A Liouville theorem on asymptotically Calabi spaces

Abstract

In this paper, we will study harmonic functions on the complete and incomplete spaces with nonnegative Ricci curvature which exhibit inhomogeneous collapsing behaviors at infinity. The main result states that any nonconstant harmonic function on such spaces yields a definite exponential growth rate which depends explicitly on the geometric data at infinity.

Appendix A: Some formulae in special functions

For developing quantitative estimates in this paper, we need to use some formulae and facts about the modified Bessel functions and the confluent hypergeometric functions. Some formulae applied in our concrete setting are in fact not completely standard in the literature, which deserves some proof. For making the paper the self-contained and for readers’ convenience, we try to summarize those results with detailed and checkable proofs in this section. Our main reference is [4].

1.1 A.2 Modified Bessel functions

Let $\nu \in {\mathbb {R}}$, we consider the following modified Bessel equation

$$\begin{aligned} y^2\cdot \frac{d^2 \mathcal {B}(y)}{dy^2}+y\cdot \frac{d\mathcal {B}(y)}{dy}-(y^2+\nu ^2)\cdot \mathcal {B}(y)=0,\ y\ge 0. \end{aligned}$$

(A.1)

First, for any $\nu \in {\mathbb {R}}$, we define

$$\begin{aligned} I_{\nu }(y)\equiv \sum \limits _{k=0}^{\infty } \frac{1}{\Gamma (k+1)\Gamma (k+\nu +1)}\Big (\frac{y}{2}\Big )^{2k+\nu }. \end{aligned}$$

(A.2)

In the special case $\nu =-\ell$ with $\ell \in {\mathbb {Z}}_+$, then the above definition can be also explained as

$$\begin{aligned} I_{\nu }(y)= \sum \limits _{k=\ell }^{\infty }\frac{1}{\Gamma (k+1) \Gamma (k-\ell +1)}\Big (\frac{y}{2}\Big )^{2k-\ell }. \end{aligned}$$

(A.3)

Immediately, for any positive integer $\ell \in {\mathbb {Z}}_+$, we have

$$\begin{aligned} I_{-\ell }(z) = I_{\ell }(z). \end{aligned}$$

(A.4)

Next we define $K_{\nu }(z)$ as follows,

$$\begin{aligned} K_{\nu }(y)\equiv {\left\{ \begin{array}{ll}\frac{\pi }{2\sin (\nu \pi )}\cdot (I_{-\nu }(y)-I_{\nu }(y)), &{} \nu \not \in {\mathbb {Z}},\\ \lim \limits _{\begin{array}{c} \nu '\rightarrow \nu \\ \nu '\not \in {\mathbb {Z}} \end{array}}K_{\nu '}(y),&\nu \in {\mathbb {Z}}. \end{array}\right. } \end{aligned}$$

(A.5)

One can check that $I_{\nu }(y)$ and $K_{\nu }(y)$ are two linearly independent solutions to (A.1). In the literature, $I_{\nu }$ and $K_{\nu }$ are usually called modified Bessel functions.

In our context, mainly we are interested in the solutions $I_{\nu }$ and $K_{\nu }$ with an index $\nu =\frac{1}{n}$ and $n\ge 2$. The simples case is $n=2$ such that both $I_{\frac{1}{2}}(y)$ and $K_{\frac{1}{2}}(y)$ have explicit formulae:

$$\begin{aligned} I_{\frac{1}{2}}(y) =\sqrt{\frac{2}{\pi y}} \sinh (y),\ K_{\frac{1}{2}}(y) =\sqrt{\frac{\pi }{2 y}} e^{-y}. \end{aligned}$$

(A.6)

The main part of this subsection is to prove the following useful integral representations for $I_{\nu }$ and $K_{\nu }$.

