In this section we aim at the generalization of the Paley–Wiener theorem for the Mellin transform which will later be applied for characterizing function spaces in terms of the distance from a Mellin–Bernstein space. The Fourier counterpart was proved in [17]. We need the following space ${H}_{c}^{\ast}$(ℍ_{a}), which lies between two Mellin-Hardy spaces.

#### Definition 3.4

Let *a*, *c* ∈ ℝ with *a* > 0 be fixed numbers. The class ${H}_{c}^{\ast}$(ℍ_{a}) comprises all functions *f* : ℍ_{a} → ℂ with the following properties:

*f* is polar-analytic on ℍ_{a};

lim_{r→0+} *r*^{c}f(*r*, 0) = lim_{r→+∞} *r*^{c}f(*r*, 0) = 0;

for every *ε* ∈ ]0, *a*[ there exists a constant *K*(*f*, *ε*) such that
$$\begin{array}{}|f(r,\theta )|\le {r}^{-c}K(f,\epsilon )\end{array}$$

for all (*r*, *θ*) ∈ ℍ_{a−ε};

for every *θ* ∈ ]−*a*, *a*[ and all *t* ∈ ℝ,
$$\begin{array}{}{\displaystyle {I}_{c}(f,\theta ,t):=\underset{R\to +\mathrm{\infty}}{lim}{\int}_{1/R}^{R}f(r,\theta ){r}^{c+it-1}dr}\end{array}$$

exists and |*I*_{c}(*f*, *θ*, *t*)| ≤ *K*(*f*) with a constant depending on *f* only.

For the subsequent considerations, we need some lemmas concerning the (*C*, 1)-summability of the *L*^{2}-Mellin inversion and a result on the inversion of the Cauchy principal value of a Mellin transform.

#### Lemma 3.1

*Let G*(*c* + *i* ⋅) *be a function in L*^{2}(*c* + *i* ℝ). *Then*, *denoting by g* ∈ ${X}_{c}^{2}$ *the inverse Mellin transform of G*, *one has*
$$\begin{array}{}{\displaystyle g(r)=\underset{\rho \to +\mathrm{\infty}}{lim}\frac{1}{2\pi}{\int}_{-\rho}^{\rho}(1-\frac{|t|}{\rho})G(c+it){r}^{-c-it}dt\phantom{\rule{2em}{0ex}}(\mathit{\text{a.e.\hspace{0.17em}}}\phantom{\rule{thinmathspace}{0ex}}r\in {\mathbb{R}}^{+}).}\end{array}$$

#### Proof

First of all, by [13, Theorem 2.9], the inverse Mellin transform *g* ∈ ${X}_{c}^{2}$ exists. Moreover, if we employ the Mellin–Fejer kernel, defined by (see [11])
$$\begin{array}{}{\displaystyle {F}_{\rho}^{c}(r):=-\frac{1}{2\pi \rho}{r}^{-c}(\frac{{r}^{\rho i/2}-{r}^{-\rho i/2}}{\mathrm{log}r}{)}^{2},\phantom{\rule{1em}{0ex}}r\ne 1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{F}_{\rho}^{c}(1)=\frac{\rho}{2\pi}\phantom{\rule{thinmathspace}{0ex}},}\end{array}$$

the Mellin convolution
$$\begin{array}{}{\displaystyle ({F}_{\rho}^{c}\ast g)(x):={\int}_{0}^{\mathrm{\infty}}{F}_{\rho}^{c}(\frac{x}{u})g(u)\frac{du}{u}}\end{array}$$

converges almost everywhere to *g*. Furthermore, using a Parseval type equation for Mellin convolutions (see [11, Theorem 9]) and the fact that the Mellin transform of ${F}_{\rho}^{c}$ is given by
$$\begin{array}{}{\displaystyle [{F}_{\rho}^{c}{]}_{{M}_{c}}^{\wedge}(c+it)=(1-\frac{|t|}{\rho}),\phantom{\rule{2em}{0ex}}0\le |t|\le \rho}\end{array}$$

and
$$\begin{array}{}{\displaystyle [{F}_{\rho}^{c}{]}_{{M}_{c}}^{\wedge}(c+it)=0,\phantom{\rule{2em}{0ex}}|t|>\rho ,}\end{array}$$

one has
$$\begin{array}{}{\displaystyle g(r)=\underset{\rho \to +\mathrm{\infty}}{lim}\frac{1}{2\pi}{\int}_{-\rho}^{\rho}(1-\frac{|t|}{\rho})G(c+it){r}^{-c-it}dt\phantom{\rule{2em}{0ex}}(\text{\hspace{0.17em}a.e.\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}r\in {\mathbb{R}}^{+}),}\end{array}$$

which is the assertion.□

The second lemma is as follows:

