{4} + \frac{i\lambda}{2}\big)}. \end{align*} $$

We have the following asymptotic estimate of $ \phi _{ \lambda } $ (see [Reference Chandra14]) for ${\mathrm {Im}}\lambda <0$ and $t\rightarrow \infty $ ,

(2-7) $$ \begin{align} \phi_\lambda(a_t)=c(\lambda)e^{(i\lambda-\rho)t}(1 + O(1)). \end{align} $$

The $ c $ -function has neither a zero nor a pole in the region $ {\mathrm {Im}} \lambda <0 $ (see [Reference Helgason16, Theorem 6.4, Ch. IV]); so it follows that for any fixed $ \lambda \in \mathbb C $ with $ {\mathrm {Im}} \lambda <0 $ ,

(2-8) $$ \begin{align} |\phi_{ \lambda}(a_t)| \asymp e^{-({\mathrm{Im}} \lambda +\rho)t}. \end{align} $$

For any $\lambda \in \mathbb C$ , the elementary spherical function $\phi _\lambda $ has the following integral representation:

(2-9) $$ \begin{align} {\phi_\lambda(x)=\int_K e^{-(i\lambda+\rho)H(xk)}\,dk \quad\text{for all } x\in G.} \end{align} $$

We now list down some well-known properties of the elementary spherical functions which are important for us [Reference Gangolli and Varadarajan11, Proposition 3.1.4 and Ch. 4, Section 4.6], [Reference Helgason17, Lemma 1.18, page 221].

1. (1) $\phi _\lambda (g)$ is K-biinvariant in $g\in G$ , $\phi _\lambda =\phi _{-\lambda }$ , $\phi _\lambda (g)=\phi _\lambda (g^{-1})$ .

2. (2) $\phi _\lambda (g)$ is $C^\infty $ in $g\in G$ and holomorphic in $\lambda \in \mathbb C$ .

3. (3) The following inequality holds:

   $$ \begin{align*} e^{-\rho t} \leq \phi_0(a_t)\leq (1+|t|)~e^{-\rho t}, \quad t\geq 0. \end{align*} $$

4. (4) $|\phi _\lambda (x)|\leq 1$ for all $x\in G$ if and only if $\lambda \in \mathbf {S}_\rho =\{\lambda \in \mathbb C : |{\mathrm {Im}}\lambda |\leq \rho \}$ .

We also have the following proposition from [Reference Helgason15, Ch. IV, Proposition 2.2].

Proposition 2.3. Let $ f $ be a complex-valued continuous function on $ G $ , not identically $ 0 $ . Then $ f $ is a spherical function if and only if

$$ \begin{align*} \int_K f(xky)\,dk = f(x) f(y) \end{align*} $$

for all $ x,y \in G $ .

The following proposition from [Reference Pusti, Ray and Sarkar20, Proposition 2.1] will be useful.

2.6 Spherical Fourier transform

The spherical transform $\widehat {f}$ of a suitable K-biinvariant function f is defined by the formula

$$ \begin{align*}\widehat{f}(\lambda)=\int_Gf(x)\phi_\lambda(x^{-1})\,dx.\end{align*} $$

It is well known that if $ f \in L^1(G//K) $ , then $ \widehat f $ is analytic on $ \mathbf {S}_{\rho }^\circ $ , continuous on $ \mathbf {S}_{\rho } $ and $ |\widehat f(\lambda )| \rightarrow 0$ as $ |\lambda |\rightarrow \infty $ in $ \mathbf {S}_{\rho } $ .

Let $C_c^\infty (G//K)$ be the set of all $C^\infty $ compactly supported K-biinvariant functions on G. Also let $PW(\mathbb C)$ be the set of all entire functions $h:\mathbb C\rightarrow \mathbb C$ such that $ h $ is of exponential type $ T $ for some $ T> 0 $ , that is, for each $N\in \mathbb N$ ,

$$ \begin{align*}\sup_{\lambda\in\mathbb C}\,(1 +|\lambda|)^N |h(\lambda)| e^{-T |{\mathrm{Im}}\lambda|}<\infty\end{align*} $$

and let $PW(\mathbb C)_e$ be the set of all even functions in $PW(\mathbb C)$ . Then we have the following Paley–Wiener theorem.

3.1 A dense subspace of $ L^1(w_{\alpha ,\,r}) $

We now construct a dense subset of $L^1(\omega _{\alpha ,r}) $ , which plays a crucial role towards proving Theorem 1.2. This collection is a suitable scalar multiple of $ \Phi _{\lambda } $ . For $\lambda \in \mathbb {C}_+: = \{ \lambda \in \mathbb C : {\mathrm {Im}} \lambda>0 \}$ , we define

(3-1) $$ \begin{align} b_\lambda(a_t):=\frac{i}{2\lambda c(-\lambda)}\Phi_\lambda(a_t) \quad \text{for }t>0. \end{align} $$

We extend $b_\lambda $ to a $ K $ -biinvariant function on $G \backslash K $ , using the Cartan decomposition

(3-2) $$ \begin{align} b_\lambda (k_1 a_t k_2) = b_\lambda(a_t) \quad\text{for all } t> 0. \end{align} $$

Hence, $b_\lambda $ is also a solution of Equation (2-5) on $G / K \setminus \{eK\}$ . Next, using the asymptotic estimates of $ \Phi _{ \lambda }(a_t) $ (see Equation (2-6)) and $\phi _{i\alpha } (a_t)$ (see Equation (2-7)) near $ t=\infty $ , we observe that if ${\mathrm {Im}} \lambda \leq \alpha $ , then the functions $ b_\lambda $ do not belong to $ L^1(\omega _{\alpha ,r}) $ . In the following lemma, we show for $ \lambda \in \mathbb C_+$ with ${\mathrm {Im}} \lambda>\alpha $ , $ b_\lambda \in L^1 (\omega _{\alpha ,r})$ , and along with this, the lemma also gives us a dense subspace of $ L^1(\omega _{\alpha ,r}) $ .

We prove this lemma step by step. First, we show that for ${\mathrm {Im}} \lambda>\alpha $ , $ b_\lambda $ belongs to $ L^1(\omega _{\alpha ,r})$ . To show this, we borrow the estimates of $ b_\lambda (a_t) $ from [Reference Pusti and Samanta21, Lemma 3.1] near $ t=0 $ and away from zero. Later in Lemma 3.5, we find the spherical transforms of $ b_\lambda $ , which are essential to prove that the collection $ \{ b_\lambda : {\mathrm {Im}} \lambda>\alpha \}$ spans a dense subspace of $ L^1(\omega _{\alpha ,r}) $ .

Lemma 3.3 [Reference Pusti and Samanta21, Lemma 3.1]

Let $\lambda \in \mathbb {C}_+$ . Then $b_\lambda (a_t)$ satisfies the following estimates near $ t =0$ and $\infty $ .

(a) There is a positive constant C such that for all $t\in (0,1/2]$ ,

$$ \begin{align*} |b_\lambda(a_t)|\leq \begin{cases}\displaystyle C (1+|\lambda|)^{N} t^{-(m_1+m_2-1)}&\text{if } m_1+m_2>1,\\ C \log \dfrac{1}{t}&\text{if }m_1+m_2=1. \end{cases} \end{align*} $$

(b) There is a positive constant C and a natural number M such that for all $t\in [1/2,\infty ]$ ,

$$ \begin{align*} |b_\lambda(a_t)|\leq C (1+|\lambda|)^Me^{-({\mathrm{Im}}\lambda+\rho)t}. \end{align*} $$

Lemma 3.4. If $ {\mathrm {Im}} \lambda> \alpha $ , then $ b_\lambda \in L^{1}(G//K,\omega _{\alpha ,r}) $ . Moreover, there exist $n, N_0\in \mathbb N $ and a constant $ C> 0 $ (independent of $ \lambda $ ) such that the following estimate holds:

$$ \begin{align*} || b_\lambda||_{L^1_{\omega_{\alpha ,r}}} \leq C \dfrac{(1+|\lambda|)^{N_0}}{({\mathrm{Im}} \lambda- \alpha)^{n+1}}. \end{align*} $$

Proof. Let $ \lambda \in \mathbb C_+ $ with $ {\mathrm {Im}} \lambda>\alpha $ and $ n=[r]+1 $ . Then using the estimates of $ b_\lambda (a_t)$ and Equation (2-8) of $ \phi _{ \alpha }(a_t) $ ,

$$ \begin{align*} \| b_\lambda \|_{L^1_{\omega_{\alpha ,r}}}&\leq \int_0^\infty |b_\lambda (a_t)| \phi_{i \alpha} (a_t)(1+t)^n \Delta(t) \,dt\\& \leq C(1+|\lambda |)^{\max \{N,M\}}\bigg( \int_{0}^{1/2} t\, \phi_{i\alpha} (a_t)(1+ t )^n dt + \int_{1/2}^\infty e^{(\alpha-{\mathrm{Im}} \lambda)t} (1+ t)^n \,dt\bigg)\\&\leq C (1+|\lambda |)^{\max \{N,M\}}\bigg(1+ \frac{1}{({\mathrm{Im}} \lambda -\alpha)^{n+1}} \bigg)\\&\leq \frac{ C (1+|\lambda |)^{N_0} }{({\mathrm{Im}} \lambda -\alpha)^{n+1}}.\\[-3.7pc] \end{align*} $$

