Here we introduce a generalized Meijer-type integral. Then we show its properties in the space of Boehmians.
Definition 10
Let \([ \frac{ ( \phi _{n} ) }{ ( \delta_{n} ) } ] \in\boldsymbol{\alpha } ( \boldsymbol{\digamma}_{\mathrm{loc}},{\Omega},\ast ,\ast ) \). Then we define the generalized Meijer-type integral transform of \([ \frac{ ( \phi_{n} ) }{ ( \delta_{n} ) } ] \) as \(\mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} [ \frac{ ( \phi_{n} ) }{ ( \delta_{n} ) } ] = [ \frac{\mathbf{l}_{\alpha_{1},\alpha_{2}} ( \phi_{n} ) }{ ( \delta_{n} ) } ] \) in \(\boldsymbol{\alpha} ( \boldsymbol {\digamma}_{\mathrm{loc}}^{\mathrm{K}},{\Omega},\ast ,\circledast ) \).
Theorem 5 shows that Definition 10 is well defined.
We discuss now some properties of \(\mathbf{l}_{\alpha _{1},\alpha_{2}}^{ge}\).
Theorem 11
Let
\(\beta_{1},\beta_{2}\in\boldsymbol{ \alpha} ( \boldsymbol{\digamma}_{\mathrm{loc}},{\Omega },\ast ,\ast ) \). Then
\(\mathbf{l}_{\alpha_{1},\alpha _{2}}^{ge} ( \beta_{1}\ast\beta_{2} ) =\mathbf{l}_{\alpha_{1},\alpha _{2}}^{ge}\beta_{1}\circledast\beta_{2}\)
in
\(\boldsymbol{\alpha } ( \boldsymbol{\digamma}_{\mathrm{loc}},{\Omega}, \ast ,\circledast ) \).
Proof
Assume the hypothesis of the theorem is satisfied for some \(\beta_{1},\beta_{2}\in\boldsymbol{\alpha} ( \boldsymbol{\digamma}_{\mathrm{loc}},{\Omega},\ast,\ast ) \). Therefore there are \(( \phi_{n} ) , ( \kappa_{n} ) \in \boldsymbol{\digamma}_{\mathrm{loc}} ( {0,\infty} ) \) and \(( \varphi_{n} ) , ( \delta_{n} ) \in\Delta\) such that \(\beta_{1}= [ \frac{ ( \phi_{n} ) }{ ( \varphi _{n} ) } ] \) and \(\beta_{2}= [ \frac{ ( \kappa _{n} ) }{ ( \delta_{n} ) } ] \). Therefore,
$$ \mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} ( \beta_{1}\ast\beta _{2} ) =\mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} \biggl( \biggl[ \frac{ ( \phi_{n} ) \ast ( \kappa_{n} ) }{ ( \varphi_{n} ) \ast ( \delta_{n} ) } \biggr] \biggr) \text{ .} $$
Definition 10 gives
$$ \mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} ( \beta_{1}\ast\beta _{2} ) = \biggl[ \frac{\mathbf{l}_{\alpha_{1},\alpha_{2}} ( ( \phi_{n} ) \ast ( \kappa_{n} ) ) }{ ( \varphi_{n} ) \ast ( \delta_{n} ) } \biggr] . $$
By the aid of Theorem 3 we obtain
$$ \mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} ( \beta_{1}\ast\beta _{2} ) = \biggl[ \frac{ ( \mathbf{l}_{\alpha_{1},\alpha _{2}}\phi_{n} ) \circledast ( \kappa_{n} ) }{ ( \varphi _{n} ) \circledast ( \delta_{n} ) } \biggr] . $$
On separating quotients this yields
$$ \mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} ( \beta_{1}\ast\beta _{2} ) = \biggl[ \frac{ ( \mathbf{l}_{\alpha_{1},\alpha _{2}}\phi_{n} ) }{ ( \varphi_{n} ) } \biggr] \circledast \biggl[ \frac{ ( \kappa_{n} ) }{ ( \delta_{n} ) } \biggr]. $$
Hence, \(\mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} ( \beta _{1}\ast \beta_{2} ) =\mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} ( \beta _{1} ) \circledast\beta_{2}\).
