Investigations into the monotonicity properties of the generalized function \({}_{a}\mathtt{B}_{b,p,c}\) hinges on the following result of Biernacki and Krzyż [19].
Lemma 1
([19])
Suppose
\(f(x)=\sum_{k=0}^{\infty }a_{k} x^{k}\)
and
\(g(x)=\sum_{k=0}^{\infty }b_{k} x^{k}\), where
\(a_{k} \in \mathbb{R}\)
and
\(b_{k} > 0\)
for all k. Further suppose that both series converge on
\(\vert x \vert < r\). If the sequence
\(\{a_{k}/b_{k}\}_{k\geq 0}\)
is increasing (or decreasing), then the function
\(x \mapsto f(x)/g(x)\)
is also increasing (or decreasing) on
\((0,r)\).
Evidently, the above lemma also holds true when both f and g are even functions, or both odd.
Theorem 2
Let
\(c \leq 0\).
-
(a)
If
\(q \geq p > -(b+1)/2\)
and
\(a \leq d\), then
\(x \mapsto ( 2^{p}x^{-p}{}_{a}\mathtt{B}_{b,p,c}(x) ) / ( 2^{q}x^{-q} {}_{d}\mathtt{B}_{b,q,c}(x) ) \)
is increasing on
\((0, \infty )\).
-
(b)
The function
\(p\mapsto {}_{a}\mathtt{B}_{b,p+a,c}(x)/{} _{a}\mathtt{B}_{b,p,c}(x)\)
is decreasing on
\((-(b+1)/2, \infty )\)
for each fixed
\(x>0\).
-
(c)
The function
\(x\mapsto x{}_{a}\mathtt{B}_{b,p,c}'(x)/{} _{a}\mathtt{B}_{b,p,c}(x)\)
is increasing on
\((0, \infty )\)
for each fixed
\(p > -{(b+1)}/{2}\).
Proof
(a) From (1) it is evident that
$$\begin{aligned} \frac{ x^{q-p}{}_{a}\mathtt{B}_{b,p,c}(x)}{2^{q-p}{}_{b}B_{b,q,c}(x)}=\frac{ \sum_{k=0}^{\infty }\alpha_{k, p,a} ( \frac{x}{2} ) ^{2k}}{ \sum_{k=0}^{\infty }\alpha_{k, q,d} ( \frac{x}{2} ) ^{2k}}, \end{aligned}$$
where
$$ \alpha_{k, p,a}=\frac{(-c)^{k}}{k! \Gamma { ( ak+p+ \frac{b+1}{2} ) }} \quad \text{and} \quad \alpha_{k, q,d}=\frac{(-c)^{k}}{k! \Gamma { ( dk+q+\frac{b+1}{2} ) }}. $$
Write \(w_{k}= \alpha_{k,p,a}/\alpha_{k,q,d}\); since \(d \geq a\) and \(q \geq p\), it follows that
$$\begin{aligned} \frac{w_{k+1}}{w_{k}} &=\frac{ \Gamma { ( ak+p+ \frac{b+1}{2} ) } \Gamma { ( dk+d+q+\frac{b+1}{2} ) }}{ \Gamma { ( dk+q+\frac{b+1}{2} ) } \Gamma { ( ak+a+p+\frac{b+1}{2} ) }}=\frac{ ( d k +q+\frac{b+1}{2} ) _{d}}{ ( a k+p+\frac{b+1}{2} ) _{a}} \geq 1. \end{aligned}$$
The result now readily follows from Lemma 1.
(b) Let \(q \geq p>-(b+1)/2\). It follows from part (a) that
$$\begin{aligned} \frac{d}{dx} \biggl( \frac{2^{p} x^{-p}{}_{a}\mathtt{B}_{b,p,c}(x)}{ {2^{q} x^{-q}{}_{a}\mathtt{B}_{b,q,c}(x)}} \biggr) \geq 0 \end{aligned}$$
on \((0,\infty )\). Thus
$$\begin{aligned} \bigl( x^{-p}{}_{a}\mathtt{B}_{b,p,c}(x) \bigr) ' \bigl( x^{-q}{}_{a} \mathtt{B}_{b,q,c}(x) \bigr) - \bigl( x^{-p}{}_{a} \mathtt{B}_{b,p,c}(x) \bigr) \bigl( x^{-q}{}_{a} \mathtt{B}_{b,q,c}(x) \bigr) '\geq 0. \end{aligned}$$
It now follows from (5) that
$$\begin{aligned} (-c)\;x^{-p-q} \biggl( \frac{x}{2} \biggr) ^{1-a} \bigl( {}_{a}\mathtt{B} _{b,p+a,c}(x)\; {}_{a} \mathtt{B}_{b,q,c}(x)-{}_{a}\mathtt{B}_{b,p,c}(x) \;{}_{a}\mathtt{B}_{b,q+a,c}(x) \bigr) \geq 0, \end{aligned}$$
whence \({}_{a} B_{b,p+a,c}/{}_{a} B_{b,p,c}\) is decreasing for \(p>-(b+1)/2\).