Lemma 3.5. Let $\lambda \in \mathbb {C}_+$ with $ {\mathrm {Im}} \lambda>\alpha $ . Then we have

$$ \begin{align*} \widehat{b}_\lambda(z)=\frac{1}{z^2-\lambda^2} \quad \text{for all } z \in \mathbf{S}_\alpha. \end{align*} $$

Proof of Lemma 3.2

We will show that $\overline {\text {span} \{ b_\lambda : {\mathrm {Im}} \lambda> \alpha \}}$ contains $C_c^\infty (G//K)$ and since $C_c^\infty (G//K)$ is dense in $L^1(\omega _{\alpha ,r})$ , the lemma will follow. Let $f\in C_c^\infty (G//K)$ , then $\widehat {f}$ is entire (see Theorem 2.5). Applying Cauchy’s integral formula for $ \widehat f $ ,

(3-3) $$ \begin{align} \widehat{f}(w)=\dfrac{1}{2\pi i } \int_{\zeta_R } \dfrac{ \widehat{f}(z)}{z-w} \,dz \quad \text{for all } w\in \mathbf{S}_\alpha , \end{align} $$

where $ \zeta _R $ is the contour consisting of a rectangle with vertices $ R+i (\alpha +1) $ , $ -R+ i (\alpha +1)$ , $-R-i(\alpha +1)$ , $R-i(\alpha +1)$ ( $R $ is sufficiently large) and the positive counterclockwise orientation. From Theorem 2.5, we get the integrals on the vertical sides of $ \zeta _R $ to go to $0$ , as $ R \rightarrow \infty. $ Therefore, Equation (3-3) gives

$$ \begin{align*}\widehat{f}(w)=\dfrac{1}{2\pi i } \int_{A } \dfrac{ \widehat{f}(z)}{z-w} \,dz + \dfrac{1}{2\pi i } \int_{B} \dfrac{ \widehat{f}(z)}{z-w} \,dz \quad\text{for }w \in \mathbf{S}_\alpha, \end{align*} $$

where $A= \mathbb R + i(\alpha +1) $ and $B= \mathbb R -i(\alpha +1) $ . We know $\widehat {f}(z)$ is an even function, and so by the change of variable $z \rightarrow -z$ in the second integral, we get for all $w \in \mathbf {S}_\alpha $ ,

$$ \begin{align*} \widehat{f}(w)= \dfrac{1}{2\pi i } \int_{A } \dfrac{ 2z \widehat{f}(z)}{z^2-w^2}\, dz. \end{align*} $$

Now for $z\in A$ , ${\mathrm {Im}} z>\alpha $ , so (by Lemma 3.4) $b_z$ is in $L^{1}(\omega _{\alpha ,r})$ . Also, from Lemma 3.5, we can write

(3-4) $$ \begin{align} \widehat{f}(w)=\frac{1}{2\pi i}\int_{A}2z\widehat{f}(z)\widehat{b}_z(w)\,dz. \end{align} $$

From the estimate of $ \|b_z \|_{\omega _{\alpha ,r}}$ and the decay condition of $ \widehat f $ , it follows that the $ L^{1}(\omega _{\alpha ,r}) $ integral

$$ \begin{align*} \frac{1}{2\pi i}\int_{A}2z\widehat{f}(z)b_z(\cdot)\,dz \end{align*} $$

converges. Equation (3-4) shows that it must converge to f. Since the Riemann sums of the integral are nothing but finite linear combinations of $b_\lambda $ , we conclude that f is in the closed subspace spanned by $\{b_\lambda : {\mathrm {Im}}\lambda>\alpha \}$ . Hence, the lemma follows.

We need to prove $ \|b_\lambda \|_{L^1_{\omega _{\alpha ,r}}} \rightarrow 0$ as $ \lambda \rightarrow \infty $ along the positive imaginary axis. For the $ r=0 $ case, Dahlner [Reference Dahlner6] proved it by using $\| b_{i\xi }\|_{L^1} =\widehat b_{i\xi }(i\alpha ) $ for large $ \xi $ . However, in general, for $ r> 0 $ , this is not true.

Proof. Let us take $ \lambda =i\xi $ , where $ \xi>0 $ is very large. Then from Equation (3-1), we have $ b_{i\xi }(a_t) $ is positive, and so

$$ \begin{align*} \| b_{i \xi}\|_{L^1_{\omega_{\alpha ,r}}} &=\int_0^\infty b_{i\xi }(a_t) \phi_{i\alpha}(a_t)(1+t)^r \Delta(t) \,dt\\ & \leq C \int_0^\infty b_{i\xi }(a_t) \phi_{2i\alpha} (a_t) \Delta(t) \,dt \quad \text{(using Proposition }{2.4}) \\ & =C \frac{1}{\xi^2- 4 \alpha^2}. \end{align*} $$

The lemma follows by sending $ \xi $ to infinity.

3.2 Resolvent transform

Let $L^1_\delta (\omega _{\alpha ,r})$ be the unitization of $L^{1}(\omega _{\alpha ,r}) $ , where $\delta $ is the $ K $ -biinvariant distribution on $ G $ , defined by $\delta (\phi )= \phi (e)$ for all $\phi \in C_c^\infty (G//K)$ . The maximal ideal space of $L^{1}_\delta (\omega _{\alpha ,r})$ is $\{L_z:z\in \mathbf {S}_\alpha \cup \{\infty \}\}$ , where the complex homomorphisms $ L_z $ on $ L^1_\delta ( \omega _{\alpha ,r}) $ are defined by the following:

$$ \begin{align*} L_z(f)= \widehat f(z), \quad z\in \mathbf{S}_\alpha, \quad L_\infty (f) =\begin{cases} 1 \quad \text{if } f=\delta\\ 0 \quad \text{if } f\in L^1(\omega_{\alpha,r}) \end{cases} \text{for all } f\in L^1_\delta(\omega_{\alpha ,r}). \end{align*} $$

Let $ I $ denote the closed ideal from the hypothesis of Theorem 1.2. Then the spherical transform of the functions in $ I $ does not have any common zero in $ \mathbf {S}_\alpha $ . Therefore, the maximal ideal space of the quotient algebra $ L^1_\delta (\omega _{\alpha ,r}) /I$ is the complex homomorphism $ \tilde {L}_\infty $ defined by

$$ \begin{align*}\tilde{ L}_\infty (f) = \widehat f (\infty) \quad\text{for all } f \in L^1_\delta(\omega_{\alpha ,r})/I. \end{align*} $$

It also follows that an element $ f +I \in L^1_\delta (\omega _{\alpha ,r} )/I $ is invertible if and only if $ \widehat f(\infty )\not = 0 $ .

Now suppose $ \lambda _0 $ is a fixed complex number with ${\mathrm {Im}} \lambda _0>\alpha. $ So by Lemma 3.4, $ b_{\lambda _0 }\in L^1(\omega _{\alpha ,r}) $ , and for $ \lambda \in \mathbb C $ , the function $\lambda \rightarrow \widehat \delta - (\lambda ^2 -\lambda _0^2) \widehat {b_{\lambda _0}}$ does not vanish at $\infty $ . Hence, $ \delta -(\lambda ^2 -\lambda _0^2) b_{\lambda _0}+I$ is invertible in the quotient algebra $ L^1_\delta (\omega _{\alpha ,r}) /I$ for all $ \lambda \in \mathbb C $ . We put

(3-5) $$ \begin{align} B_\lambda := (\delta -(\lambda ^2-\lambda_0^2)b_{\lambda_0} +I)^{-1}*(b_{\lambda_0}+I). \end{align} $$

For each $ g \in L^\infty (G//K,1/\omega _{\alpha ,r}) $ that annihilates the closed ideal $ I $ , we associate its resolvent transform

$$ \begin{align*} \mathcal{R}[g](\lambda )= \langle B_\lambda , g \rangle \quad\text{for } \lambda \in \mathbb C. \end{align*} $$

3.3 Properties of $\mathcal {R}[g](\lambda )$

We first find the representative of $ B_\lambda $ in $ L^1(G//K,\omega _{\alpha ,r}) $ that gives the explicit expression of $ \mathcal {R}[g](\lambda ) $ . Let $ \lambda \in \mathbb C $ be such that $ 0<{\mathrm {Im}} \lambda <\alpha $ and $ f \in L^1(\omega _{\alpha ,r}) $ . Then we define

(3-6) $$ \begin{align} T_\lambda f(a_t)=b_\lambda(a_t)\int_t^\infty f(a_s)\phi_{\lambda}(a_s)\Delta(s)\,ds-\phi_{\lambda}(a_t) \int_t^\infty f(a_s)b_\lambda(a_s)\Delta(s)\,ds ,\quad t>0, \end{align} $$

and extend it to a $ K$ -biinvariant function on $ G\setminus K $ .