#### Lemma 3.2

*Suppose that ϕ* : ℝ^{+} → ℂ *is integrable on every compact subinterval and*
$$\begin{array}{}{\displaystyle G(c+it):=\underset{R\to +\mathrm{\infty}}{lim}{\int}_{1/R}^{R}\varphi (r){r}^{c-1+it}dr}\end{array}$$

*exists for each t* ∈ ℝ *and is integrable on compact subintervals*. *Then*,
$$\begin{array}{}{\displaystyle \varphi (r)=\underset{\rho \to +\mathrm{\infty}}{lim}\frac{1}{2\pi}{\int}_{-\rho}^{\rho}(1-\frac{|t|}{\rho})G(c+it){r}^{-c-it}dt}\end{array}$$

*almost everywhere for r* ∈ ℝ^{+}.

In the Fourier case this lemma is stated and proved in [28, Theorem 120] and in [33, Theorem 10.3]. Both proofs are quite long with several intermediate results. For this reason, we made no attempts to design an intrinsic Mellin type proof, but leave the deduction of Lemma 3.2 to the reader, using the classical substitution from the Fourier to the Mellin instance. (Note: The modern literature on integral transforms prefers an *L*^{p}-frame. However, as soon as contour integration of analytic functions is involved, Cauchy principal values occur in a natural way. Therefore it would be desirable to find a new and short approach to Lemma 3.2 (without substitutions) or at least to its Fourier or Laplace version.)

Combining Lemma 3.1 and Lemma 3.2, we obtain the following basic lemma whose Fourier version was given in [17, Lemma 2.1].

#### Lemma 3.3

*Let ϕ* : ℝ^{+} → ℂ *be integrable on every compact subinterval of* ℝ^{+}, *and suppose that*
$$\begin{array}{}{\displaystyle {M}_{c}^{\ast}[\varphi ](c+it):=\underset{R\to +\mathrm{\infty}}{lim}{\int}_{1/R}^{R}\varphi (r){r}^{c+it-1}dr}\end{array}$$

*exists for all t* ∈ ℝ. *If* ${M}_{c}^{\ast}$ [*ϕ*] ∈ *L*^{2}(*c* + *i* ℝ), *then ϕ* ∈ ${X}_{c}^{2}$ *and* ${M}_{c}^{\ast}$[*ϕ*](*c* + *it*) = ${M}_{c}^{2}$[*ϕ*](*c* + *it*) *almost everywhere for t* ∈ ℝ.

#### Proof

If ${M}_{c}^{\ast}$[*ϕ*] ∈ *L*^{2}(*c* + *i* ℝ), then, by a Plancherel type theorem for the Mellin transform (see [13, Theorem 2.9]), there exists a function *ψ* ∈ ${X}_{c}^{2}$ such that ${M}_{c}^{2}$[*ψ*] = ${M}_{c}^{\ast}$[*ϕ*]. Employing Lemma 3.1, we conclude that
$$\begin{array}{}{\displaystyle \psi (r)=\underset{\rho \to +\mathrm{\infty}}{lim}\frac{1}{2\pi}{\int}_{-\rho}^{\rho}(1-\frac{|t|}{\rho}){M}_{c}^{\ast}[\varphi ](c+it){r}^{-c-it}dt\phantom{\rule{2em}{0ex}}(\text{a.e.\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}r\in {\mathbb{R}}^{+}).}\end{array}$$

On the other hand, by Lemma 3.2, we have
$$\begin{array}{}{\displaystyle \varphi (r)=\underset{\rho \to +\mathrm{\infty}}{lim}\frac{1}{2\pi}{\int}_{-\rho}^{\rho}(1-\frac{|t|}{\rho}){M}_{c}^{\ast}[\varphi ](c+it){r}^{-c-it}dt\phantom{\rule{2em}{0ex}}(\text{a.e.\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}r\in {\mathbb{R}}^{+}).}\end{array}$$

Hence *ϕ*(*r*) = *ψ*(*r*) almost everywhere for *r* ∈ ℝ^{+}. Now the assertion follows immediately.□

With the help of Lemma 3.3, we can easily see that ${H}_{c}^{\ast}$(ℍ_{a}) constitutes a normed linear space.