(c) Let \(\beta_{k,p,a}:=(2k+p)\alpha_{k, p,a}\). Then the quotient \(x {}_{a}\mathtt{B}_{b,p,c}'/{}_{a}\mathtt{B}_{b,p,c}\) can be written as
$$\begin{aligned} \frac{x {}_{a}B'_{b,p,c}(x)}{{}_{a}\mathtt{B}_{b,p,c}(x)}= \frac{ \sum_{k=0}^{\infty }{\beta_{k,p,a} ( \frac{x}{2} ) ^{2k}}}{ \sum_{k=0}^{\infty }{\alpha_{k,p,a} ( \frac{x}{2} ) ^{2k}}}. \end{aligned}$$
Clearly, the sequence \(\{\beta_{k,p,a}/\alpha_{k,p,a}\}_{k \geq 0}=\{2k+p \}_{k\geq 0}\) is increasing, and hence Lemma 1 shows that the function \(x \mapsto x {}_{a}\mathtt{B}_{b,p,c}'/{}_{a}\mathtt{B} _{b,p,c}\) is increasing on \(( 0, \infty ) \). □
Next consider the normalized function
$$\begin{aligned} {}_{a}\mathcal{B}_{b,p,c}(x)&=2^{p}x^{-p} \Gamma { \biggl( p+\frac{b+1}{2} \biggr) } {}_{a}\mathtt{B}_{b, p,c}(x) \\ &= \Gamma { \biggl( p+\frac{b+1}{2} \biggr) }\sum_{k=0}^{\infty }\frac{(-c)^{k}}{k! \Gamma { ( ak+p+\frac{b+1}{2} ) }} \biggl( \frac{x}{2} \biggr) ^{2k}. \end{aligned}$$
(14)
Also let \({}_{1}\Phi_{1}\) be the confluent hypergeometric function
$$ {}_{1}\Phi_{1}(\alpha ; \beta ; x)= \sum _{k=0}^{\infty }\frac{ ( \alpha )_{k} x^{k}}{(\beta )_{k} k!}. $$
The next result discusses the monotonicity property of rational functions involving \({}_{a}\mathcal{B}_{b,p,c}\).
Theorem 3
Let
\(c \leq 0\).
-
(a)
If
\(\alpha \geq \beta >0\), then the function
\(x \mapsto {}_{a}\mathcal{B}_{b,p,c}(x)/{}_{1}\Phi_{1}(\alpha ; \beta ; -c x^{2}/4)\)
is decreasing on
\(\mathbb{R}\)
for each fixed
\(p > -{(b+1)}/ {2}\).
-
(b)
If
\(0< \beta \leq (2p+b+1)/(2a)\), then the function
\(x \mapsto {}_{a}\mathcal{B}_{b,p,c}(x)/F_{a}(\beta ; -c x^{2}/4)\)
is decreasing on
\(\mathbb{R}\)
for each fixed
\(p > -{(b+1)}/{2}\), where
$$ F_{a}(\beta , x):= {}_{0}F_{a} \biggl( -; \beta , \beta +\frac{1}{a}, \ldots , \beta +\frac{a-1}{a}; x \biggr) . $$
Proof
(a) It follows from (14) that
$$ \frac{{}_{a}\mathcal{B}_{b, p, c}(x)}{{}_{1}\Phi_{1} ( \alpha ; \beta ; -c\frac{x^{2}}{4} ) }=\frac{\sum_{k=0}^{\infty }\delta _{1}(k) ( \frac{x}{2} ) ^{2k}}{\sum_{k=0}^{\infty }\delta _{2}(k) ( \frac{x}{2} ) ^{2k}} $$
with
$$ \delta_{1}(k)=\frac{(-c)^{k}\Gamma ( p+\frac{b+1}{2} ) }{k! \Gamma ( ak+p+\frac{b+1}{2} ) } \quad \text{and} \quad \delta _{2}(k)=\frac{(-c)^{k} (\alpha )_{k}}{k! (\beta )_{k}}. $$
Set \(w_{k}:= \delta_{1}(k)/\delta_{2}(k)\). Since \(\alpha \geq \beta \),
$$ \frac{w_{k+1}}{w_{k}}= \frac{(\beta +k)\Gamma ( ak+p+ \frac{b+1}{2} ) }{(\alpha +k)\Gamma ( ak+a+p+\frac{b+1}{2} ) } \leq 1. $$
Thus \(\{w_{k}\}_{k}\) is decreasing, and the result follows from Lemma 1.