Lemma A.1

Given $\nu \in {\mathbb {R}}$, then the following integral formulae hold for each $y>0$,

$$\begin{aligned} I_{\nu }(y)&=\frac{1}{\pi }\int _{0}^{\pi }e^{y\cos \theta } \cos (\nu \theta ) d\theta - \frac{\sin (\nu \pi )}{\pi }\int _0^{\infty }e^{-y\cosh t - \nu t} dt,\end{aligned}$$

(A.7)

$$\begin{aligned} K_{\nu }(y)&= \int _{0}^{\infty }e^{-y\cosh t}\cosh (\nu t)dt. \end{aligned}$$

(A.8)

Proof

First, we prove the integral formula for $I_{\nu }$. The idea of the proof was originally inspired by Hankel’s representation formula for the reciprocal gamma function. In fact, let $\mathcal {L}\subset {\mathbb {C}}$ be a contour winding around the negative Ox-axis. In our particular case, $\mathcal {L}=\mathcal {L}_1 + \mathcal {L}_2+\mathcal {L}_3$, where $\mathcal {L}_1$ and $\mathcal {L}_3$ are two rays parallel to Ox and $\mathcal {L}_2$ is an arc of the unit circle centered at the origin (See Fig. 2). So Hankel’s representation formula gives that

$$\begin{aligned} \frac{1}{\Gamma (k+\nu +1)}=\frac{1}{2\pi \sqrt{-1}}\int _{\mathcal {L}}e^{w} w^{-(k+\nu +1)}dw,\ w\in {\mathbb {C}}. \end{aligned}$$

(A.9)

By the power series definition of $I_{\nu }$,

$$\begin{aligned} I_{\nu }(y)= & {} \sum \limits _{k=0}^{\infty }\frac{1}{\Gamma (k+1) \Gamma (k+\nu +1)}\Big (\frac{y}{2}\Big )^{2k+\nu }\nonumber \\= & {} (\frac{y}{2})^{\nu }\frac{1}{2\pi \sqrt{-1}}\int _{\mathcal {L}}e^w w^{-\nu -1}\sum \limits _{k=0}^{\infty }\frac{(\frac{y^2}{4w})^{k}}{k!}dw\nonumber \\= & {} (\frac{y}{2})^{\nu }\frac{1}{2\pi \sqrt{-1}}\int _{\mathcal {L}} e^{w+\frac{y^2}{4w}} w^{-\nu -1}dw. \end{aligned}$$

(A.10)

For every $y>0$, we make change of variables for each $w\in {\mathbb {C}}$,

$$\begin{aligned} w=\frac{y\cdot e^{\zeta }}{2}=\frac{ye^t}{2} \cdot e^{\sqrt{-1}\theta },\ 0< t<\infty ,\ 0\le \theta \le 2\pi . \end{aligned}$$

(A.11)

Letting $\mathcal {L}_1$ and $\mathcal {L}_3$ tend to each other, then in terms of the variables $(t,\theta )$,

$$\begin{aligned} \int _{\mathcal {L}} e^{w+\frac{y^2}{4w}} w^{-\nu -1}dw =\frac{1}{\pi }\int _{0}^{\pi }e^{y\cos \theta } \cos (\nu \theta ) d\theta - \frac{\sin (\nu \pi )}{\pi }\int _0^{\infty }e^{-y\cosh t - \nu t} dt. \end{aligned}$$

(A.12)

The integral formula for $K_{\nu }$ follows easily from the above integral representation for $I_{\nu }$ and the definition

$$\begin{aligned} K_{\nu }(y)=\frac{\pi (I_{-\nu }(y)-I_{\nu }(y))}{2\sin (\nu \pi )}. \end{aligned}$$

(A.13)

Fig. 2

The contour $\mathcal {L}=\mathcal {L}_1+\mathcal {L}_2 + \mathcal {L}_3$ for the integral (A.9)