#### Proposition 3.1

*Let f* ∈ ${H}_{c}^{\ast}$(ℍ_{a}). *By introducing*
$$\begin{array}{}{\displaystyle \parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}:=sup\{|{I}_{c}(f,\theta ,t)|:\theta \in ]-a,a[,t\in \mathbb{R}\},}\end{array}$$

*the class* ${H}_{c}^{\ast}$(ℍ_{a}) *becomes a normed linear space*.

#### Proof

As the only non-trivial assertion, we have to show that if *f* ∈ ${H}_{c}^{\ast}$(ℍ_{a}) and $\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}$ = 0, then *f*(*r*, *θ*) = 0 for all (*r*, *θ*) ∈ ℍ_{a}. Now, if $\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}$ = 0, then, clearly, *I*_{c}(*f*, *θ*, *t*) = 0 for all *t* ∈ ℝ. In view of Lemma 3.3, we obtain that ${M}_{c}^{2}$ [*f*(⋅, *θ*)](*c* + *it*) = 0 almost everywhere for *t* ∈ ℝ, and application of the the Mellin–Plancherel theorem (see [13, Lemma 2.6]) shows that *f*(*r*, *θ*) = 0 almost everywhere for *r* ∈ ℝ^{+}. Since a polar-analytic function is continuous, we conclude that *f*(*r*, *θ*) = 0 for all (*r*, *θ*) ∈ ℍ_{a}.□

Next we want to show that in Definition 3.4, statement (b) can be generalized by employing (a) and (c).

#### Proposition 3.2

*Let f* : ℍ_{a} → ℂ *satisfy conditions (a)*-*(c) of Definition 3.4*. *Then*,
$$\begin{array}{}{\displaystyle \underset{r\to {0}^{+}}{lim}{r}^{c}f(r,\theta )=\underset{r\to +\mathrm{\infty}}{lim}{r}^{c}f(r,\theta )=0}\end{array}$$

*uniformly with respect to* *θ* *on all compact subintervals of* ] − *a*, *a* [.

#### Proof

For *x* ∈ ℝ and *y* ∈ ]−*a*, *a*[, we define
$$\begin{array}{}F(x+iy):={e}^{c(x+iy)}f({e}^{x},y).\end{array}$$

Then (a) of Definition 3.4 implies that *F* is analytic in the strip *S*_{a} := {*z* ∈ ℂ : |ℑ *z*|< *a*} and (c) shows that for every *ε* ∈]0, *a* [ it is bounded on *S*_{a−ε}. Furthermore, by (b) we have *F*(*x*) → 0 as *x* → ± ∞. Then, by a theorem of Montel (see [9, Theorem 1.4.9]), it follows that *F*(*x* + *iy*) → 0 as *x* → ± ∞ uniformly with respect to *y* on compact subintervals of ]−*a*, *a*[. Expressing this conclusion in terms of *f*, we arrive at the desired result.□

Now we want to compare ${H}_{c}^{\ast}$(ℍ_{a}) with ${H}_{c}^{1}$(ℍ_{a}).

#### Proposition 3.3

*We have*
$$\begin{array}{}{H}_{c}^{1}({\mathbb{H}}_{a})\u2acb{H}_{c}^{\ast}({\mathbb{H}}_{a})\phantom{\rule{1em}{0ex}}\mathit{\text{\hspace{0.17em}and\hspace{0.17em}}}\phantom{\rule{1em}{0ex}}\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}\le 2\parallel f{\parallel}_{{H}_{c}^{1}({\mathbb{H}}_{a})}.\end{array}$$(3.2)

#### Proof

Let *f* ∈ ${H}_{c}^{1}$(ℍ_{a}). First we have to show that *f* satisfies conditions (a)–(*d*) of Definition 3.4. Clearly, (a) is guaranteed by the definition of ${H}_{c}^{1}$(ℍ_{a}). Furthermore, (b) and (c) follow from Propositon 2 and Proposition 4 in [8], respectively. As regards (*d*), we note that *I*_{c}(*f*, *θ*, *t*) exists by the hypotheses of ${H}_{c}^{1}$(ℍ_{a}) and
$$\begin{array}{}{\displaystyle |{I}_{c}(f,\theta ,t)|\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}\underset{R\to +\mathrm{\infty}}{lim}{\int}_{1/R}^{R}|f(r,\theta )|{r}^{c-1}dr=\parallel f(\cdot ,\theta ){\parallel}_{{X}_{c}^{1}}\le 2\parallel f{\parallel}_{{H}_{c}^{1}({\mathbb{H}}_{a})}.}\end{array}$$

This not only verifies (*d*) but proves (3.2) as well.