(b) From Proposition 1, a representation of \({}_{a} \mathcal{B}_{b, p, c}\) by the generalized hypergeometric function is
$$ {}_{a}\mathcal{B}_{b, p, c}(x)=\sum _{k=0}^{\infty }\sigma_{1}(k) \biggl( \frac{x}{2} \biggr) ^{2k}, $$
where
$$ \sigma_{1}(k):=\frac{(-c)^{k} }{a^{ak}k! \prod_{j=1}^{a} ( \frac{2p+b+2j-1}{2a} ) _{k}}. $$
Let
$$ \sigma_{2}(k):=\frac{(-c)^{k} }{k! \prod_{j=1}^{a} ( \beta + \frac{j-1}{a} ) _{k}}. $$
Then
$$ \frac{{}_{a}\mathcal{B}_{b, p, c}(x)}{F_{a} ( \beta ,-c\frac{x ^{2}}{4} ) } = \frac{\sum_{k=0}^{\infty }\sigma_{1}(k) ( \frac{x}{2} ) ^{2k}}{\sum_{k=0}^{\infty }\sigma_{2}(k) ( \frac{x}{2} ) ^{2k}}. $$
With \(\tau_{k}:= \sigma_{1}(k)/\sigma_{2}(k)\), a computation shows that
$$ \frac{\tau_{k+1}}{\tau_{k}}= \frac{1}{a^{a}} \prod_{j=1}^{a} \frac{ ( \frac{2p+b+2j-1}{2a} ) _{k}}{ ( \frac{2p+b+2j-1}{2a} ) _{k+1}} \frac{ ( \beta +\frac{j-1}{a} ) _{k+1}}{ ( \beta + \frac{j-1}{a} ) _{k}} =\frac{1}{a^{a}} \prod _{j=1}^{a} \frac{ \beta +\frac{j-1}{a}+k}{\frac{2p+b+2j-1}{2a}+k}. $$
Now
$$ \frac{2p+b+2j-1}{2a} \geq \beta +\frac{j-1}{a} $$
for each fixed j, \(1 \leq j \leq a\), provided \(0 < \beta \leq (2p+b+1)/(2a)\). Hence \(\tau_{k+1} \leq \tau_{k}\), and the result follows from Lemma 1. □
Another function of interest is that given by
$$ {}_{a}\mathcal{B}^{d}_{b,p, c}(x):= \sum_{k=0}^{\infty }\frac{(-c/4)^{k}\Gamma { ( p+\frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \Gamma { ( ak+p+ \frac{b+1}{2} ) }} \frac{(d)_{k}}{k!}x^{k} =\sum_{k=0}^{\infty } \frac{(-c/4)^{k}(d)_{k}}{\Gamma{ ( k+1 ) } ( p+\frac{b+1}{2} ) _{ak} k!}x ^{k}. $$
(15)
Note that \({}_{a}\mathcal{B}^{1}_{b,p,c}(x)={}_{a}\mathcal{B}_{b,p,c}( \sqrt{x})\), where \({}_{a}\mathcal{B}_{b,p,c}\) is given by (14). The following result by Karp and Sitnik [20, Theorem 1] is required to deduce the log-concavity of \({}_{a}\mathcal{B}^{d}_{b,p, c}\) in terms of the parameter d.
Lemma 2
([20])
Let
$$ f(d,x)=\sum_{k=0}^{\infty }f_{k} \frac{(d)_{k}}{k!} x^{k}, $$
where
\(f_{k}>0\) (and is independent of
d). Suppose
\(e>d>0\), \(\delta >0\). Then the function
$$ \phi_{d, e, \delta }(x)=f(d+\delta ,x)f(e,x)-f(e+\delta ,x)f(d,x)= \sum _{m=2}^{\infty }\phi_{m} x^{m} $$
has positive power series coefficients
\(\phi_{m} >0\)
so that
\(d \mapsto f(d,x)\)
is strictly log-concave for
\(x>0\)
whenever the sequence
\(\{ f_{k}/f_{k-1} \} \)
is decreasing. On the other hand, \(\phi_{d, e, \delta }(x)\)
has negative power series coefficients
\(\phi_{m} <0\)
so that
\(d \mapsto f(d,x)\)
is strictly log-convex for
\(x>0\)
whenever the sequence
\(\{ f_{k}/f_{k-1} \} \)
is increasing.
Theorem 4
Let
\(c \leq 0\)
and
\(d>0\).
-
(a)
The function
$$ p \mapsto {}_{a}\mathcal{B}^{d}_{b,p, c}(x) $$
given by (15) is decreasing and log-convex on
\((-(b+1)/2, \infty )\)
for each fixed
\(x>0\)
and
\(d>0\).
-
(b)
The function
$$ p \mapsto {}_{a}\mathcal{B}_{b,p+1, c}^{d}(x)/{}_{a} \mathcal{B}_{b,p, c}^{d}(x) $$
is increasing on
\((-(b+1)/2, \infty )\), that is, for
\(q \geq p >-(b+1)/2\), the inequality
$$ {}_{a}\mathcal{B}_{b, q+1, c}^{d}(x){}_{a} \mathcal{B}_{b,p, c}^{d}(x) \geq {}_{a} \mathcal{B}_{b, q, c}^{d}(x){}_{a}\mathcal{B}_{b,p+1, c} ^{d}(x) $$
(16)
holds for each fixed
\(x>0\)
and
\(d>0\).
-
(c)
The function
\(d \mapsto {}_{a}\mathcal{B}^{d}_{b,p, c}(x)\)
is log-concave on
\((0, \infty )\)
for each fixed
\(x>0\)
and
\(p >(2a-b-1)/2\).
Proof
(a) Let \(q\geq p > -(b+1)/2\). Then \((q+(b+1)/2)_{ak} > (p+(b+1)/2)_{ak}\) for all \(k \in \{0, 1, 2, \ldots \}\). Thus
$$\begin{aligned} \gamma_{k}(q, d):=\frac{(-c)^{k}(d)_{k}}{4^{k} \Gamma { ( k+1 ) } ( q+\frac{b+1}{2} ) _{ak}k!} \leq \frac{(-c)^{k} (d)_{k}}{4^{k}\Gamma { ( k+1 ) } ( p+\frac{b+1}{2} ) _{ak}k!}:= \gamma_{k}(p, d). \end{aligned}$$
Since
$$ {}_{a}\mathcal{B}_{b,q, c}^{d}(x)=\sum _{k=0}^{\infty } \gamma_{k}(q, d) x^{k}, $$
we deduce that
$$\begin{aligned} {}_{a}\mathcal{B}_{b,q, c}^{d}(x) \leq {}_{a}\mathcal{B}_{b,p, c}^{d}(x) \end{aligned}$$
for each fixed \(x >0\) and \(d>0\). Therefore \(p \mapsto {}_{a} \mathcal{B}_{b,p, c}^{d}\) is decreasing for \(p > -(b+1)/2\).