1.2 A. 2 The confluent hypergeometric functions

Now we summarize some results regarding the confluent hypergeometric functions which are used in this paper. Given $\alpha ,\beta \in {\mathbb {R}}$ such that $\alpha >\beta$ and $\alpha$ is not a negative integer, we consider the following confluent hypergeometric equation

$$\begin{aligned} y\cdot \frac{d^2 \mathcal {J}(y)}{dy^2}+(\alpha -y)\cdot \frac{d\mathcal {J}(y)}{dy}-\beta \cdot \mathcal {J}(y)=0. \end{aligned}$$

(A.14)

Let

$$\begin{aligned} \Phi ^{\sharp }(\beta ,\alpha ,y)\equiv \sum \limits _{k=0}^{\infty } \frac{(\beta )_k}{(\alpha )_k}\cdot \frac{y^k}{k!}, \end{aligned}$$

(A.15)

where we define the notation $(x)_k\equiv \prod \limits _{m=1}^{k}(x+m-1)$ and $(x)_0=1$. So the power series $\Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},z)$ is always well-defined for all $\beta \in {\mathbb {C}}$, $z\in {\mathbb {C}}$ and ${{\,\mathrm{\alpha }\,}}\in {\mathbb {C}}\setminus \{0,-1,-2,\ldots \}$. Moreover, for any fixed $z\in {\mathbb {C}}$, the function $\Phi ^{\sharp }$ is entire in $\beta$ and meromorphic in ${{\,\mathrm{\alpha }\,}}$ with simple poles at negative integers.

It is by straightforward calculations that the function $\Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)$ is a solution to (A.14). In the literature, $\Phi ^{\sharp }$ is called Kummer’s (confluent hypergeometric) function. Moreover, when $y>0$, one can directly check that the function $\widehat{\Phi ^{\sharp }}(\beta ,{{\,\mathrm{\alpha }\,}},y)\equiv y^{1-{{\,\mathrm{\alpha }\,}}}\cdot \Phi ^{\sharp }(1+\beta -{{\,\mathrm{\alpha }\,}},2-{{\,\mathrm{\alpha }\,}}, y)$, which is linearly independent of $\Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)$, also solves (A.14). Therefore, the general solution of (A.14) for $y>0$ is

$$\begin{aligned} \mathcal {J}(y) = C\cdot \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y) + C^* \cdot y^{1-{{\,\mathrm{\alpha }\,}}}\cdot \Phi ^{\sharp }(1+\beta -{{\,\mathrm{\alpha }\,}},2-{{\,\mathrm{\alpha }\,}}, y). \end{aligned}$$

(A.16)

The power series definition of $\Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)$ immediately gives the following integral representation formula which is well known in the literature. We include a short proof just for the convenience of the readers.

Lemma A.2

For any ${{\,\mathrm{\alpha }\,}}>\beta >0$, then for each $y\in {\mathbb {R}}$,

$$\begin{aligned} \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y) = \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma (\beta )\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )} \int _0^1 e^{yt} t^{\beta -1}(1-t)^{{{\,\mathrm{\alpha }\,}}-\beta -1}dt. \end{aligned}$$

(A.17)

Proof

Given $p,q>0$, let B(p, q) be the beta function which is defined by

$$\begin{aligned} B(p,q) \equiv \int _0^{1}t^{p-1}(1-t)^{q-1}dt. \end{aligned}$$

(A.18)

Then the beta function satisfies $B(p,q)=\frac{\Gamma (p)\Gamma (q)}{\Gamma (p+q)}$. The above formulae imply that

$$\begin{aligned} \frac{(\beta )_k}{({{\,\mathrm{\alpha }\,}})_k}= & {} \frac{\Gamma (\beta +k)}{\Gamma (\beta )}\cdot \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma ({{\,\mathrm{\alpha }\,}}+k)} \nonumber \\= & {} \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma (\beta )}\cdot \frac{B(\beta +k,{{\,\mathrm{\alpha }\,}}-\beta )}{\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )}\nonumber \\= & {} \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma (\beta )\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )} \int _0^1t^{\beta +k-1}(1-t)^{{{\,\mathrm{\alpha }\,}}-\beta -1}dt. \end{aligned}$$