This completes the proof of the theorem. □
Theorem 12
\(\mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge}\)
defines a linear mapping from
\(\boldsymbol{\alpha} ( \boldsymbol{\digamma}_{\mathrm{loc}},{\Omega},\ast,\ast ) \)
into
\(\boldsymbol{\alpha} ( \boldsymbol{\digamma}_{\mathrm {loc}},{\Omega},\ast,\circledast ) \).
The proof is straightforward. Details are therefore omitted.
Theorem 13
Let
\([ \frac{ ( \phi_{n} ) }{ ( \delta_{n} ) } ] \in\boldsymbol{\alpha} ( \boldsymbol{\digamma}_{\mathrm{loc}},{\Omega},\ast,\ast ) \)
and
\(\delta\in{\Omega} ( {0,\infty} ) \). Then we have
$$ \mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} \biggl( \biggl[ \frac{ ( \phi_{n} ) }{ ( \delta_{n} ) } \biggr] \ast\delta \biggr) = \biggl[ \frac{ ( \mathbf{l}_{\alpha_{1},\alpha_{2}}\phi _{n} ) }{ ( \delta_{n} ) } \biggr] \circledast\delta. $$
Proof
Let \([ \frac{ ( \phi _{n} ) }{ ( \delta_{n} ) } ] \in\boldsymbol{\alpha } ( \boldsymbol{\digamma}_{\mathrm{loc}},{\Omega},\ast ,\ast ) \) and \(\delta\in{\Omega} ( {0,\infty } ) \). By virtue of Definition 10 we write
$$ \mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} \biggl( \biggl[ \frac{ ( \phi_{n} ) }{ ( \delta_{n} ) } \biggr] \ast\delta \biggr) = \biggl[ \frac{\mathbf{l}_{\alpha_{1},\alpha_{2}} ( ( \phi _{n} ) \ast\delta ) }{ ( \delta_{n} ) } \biggr] . $$
Once again, Definition 10 and Theorem 3 give
$$ \mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} \biggl( \biggl[ \frac{ ( \phi_{n} ) }{ ( \delta_{n} ) } \biggr] \ast\delta \biggr) = \biggl[ \frac{ ( \mathbf{l}_{\alpha_{1},\alpha_{2}}\phi _{n} ) \circledast\delta}{ ( \delta_{n} ) } \biggr] = \biggl[ \frac{ ( \mathbf{l}_{\alpha_{1},\alpha_{2}}\phi_{n} ) }{ ( \delta_{n} ) } \biggr] \circledast\delta. $$
This completes the proof of the theorem. □
Theorem 14
The transform
\(\mathbf{l}_{\alpha _{1},\alpha_{2}}^{ge}\)
is consistent with
\(\mathbf{l}_{\alpha _{1},\alpha_{2}}^{ge}:\boldsymbol{\digamma}_{\mathrm{loc}} ( {0,\infty} ) \rightarrow\boldsymbol{\digamma}_{\mathrm {loc}} ( {0,\infty} ) \).
Proof
For every \(\phi\in\boldsymbol {\digamma}_{\mathrm{loc}} ( {0,\infty} ) \), let \(\beta\in \boldsymbol{\alpha} ( \boldsymbol{\digamma}_{\mathrm {loc}},{\Omega},\ast,\ast ) \) be the representative in the space \(\boldsymbol{\alpha} ( \boldsymbol{\digamma}_{\mathrm {loc}},{\Omega},\ast,\ast ) \), then \(\forall n\in\mathbb{N}\), \(( \delta_{n} ) \in\Delta\), \(\beta= [ \frac{\phi\ast ( \delta _{n} ) }{ ( \delta_{n} ) } ] \). For all \(n\in \mathbb{N}\) it is clear that \(( \delta_{n} ) \) is independent from the choice of the representative.