To show log-convexity of \({}_{a}\mathcal{B}_{b,p, c}^{d}\), it suffices to show that \(p \mapsto \gamma_{k}(p, d)\) is log-convex for all \(k \in \{0,1, 2, 3, \ldots \}\) and fixed \(d>0\). Then the result follows from the fact that sums of log-convex functions are also log-convex.
Let Ψ be the digamma function given by \(\Psi (p)= \Gamma '{(p)}/ \Gamma {(p)}\). Then evidently
$$\begin{aligned} \frac{\partial^{2}}{\partial {p^{2}}} \bigl(\log \bigl(\gamma_{k}( p,d) \bigr) \bigr)= \Psi ' \biggl( p+\frac{b+1}{2} \biggr) -\Psi ' \biggl( ak+p+\frac{b+1}{2} \biggr) . \end{aligned}$$
Note that [16, p. 260] \(\Psi '\) has the explicit form
$$\begin{aligned} \Psi '(t)=\sum_{n=0}^{\infty } \frac{1}{(t+n)^{2}}, \quad t \in \mathbb{R}\setminus \{0,-1,-2,\ldots \}. \end{aligned}$$
This implies that
$$\begin{aligned} \frac{\partial^{2}}{\partial {p^{2}}} \bigl(\log \bigl(\gamma_{k}(p, d) \bigr) \bigr)= \sum_{n=0}^{\infty }\frac{ak(ak+2p+(b+1)+2n)}{ ( p+\frac{b+1}{2}+n ) ^{2} ( ak+p+\frac{b+1}{2}+n ) ^{2}} \geq 0 \end{aligned}$$
for all \(k \in \{0, 1, 2, \ldots \}\) and \(p > -(b+1)/2\). Thus \(p \mapsto \gamma_{k}( p, d)\) is log-convex on \((-(b+1)/2, \infty )\), and consequently \({}_{a}\mathcal{B}_{b,p, c}^{d}\) is log-convex for each fixed \(x>0\).
(b) It is clear that (16) is equivalent to showing
$$\begin{aligned} \Biggl( \sum_{k=0}^{\infty } \gamma_{k}(q+1, d)x^{k} \Biggr) \Biggl( \sum _{k=0}^{\infty }\gamma_{k}(p, d)x^{k} \Biggr) \geq \Biggl( \sum_{k=0} ^{\infty } \gamma_{k}(q, d)x^{k} \Biggr) \Biggl( \sum _{k=0}^{\infty } \gamma_{k}(p+1, d)x^{k} \Biggr) , \end{aligned}$$
which holds whenever
$$\begin{aligned} \gamma_{i}(q+1, d)\gamma_{j}(p, d)+ \gamma_{j}(q+1, d) \gamma_{i}(p, d) \geq \gamma_{j}(q, d)\gamma_{i}(p+1, d)+\gamma_{i}(q, d) \gamma_{j}(p+1, d) \end{aligned}$$
(17)
for all \(i, j \in \mathbb{N}\).
Let
$$ \lambda_{1}= \Gamma { \biggl( ai+q+\frac{b+3}{2} \biggr) } \Gamma { \biggl( aj+p+\frac{b+3}{2} \biggr) } $$
and
$$ \lambda_{2}= \Gamma { \biggl( aj+q+\frac{b+3}{2}\biggr) } \Gamma { \biggl( ai+p+\frac{b+3}{2} \biggr) }. $$
Then
$$\begin{aligned} &\gamma_{i}(q+1, d)\gamma_{j}(p, d)+\gamma_{j}(q+1, d) \gamma_{i}(p, d) \\ &\quad = \frac{(-c/4)^{i+j} \Gamma ( p+\frac{b+1}{2} ) \mathrm{ \Gamma } ( q+\frac{b+3}{2} ) (d)_{i} (d)_{j}}{ \Gamma ( i+1 ) \Gamma ( j+1 ) i! j!} \biggl[ \frac{aj+p+ \frac{b+1}{2}}{\lambda_{1}} +\frac{ai+q+\frac{b+1}{2}}{\lambda_{2}} \biggr] . \end{aligned}$$
(18)
Similarly,
$$\begin{aligned} &\gamma_{j}(q, d)\gamma_{i}(p+1, d)+\gamma_{i}(q, d) \gamma_{j}(p+1, d) \\ &\quad = \frac{ ( -c/4 ) ^{i+j} \Gamma ( q+ \frac{b+1}{2} ) \Gamma ( p+\frac{b+3}{2} ) (d)_{i} (d)_{j}}{ \Gamma ( i+1 ) \Gamma ( j+1 ) i! j!} \biggl[ \frac{ai+q+\frac{b+1}{2}}{ \lambda_{1}} +\frac{aj+p+\frac{b+1}{2}}{\lambda_{2}} \biggr] . \end{aligned}$$
(19)
Next, for \(i \geq j\), relations (18) and (19) show that inequality (17) is equivalent to
$$\begin{aligned} & \biggl( q+\frac{b+1}{2} \biggr) \biggl[ \biggl( aj+p+\frac{b+1}{2} \biggr) \lambda_{2} + \biggl( ai+q+\frac{b+1}{2} \biggr) \lambda_{1} \biggr] \\ &\quad \geq \biggl( p+\frac{b+1}{2} \biggr) \biggl[ \biggl( aj+p+ \frac{b+1}{2} \biggr) \lambda_{1}+ \biggl( ai+q+\frac{b+1}{2} \biggr) \lambda_{2} \biggr] . \end{aligned}$$
Since \(q\geq p\), this can be further simplified to showing
$$\begin{aligned} a ( \lambda_{1}-\lambda_{2} ) (i-j) \biggl( p+\frac{b+1}{2} \biggr) \geq 0. \end{aligned}$$
The latter inequality clearly holds true whenever \(\lambda_{1} \geq \lambda_{2}\).