We first show that for a given $ f \in L^1(\omega _{\alpha ,r}) $ , $ T_\lambda f \in L^1(\omega _{\alpha ,r}) $ , and find a good quantitative bound of $ \|T_\lambda f\|_{L^1_{\omega _{\alpha ,r}}} $ (in the next lemma). Then (in Lemma 3.9), we find its spherical Fourier transform. This is essential to finding the representative of $ B_\lambda $ in terms of $T_\lambda f $ for $ 0<{\mathrm {Im}} \lambda <\alpha $ .

Lemma 3.7. Let $0<{\mathrm {Im}}\lambda <\alpha $ and $f\in L^1(\omega _{\alpha ,r})$ . Then, $T_\lambda f\in L^1(\omega _{\alpha ,r})$ . Moreover, if $\lambda \notin B_\alpha (0)$ ,

$$ \begin{align*}||T_\lambda f||_{L^1_{\omega_{\alpha,r}}}\leq C||f||_{L^1_{\omega_{\alpha,r}}}(1+|\lambda|)^Ld(\lambda,\partial \mathbf{S}_\alpha)^{-1}\end{align*} $$

for some nonnegative integer L, where $d(\lambda ,\partial \mathbf {S}_\alpha )$ denotes the Euclidean distance of $\lambda $ from the boundary $\partial \mathbf {S}_\alpha $ of the strip $ \mathbf {S}_\alpha $ and $ B_\alpha (0) $ denotes the closed ball in $ \mathbb C $ of radius $ \alpha $ centred at zero.

To find the estimate of $ \|T_\lambda f \|_{L^1(\omega _{\alpha , r})}$ , we need the following property of spherical functions.

Proof. By the Cartan decomposition, it is enough to show that $\phi _{i\nu } (a_t) \leq \phi _{i\alpha }(a_t) $ for all $ t>0 $ , when $ 0<\nu <\alpha $ . For any two smooth functions $ f $ and $ g $ on $ (0,\infty ) $ , we define

$$ \begin{align*} [f,g](t):=( f^\prime(t)g(t) - f(t)g^\prime(t))\,\Delta(t),\,\,\,t>0 .\end{align*} $$

An easy calculation shows that

(3-7) $$ \begin{align} [f,g]^\prime(t)=( L f\cdot g-f \cdot L g)(t)\Delta(t), \end{align} $$

where we recall $ L $ is the Laplace–Beltrami operator on $ G/K $ and

$$ \begin{align*} \Delta(t) = (2\sinh t)^{m_1+m_2}(2\cosh t)^{m_2} .\end{align*} $$

Now putting $ f= \phi _{i\nu }$ and $ g= \phi _{i\alpha } $ in Equation (3-7),

$$ \begin{align*} [\phi_{i\nu},\phi_{i\alpha}]^\prime (a_t) &= ((\nu^2-\rho^2)\phi_{i\nu}\phi_{i\alpha} -\phi_{i\nu}\phi_{i\alpha}(\alpha^2-\rho^2)) \Delta(t)\\ & = (\nu^2-\alpha^2)\phi_{i\nu}\phi_{i\alpha}\, \Delta(t). \end{align*} $$

Hence, $[\phi _{i\nu },\phi _{i\alpha }] (a_t) \leq 0$ for all $ t>0 $ . So we have

$$ \begin{align*} \bigg(\frac{\phi_{i\nu}}{\phi_{i\alpha}}\bigg)^\prime(a_t)&= \bigg(\dfrac{\phi_{i\alpha}(a_t) \phi_{i\nu}^\prime(a_t)-\phi_{i\nu}(a_t)\phi_{i\alpha}^\prime(a_t)}{\phi_{i\alpha}^2(a_t)}\bigg)\leq 0. \end{align*} $$

Therefore, $ \phi _{i\nu }(a_t)\leq \phi _{i\alpha }(a_t) $ for all $ t>0 $ .

Proof of Lemma 3.7

We have

$$ \begin{align*} \| T_\lambda f\|_{L^1_{\omega_{\alpha ,r}}} &= \int_{0}^\infty | T_\lambda f(a_t)| {\phi_{i\alpha}(a_t)\Delta(t)}{(1+t)^r}\,dt. \end{align*} $$

Let

$$ \begin{align*} r_\lambda(t) = \begin{cases}C(1+|\lambda|)^Nt^{(1-m_1-m_2)}&\text{if } m_1+m_2>1,\\ C \log \dfrac{1}{t}&\text{if } m_1+m_2=1. \end{cases} \end{align*} $$

We observe the following properties of $ r_\lambda (t) $ : :

1. (i) $r_\lambda $ is a decreasing function;

2. (ii) $\int _0^{1/2}r_\lambda (t)\Delta (t)\,dt\leq C(1+|\lambda |)^N$ ;

3. (iii) $ |b_\lambda (a_t)| \leq r_\lambda (t) ,\ t\in (0,1/2] .$

We write

$$ \begin{align*} \int_{0}^\infty | T_\lambda f(a_t)| {\phi_{i\alpha}(a_t)\Delta(t)}{(1+t)^r}\,dt\leq I_1 +I_2+I_3+I_4, \end{align*} $$

where

$$ \begin{align*}& I_1= \int_0^{1/2}|b_\lambda(a_t)| {\phi_{i\alpha}(a_t)\Delta(t)}{(1+t)^r} \bigg(\int_t^{\infty}|f(a_s)||\phi_\lambda(a_s)|\Delta(s)\,ds\bigg) \,dt,\\ & I_2 =\int_0^{1/2}|\phi_\lambda(a_t)| {\phi_{i\alpha}(a_t)\Delta(t)}{(1+t)^r} \bigg(\int_t^\infty|f(a_s)||b_\lambda(a_s)|\Delta(s)\,ds\bigg) \,dt,\\ &I_3 = \int_{1/2}^\infty|b_\lambda(a_t)|| {\phi_{i\alpha}(a_t)\Delta(t)}{(1+t)^r} \bigg(\int_t^{\infty}|f(a_s)||\phi_{\lambda}(a_s)|\Delta(s)\,ds\bigg) \,dt,\\ & I_4 =\int_{1/2}^\infty|\phi_{\lambda}(a_t)| {\phi_{i\alpha}(a_t)\Delta(t)}{(1+t)^r} \bigg(\int_t^\infty|f(a_s)||b_\lambda(a_s)|\Delta(s)\,ds\bigg) \,dt. \end{align*} $$

Then,

$$ \begin{align*} I_1 &\leq C \int_0^{1/2}r_\lambda(t) {\phi_{i\alpha}(a_t)\Delta(t)}{(1+t)^r} \bigg(\int_0^{\infty}|f(a_s)|\phi_{i\alpha}(a_s)\Delta(s)\,ds\bigg) \,dt \quad\text{(using Lemma } {3.8})\\ & \leq C (1+|\lambda|)^N \|f\|_{L^1_{\omega_{\alpha,r}}},\\ I_2 &\leq \int_0^{1/2}\phi_{i{\mathrm{Im}}\lambda}(a_t){\phi_{i\alpha}(a_t)\Delta(t)}{(1+t)^r}\bigg(\int_t^{{1}/{2}}|f(a_s)||b_\lambda(a_s)|\Delta(s)\,ds\bigg) \,dt\\ &\quad+\int_0^{1/2}\phi_{i{\mathrm{Im}}\lambda}(a_t){\phi_{i\alpha}(a_t)\Delta(t)}{(1+t)^r}\bigg(\int_{{1}/{2}}^\infty|f(a_s)||b_\lambda(a_s)|\Delta(s)\,ds\bigg) \,dt.\\ & \leq C \int_0^{1/2} r_\lambda(t)\Delta(t) \,dt\int_0^{{1}/{2}}|f(a_s)|{\phi_{i\alpha}(a_s)\Delta(s)}{(1+s)^r}\Delta(s)\,ds \\ &\quad+ C (1+|\lambda|)^M \int_{{1}/{2}}^\infty|f(a_s)|e^{(\alpha -\rho)s}\Delta(s)\,ds \\ &\leq C (1+|\lambda|)^{\max(M,N)}\bigg(\int_0^{{1}/{2}}|f(a_s)|{\phi_{i\alpha}(a_s)\Delta(s)}{(1+s)^r}\,ds\\ &\quad + \int_{{1}/{2}}^{\infty}|f(a_s)|{\phi_{i\alpha}(a_s)\Delta(s)}{(1+s)^r}\,ds \bigg)\quad (\text{using Equation}\ ({2}-{7}))\\ & \leq C (1+|\lambda|)^{\max(M,N)}\|f\|_{L^1_{\omega_{\alpha,r}}}. \end{align*} $$

