# Laguerre polynomials and Perron-Frobenius operators

Jordan Bell
April 19, 2016

## 1 Laguerre polynomials

### 1.1 Definition and generating functions

Let $D=\frac{d}{dx}$. For $\alpha>-1$ and $n\geq 0$ let

 $L_{n}^{\alpha}(x)=e^{x}\frac{x^{-\alpha}}{n!}D^{n}(e^{-x}x^{n+\alpha}),$

called the Laguerre polynomials. Using the Leibniz rule for $D^{n}(f\cdot g)$ yields

 $L_{n}^{\alpha}(x)=\sum_{k=0}^{n}\frac{\Gamma(n+\alpha+1)}{\Gamma(k+\alpha+1)}% \frac{(-x)^{k}}{k!(n-k)!}.$

The generating function for the Laguerre polynomials is11 1 N. N. Lebedev, Special Functions and Their Applications, p. 77, §4.17.

 $w(x,z)=(1-z)^{-\alpha-1}e^{-xz/(1-z)}=\sum_{n=0}^{\infty}L_{n}^{\alpha}(x)z^{n% },\qquad|z|<1.$

Define

 $W(x,y,z)=(1-z)^{-1}e^{-(x+y)z/(1-z)}(xyz)^{-\alpha/2}I_{\alpha}\left(\frac{2% \sqrt{xyz}}{1-z}\right),\qquad|z|<1,$

where

 $I_{\alpha}(x)=i^{-\alpha}J_{\alpha}(ix)=\sum_{m=0}^{\infty}\frac{1}{m!\Gamma(m% +\alpha+1)}(x/2)^{2m+\alpha}$

and

 $J_{\alpha}(x)=\sum_{m=0}^{\infty}\frac{(-1)^{m}}{m!\Gamma(m+\alpha+1)}(x/2)^{2% m+\alpha}.$

$W$ satisfies

 $W(x,y,z)=\sum_{n=0}^{\infty}\frac{n!L_{n}^{\alpha}(x)L_{n}^{\alpha}(y)}{\Gamma% (n+\alpha+1)}z^{n}.$

### 1.2 Differential equations satisfied by Laguerre polynomials

$w$ satisfies the ordinary differential equation

 $(1-z^{2})\partial_{z}w+(x-(1-z)(1+\alpha))w=0.$

This yields, for $n\geq 1$,

 $(n+1)L_{n+1}^{\alpha}(x)+(x-\alpha-2n-1)L_{n}^{\alpha}(x)+(n+\alpha)L_{n-1}^{% \alpha}(x)=0.$ (1)

$w$ also satisfies the ordinary differential equation

 $(1-t)\partial_{x}w+tw=0,$

which yields, for $n\geq 1$,

 $DL_{n}^{\alpha}-DL_{n-1}^{\alpha}+L_{n-1}^{\alpha}=0.$ (2)

Using (1) and (2) gives

 $xDL_{n}^{\alpha}=nL_{n}^{\alpha}-(n+\alpha)L_{n-1}^{\alpha},\qquad n\geq 1.$ (3)

Using (3) and (2) we get, for $n\geq 0$,

 $xD^{2}L_{n}^{\alpha}(x)+(\alpha+1-x)DL_{n}^{\alpha}(x)+nL_{n}^{\alpha}(x)=0.$ (4)

### 1.3 Integral formulas for Laguerre polynomials

For $\nu>-1$, $a>0$, $b>0$, using the series for $J_{\nu}$ one calculates22 2 N. N. Lebedev, Special Functions and Their Applications, p. 132, §5.15, Example 2.

 $\int_{0}^{\infty}e^{-a^{2}x^{2}}J_{\nu}(bx)x^{\nu+1}dx=\frac{b^{\nu}}{(2a^{2})% ^{\nu+1}}e^{-\frac{b^{2}}{4a^{2}}}.$ (5)

Applying this with $\nu=n+\alpha$, $a=1$, $b=2\sqrt{x}$, $x=\sqrt{t}$ yields

 $\int_{0}^{\infty}e^{-t}J_{n+\alpha}(2\sqrt{xt})(\sqrt{t})^{n+\alpha+1}\cdot% \frac{1}{2\sqrt{t}}dt=\frac{(2\sqrt{x})^{n+\alpha}}{2^{n+\alpha+1}}e^{-x},$

i.e.