To see that this is indeed the case, for \(q \geq p\), let
$$\begin{aligned} \phi (x):= \frac{ \Gamma { ( ax+ q +\frac{b+3}{2} ) }}{\mathrm{ \Gamma }{ ( ax+ p +\frac{b+3}{2} ) }}. \end{aligned}$$
Since \(x \mapsto \Gamma {(ax+y)}\) is log-convex, it follows that \(\phi '(x) \geq 0\). Thus \(\phi (i) \geq \phi (j)\) for \(i \geq j\), and consequently \(\lambda_{1} \geq \lambda_{2}\). This validates inequality (16).
(c) To apply Lemma 2, let
$$ f_{k}:=\frac{(-c/4)^{k} {\Gamma ( p+\frac{b+1}{2} ) }}{ \Gamma (k+1) \Gamma ( ak+p+\frac{b+1}{2} ) }. $$
We shall show that the sequence \(b_{k} = \{ f_{k}/f_{k-1} \} \) is decreasing. A calculation gives
$$ b_{k}= -\frac{c}{4} \frac{\Gamma ( ak+p+\frac{b+1}{2}-a ) }{k \; \Gamma ( ak+p+\frac{b+1}{2} ) }, $$
and so we need to show that the function \(\xi :(0,\infty )\to \mathbb{R}\) given by
$$ \xi (x)= \frac{\Gamma ( ax+p+\frac{b+1}{2}-a ) }{x\; \Gamma ( ax+p+\frac{b+1}{2} ) } $$
is decreasing for \(p>(2a-b-1)/2\). Logarithmic differentiation gives
$$\begin{aligned} \frac{\xi '(x)}{\xi (x)}= a \Psi \biggl( ax+p+\frac{b+1}{2}-a \biggr) - a \Psi \biggl( ax+p+\frac{b+1}{2} \biggr) -\frac{1}{x}. \end{aligned}$$
Since the digamma function is known to be increasing on \((0, \infty )\) for \(p>(2a-b-1)/2\) and \(x>0\), it follows that
$$\begin{aligned} \frac{\xi '(x)}{\xi (x)} < 0. \end{aligned}$$
Thus ξ is indeed decreasing, and Lemma 2 shows that the function
$$ d\mapsto {}_{a}\mathcal{B}_{b,p, c}^{d}(x) =\sum _{k=0}^{\infty } f _{k} \frac{(d)_{k}}{k!} x^{k} $$
is log-concave. □
The results of parts (a) and (b) in Theorem 4 in the case \(d=1\) were also obtained by Baricz [1, Theorem 3, Theorem 4].
Remark 5
Theorem 4 has interesting consequences, among which is the Tur\(\acute{\text{a}}\)n-type inequality for the function \({}_{a}\mathcal{B}_{b,p, c}^{d}\) given by (15). From the definition of log-convexity, it follows from Theorem 4(a) that
$$\begin{aligned} {}_{a}\mathcal{B}_{b, \alpha p_{1}+(1-\alpha )p_{2}, c}^{d}(x) \leq \bigl( {}_{a}\mathcal{B}_{b,p_{1}, c}^{d}(x) \bigr) ^{\alpha } \bigl( {} _{a}\mathcal{B}_{b,p_{2}, c}^{d}(x) \bigr) ^{1-\alpha }, \end{aligned}$$
where \(\alpha \in [0,1]\), \(p_{1}, p_{2} > -(b+1)/2\), and \(x >0\). Choosing \(\alpha =1/2\), \(p_{1}=p-\nu \) and \(p_{2}=p+\nu \), the above inequality yields
$$\begin{aligned} \bigl({}_{a}\mathcal{B}_{b,p,c}^{d}(x) \bigr)^{2} - {}_{a}\mathcal{B} _{b,p+\nu , c}^{d}(x) {}_{a}\mathcal{B}_{b,p-\nu }^{d}(x)\leq 0. \end{aligned}$$
(20)
On the other hand, the log-concavity of \(d \mapsto {}_{a}\mathcal{B} _{b,p, c}^{d}\) implies that
$$\begin{aligned} {}_{a}\mathcal{B}_{b, p, c}^{td_{1}+(1-t)d_{2}}(x) \geq \bigl( {}_{a} \mathcal{B}_{b,p, c}^{d_{1}}(x) \bigr) ^{t} \bigl( {}_{a}\mathcal{B} _{b,p, c}^{d_{2}}(x) \bigr) ^{1-t} \end{aligned}$$