(A.19)

Now we return to the definition of $\Phi ^{\sharp }$, combining the above summation,

$$\begin{aligned} \Phi ^{\sharp }(\beta ,\alpha ,y)= & {} \sum \limits _{k=0}^{\infty }\frac{(\beta )_k}{(\alpha )_k} \cdot \frac{y^k}{k!}\nonumber \\= & {} \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma (\beta )\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )} \int _0^1t^{\beta -1}(1-t)^{{{\,\mathrm{\alpha }\,}}-\beta -1}\sum \limits _{k=0}^{\infty } \frac{(yt)^{k-1}}{k!}dt\nonumber \\= & {} \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma (\beta )\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )} \int _0^1e^{yt}t^{\beta -1}(1-t)^{{{\,\mathrm{\alpha }\,}}-\beta -1}dt. \end{aligned}$$

(A.20)

The proof is done.

Given $\beta >0$ and $y>0$, we define the function

$$\begin{aligned} \mathcal {U}(\beta ,{{\,\mathrm{\alpha }\,}}, y) \equiv \frac{1}{\Gamma (\beta )}\int _0^{\infty }e^{-yt}t^{\beta -1}(1+t)^{{{\,\mathrm{\alpha }\,}}-\beta -1}dt. \end{aligned}$$

(A.21)

Quick computations show that for each $\beta >0$, the function $\mathcal {U}(\beta ,{{\,\mathrm{\alpha }\,}}, y)$ is a solution to the confluent hypergeometric equation (A.14) on the positive real axis ${\mathbb {R}}_+$. Now let $\beta >0$ and ${{\,\mathrm{\alpha }\,}}\in {\mathbb {R}}\setminus \{0,-1,-2,-3,\ldots \}$, thanks to (A.16), the function $\mathcal {U}(\beta ,{{\,\mathrm{\alpha }\,}}, y)$ can be written in terms of Kummer’s function $\Phi ^{\sharp }$. Evaluating those functions and their derivatives at $y=0$, one can easily obtain

$$\begin{aligned} \mathcal {U}(\beta ,{{\,\mathrm{\alpha }\,}},y) = \frac{\Gamma (1-{{\,\mathrm{\alpha }\,}})}{\Gamma (1+\beta -{{\,\mathrm{\alpha }\,}})}\cdot \Phi ^{\sharp }(\beta , {{\,\mathrm{\alpha }\,}}, y) + \frac{\Gamma ({{\,\mathrm{\alpha }\,}}-1)}{\Gamma (\beta )}\cdot y^{1-{{\,\mathrm{\alpha }\,}}}\cdot \Phi ^{\sharp }(1+\beta -{{\,\mathrm{\alpha }\,}},2-{{\,\mathrm{\alpha }\,}}, y). \end{aligned}$$

(A.22)

Notice that, the above relation is well-defined for each $y\ge 0$ and non-integral $\alpha$. Moreover, if $\alpha \rightarrow n+1\in {\mathbb {Z}}_+$, then the right hand side of (A.22) will tend to a definite limit. The function $\mathcal {U}(\beta , {{\,\mathrm{\alpha }\,}},y )$ is usually called Tricomi’s (confluent hypergeometric) function. In our context, we are also interested in the case $y<0$. It can be directly verified that, if $y<0$, the function

$$\begin{aligned} \Psi ^{\flat }(\beta , {{\,\mathrm{\alpha }\,}}, y) \equiv e^{y} \cdot \mathcal {U}({{\,\mathrm{\alpha }\,}}-\beta , {{\,\mathrm{\alpha }\,}}, -y) \end{aligned}$$

(A.23)

solves equation (A.14). Moreover, it immediately follows from the integral representation of $\mathcal {U}$ that for any $y<0$,

$$\begin{aligned} \Psi ^{\flat }(\beta ,\alpha ,y)= \frac{e^y}{\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )}\int _0^{\infty }e^{yt} t^{{{\,\mathrm{\alpha }\,}}-\beta -1}(1+t)^{\beta -1}dt. \end{aligned}$$

(A.24)

In summary, if $y<0$, the equation (A.14) has two linearly independent solutions $\Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}}, y)$ and $\Psi ^{\flat }(\beta , {{\,\mathrm{\alpha }\,}}, y)$.