Next, using the estimates of $ b_\lambda (a_t)$ and $ \phi _{ \lambda }(a_t) $ (see Equation (2-7) and Lemma 3.3) and changing the order of integration,

$$ \begin{align*} I_3 & \leq C (1 + |\lambda|)^M |c(-\lambda)| |c(-\alpha)|\int_{1/2}^\infty e^{(\alpha -{\mathrm{Im}} \lambda)t } (1+t)^r\bigg(\int_t^\infty |f(a_s)|e^{({\mathrm{Im}}\lambda-\rho)s}\Delta(s)\,ds\bigg)\,dt\\ & \leq C (1 + |\lambda|)^M |c(-\lambda)| |c(-\alpha)| \int_{1/2}^\infty |f(a_s)|e^{(\alpha-{\mathrm{Im}}\lambda)s} \Delta(s)\bigg(\int_0^s e^{(\alpha -{\mathrm{Im}} \lambda)t}(1+t)^rdt \bigg)\,ds\\ &\leq C (1 + |\lambda|)^M |c(-\lambda)| |c(-\alpha)| \!\int_{1/2}^\infty \!|f(a_s)|e^{({\mathrm{Im}}\lambda-\rho)s} (1+s)^r \Delta(s)\bigg( \dfrac{e^{(\alpha-{\mathrm{Im}} \lambda)s}-e^{(\alpha-{\mathrm{Im}} \lambda){1}/{2} }}{\alpha- {\mathrm{Im}} \lambda} \bigg)\,ds\\ &\leq { C (1 + |\lambda|)^M |c(-\lambda)|} \|f\|_{L^1_{\omega_{\alpha,r}}}d(\lambda,\partial \mathbf{S}_\alpha)^{-1}. \end{align*} $$

Simliarly, we can prove

$$ \begin{align*} I_4 &\leq { C (1 + |\lambda|)^M |c(-\lambda)|} \|f\|_{L^1_{\omega_{\alpha,r}}}\,d(\lambda,\partial \mathbf{S}_\alpha)^{-1}. \end{align*} $$

Since

$$ \begin{align*} c(-\lambda)=\frac{2^{\rho -i\lambda} \Gamma(\frac{m_1 + m_2 +1}{2})\Gamma(i\lambda)}{\Gamma(\frac{\rho+i\lambda}{2}) \Gamma(\frac{m_1+ 2}{4}+ \frac{i\lambda}{2})}=\frac{2^{\rho -i\lambda} \Gamma(\frac{m_1 + m_2 +1}{2})\Gamma(1+i\lambda)}{i\lambda\, \Gamma(\frac{\rho+i\lambda}{2}) \Gamma(\frac{m_1+ 2}{4} + \frac{i\lambda}{2})},\end{align*} $$

by the polynomial approximation of gamma functions,

$$ \begin{align*}|c(-\lambda)|\leq \frac{C}{|\lambda|(1 + |\lambda|)^{(m_1 + m_2 -2)/2}}.\end{align*} $$

Now $ \lambda \notin B_\alpha (0) $ , so we can dominate $ |c(-\lambda )| $ by a polynomial. Finally, from the estimates $I_1,I_2, I_3$ and $ I_4 $ , the lemma follows.

We will use the inverse spherical transform to show $ T_\lambda f $ is a representative of $ B_\lambda $ for $0 < {\mathrm {Im}}\lambda < \alpha $ . To apply an inverse spherical transform, we need the following lemma.

Lemma 3.9. Let $0 < {\mathrm {Im}}\lambda < \alpha $ and $f \in L^1(G//K,\omega _{\alpha ,r})$ . Then,

$$ \begin{align*} &\widehat{T_\lambda f}( z)=\frac{\widehat{f}(\lambda)-\widehat{f}(z)}{z^2-\lambda^2 }\quad\text{for all } z\in \mathbf{S}_\alpha\setminus\{ \pm \lambda\}. \end{align*} $$

Proof. Using the definition of $ T_\lambda f $ in Equation (3-6),

$$ \begin{align*} { \widehat{T_\lambda f}( z)}&= \int_0^\infty b_\lambda(a_t) \phi_{ z}(a_t) \bigg(\int_t^\infty f(a_s)\phi_{\lambda}(a_s)\Delta(s)\,ds \bigg) \Delta(t)\,dt\\ & \quad- \int_0^\infty \phi_{\lambda}(a_t) \phi_{ z}(a_t) \bigg( \int_t^\infty f(a_s)b_\lambda(a_s)\Delta(s)\,ds\bigg) \Delta(t)\;dt. \end{align*} $$

By changing the order of integration,

$$ \begin{align*} \widehat{T_\lambda f}( z) &=\!\int_0^\infty \!f(a_s) \bigg( \phi_{\lambda}(a_s) \!\int_0^s\! b_\lambda(a_t) \phi_{ z}(a_t) \Delta(t) \,dt - b_{\lambda}(a_s)\!\int_0^s\! \phi_\lambda(a_t) \phi_{ z}(a_t) \Delta(t) \,dt \bigg)\Delta(s) \,ds. \end{align*} $$

Putting $ f =\Phi _{ \lambda } $ , $ g = \phi _{ z}$ in Equation (3-7), we get for any $ 0<r <s $ ,

(3-8) $$ \begin{align} \int_{r}^{s} \Phi_{ \lambda}(a_t)\phi_{ z}(a_t)\Delta(t) \,dt =\frac{1}{z^2-\lambda^2}( [\Phi_\lambda,\phi_{ z}](a_s)-[\Phi_{ \lambda},\phi_{ z}](a_r) ). \end{align} $$

Now sending $ r \rightarrow 0 $ and using the asymptotic behaviour of $ \Phi _{ \lambda }(a_t) $ near $ t=0 $ , we get (see [Reference Pusti and Samanta21, Lemma 8.1])

$$ \begin{align*} \lim\limits_{r\rightarrow 0^+} [\Phi_{ \lambda},\phi_{ z}](a_r) =2i\lambda c(-\lambda). \end{align*} $$

Consequently, Equation (3-8) becomes

$$ \begin{align*} \int_0^s b_\lambda(a_t) \phi_{ z}(a_t) \Delta(t) \,dt =\dfrac{1}{z^2-\lambda^2} ( [b_\lambda,\phi_{ z}](a_s) + 1 ). \end{align*} $$

Similarly, putting $ f =\Phi _{ \lambda } $ , $ g = \phi _{ \lambda }$ in Equation (3-7), we get $ [\phi _{ \lambda },\Phi _{ \lambda }](\cdot ) $ is constant on $ (0,\infty )$ . Next, using the asymptotic behaviour of $ \Delta (t) $ and $ \Phi _{ \lambda }(a_t) $ near $ t =\infty $ , we get (see [Reference Pusti and Samanta21, Lemma 8.1])

$$ \begin{align*} [\phi_{ \lambda},\Phi_{ \lambda}](\cdot) = \lim\limits_{t \rightarrow \infty }[\phi_{ \lambda},\Phi_{ \lambda}](a_t)= -2i\lambda c(-\lambda). \end{align*} $$

Next, from the equations above, it follows that

$$ \begin{align*} I(s)&:= \phi_{\lambda}(a_s) \Delta(s) \int_0^s b_\lambda(a_t) \phi_{ z}(a_t) \Delta(t) \,dt - b_{\lambda}(a_s) \Delta(s)\int_0^s \phi_\lambda(a_t) \phi_{ z}(a_t) \Delta(t) \,dt \\ &=\dfrac{1}{z^2-\lambda^2} ( \phi_{\lambda}(a_s) \Delta(s) ( [b_\lambda,\phi_{ z}](a_s) + 1 ) - b_{\lambda}(a_s) \Delta(s) [\phi_{z},\phi_\lambda](a_s) )\\ &= \dfrac{1}{z^2-\lambda^2} ( \phi_{\lambda}(a_s) - \phi_z(a_s) ) \Delta(s) \text{ (here we use the fact } [\phi_{ \lambda},b_\lambda](\cdot) = 1\text{)}. \end{align*} $$

Therefore,

$$ \begin{align*} \widehat{T_\lambda f}( z)= \frac{\widehat{f}(\lambda)-\widehat{f}(z)}{z^2-\lambda^2}\quad\text{for all } z\in \mathbf{S}_\alpha\setminus\{\pm \lambda\}.\\[-42pt] \end{align*} $$

We are now equipped with all the tools to find out the explicit formula of $\mathcal {R}[g]$ . The relevancy of the functions $ b_\lambda $ and $ T_\lambda f $ is made clear in the following lemma, where we summarize the necessary properties of the resolvent transform.

Proof. (a) This follows from the definition of $ B_\lambda $ (see Equation (3-5)).