for \(t \in [0, 1]\), \(d_{2} > d_{1} >0\), \(p> (2a-b-1)/2\) and \(x>0\). Choosing \(t=1/2\), \(d_{1}=d-\mu \), and \(d_{2}=d+\mu \), \(\mu \in \mathbb{R}\), the inequality reduces to
$$\begin{aligned} \bigl( {}_{a}\mathcal{B}_{b, p, c}^{d}(x) \bigr) ^{2} \geq {}_{a} \mathcal{B}_{b,p, c}^{d+\mu }(x) {}_{a}\mathcal{B}_{b,p, c}^{d-\mu }(x). \end{aligned}$$
(21)
Thus (20) and (21) lead to direct and reverse Tur\(\acute{\text{a}}\)n-type inequalities
$$\begin{aligned} {}_{a}\mathcal{B}_{b,p, c}^{d+\mu }(x) {}_{a} \mathcal{B}_{b,p, c}^{d- \mu }(x)\leq \bigl( {}_{a} \mathcal{B}_{b, p, c}^{d}(x) \bigr) ^{2} \leq {}_{a}\mathcal{B}_{b,p+\nu , c}^{d}(x) {}_{a} \mathcal{B}_{b,p- \nu }^{d}(x). \end{aligned}$$
Remark 6
For \(d=2\), it follows from (15) that
$$\begin{aligned} {}_{a}\mathcal{B}^{2}_{b,p, c}(x) &=\sum_{k=0}^{\infty }\frac{(-c/4)^{k}\mathrm{ \Gamma }{ ( p+\frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \Gamma { ( ak+p+ \frac{b+1}{2} ) }} \frac{(2)_{k}}{k!}x^{k} \\ &=\sum_{k=0}^{\infty }\frac{(-c/4)^{k} \Gamma { ( p+ \frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \mathrm{ \Gamma }{ ( ak+p+\frac{b+1}{2} ) }} \frac{(k+1)!}{k!}x^{k} \\ &=x \sum_{k=1}^{\infty } \frac{(-c/4)^{k} \Gamma { ( p+ \frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \mathrm{ \Gamma }{ ( ak+p+\frac{b+1}{2} ) }} kx^{k-1} \\ &\quad {}+ \sum_{k=0}^{ \infty } \frac{(-c/4)^{k} \Gamma { ( p+\frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \Gamma { ( ak+p+\frac{b+1}{2} ) }} x^{k} \\ &= x \frac{d}{dx} \bigl( {}_{a}\mathcal{B}_{b,p, c} ( \sqrt{x} ) \bigr) + {}_{a}\mathcal{B}_{b,p, c} ( \sqrt{x} ) , \end{aligned}$$
(22)
where \({}_{a}\mathcal{B}_{b,p, c}\) is given by (14). With \(d=1\) and \(\mu =1\) in (21), then \({}_{a}\mathcal{B} _{b,p, c}^{1}(x)={}_{a}\mathcal{B}_{b,p, c} ( \sqrt{x} ) \). Thus (22) shows that
$$\begin{aligned} \bigl( {}_{a}\mathcal{B}_{b,p, c} ( \sqrt{x} ) \bigr) ^{2} \geq x \frac{d}{dx} \bigl( {}_{a} \mathcal{B}_{b,p, c} ( \sqrt{x} ) \bigr) + {}_{a} \mathcal{B}_{b,p, c} ( \sqrt{x} ) . \end{aligned}$$
Remark 7
Inequality (16) leads to a generalization of the Turán-type inequality
$$ \bigl( {}_{a}\mathcal{B}_{b, p+1,c}^{d}(x) \bigr) ^{2}\leq {}_{a} \mathcal{B}_{b, p,c}^{d}(x) \;{}_{a}\mathcal{B}_{b, p+2,c}^{d}(x). $$
Inequality (16) also yields
$$\begin{aligned} \frac{{}_{a}\mathcal{B}_{b, q,c}^{d}(x)}{{}_{a}\mathcal{B}_{b,p,c} ^{d}(x)}\leq \frac{{}_{a}\mathcal{B}_{b, q+1,c}^{d}(x)}{{}_{a} \mathcal{B}_{b,p+1,c}^{d}(x)} \leq \frac{{}_{a}\mathcal{B}_{b, q+2,c} ^{d}(x)}{{}_{a}\mathcal{B}_{b,p+2,c}^{d}(x)}\leq \cdots \leq \frac{ {}_{a}\mathcal{B}_{b, q+a,c}^{d}(x)}{{}_{a}\mathcal{B}_{b,p+a,c}^{d}(x)}. \end{aligned}$$
Thus
$$ \frac{{}_{a}\mathcal{B}_{b, q,c}^{d}(x)}{{}_{a}\mathcal{B}_{b, p,c} ^{d}(x)}\leq \frac{{}_{a}\mathcal{B}_{b, q+a,c}^{d}(x)}{{}_{a} \mathcal{B}_{b, p+a,c}^{d}(x)}. $$
The next result gives a dominant function for \({}_{a}\mathtt{B}_{b,p, -\alpha^{2}}\).