The asymptotic behavior of $\Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)$, $\mathcal {U}(\beta ,{{\,\mathrm{\alpha }\,}},y)$ and $\Psi ^{\flat }(\beta , {{\,\mathrm{\alpha }\,}}, y)$ can be easily seen from the above integral formulae. In fact, we have the following

Lemma A.3

The following asymptotics hold:

• (1) Let ${{\,\mathrm{\alpha }\,}}\in {\mathbb {R}}\setminus \{0,-1,-2,-3,\ldots \}$ and $\beta >0$ satisfy ${{\,\mathrm{\alpha }\,}}>\beta +1$, then

  $$\begin{aligned} \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y) \sim {\left\{ \begin{array}{ll} \frac{\Gamma (\alpha )}{\Gamma (\alpha -\beta )}\cdot (-y)^{-\beta }, &{} y\rightarrow -\infty , \\ \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma (\beta )}\cdot e^y\cdot y^{\beta -{{\,\mathrm{\alpha }\,}}}, &{} y\rightarrow +\infty . \end{array}\right. } \end{aligned}$$

  (A.25)

• (2) Let $\beta >0$, then

  $$\begin{aligned} \mathcal {U}(\beta ,{{\,\mathrm{\alpha }\,}}, y) \sim y^{-\beta },\ y\rightarrow +\infty . \end{aligned}$$

  (A.26)

• (3) Let $\alpha >\beta$, then

  $$\begin{aligned} \Psi ^{\flat }(\beta ,{{\,\mathrm{\alpha }\,}}, y)\sim e^y\cdot (-y)^{\beta -{{\,\mathrm{\alpha }\,}}},\ y\rightarrow -\infty . \end{aligned}$$

  (A.27)

Proof

The proof is straightforward. For example, we only prove

$$\begin{aligned} \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y) \sim \frac{\Gamma (\alpha )}{\Gamma (\alpha -\beta )}\cdot (-y)^{-\beta } \end{aligned}$$

(A.28)

as $y\rightarrow -\infty$. The calculations of the remaining cases are the same. We make change of variables and let $u=-yt$, then

$$\begin{aligned} \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)&= \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma (\beta )\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )} \int _0^1 e^{yt} t^{\beta -1}(1-t)^{{{\,\mathrm{\alpha }\,}}-\beta -1}dt\nonumber \\&=\frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma (\beta )\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )}\cdot (-y)^{-\beta }\cdot \int _0^{-y}e^{-u}u^{\beta -1}\Big (1+\frac{u}{y}\Big )^{{{\,\mathrm{\alpha }\,}}-\beta -1}du. \end{aligned}$$

(A.29)

Since ${{\,\mathrm{\alpha }\,}}-\beta -1>0$ and $-1\le \frac{u}{y}\le 0$, it is obvious $(1+\frac{u}{y})^{{{\,\mathrm{\alpha }\,}}-\beta -1}\le 1$. Hence dominated convergence theorem implies

$$\begin{aligned} \lim \limits _{y\rightarrow -\infty }\int _0^{-y}e^{-u} u^{\beta -1}\Big (1+\frac{u}{y}\Big )^{{{\,\mathrm{\alpha }\,}}-\beta -1} du=\int _0^{\infty }e^{-u}u^{\beta -1}du = \Gamma (\beta ). \end{aligned}$$

(A.30)

Therefore, as $y\rightarrow -\infty$,

$$\begin{aligned} \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)\sim \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )}\cdot (-y)^{-\beta }. \end{aligned}$$

(A.31)

Next we introduce some recurrence formulae for Kummer’s function.