(b) Suppose $ \lambda _0 $ be a fixed complex number with $ {\mathrm {Im}} \lambda _0>\alpha $ . We recall from Lemma 3.4 that for $ {\mathrm {Im}} \lambda>\alpha $ , we have $ b_\lambda \in L^1(\omega _{\alpha ,r}) $ . Then from Lemma 3.5, for all $ z \in \mathbf {S}_\alpha $ ,

$$ \begin{align*} (1- (\lambda^2-\lambda_0^2)\widehat b_{\lambda_0}(z)) \widehat {b_\lambda}(z) = \widehat {b_{\lambda_0}}(z),\quad \text{for all }z \in \mathbf{S}_\alpha. \end{align*} $$

Hence, by the inverse spherical transform,

$$ \begin{align*} (\delta - (\lambda^2-\lambda_0^2) b_{\lambda_0}(\cdot))* b_\lambda(\cdot) = b_{\lambda_0}(\cdot) \end{align*} $$

as $ L^{1}_\delta (\omega _{\alpha ,r}) $ functions. Therefore, in the quotient algebra $ L^{1}_\delta (\omega _{\alpha ,r}) /I$ ,

(3-9) $$ \begin{align} (\delta -(\lambda^2-\lambda_0^2) b_{\lambda_0} +I)*(b_\lambda +I)= b_{\lambda_0}+I. \end{align} $$

Since $ (\delta - (\lambda ^2-\lambda _0^2) b_{\lambda _0}^{\lambda } +I) $ is invertible in $ L^{1}_\delta (\omega _{\alpha ,r})/I $ , from the definition of $ B_\lambda $ (see Equation (3-5)) and Equation (3-9), we get $ B_\lambda =b_\lambda +I. $ Accordingly,

(3-10) $$ \begin{align} \mathcal{R}[g](\lambda ) =\langle b_\lambda ,g \rangle. \end{align} $$

(c) Let $ f \in I $ and $ \lambda \in \mathbb C$ with $0<{\mathrm {Im}}\lambda < \alpha $ , then by Lemma 3.7, $T_\lambda f$ is in $L^1(\omega _{\alpha ,r})$ . Now if $ \lambda $ is such that $ \widehat f(\lambda ) \not = 0 $ , then, similarly as in the previous case, we have from Lemma 3.9 and the inverse spherical transform,

$$ \begin{align*} (\delta - (\lambda^2-\lambda_0^2) b_{\lambda_0} (\cdot))* \bigg( \frac{T_\lambda f (\cdot)}{\widehat f(\lambda)}\bigg)= b_{\lambda_0}(\cdot)-\frac{ b_{\lambda_0}(\cdot)*f(\cdot)}{ \widehat f(\lambda)} \end{align*} $$

as $ L^1_\delta (\omega _{\alpha ,r}) $ functions. Since $ f \in I $ ,

$$ \begin{align*} (\delta - (\lambda^2-\lambda_0^2) b_{\lambda_0} +I )* \bigg( \frac{T_\lambda f }{\widehat f(\lambda)}+I\bigg)= b_{\lambda_0}+I. \end{align*} $$

From the definition in Equation (3-5) of $ B_\lambda $ , we get $ B_\lambda = {T_\lambda f }/{\widehat f(\lambda )} +I $ . Hence, from the equation above and the fact that the spherical transforms of functions in $ I $ do not have a common zero, we get the expression

$$ \begin{align*} \mathcal{R}[g](\lambda )= \bigg\langle \frac{T_\lambda f }{\widehat f(\lambda)},g\bigg\rangle \end{align*} $$

provided $ \lambda \notin Z(\widehat f) =\{ z \in S_\alpha : \widehat f(z)=0\} $ .

(d) From item (a), we have $ \mathcal {R}[g] $ is an even function, so it is enough to consider the case ${\mathrm {Im}} \lambda>\alpha $ . Now for $ {\mathrm {Im}} \lambda>\alpha $ , we have from Equation (3-10) and Lemma 3.4,

$$ \begin{align*} | \mathcal{R}[g](\lambda )| \leq C \|g\|_{L^\infty_{1/\omega_{\alpha ,r}}}{ \dfrac{(1+|\lambda |)^N} {d(\lambda,\partial S_\alpha)^n}} \quad \text{for some } C>0. \end{align*} $$

(e) From the estimate of $ \| T_\lambda f\|_{L^1_{\omega _{\alpha ,r}}} $ (Lemma 3.7), we get for $ 0<{\mathrm {Im}} \lambda <\alpha $ , $ \lambda \notin B_{\alpha }(0) $ ,

(3-11) $$ \begin{align} |\widehat f(\lambda)\mathcal{R}[g](\lambda)|\leq C \|f\|_{L^1_{\omega_{\alpha ,r}}}\|g\|_{L^\infty_{1/\omega_{\alpha ,r}}}\frac {(1+\lambda)^L}{d(\lambda,\partial S_\alpha)}. \end{align} $$

As $ \widehat f(\lambda ) \mathcal {R}[g](\lambda ) $ is an even function on $ S_\alpha $ , we get the same estimate for $ 0<|{\mathrm {Im}} \lambda |<\alpha $ , $ \lambda \notin B_{\alpha }(0) $ . Now from continuity of $ \mathcal {R}[g](\lambda ) $ and using Hölder’s inequality for $ \lambda \in B_{\alpha }(0) $ ,

$$ \begin{align*} |\widehat f(\lambda ) \mathcal{R}[g](\lambda)| \leq C \|f\|_{L^1_{\omega_{\alpha ,r}}}, \end{align*} $$

where $ C $ is a constant independent of $ f $ and $ \lambda $ . Hence, using continuity of $ \mathcal {R}[g](\lambda ) $ and $ \widehat f(\lambda ) $ again, we can find a constant $ C>0 $ such that for all $ \lambda \in \mathbb C $ with $ 0\leq {\mathrm {Im}} \lambda <\alpha $ , Equation (3-11) holds.

Proof of Theorem 1.2

Since the ideal generated by $\{f_\beta : \beta \in \Lambda \}$ is the same as the ideal generated by the elements $\{{f_\beta }/{\|f_\beta \|_{L^1_{\omega _{\alpha ,r}}}}: \beta \in \Lambda \}$ and $\delta ^{\alpha \pm }_\infty (\widehat {f})= \delta ^{\alpha +}_\infty ( {\widehat {f}}/{\|f \|_{L^1_{\omega _{\alpha ,r}}}})$ , we assume that the functions $f_\beta $ are of unit $L^1({\omega _{\alpha ,r}})$ norm. Let $g\in L^{\infty }(G//K,1/\omega _{\alpha ,r})$ annihilate the closed ideal I generated by $\{f_\beta : \beta \in \Lambda \}$ . We will show that $g = 0 $ . Then by an application of the Hahn–Banach theorem, it will follow that $I = L^{1}(G//K,\omega _{\alpha ,r})$ . From the hypothesis,

$$ \begin{align*} \inf_{\beta\in\Lambda}\delta^{\alpha+}_\infty(\widehat{f_\beta})=\inf_{\beta\in\Lambda}\delta^{\alpha-}_\infty(\widehat{f_\beta})=0. \end{align*} $$

By Lemma 3.10, the entire function $\mathcal {R}[g]$ satisfies the following estimates:

$$ \begin{align*} |\mathcal{R}[g](z)|&\leq C(1+|z|)(d(z,\partial \mathbf{S}_\alpha))^{-n-1}, \hspace{3mm}z\in\mathbb{C}\backslash \mathbf{S}_\alpha,\\ |\widehat{f_\beta}(z)\mathcal{R}[g](z)|&\leq C(1+|z|)(d(z,\partial \mathbf{S}_\alpha))^{-1}, \hspace{3mm}z\in \mathbf{S}_\alpha^0, \end{align*} $$

for all $\beta \in \Lambda $ , where C is a constant and we choose it to be greater than e and we recall $ n=[r]+1 $ . Let $M:(0,\infty )\rightarrow (e,\infty )$ be a continuously differentiable decreasing function such that $M(t)=\frac {C}{t^{n+1}}$ for $0<t< 1$ , and $\int _1^\infty \log \log M(t)\,dt<\infty $ . With this definition of M,

$$ \begin{align*} |\mathcal{R}[g](z)|&\leq (1+|z|)M(d(z,\partial \mathbf{S}_\alpha)) \hspace{3mm}z\in\mathbb{C}\backslash \mathbf{S}_\alpha,\\ |\widehat{f_\beta}(z)\mathcal{R}[g](z)|&\leq (1+|z|)M(d(z,\partial \mathbf{S}_\alpha)) \hspace{3mm}z\in \mathbf{S}_\alpha^0, \quad\text{for all }\beta \in \Lambda. \end{align*} $$

Therefore, applying Theorem 2.2, we get $\mathcal {R}[g](z)$ is a polynomial. Now from Lemma 3.6, we get $ \|b_z\|_{L^1_{\omega _{\alpha ,r}}} \rightarrow 0$ as $ z $ goes to infinity along the positive imaginary axis. However, Lemma 3.10 gives

$$ \begin{align*}|\mathcal{R}[g](z)| \leq \|b_z\|_{L^1_{\omega_{\alpha ,r}}} \|g\|_{L^\infty_{1/\omega_{\alpha ,r}}} \quad\text{for } {\mathrm{Im}} z>\alpha,\end{align*} $$

so we get $\mathcal {R}[g](z)\rightarrow 0$ , when $z \rightarrow \infty $ along the positive imaginary axis. Therefore, $\mathcal {R}[g]$ is the zero polynomial. Hence, $\langle b_\lambda ,g\rangle =0$ whenever ${\mathrm {Im}} \lambda>\alpha $ , but the collection $\{ b_\lambda : \, {\mathrm {Im}} \lambda>\alpha \}$ spans a dense subset of $L^{1}(G//K,\omega _{\alpha ,r})$ . So $g= 0$ , and the theorem is proved.