Theorem 5
Let
\(p\geq -(b+1)/2\)
and
\(x \geq 0\). Then
$$ {}_{a}\mathtt{B}_{b,p, -\alpha^{2}}(x) \leq \frac{1}{\Gamma ( p+ \frac{b+1}{2} ) } \biggl( \frac{x}{2} \biggr) ^{p}\exp \biggl( \frac{ \alpha^{2} x^{2}}{4 ( p+\frac{b+1}{2} ) ^{a}} \biggr) . $$
Proof
Clearly the estimate trivially holds for \(x=0\). Let \(x>0\). It is readily established by mathematical induction that \(\Gamma (m+x) \geq x^{m} \Gamma (x)\) for \(m \in \mathbb{N}\) and \(x\geq 0\). Then
$$ \Gamma \biggl( a k + p+\frac{b+1}{2} \biggr) \geq \biggl( p+ \frac{b+1}{2} \biggr) ^{a k} \Gamma \biggl( p+\frac{b+1}{2} \biggr) , $$
and thus
$$\begin{aligned} {}_{a}\mathtt{B}_{b,p, -\alpha^{2}}(x) &\leq \frac{1}{\Gamma ( p+ \frac{b+1}{2} ) }\sum _{k=0}^{\infty }\frac{\alpha^{2k}}{ ( p+ \frac{b+1}{2} ) ^{ak} k!} \biggl( \frac{x}{2} \biggr) ^{2k+p} \\ &=\frac{1}{\Gamma ( p+\frac{b+1}{2} ) } \biggl( \frac{x}{2} \biggr) ^{p}\exp \biggl( \frac{\alpha^{2} x^{2}}{4 ( p+\frac{b+1}{2} ) } \biggr) ^{a} . \end{aligned}$$
□
For \(\alpha =\pm 1\), \(b=a=1\), Theorem 5 leads to a dominant for the modified Bessel function
$$ I_{p}(x) \leq \frac{1}{\Gamma ( p+1 ) } \biggl( \frac{x}{2} \biggr) ^{p}e^{\frac{ x^{2}}{4 ( p+1 ) }} $$
obtained by Baricz in [21].
The final result uses the Chebyshev integral inequality [22, p. 40]: Suppose f and g are two integrable functions monotonic in the same sense (either both decreasing or both increasing). Let \(q: (a, b) \to \mathbb{R}\) be a positive integrable function. Then
$$\begin{aligned} \biggl( \int_{a}^{b} q(t) f(t)\,dt \biggr) \biggl( \int_{a}^{b} q(t) g(t)\,dt \biggr) \leq \biggl( \int_{a}^{b} q(t)\,dt \biggr) \biggl( \int_{a}^{b} q(t) f(t) g(t)\,dt \biggr) . \end{aligned}$$
(23)
The inequality in (23) is reversed if f and g are monotonic of the opposite sense.
Theorem 6
Let
\(p > -(b-1)/2\), \(\alpha \in \mathbb{R}\backslash \{0\}\), and
\(x \in (0, \pi /\vert \alpha \vert )\). Then
$$\begin{aligned}& {}_{2}\mathtt{B}_{b, p, \alpha^{2}}(x) \geq \frac{ \pi \alpha^{1-p-\frac{b}{2}}x^{1-\frac{b}{2}} ( \Gamma ( \frac{2p+b}{4} ) ) ^{2}}{2^{\frac{5-b}{2}} \Gamma (\frac{2p+b-1}{4})\Gamma ( \frac{2p+b+1}{4})}\; \biggl( \mathtt{J}_{\frac{2p+b-2}{4}} \biggl( \frac{ \alpha x}{2} \biggr) \biggr) ^{2}, \\& {}_{2}\mathtt{B}_{b, p, -\alpha^{2}}(x) \geq \frac{ \pi \alpha^{1-p-\frac{b}{2}}x^{1-\frac{b}{2}} ( \Gamma ( \frac{2p+b}{4} ) ) ^{2}}{2^{\frac{5-b}{2}} \Gamma (\frac{2p+b-1}{4})\Gamma ( \frac{2p+b+1}{4})}\; \biggl( \mathtt{I}_{\frac{2p+b-2}{4}} \biggl( \frac{ \alpha x}{2} \biggr) \biggr) ^{2}. \end{aligned}$$
Proof
Putting \(a=2\) in Remark 4, the integral form for \({}_{2}\mathtt{B}_{b, p, \alpha^{2}}(x)\) is
$$\begin{aligned} {}_{2}\mathtt{B}_{b, p, \alpha^{2}}(x) & = \frac{x^{p}}{2^{2p+ \frac{b-3}{2}} { \Gamma { ( \frac{2p+b-1}{4} ) }} { \Gamma { ( \frac{2p+b+1}{4} ) }}} \biggl( \int_{0} ^{1} \bigl(1-t^{2} \bigr)^{\frac{2p+b-5}{4}} \cos \biggl( \frac{\alpha xt}{2} \biggr)\,dt \biggr) \\ & \quad {} \times \biggl( \int_{0}^{1} \bigl(1-t^{2} \bigr)^{\frac{2p+b-3}{4}} \cos \biggl( \frac{ \alpha xt}{2} \biggr)\,dt \biggr) . \end{aligned}$$
To establish the subordinant for \({}_{2}\mathtt{B}_{b, p, \alpha^{2}}\), let
$$ q(t) = \bigl(1-t^{2} \bigr)^{\frac{2p+b-3}{4}} \cos \biggl( \frac{\alpha x t}{2} \biggr) ; \quad \quad f(t)= g(t) := \bigl(1-t^{2} \bigr)^{-\frac{1}{4}}, \quad 0 < t < 1. $$