Lemma A.4

Let ${{\,\mathrm{\alpha }\,}}\in {\mathbb {R}}\setminus \{0,-1,-2,-3,\ldots \}$ and $\beta \in {\mathbb {R}}$, then for each $y\in {\mathbb {R}}$,

$$\begin{aligned} \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}}, y)&= \Phi ^{\sharp }(\beta +1,{{\,\mathrm{\alpha }\,}},y) - \frac{y}{{{\,\mathrm{\alpha }\,}}}\Phi ^{\sharp }(\beta +1,{{\,\mathrm{\alpha }\,}}+1,y), \end{aligned}$$

(A.32)

$$\begin{aligned} \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)&= \frac{{{\,\mathrm{\alpha }\,}}+y}{{{\,\mathrm{\alpha }\,}}}\cdot \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}}+1,y) - \frac{{{\,\mathrm{\alpha }\,}}-\beta +1}{{{\,\mathrm{\alpha }\,}}({{\,\mathrm{\alpha }\,}}+1)}\cdot y\cdot \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}}+1,y). \end{aligned}$$

(A.33)

Proof

The formula can be quickly verified by applying the power series definition of $\Phi ^{\sharp }$.

With the above recurrence formula, we can extend the domain of indices in Lemma A.3 for Kummer’s function.

Lemma A.5

For any ${{\,\mathrm{\alpha }\,}}\in {\mathbb {R}}\setminus \{0,-1,-2,-3,\ldots \}$ and $\beta \in {\mathbb {R}}$ such that ${{\,\mathrm{\alpha }\,}}>\beta$, then

$$\begin{aligned} \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y) \sim {\left\{ \begin{array}{ll} \frac{\Gamma (\alpha )}{\Gamma (\alpha -\beta )}\cdot (-y)^{-\beta }, &{} y\rightarrow -\infty , \\ \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma (\beta )}\cdot e^y\cdot y^{\beta -{{\,\mathrm{\alpha }\,}}}, &{} y\rightarrow +\infty . \end{array}\right. } \end{aligned}$$

(A.34)

Proof

We start with the initial step by assuming $\alpha -\beta > 1$ and $\beta >1$. Then Lemma A.3 in this case shows that the desired asymptotics hold in this case.

Applying the recurrence formula (A.33), we can extend the domain of indices to $\alpha -\beta >0$ and $\beta >1$. Then applying (A.32), one can obtain the desired asymptotics for all $\beta \in {\mathbb {R}}$. The proof is done.

Lemma A.6

(Kummer’s transformation law) Let ${{\,\mathrm{\alpha }\,}}\in {\mathbb {R}}\setminus \{0,-1,-2,-3,\ldots \}$ and $\beta \in {\mathbb {R}}$, then for any $y\in {\mathbb {R}}$,

$$\begin{aligned} \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)=e^y \cdot \Phi ^{\sharp }({{\,\mathrm{\alpha }\,}}-\beta ,{{\,\mathrm{\alpha }\,}},-y). \end{aligned}$$

(A.35)

Proof

First, we temporarily assume ${{\,\mathrm{\alpha }\,}}>\beta >0$. By Lemma A.2,

$$\begin{aligned} e^y\cdot \Phi ^{\sharp }({{\,\mathrm{\alpha }\,}}-\beta ,{{\,\mathrm{\alpha }\,}},-y)= & {} \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )\Gamma (\beta )} \int _0^1 e^{y(1-t)} t^{{{\,\mathrm{\alpha }\,}}-\beta -1}(1-t)^{\beta -1}dt\nonumber \\= & {} \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )\Gamma (\beta )} \int _0^1 e^{ys} (1-s)^{{{\,\mathrm{\alpha }\,}}-\beta -1}s^{\beta -1}ds\nonumber \\= & {} \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}}, y). \end{aligned}$$

(A.36)