4.1 Spherical functions

The spherical functions on $ G $ with respect to $ K $ have the following formula (see [Reference Graczyk and Lœb12, Equation (2.2)] and [Reference Helgason16, page 432, Theorem 5.7]):

$$ \begin{align*} \phi_\lambda(a_t) = i c(\lambda) \dfrac{ \sin \lambda t}{\sinh t}, \end{align*} $$

where

$$ \begin{align*} c(\lambda)& =\dfrac{ 2^{(1-i\lambda)} \Gamma(\frac{3}{2})\Gamma(i\lambda) }{\Gamma(1+ \frac{i\lambda}{2}) \Gamma(\frac{1+i\lambda}{2})} =\dfrac { \Gamma(\frac{3}{2})\Gamma(\frac{i\lambda}{2})\Gamma(\frac{1+i\lambda}{2})}{\sqrt{\pi}\,\Gamma(1+ \frac{i\lambda}{2}) \Gamma(\frac{1+i\lambda}{2})} =\dfrac{1}{i\lambda}. \end{align*} $$

The spherical transform $ \widehat f $ of a function $ f \in L^{p,1}(G//K) $ is defined as

$$ \begin{align*} \widehat f (\lambda)= \int_G f(x) \phi_{- \lambda}(x^{-1}) \,dx\quad \text {for all } \lambda \in \mathbf{S}_{\gamma_p}. \end{align*} $$

As in the rank-one real semisimple Lie group, similar properties of spherical functions and spherical Fourier transforms hold for the rank-one complex semisimple Lie group. We refer the reader to [Reference Flensted-Jensen10] for further details.

4.2 A dense subspace of $ L^{p,1}(G//K)$

Now, to prove Theorem 1.3, we follow a similar strategy as in Theorem 1.2. First, we construct a dense subspace of $ L^{p,1}(G//K)$ . We define for $ \lambda \in \mathbb C_+$ ,

$$ \begin{align*} b_\lambda(a_t)= \dfrac{i e^{i\lambda t}}{2\lambda c(-\lambda)J(t)} = \dfrac{e^{i\lambda t}}{2J(t)} \quad \text{for all } t>0, \end{align*} $$

where $ J(t)= (e^{t }-e^{- t}).$ We extend $ b_\lambda (\cdot ) $ as a $ K $ -biinvariant function in $ G/K\setminus \{eK\} $ using the Cartan decomposition of $ G$ .

To prove the lemma above, we first show that for $ {\mathrm {Im}} \lambda>\gamma _p $ , $ b_\lambda \in L^{p,1}(G//K) $ . After that, we show that the spherical transform of $ b_\lambda $ is $ \widehat {b_\lambda }(z) = 1/(z^2-\lambda ^2)$ for all $ z \in \mathbf {S}_{\gamma _p} $ , and then the proof of the lemma above follows exactly as in Lemma 3.2.

Lemma 4.2. Let $ 1\leq p<2 $ and $ \lambda \in \mathbb C_+ $ with $ {\mathrm {Im}} \lambda>\gamma _p$ , then $ b_\lambda \in L^{p,1}(G//K)$ . Moreover, there exists a constant $ C>0$ (independent of $ \lambda $ ) such that the following estimate holds:

(4-1) $$ \begin{align} \|b_\lambda\|_{p,1} \leq C \frac{1+|\lambda |}{ {\mathrm{Im}} \lambda -\gamma_p}. \end{align} $$

Proof. Suppose $ \lambda \in \mathbb C_+ $ . Then using an asymptotic estimate of $ J(t) $ , we get

(4-2) $$ \begin{align} |b_\lambda(a_t)| \leq C(f(t)+ g(t)), \end{align} $$

where

$$ \begin{align*} { f(t)= \chi_{(0,1]} (t)t^{-1} \quad\text{and}\quad g(t) = \chi_{[1,\infty)}(t)e^{-({\mathrm{Im}} \lambda +1)t} \quad\text{for } t\geq0 .} \end{align*} $$

We extend $ f $ and $ g $ as $ K $ biinvariant functions in $ G $ similarly as in Equation (3-2). The proof will be completed if we show that $ \|f\|_{p,1}+{\| g\|_{p,1}} $ is dominated by the right-hand side of Equation (4-1). We have, for $ \alpha>0 $ ,

$$ \begin{align*} d_f(\alpha)& = m\{ t \in [0,\infty] : |f(t)|> \alpha \}= m\{ t \in (0,1] : t^{-1} > \alpha \}= m\bigg\{ t \in (0,1] : t <\frac{1} {\alpha^{-1}} \bigg\}, \end{align*} $$

where $ m $ is the Haar measure on $ G $ in the Cartan decomposition. We observe that for $ \alpha <1 $ , $ d_f(\alpha ) $ is constant. For $ \alpha>1 $ , we have

$$ \begin{align*} d_f(\alpha) = \int_{0}^{\alpha^{-1}} \widetilde{\Delta}(t) \,dt = \int_{0}^{\alpha^{-1}} (e^{ t }-e^{- t})^2 \,dt \asymp \int_{0}^{\alpha^{-1}} t^{2} \,dt \asymp \alpha ^{-3}. \end{align*} $$

Now,

(4-3) $$ \begin{align} \begin{aligned} \|f\|_{p,1}& = p\bigg( \int_0^\infty d_f(\alpha)^{{1}/{p}}d\alpha\bigg)\leq p\bigg( C+ \int_1^\infty \alpha^{-{3}/{p}} d\alpha\bigg)=C \text{ (since } p< 2 \text{).}\ \end{aligned} \end{align} $$

For the function $ g $ ,

$$ \begin{align*} d_g(\alpha)& = m\{ t \in [1/2,\infty): e^{-({\mathrm{Im}}\lambda+1)t}> \alpha \} = m\bigg\{ t \in [1/2,\infty) : t < \frac{1}{({\mathrm{Im}}\lambda+1)}\log \frac{1}{\alpha} \bigg\}. \end{align*} $$

Therefore, $ d_g(\alpha ) = 0 $ whenever $ \alpha> e^{{-({\mathrm {Im}}\lambda +1)}/{2}} $ and for the range $ 0<\alpha < e^{{-({\mathrm {Im}}\lambda +1)}/{2}} $ ,

$$ \begin{align*} d_g(\alpha) \asymp \int_{{1}/{2}}^{{1}/{({\mathrm{Im}}\lambda+1)}\log {{1}/{\alpha}}} e^{2 t}\,dt \leq \dfrac{1}{\alpha^{{2 }/({{\mathrm{Im}} \lambda + 1})}}. \end{align*} $$

Then,

(4-4) $$ \begin{align} \|g\|_{p,1}=p\bigg( \int_0^\infty d_g(\alpha)^{{1}/{p}}d\alpha\bigg) \leq p\bigg( \int_0^1 \alpha^{-{2}/{p({\mathrm{Im}} \lambda +1)}}d\alpha\bigg) = \dfrac{1}{1 -\frac{2}{ p({\mathrm{Im}} \lambda +1)}}. \end{align} $$

The integral in Equation (4-4) converges since $ {2 }/{({\mathrm {Im}} \lambda + 1)p} < 1 $ for $ {\mathrm {Im}} \lambda> \gamma _p. $ Next, from Equations (4-3) and (4-4),

$$ \begin{align*} \|b_\lambda\|_{p,1} \leq C+ \dfrac{1}{1 -\frac{2}{ p({\mathrm{Im}} \lambda +1)}} \leq C \frac{1+|\lambda |}{ {\mathrm{Im}} \lambda -\gamma_p}. \end{align*} $$

This completes the proof of the lemma.

Lemma 4.3. For $ \lambda \in \mathbb C_+ $ with $ {\mathrm {Im}} \lambda>\gamma _p,$

$$ \begin{align*} \widehat b_\lambda(z) = \frac{1}{z^2-\lambda^2}, \quad \text{for all } z\in \mathbf{S}_{\gamma_p}. \end{align*} $$

Proof. Suppose $ z \in \mathbf {S}_{\gamma _p},$ then,

$$ \begin{align*} \widehat b_\lambda (z)& = \dfrac{1}{2z}\int_{0}^{\infty}\dfrac{ e^{i\lambda t} \sin(z t)}{J(t) \sinh t} (e^{t }-e^{- t})^2\,dt\\ &= \dfrac{1}{2iz}\int_{0}^{\infty} (e^{i(\lambda +z)t}-e^{i(\lambda -z)t})\,dt\\ & = \dfrac{1}{-2z} \dfrac{2z}{ (\lambda^2 -z^2)} =\dfrac{1}{(z^2-\lambda^2)}.\\[-3.5pc] \end{align*} $$

From the pointwise estimate in Equation (4-2) of $ b_\lambda (a_t) $ , we observe that if $ {\mathrm {Im}} (\lambda ) $ is sufficiently large, then $ b_\lambda \in L^{p,1}(G//K)$ for any small $ p $ . In particular, if $ {\mathrm {Im}} \lambda>\gamma _p+1 $ , then $ \widehat b_\lambda $ exists at the point $ i $ and $\widehat {b}_{\lambda }(i)={1}/{-(\lambda ^2 +1)} $ . Using this along with the estimate of $ b_\lambda (a_t) $ , we show that for all $ p \in [1,2) $ , $ \|b_\lambda \|_{p,1} \rightarrow 0$ whenever $ \lambda \rightarrow \infty $ along the positive imaginary axis.