The inclusion in (3.2) is strict since it can be shown that
$$\begin{array}{}{\displaystyle f(r,\theta ):=\frac{{r}^{-c}{e}^{ic\theta}}{a+i(\mathrm{log}r+i\theta )}}\end{array}$$

belongs to ${H}_{c}^{\ast}$(ℍ_{a}) but *f* ∉ ${H}_{c}^{1}$(ℍ_{a}).□

We can also compare ${H}_{c}^{\ast}$(ℍ_{a}) with a corresponding Hardy type space for *L*^{2}-norms. The following proposition implies that ${H}_{c}^{\ast}$(ℍ_{a}) is a subspace of ${H}_{c}^{2}$(ℍ_{a−ε}) for every *ε* ∈ ]0, *a*[.

#### Proposition 3.4

*Let f* ∈ ${H}_{c}^{\ast}$(ℍ_{a}). *Then*, *for all* *α* ∈ ]−*a*, *a*[, *we have*
$$\begin{array}{}|{M}_{c}^{\ast}[f(\cdot ,\alpha )](c+it)|\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}{e}^{-|t|(a-|\alpha |)}\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})},\end{array}$$(3.3)

*f*(⋅, *α*) ∈ ${X}_{c}^{2}$ *and*
$$\begin{array}{}{\displaystyle \parallel f(\cdot ,\alpha ){\parallel}_{{X}_{c}^{2}}\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}\frac{1}{\sqrt{2\pi (a-|\alpha |)}}\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}.}\end{array}$$(3.4)

#### Proof

We proceed in a way similar to the proof of [8, Theorem 5]. For *f* ∈ ${H}_{c}^{\ast}$(ℍ_{a}) and *t* ∈ ℝ, we consider the function *g* defined by
$$\begin{array}{}{\displaystyle g(r,\theta ):={e}^{(c-1+it)(\mathrm{log}r+i\theta )}f(r,\theta ),}\end{array}$$

which is polar-analytic on ℍ_{a}. For *R* > 1 and *α* ∈ ]−*a*, *a*[\ {0}, let *γ* be the positively oriented rectangular curve with vertices at (1/*R*, 0), (*R*, 0), (1/*R*, *α*) and (*R*, *α*). By an integral theorem for polar-analytic functions (see [8, Proposition 1]), we have
$$\begin{array}{}{\displaystyle {\int}_{\gamma}g(r,\theta ){e}^{i\theta}(dr+ird\theta )=0.}\end{array}$$

This equation may be written in terms of ordinary integrals as
$$\begin{array}{}{\displaystyle {\int}_{1/R}^{R}g(r,0)dr={\int}_{1/R}^{R}g(r,\alpha ){e}^{i\alpha}dr+I(1/R,t)-I(R,t),}\end{array}$$(3.5)

where, for any *r* > 0,
$$\begin{array}{}{\displaystyle I(r,t)={\int}_{0}^{\alpha}g(r,\theta )ir{e}^{i\theta}d\theta =i{r}^{c+it}{\int}_{0}^{\alpha}{e}^{-(t-ic)\theta}f(r,\theta )d\theta .}\end{array}$$

Using statement (*d*) of Definition 3.4, we find that the first two integrals in (3.5) converge as *R* → +∞ while the last two expressions approach 0 as a consequence of Proposition 3.2. More precisely, in the notation of Definition 3.4, the limit *R* → +∞ in (3.5) leads to
$$\begin{array}{}{I}_{c}(f,0,t)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{e}^{i\alpha (c+it)}{I}_{c}(f,\alpha ,t)\end{array}$$

and so
$$\begin{array}{}|{I}_{c}(f,0,t)|\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{e}^{-\alpha t}|{I}_{c}(f,\alpha ,t)|.\end{array}$$(3.6)