We have
$$ \mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} ( \beta ) =\mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge} \biggl( \biggl[ \frac{\phi \ast ( \delta_{n} ) }{ ( \delta_{n} ) } \biggr] \biggr) = \biggl[ \frac{\mathbf{l}_{\alpha_{1},\alpha_{2}} ( \phi\ast ( \delta_{n} ) ) }{ ( \delta_{n} ) } \biggr] = \biggl[ \frac{\mathbf{l}_{\alpha_{1},\alpha_{2}}\phi\ast ( \delta_{n} ) }{ ( \delta_{n} ) } \biggr], $$
which is the representative of \(\mathbf{l}_{\alpha_{1},\alpha _{2}}^{ge}\phi\) in the space \(\boldsymbol{\digamma}_{\mathrm {loc}} ( {0,\infty} ) \).
Hence the proof is completed. □
Theorem 15
The necessary and sufficient condition for
\([ \frac{ ( g_{n} ) }{ ( \psi_{n} ) } ] \in \boldsymbol{\alpha} ( \boldsymbol{\digamma}_{\mathrm {loc}},{\Omega},\ast,\circledast ) \)
to be in the range of
\(\mathbf {l}_{\alpha_{1},\alpha_{2}}^{ge}\)
is that
\(g_{n}\)
belongs to the range of
\(\mathbf{l}_{\alpha_{1},\alpha_{2}}\)
for every
\(n\in\mathbb{N}\).
Proof
Let \([ \frac{ ( g_{n} ) }{ ( \psi_{n} ) } ] \) be in the range of \(\mathbf{l}_{\alpha_{1},\alpha_{2}}^{ge}\). Then of course \(g_{n}\) belongs to the range of \(\mathbf{l}_{\alpha_{1},\alpha_{2}}\), \(\forall n\in\mathbb {N}\). To establish the converse, let \(g_{n}\) be in the range of \(\mathbf{l}_{\alpha_{1},\alpha_{2}}\), \(\forall n\in\mathbb{N}\). Then there is \(\phi _{n}\in\boldsymbol{\digamma}_{\mathrm{loc}} ( {0,\infty} ) \) such that \(\mathbf{l}_{\alpha_{1},\alpha_{2}}\phi _{n}=g_{n}\), \(n\in\mathbb{N}\).
Since \([ \frac{ ( g_{n} ) }{ ( \psi _{n} ) } ] \in\boldsymbol{\alpha} ( \boldsymbol{\digamma}_{\mathrm {loc}},{\Omega},\ast,\circledast ) \) we get \(g_{n}\circledast \psi _{m}=g_{m}\circledast\psi_{n}\), \(\forall m,n\in\mathbb{N}\). Therefore, Theorem 3 yields
$$ \mathbf{l}_{\alpha_{1},\alpha_{2}} ( \phi_{n}\ast\delta _{m} ) = \mathbf{l}_{\alpha_{1},\alpha_{2}} ( \phi _{m}\ast \delta_{n} ) , \quad \forall m,n\in\mathbb{N}, $$
where \(\phi_{n}\in\boldsymbol{\digamma}_{\mathrm{loc}} ( {0,\infty} ) \) and \(( \delta_{n} ) \in\Delta\), \(\forall n\in\mathbb{N}\). Thus \(\phi_{n}\ast\delta_{m}=\phi_{m}\ast\delta _{n}\), \(m, \forall n\in\mathbb{N}\). Hence,
$$ \biggl[ \frac{ ( \phi_{n} ) }{ ( \delta_{n} ) } \biggr] \in\boldsymbol{\alpha} ( \boldsymbol{ \digamma}_{\mathrm{loc}},{\Omega},\ast,\ast ) \quad \text{and}\quad \mathbf{l}_{\alpha _{1},\alpha_{2}}^{ge} \biggl( \biggl[ \frac{ ( \phi_{n} ) }{ ( \delta_{n} ) } \biggr] \biggr) = \biggl[ \frac{ ( g_{n} ) }{ ( \psi_{n} ) } \biggr] . $$
The theorem is therefore completely proved. □
Theorem 16
The mappings
\(\mathbf{l}_{\alpha _{1},\alpha_{2}}^{ge}\)
are continuous with respect to
δ
and Δ-convergence.
The proof of this theorem is given by various papers of many authors. A detailed proof is therefore omitted.