Proof. We are going to show for $ \lambda =\zeta + i\xi $ , $ \| b_\lambda \|_{p,1} \rightarrow 0 $ as $ \xi \rightarrow \infty $ , for any fixed $ \zeta \in \mathbb R $ . Suppose $ \lambda =\zeta + i\xi $ and $ \xi $ is a large positive real number and $ q \in [1,2) $ . Then, $ b_{i\xi } (a_t) $ is positive and for a fixed $ \zeta $ , $ |b_\lambda (a_t)| \leq C b_{i\xi }(a_t) $ for all $ t> 0 $ . Hence, for large $ \xi>0$ ,

$$ \begin{align*} ||b_{i\xi}||_1 = \int_{\mathbb R_+} b_{i\xi} (a_t) \widetilde{\Delta}(t) \,dt = \widehat b_{i\xi}(i)=\frac{1}{\xi^2-1} \end{align*} $$

goes to zero as $ \xi \rightarrow \infty. $ Next, we show that for any $ q \in [1,2)$ ,

(4-5) $$ \begin{align} \lim\limits_{\xi \rightarrow \infty} \| b_{i\xi}\|_{q} =0. \end{align} $$

We have

$$ \begin{align*} \| b_{i\xi}\|_{q}^q & \leq C\int_0^{1}\bigg(\frac{1}{t}\bigg)^q \widetilde{\Delta} (t) \,dt +C\int_{1}^\infty e^{-( \xi +1)qt+2t} \, dt\\&\leq C + C e^{1-{( \xi +1)q}/{2}}. \end{align*} $$

Let $ 1<p<q $ , then there exists $ \theta \in (0,1)$ such that $ 1/p= (1-\theta )+\theta /p $ . By Hölder’s inequality,

$$ \begin{align*} \lim\limits_{\xi \rightarrow \infty } \| b_{i\xi}\|_{p} \leq \lim\limits_{\xi \rightarrow \infty } \| b_{i\xi}\|_{1}^{1-\theta} \lim\limits_{\xi \rightarrow \infty } \| b_{i\xi}\|_{q}^\theta = 0 \end{align*} $$

as $ \lim \nolimits _{\xi \rightarrow \infty } \| b_{i\xi }\|_{1} = 0 $ (see [Reference Pusti and Samanta21, Lemma 3.5]). Since $ p,q $ are arbitrary, we get Equation (4-5). Now for $ p_1 <p <p_2 $ , we have from Lemma 2.1,

$$ \begin{align*} \|b_{i\xi }\|_{p,1}\leq C ( \|b_{i\xi}\|_{p_1} +\|{b_{i\xi}}\|_{p_2}). \end{align*} $$

This implies $ \|b_\lambda \|_{p,1} \rightarrow 0 $ as $ \lambda \rightarrow \infty $ through any vertical line.

4.3 Resolvent transform

As before, here also we define a resolvent transform associated to each $ g \in L^{p',\infty }(G//K) $ that annihilates the closed ideal $ I $ generated by the functions $ \{ f_\beta : \beta \in \Lambda \} $ from the hypothesis of Theorem 1.3. Let $ L^{p,1}_\delta (G//K) $ be the unitization of $ L^{p,1}(G//K) $ and $ \delta $ . The maximal ideal space of $ L^{p,1}_\delta (G//K) $ is $ \{ L_z :z \in \mathbf {S}_{\gamma _p} \cup \infty \} $ , where $ L_z $ are complex homomorphisms on $ L^{p,1}_\delta (G//K) $ defined similarly as in Equation (3-2). We have that the collection $ \{ \widehat f_\beta : \beta \in \Lambda \} $ does not have any common zero in $ \mathbf {S}_{\gamma _p} $ . So by Banach algebra theory, we get that the maximal ideal space of $ L^{p,1}_\delta (G//K)/I $ is the complex homomorphism $ \tilde L_\infty $ , defined by

$$ \begin{align*} \tilde L_\infty (f+I) = \widehat f(\infty) \quad\text{for all } f\in L^{p,1}_\delta(G//K)/I. \end{align*} $$

For each $ g \in L^{p',\infty }(G//K) $ that annihilates the closed ideal $ I $ , we associate its resolvent transform

$$ \begin{align*} \mathcal{R}[g] (\lambda )= \langle b_\lambda, g\rangle, \quad {\mathrm{Im}} \lambda>\gamma_p. \end{align*} $$

Now suppose $ \lambda _0 $ is a fixed complex number with ${\mathrm {Im}} \lambda _0>\gamma _p. $ Then, using an analogous argument as in Equation (3-5), we define the resolvent transform

$$ \begin{align*} \mathcal{R}[g](\lambda)= \langle B_\lambda,g \rangle \quad \lambda \in \mathbb C, \end{align*} $$

which is analytic on the entire complex plane, where

$$ \begin{align*} B_\lambda = (\delta - (\lambda ^2-\lambda_0^2)b_{\lambda_0} +I)^{-1}*(b_{\lambda_0}+I). \end{align*} $$

Our next objective is to find an explicit formula for $ \mathcal {R}[g](\lambda ) $ by acquiring representatives of $ B_\lambda $ in $ L^{p,1}(G//K) /I$ . We will show, for $ {\mathrm {Im}} \lambda>\gamma _p$ , a representative of $ B_\lambda $ is $b_\lambda $ , which is in $ L^{p,1}_\delta (G//K) $ . Before that, we find a representative of $ B_\lambda $ in $ L^{p,1}_\delta (G//K) /I $ for $ 0<{\mathrm {Im}} \lambda <\gamma _p $ . Suppose $ \lambda \in \mathbb C $ with $ 0<{\mathrm {Im}} \lambda <\gamma _p$ and $ f \in L^{p,1}(G//K) $ . Then for all $ t>0 $ , we define

(4-6) $$ \begin{align} T_\lambda f(a_t) = b_\lambda(a_t) \int_{t}^{\infty} f(a_s)\phi_\lambda(a_s)\widetilde{\Delta}(s) \,ds-\phi_\lambda(a_t) \int_{t}^{\infty} f(a_s) b_\lambda(a_s)\widetilde{\Delta}(s) \,ds, \end{align} $$

and extend it as a $ K $ biinvariant function on $ G. $ Next, using the estimate of $ b_\lambda (a_t) $ , we find a quantitative estimate of $ \|T_\lambda f\|_{p,1} $ in the following lemma.

Proof. Suppose $ h \in L^{p^{\prime },\infty }(G//K)$ with $ \| h\|_{p^{\prime },\infty } \leq 1$ , then from Equation (4-6),

$$ \begin{align*} \bigg| \int_G T_\lambda f(x) h(x) \,dx \bigg| & = \bigg| \int_0^\infty T_\lambda f(a_t) h(a_t) \widetilde{\Delta}(t) \,dt \bigg|\\ & = \bigg| \int_0^\infty\bigg( b_\lambda(a_t)\int_t^\infty f(a_s)\phi_{\lambda}(a_s)\widetilde{\Delta}(s)\,ds \\ & \quad -\phi_{\lambda}(a_t) \int_t^\infty f(a_s)b_\lambda(a_s)\widetilde{\Delta}(s)\,ds \bigg)h(t)\widetilde{\Delta}(t) \,dt\bigg|. \end{align*} $$

Now we divide the integral into four parts to use the estimates of $ b_\lambda (a_t) $ :

$$ \begin{align*} \bigg|\int_G T_\lambda f (x)h(x) \,dx \bigg| \leq I_1 +I_2 + I_3+I_4, \end{align*} $$

where

$$ \begin{align*} I_1 & =\int_0^{1}|b_\lambda (a_t)||h(a_t)|\bigg(\int_t^{\infty}|f(a_s)||\phi_{\lambda}(a_s)|\widetilde{\Delta}(s)\,ds\bigg) \widetilde{\Delta}(t)\,dt, \\ I_2 & = \int_0^{1}|\phi_{\lambda}(a_t)||h(a_t)|\bigg(\int_t^\infty|f(a_s)||b_\lambda(a_s)|\widetilde{\Delta}(s)\,ds\bigg) \widetilde{\Delta}(t) \,dt,\\ I_3 & = \int_{1}^\infty|b_\lambda (a_t)||h(a_t)|\bigg(\int_t^{\infty}|f(a_s)||\phi_{\lambda}(a_s)|\widetilde{\Delta}(s)\,ds\bigg) \widetilde{\Delta}(t)\,dt\\ \quad\text{and}\quad\\ I_4 & =\int_{1}^\infty|\phi_{\lambda}(a_t)||h(a_t)|\bigg(\int_t^\infty|f(a_s)||b_\lambda(a_s)|\widetilde{\Delta}(s)\,ds\bigg) \widetilde{\Delta}(t) \,dt. \end{align*} $$