Given *t* ≠ 0, we may choose in (3.6) the sign of *α* such that *α* *t* > 0. Noting that |*I*_{c}(*f*, *α*, *t*)| ≤ $\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}$ and letting |*α*| → *a*^{−}, we obtain
$$\begin{array}{}|{I}_{c}(f,0,t)|\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}{e}^{-a|t|}\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}.\end{array}$$

Combining this with (3.6), we arrive at
$$\begin{array}{}|{I}_{c}(f,\alpha ,t)|\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}{e}^{-|t|(a-|\alpha |)}\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}.\end{array}$$(3.7)

Since *I*_{c}(*f*, *α*, *t*) = ${M}_{c}^{\ast}$[*f*(⋅, *α*)](*c* + *it*), we have proved (3.3).

Using Lebesgue’s theorem of dominated convergence, we conclude from (3.7) that *I*_{c}(*f*, *α*, ⋅) ∈ *L*^{2}(ℝ) and
$$\begin{array}{}{\displaystyle \parallel {I}_{c}(f,\alpha ,\cdot ){\parallel}_{{L}^{2}(\mathbb{R})}\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}\frac{1}{\sqrt{a-|\alpha |}}\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}.}\end{array}$$

Now Lemma 3.3 implies that *f*(⋅, *α*) ∈ ${X}_{c}^{2}$ for |*α*| < *a*. Finally, the same lemma and a Parseval type theorem for Mellin transform (see [13, Lemma 2.6]) lead us to inequality (3.4).□

Now we are ready for the announced generalization of the Paley–Wiener theorem for the Mellin transform.

#### Theorem 3.2

*A continuous function ϕ* : ℝ^{+} → ℂ *is the restriction to* ℝ^{+} *of a function f* ∈ ${H}_{c}^{\ast}$(ℍ_{a}) *if and only if* ${M}_{c}^{\ast}$[*ϕ*] *exists and*
$$\begin{array}{}{\displaystyle |{M}_{c}^{\ast}[\varphi ](c+it)|\le C{e}^{-a|t|}\phantom{\rule{2em}{0ex}}(t\in \mathbb{R}),}\end{array}$$(3.8)

*with a constant C that may be taken to be* $\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}$.

#### Proof

If *ϕ* is the restriction to ℝ^{+} of a function *f* ∈ ${H}_{c}^{\ast}$(ℍ_{a}), then Proposition 3.4 applies and (3.3) for *α* = 0 shows that (3.8) holds with *C* = $\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}$.

Conversely, suppose that ${M}_{c}^{\ast}$[*ϕ*] exists and (3.8) holds with *C* = $\parallel f{\parallel}_{{H}_{c}^{\ast}({\mathbb{H}}_{a})}$. Then, as a function of *t*, the expression ${M}_{c}^{\ast}$[*ϕ*](*c* + *it*)*e*^{iαt} belongs to *L*^{1}(ℝ)∩ *L*^{2}(ℝ) whenever |*α*| < *a*. Therefore,
$$\begin{array}{}{\displaystyle f(r,\theta ):=\frac{1}{2\pi}{\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}{M}_{c}^{\ast}[\varphi ](c+it)(r{e}^{i\theta}{)}^{-c-it}dt}\end{array}$$

exists for (*r*, *θ*) ∈ ℍ_{a}. By Lemma 3.3 and the Mellin inversion theorem in ${X}_{c}^{2}$ (see [13, Theorem 2.9]), we obtain that *f*(*r*, 0) = *ϕ* (*r*) almost everywhere on ℝ^{+}. Since ${M}_{c}^{\ast}$[*ϕ*](*c* + *i* ⋅) ∈ *L*^{1}(ℝ), we conclude that *f*(⋅, 0) is continuous. Hence *f*(*r*, 0) = *ϕ*(*r*) for every *r* ∈ ℝ^{+}.