Now we prove the general case. Since both $\frac{e^y\cdot \Phi ^{\sharp }({{\,\mathrm{\alpha }\,}}-\beta ,{{\,\mathrm{\alpha }\,}},-y)}{\Gamma ({{\,\mathrm{\alpha }\,}})}$ and $\frac{\Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)}{\Gamma ({{\,\mathrm{\alpha }\,}})}$ are entire functions in ${\mathbb {C}}$, so the standard analytic continuation theorem implies that $\Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)=e^y \cdot \Phi ^{\sharp }({{\,\mathrm{\alpha }\,}}-\beta ,{{\,\mathrm{\alpha }\,}},-y)$ holds for any arbitrary $\beta \in {\mathbb {R}}$ and $\alpha \in {\mathbb {R}}\setminus \{0,-1,-2,-3,\ldots \}$.

Next we give another integral representation for Kummer’s function $\Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)$ in the case $y\le 0$, which has a crucial role in proving the uniform estimates in Sect. 4.

Lemma A.7

Assume that ${{\,\mathrm{\alpha }\,}}>\beta$ and $y\le 0$, then it holds that

$$\begin{aligned} \Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y)=\frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )}\cdot e^{y}(-y)^{\frac{1-{{\,\mathrm{\alpha }\,}}}{2}}\cdot \int _0^{\infty }e^{-t}\cdot t^{\frac{{{\,\mathrm{\alpha }\,}}-1}{2}-\beta }\cdot I_{{{\,\mathrm{\alpha }\,}}-1}(2\sqrt{-yt})dt. \end{aligned}$$

(A.37)

Proof

By definition,

$$\begin{aligned} I_{{{\,\mathrm{\alpha }\,}}-1}(2\sqrt{-yt})=\sum \limits _{k=0}^{\infty } \frac{(-yt)^{k+\frac{{{\,\mathrm{\alpha }\,}}-1}{2}}}{k!\cdot \Gamma (k+{{\,\mathrm{\alpha }\,}})}. \end{aligned}$$

(A.38)

Integrating the above expansion, it follows that

$$\begin{aligned}&\frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )}\cdot \int _0^{\infty }e^{-t}\cdot t^{\frac{{{\,\mathrm{\alpha }\,}}-1}{2}-\beta }\cdot I_{{{\,\mathrm{\alpha }\,}}-1}(2\sqrt{-yt})dt\nonumber \\&\quad = \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )}\cdot (-y)^{\frac{\alpha -1}{2}} \cdot \sum \limits _{k=0}^{\infty } \frac{(-y)^k}{k!\cdot \Gamma (k+{{\,\mathrm{\alpha }\,}})}\cdot \int _0^{\infty }e^{-t}\cdot t^{{{\,\mathrm{\alpha }\,}}-\beta +k-1}dt\nonumber \\&\quad = \frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )}\cdot (-y)^{\frac{\alpha -1}{2}}\cdot \sum \limits _{k=0}^{\infty }\frac{(-y)^k\cdot \Gamma (\alpha -\beta +k )}{k!\cdot \Gamma (k+{{\,\mathrm{\alpha }\,}})}. \end{aligned}$$

(A.39)

By the recursive formula of the Gamma function, $\frac{\Gamma (\alpha -\beta +k)}{\Gamma (k+\alpha )} = \frac{(\alpha -\beta )_k \cdot \Gamma (\alpha -\beta )}{(\alpha )_k \cdot \Gamma (\alpha )}$, so it follows that

$$\begin{aligned}&\frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )}\cdot \int _0^{\infty }e^{-t}\cdot t^{\frac{{{\,\mathrm{\alpha }\,}}-1}{2}-\beta }\cdot I_{{{\,\mathrm{\alpha }\,}}-1}(2\sqrt{-yt})dt\nonumber \\&\quad = (-y)^{\frac{\alpha -1}{2}}\cdot \sum \limits _{k=0}^{\infty } \frac{(\alpha -\beta )_k (-y)^k}{(\alpha )_k \cdot k!}\nonumber \\&\quad =(-y)^{\frac{\alpha -1}{2}}\cdot \Phi ^{\sharp }({{\,\mathrm{\alpha }\,}}-\beta ,{{\,\mathrm{\alpha }\,}},-y). \end{aligned}$$