Then, using Hölder's inequality and our estimates of $ b_\lambda $ ,

$$ \begin{align*} I_1 & \leq\int_0^{1}|b_\lambda(a_t)||h(a_t)|\bigg(\int_t^{\infty}|f(a_s)||\phi_{ \lambda}(a_s)|\widetilde{\Delta}(s)\,ds\bigg) \widetilde{\Delta}(t)\,dt\\ & \leq C \|f\|_{p,1} \| \phi_{ \lambda}\|_{p^{\prime},\infty } \|h\|_{p^{\prime},\infty }\\ I_2 & \leq \int_0^{1}|\phi_{ \lambda}(a_t)||h(a_t)|\bigg(\int_t^\infty|f(a_s)||b_\lambda(a_s)|\widetilde{\Delta}(s)\,ds\bigg) \widetilde{\Delta}(t) \,dt\\ & \leq C \int_0^{1/2}|\phi_{ \lambda}(a_t)||h(a_t)| \|f\|_{p,1}( C+ \| \chi_{[1,\infty)}(s) e^{-({\mathrm{Im}} \lambda + 1)s}\|_{p^{\prime},\infty} )\, dt\\ & \leq C \|f\|_{p,1} \|h\|_{p^{\prime},\infty.} \end{align*} $$

Since $ \lambda $ is inside the strip $ \mathbf {S}_{\gamma _p} $ , that is, $ 0<{\mathrm {Im}} \lambda < \gamma _p $ , by a similar calculation as in Equation (4-2), we can show $ \|\phi _{ \lambda } \|_{p^{\prime },\infty }$ and $ \| \chi _{[1,\infty )}(s)b_\lambda (a_s) \|_{p^{\prime },\infty }$ is bounded by a constant independent of $ \lambda $ .

Before we estimate $ I_3 $ and $ I_4 $ , we need to find an $ L^{p,1} $ -norm estimate of the following $ K $ biinvariant function on $ G $ , defined by

$$ \begin{align*} g(t)=e^{({\mathrm{Im}} \lambda -1)t} \chi_{[1,s ]}(t), \quad \text{where } s \geq 1\text{ is fixed}. \end{align*} $$

Then the distribution function is

$$ \begin{align*} d_g(\alpha) = m\bigg\{ t \in [1,s] : t < \frac{1}{(1-{\mathrm{Im}}\lambda)}\log \frac{1}{\alpha} \bigg\}.\end{align*} $$

Recall m is the Haar measure on G in the Cartan decomposition. Now we observe, unless $ 0< \alpha <e^{{({\mathrm {Im}} \lambda -1)}}$ , $ d_g(\alpha ) =0 $ . Furthermore, when $ 0 < \alpha < e^{{({\mathrm {Im}} \lambda -1)}s} $ ,

$$ \begin{align*} d_g(\alpha) \asymp \int_{1}^s e^{2 t}\,dt = \dfrac{e^{2 s}- e^{2}}{2} \end{align*} $$

and for $ e^{{({\mathrm {Im}} \lambda -1)}s}< \alpha < e^{{({\mathrm {Im}} \lambda -1)}} $ , we get

$$\begin{align*}d_g(\alpha) \asymp \int_{1}^{({\log{1}/{\alpha}})/({1-{\mathrm{Im}} \lambda})} e^{2 t}\,dt = \dfrac{1}{2\alpha^{{2}/({1-{\mathrm{Im}} \lambda})}} -\dfrac{e^{2 }}{2}. \end{align*}$$

Therefore,

$$ \begin{align*} \|g\|_{p,1}= p \int_0^\infty d_g(\alpha)^{{1}/{p}} d\alpha & = p \int_0^{e^{{({\mathrm{Im}} \lambda -1)}s}} d_g(\alpha)^{{1}/{p}} d\alpha + p \int_{e^{{({\mathrm{Im}} \lambda -1)}s}}^{e^{{({\mathrm{Im}} \lambda -1)}}}d_g(\alpha)^{{1}/{p}} d\alpha \notag \\[5pt] & \leq C e^{({\mathrm{Im}} \lambda +{2}/{p}-1)s} + \dfrac{e^{{{\mathrm{Im}} \lambda -1}+{2}/{p}} - e^{({\mathrm{Im}} \lambda -1 +{2 }/{p})s} }{1 -\frac{2 }{p( -{\mathrm{Im}} \lambda)}} \notag \\[5pt] & \leq C (1+|\lambda|) \dfrac{e^{({\mathrm{Im}} \lambda +{2 }/{p}- 1)s}}{(\frac{2}{p}-1) 1 +{\mathrm{Im}} \lambda} \leq C (1+|\lambda|) \dfrac{e^{({2 }/{p}-1 + {\mathrm{Im}} \lambda )s}}{(\frac{2}{p}-1) -{\mathrm{Im}} \lambda}. \end{align*} $$

Similarly, if we take $ \tilde g(t) = e^{-({\mathrm {Im}} \lambda +1 )t} \chi _{[1,s]}(t), $ for a fixed $ s\geq 1 $ ,

(4-7) $$ \begin{align} \|\tilde g\|_{p,1} \leq C (1+|\lambda|) \dfrac{e^{({2 }/{p}- 1 - {\mathrm{Im}} \lambda )s}}{(\frac{2}{p}-1) -{\mathrm{Im}} \lambda}. \end{align} $$

Now, using the estimates of $ b_\lambda $ , $ \phi _{\lambda }$ and changing the order of integration,

$$ \begin{align*} I_3 & \leq \dfrac{C}{|\lambda|} \int_{1}^\infty e^{-({\mathrm{Im}} \lambda +1)t} |h(t)| \bigg( \int_t^\infty |f(a_s)| e^{({\mathrm{Im}} \lambda -1)s} \widetilde{\Delta}(s) \,ds\bigg) \widetilde{\Delta}(t) \,dt\\[5pt] & \leq \dfrac{C}{|\lambda|} \int_{1}^\infty |f(a_s)| e^{({\mathrm{Im}} \lambda -1)s} \bigg( \int_{1}^s e^{-({\mathrm{Im}} \lambda +1)t} |h(t)| \widetilde{\Delta}(t)\,dt \bigg) \widetilde{\Delta}(s) \,ds\\[5pt] & \leq \dfrac{C}{|\lambda|} (1+|\lambda|) \!\int_{1}^\infty \kern-1pt\!|f(a_s)| e^{({\mathrm{Im}} \lambda -1)s} \dfrac{e^{({2 }/{p}-1 - {\mathrm{Im}} \lambda )s}}{(\frac{2}{p}-1) -{\mathrm{Im}} \lambda} \|h\|_{p^{\prime},\infty} \widetilde{\Delta}(s)\,ds\ \text{ (using Equation}\ ({4\hbox{-}7})) \\& \leq \dfrac{C (1+|\lambda|) |}{|\lambda|(\frac{2 }{p}- 1-{\mathrm{Im}} \lambda)} \int_{1}^\infty |f(a_s)| e^{-{2 }/{p'}s} \|h\|_{p^{\prime},\infty} \widetilde{\Delta}(s)\,ds\\[5pt] &\leq \dfrac{C(1+|\lambda|)}{|\lambda|} d(\lambda, \partial \mathbf{S}_{\gamma_p})^{-1} \|f\|_{p,1} \|h\|_{p^{\prime},\infty}. \end{align*} $$

Similarly, we can prove

$$ \begin{align*} I_4 &\leq \dfrac{C (1+|\lambda|)}{|\lambda|} d(\lambda, \partial \mathbf{S}_{\gamma_p})^{-1} \|f\|_{p,1} \|h\|_{p^{\prime},\infty}. \end{align*} $$

Since $ \lambda \notin B_{\gamma _p}(0) $ , ${1}/{|\lambda |}\leq C$ , and so adding the estimates of $ I_1,I_2,I_3 $ and $ I_4 $ , the lemma follows.

Now that we have $ T_\lambda f \in L^{p,1}(G//K) $ for $ 0<{\mathrm {Im}} \lambda <\gamma _p $ , next we need the spherical transform of $ T_\lambda f $ to prove that $T_{\lambda} f$ is a representative of $ B_\lambda $ in $ L^{p,1}(G//K) $ . We find the spherical transform of $T_\lambda f $ , using the same calculation in Lemma 3.9 to prove the following lemma.

Lemma 4.6. Suppose $0<{\mathrm {Im}}\lambda <\gamma _p$ and $f \in L^{p,1}(G//K)$ . Then,

$$ \begin{align*} \widehat{T_\lambda f}(z)=\frac{\widehat{f}(\lambda)-\widehat{f}(z)}{z^2-\lambda^2}\quad \text{for all } z\in \mathbf{S}_{\gamma_p} \ {\setminus}\ \{\pm \lambda\}. \end{align*} $$