Now we want to show that *f* satisfies (a)–(*d*) of Definition 3.4. By a refinement of the proof of [13, Theorem 4], one can verify that *f* is polar-analytic in ℍ_{a}. For the reader’s convenience we give details. Let (*r*_{0}, *θ*_{0}) be any point in ℍ_{a}. Denote by *Q* a sufficiently small closed rectangular domain centered at (*r*_{0}, *θ*_{0}) such that *Q* ⊂ ℍ_{a}. We consider the difference quotient
$$\begin{array}{}{\displaystyle \frac{f(r,\theta )-f({r}_{0},{\theta}_{0})}{r{e}^{i\theta}-{r}_{0}{e}^{i{\theta}_{0}}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\frac{1}{2\pi}{\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}{M}_{c}^{\ast}[\varphi ](c+it)\frac{(r{e}^{i\theta}{)}^{-c-it}-({r}_{0}{e}^{i{\theta}_{0}}{)}^{-c-it}}{r{e}^{i\theta}-{r}_{0}{e}^{i{\theta}_{0}}}dt.}\end{array}$$(3.9)

For (*r*, *θ*) ∈ *Q*, the limit (*r*, *θ*) → (*r*_{0}, *θ*_{0}) carried out inside the integral leads to an ordinary differentiation of *z*^{−c−it} with respect to *z* at the point *z*_{0} := *r*_{0} *e*^{iθ0} and the integral with the derivative in place of the difference quotient exists. However, we have to justify that the limit and the integration can be interchanged. For this purpose, we first note that the difference quotient on the right-hand side of (3.9) can be expressed as
$$\begin{array}{}{\displaystyle \frac{(r{e}^{i\theta}{)}^{-c-it}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}({r}_{0}{e}^{i{\theta}_{0}}{)}^{-c-it}}{r{e}^{i\theta}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}{r}_{0}{e}^{i{\theta}_{0}}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}-(c+it)\phantom{\rule{negativethinmathspace}{0ex}}{\int}_{0}^{1}\phantom{\rule{negativethinmathspace}{0ex}}({r}_{0}{e}^{i{\theta}_{0}}\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}s(r{e}^{i\theta}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}{r}_{0}{e}^{i{\theta}_{0}})\phantom{\rule{negativethinmathspace}{0ex}}{)}^{-c-1-it}ds,}\end{array}$$

which implies that
$$\begin{array}{}{\displaystyle \left|\frac{(r{e}^{i\theta}{)}^{-c-it}-({r}_{0}{e}^{i{\theta}_{0}}{)}^{-c-it}}{r{e}^{i\theta}-{r}_{0}{e}^{i{\theta}_{0}}}\right|}& \le & \sqrt{{c}^{2}+{t}^{2}}\underset{(r,\theta )\in Q}{sup}\left|(r{e}^{i\theta}{)}^{-c-1+it}\right|\\ & \le & {\displaystyle \frac{\sqrt{{c}^{2}+{t}^{2}}}{{r}_{1}^{c+1}}{e}^{{\theta}_{1}|t|}}\end{array}$$

for some *r*_{1} > 0 and *θ*_{1} ∈ ]0, *a*[. Thus, for (*r*, *θ*) ∈ *Q* the absolute value of the integrand in (3.9) is bounded by
$$\begin{array}{}{\displaystyle \frac{C\sqrt{{c}^{2}+{t}^{2}}}{{r}_{1}^{c+1}}{e}^{-(a-{\theta}_{1})|t|},}\end{array}$$

which is integrable with respect to *t* over ℝ. Hence the desired interchange is guaranteed by Lebesgue’s theorem of dominated convergence. This completes the proof of property (a).

Next we note that (b) is a consequence of the Riemann–Lebesgue lemma. Furthermore, using (3.8), we see that
$$\begin{array}{}|f(r,\theta )|& \le & {\displaystyle \frac{1}{2\pi}{\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}|{M}_{c}^{\ast}[\varphi ](c+it)|{r}^{-c}{e}^{\theta t}dt}\\ & \le & {\displaystyle \frac{C{r}^{-c}}{2\pi}{\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}{e}^{-|t|(a-|\theta |)}dt={r}^{-c}\frac{C}{\pi (a-|\theta |)}\phantom{\rule{thinmathspace}{0ex}},}\end{array}$$

which shows that (c) also holds.

It remains to verify (d). By the hypothesis *I*_{c}(*f*, 0, *t*) = ${M}_{c}^{\ast}$[*ϕ*](*c* + *it*) exists. Now performing a contour integration as in the proof of Proposition 3.4 and noting that Proposition 3.2 is applicable to *f*, we conclude that *I*_{c}(*f*, *α*, *t*) also exists for |*α*| < *a* and (3.5) holds. Combined with (3.8), this yields
$$\begin{array}{}|{I}_{c}(f,\alpha ,t)|\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}C{e}^{-|t|(a-|\alpha |)},\end{array}$$

which shows that (*d*) holds with *K*(*f*) = C.□

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