(A.40)

Therefore,

$$\begin{aligned}&\frac{\Gamma ({{\,\mathrm{\alpha }\,}})}{\Gamma ({{\,\mathrm{\alpha }\,}}-\beta )}\cdot e^{y}(-y)^{\frac{1-{{\,\mathrm{\alpha }\,}}}{2}}\cdot \int _0^{\infty }e^{-t}\cdot t^{\frac{{{\,\mathrm{\alpha }\,}}-1}{2}-\beta }\cdot I_{{{\,\mathrm{\alpha }\,}}-1}(2\sqrt{-yt})dt\nonumber \\&\quad = e^y \cdot \Phi ^{\sharp }({{\,\mathrm{\alpha }\,}}-\beta ,{{\,\mathrm{\alpha }\,}},-y)\nonumber \\&\quad =\Phi ^{\sharp }(\beta ,{{\,\mathrm{\alpha }\,}},y). \end{aligned}$$

(A.41)

The last equality follows from Kummer’s transformation law.

Lemma A.8

Let $\nu >0$, then for all $y>0$

$$\begin{aligned} I_{\nu }(y)&= \frac{(\frac{y}{2})^{\nu }e^{-y}}{\Gamma (\nu +1)} \Phi ^{\sharp }(\nu +\frac{1}{2},2\nu +1,2y), \end{aligned}$$

(A.42)

$$\begin{aligned} K_{\nu }(y)&= \sqrt{\pi }(2y)^{\nu }e^{-y}\mathcal {U}(\nu + \frac{1}{2},2\nu +1,2y). \end{aligned}$$

(A.43)

Proof

The relation (A.42) can be verified by the power series definition of $I_{\nu }$ and $\Phi ^{\sharp }(\nu +\frac{1}{2},2\nu +1,2y)$, so we just omit the computations.

To prove (A.43), first we assume $\nu$ is not an integer. Combining the definition

$$\begin{aligned} K_{\nu }(y) = \frac{\pi }{\sin (\nu \pi )}\cdot \frac{I_{-\nu }(y) - I_{\nu }(y)}{2} \end{aligned}$$

(A.44)

and the relation

$$\begin{aligned} \mathcal {U}(\nu +\frac{1}{2},2\nu +1,y) = \frac{\Gamma (-2\nu )}{\Gamma (\frac{1}{2}-\nu )} \cdot \Phi ^{\sharp }(\nu +\frac{1}{2},2\nu +1,y) + \frac{\Gamma (2\nu )}{\Gamma (\nu +\frac{1}{2})}\cdot y^{-2\nu }\cdot \Phi ^{\sharp }(\frac{1}{2}-\nu ,1-2\nu , y), \end{aligned}$$

(A.45)

which is given by (A.22). If $\nu$ is an integer, the relation (A.43) can be obtained by the limiting definition of $K_{\nu }$ and the continuity argument for $\nu$.

The following corollary shows the asymptotic behavior of $I_{\nu }(y)$ and $K_{\nu }(y)$ as $y\rightarrow +\infty$.

Corollary A.8.1

Let $\nu >0$, then we have

$$\begin{aligned} \lim \limits _{y\rightarrow +\infty } \frac{I_{\nu }(y)}{\frac{e^y}{\sqrt{2\pi y}}}=1 \end{aligned}$$

(A.46)

and

$$\begin{aligned} \lim \limits _{y\rightarrow +\infty } \frac{K_{\nu }(y)}{\sqrt{\frac{\pi }{2y}} \cdot e^{-y}}=1. \end{aligned}$$

(A.47)

Proof

The proof follows from Lemma A.3, Lemma A.5 and Lemma A.8.