Then
$$\begin{aligned} \int_{0}^{1} q(t)f(t)\,dt= \int_{0}^{1} q(t)g(t)\,dt &= \int_{0}^{1} \bigl(1-t ^{2} \bigr)^{\frac{2p+b-4}{4}} \cos \biggl( \frac{\alpha x t}{2} \biggr)\,dt. \end{aligned}$$
It is known that for \(\operatorname{Re} \nu >-1/2\), the classical Bessel function \(\mathtt{J}_{\nu }\) has the integral representation
$$\begin{aligned} \mathtt{J}_{\nu }(y)= \frac{2^{1-\nu }y^{\nu }}{\sqrt{\pi } \mathrm{ \Gamma }{ ( \nu +\frac{1}{2} ) }} \int_{0}^{1} \bigl(1-t^{2} \bigr)^{ \nu -\frac{1}{2}} \cos ( y t )\,dt. \end{aligned}$$
Replacing y by \((\alpha x)/2\) and ν by \((2p+b-2)/4\), we obtain
$$\begin{aligned} \biggl( \int_{0}^{1}q(t) f(t)\,dt \biggr) \biggl( \int_{0}^{1}q(t) g(t)\,dt \biggr) = 2^{2p+b-4} \pi (\alpha x)^{1-p-\frac{b}{2}} \biggl( \Gamma \biggl( \frac{2p+b}{4} \biggr) \biggr) ^{2} \mathtt{J}^{2}_{ \frac{2p+b-2}{4}} \biggl( \frac{\alpha x}{2} \biggr) . \end{aligned}$$
Since f and g both are increasing on \((0, 1)\), it is evident from (23) that
$$\begin{aligned} {}_{2}\mathtt{B}_{b, p, \alpha^{2}}(x) & =\frac{x^{p}}{2^{2p+ \frac{b-3}{2}} { \Gamma { ( \frac{2p+b-1}{4} ) }} { \Gamma { ( \frac{2p+b+1}{4} ) }}} \biggl( \int_{0} ^{1} \bigl(1-t^{2} \bigr)^{\frac{2p+b-5}{4}} \cos \biggl( \frac{\alpha xt}{2} \biggr)\,dt \biggr) \\ & \quad {} \times \biggl( \int_{0}^{1} \bigl(1-t^{2} \bigr)^{\frac{2p+b-3}{4}} \cos \biggl( \frac{ \alpha xt}{2} \biggr)\,dt \biggr) \\ &=\frac{x^{p}}{2^{2p+\frac{b-3}{2}} { \Gamma { ( \frac{2p+b-1}{4} ) }} { \Gamma { ( \frac{2p+b+1}{4} ) }}} \biggl( \int_{0} ^{1}p(t)f(t) g(t)\,dt \biggr) \biggl( \int_{0}^{1}p(t)\,dt \biggr) \\ &\geq \frac{ \pi \alpha^{1-p-\frac{b}{2}}x^{1-\frac{b}{2}} ( \Gamma ( \frac{2p+b}{4} ) ) ^{2}}{2^{\frac{5-b}{2}} \Gamma ( \frac{2p+b-1}{4})\Gamma (\frac{2p+b+1}{4})}\; \mathtt{J}^{2}_{ \frac{2p+b-2}{4}} \biggl( \frac{\alpha x}{2} \biggr) . \end{aligned}$$
The subordinant for \({}_{2}\mathtt{B}_{b, p, - \alpha^{2}}\) is readily established in a similar manner by choosing
$$ q(t) = \bigl(1-t^{2} \bigr)^{\frac{2p+b-3}{4}} \cosh \biggl( \frac{\alpha x t}{2} \biggr) ; \quad \quad f(t)= g(t) := \bigl(1-t^{2} \bigr)^{-\frac{1}{4}}, \quad 0 < t < 1. $$
□
Remark 8
As a final application, choose \(b=3\) and \(\alpha =1\) in Theorem 6. Then
$$ {}_{2}\mathtt{B}_{3, p, 1}(x) \geq \frac{ \pi x^{-\frac{1}{2}} ( \Gamma ( \frac{2p+3}{4} ) ) ^{2}}{2 \Gamma (\frac{2p+2}{4}) \Gamma (\frac{2p+4}{4})}\; \biggl( \mathtt{J}_{\frac{2p+1}{4}} \biggl( \frac{ x}{2} \biggr) \biggr) ^{2}. $$
Remark 1 now shows that
$$ J_{\frac{p}{2}} \biggl( \frac{x}{ 2} \biggr) J_{\frac{p+1}{2}} \biggl( \frac{x}{ 2} \biggr) \geq \frac{ \sqrt{\pi } ( \Gamma ( \frac{2p+3}{4} ) ) ^{2}}{2 \Gamma (\frac{2p+2}{4})\Gamma (\frac{2p+4}{4})}\; \biggl( \mathtt{J} _{\frac{2p+1}{4}} \biggl( \frac{ x}{2} \biggr) \biggr) ^{2}. $$
Similarly,
$$ {}_{2}\mathtt{B}_{3, p, -1}(x) \geq \frac{ \pi x^{-\frac{1}{2}} ( \Gamma ( \frac{2p+3}{4} ) ) ^{2}}{2 \Gamma ( \frac{2p+2}{4})\Gamma (\frac{2p+4}{4})}\; \biggl( \mathtt{I}_{ \frac{2p+1}{4}} \biggl( \frac{ x}{2} \biggr) \biggr) ^{2}, $$
and thus
$$ I_{\frac{p}{2}} \biggl( \frac{x}{ 2} \biggr) I_{\frac{p+1}{2}} \biggl( \frac{x}{ 2} \biggr) \geq \frac{ \sqrt{\pi } ( \Gamma ( \frac{2p+3}{4} ) ) ^{2}}{2 \Gamma (\frac{2p+2}{4})\Gamma (\frac{2p+4}{4})}\; \biggl( \mathtt{I} _{\frac{2p+1}{4}} \biggl( \frac{ x}{2} \biggr) \biggr) ^{2}. $$