 $\int_{0}^{\infty}e^{-t}J_{n+\alpha}(2\sqrt{xt})(\sqrt{xt})^{n+\alpha}dt=e^{-x}% x^{n+\alpha}.$ (6)

Now, it is a fact that

 $\frac{d}{du}u^{\nu/2}J_{\nu}(2\sqrt{u})=u^{(\nu-1)/2}J_{\nu-1}(2\sqrt{u}),$

and using this and (6), we get that for $\alpha>1$ and $n\geq 0$,

 $L_{n}^{\alpha}(x)=\frac{e^{x}x^{-\alpha/2}}{n!}\int_{0}^{\infty}t^{n+\alpha/2}% J_{\alpha}(2\sqrt{xt})e^{-t}dt.$ (7)

We remind ourselves that for $\alpha>-1$ and $|z|<1$,

 $(1-z)^{-\alpha-1}e^{-yt/(1-t)}=\sum_{n=0}^{\infty}L_{n}^{\alpha}(y)z^{n}.$

For $|z|<\frac{1}{3}$, using this and $e^{-\frac{yt}{1-t}-\frac{y}{2}}=e^{-\frac{2yt+y-yt}{2(1-t)}}=e^{-\frac{y(1+t)}% {2(1-t)}}$ one checks that

 $\begin{split}&\displaystyle(1-z)^{-\alpha-1}\int_{0}^{\infty}e^{-\frac{y(1+t)}% {2(1-t)}}y^{\alpha/2}J_{\alpha}(\sqrt{xy})dy\\ \displaystyle=&\displaystyle\sum_{n=0}^{\infty}z^{n}\int_{0}^{\infty}e^{-y/2}y% ^{\alpha/2}J_{\alpha}(\sqrt{xy})L_{n}^{\alpha}(y)dy.\end{split}$

Then one gets, for $|z|<1$,

 $2e^{-x/2}x^{\alpha/2}\sum_{n=0}^{\infty}(-1)^{n}L_{n}^{\alpha}(x)z^{n}=\sum_{n% =0}^{\infty}z^{n}\int_{0}^{\infty}e^{-y/2}y^{\alpha/2}J_{\alpha}(\sqrt{xy})L_{% n}^{\alpha}(y)dy.$

Therefore for $\alpha>-1$ and $n\geq 0$,

 $e^{-x/2}x^{\alpha/2}L_{n}^{\alpha}(x)=\frac{(-1)^{n}}{2}\int_{0}^{\infty}J_{% \alpha}(\sqrt{xy})e^{-y/2}y^{\alpha/2}L_{n}^{\alpha}(y)dy.$ (8)

### 1.4 Orthogonality of Laguerre polynomials

Let

 $\rho_{\alpha}(x)=e^{-x}x^{\alpha}.$

Let

 $u_{n}=\rho_{\alpha}^{1/2}L_{n}^{\alpha},\qquad n\geq 0.$

$u_{n}$ satisfies the differential equation

 $(xu_{n}^{\prime})^{\prime}+\left(n+\frac{\alpha+1}{2}-\frac{x}{4}-\frac{\alpha% ^{2}}{4x}\right)u_{n}=0.$

Using this we get

 $x(u_{n}^{\prime}u_{m}-u_{m}^{\prime}u_{n})\bigg{|}_{0}^{\infty}+(n-m)\int_{0}^% {\infty}u_{m}u_{n}dx=0.$

Then

 $(n-m)\int_{0}^{\infty}u_{m}u_{n}dx=0.$ (9)

Using (1) yields for $n\geq 2$,

 $\begin{split}&\displaystyle n(L_{n}^{\alpha})^{2}-(n+\alpha)(L_{n-1}^{\alpha})% ^{2}-(n+1)L_{n+1}^{\alpha}L_{n-1}^{\alpha}+2L_{n}^{\alpha}L_{n-1}^{\alpha}+(n+% \alpha-1)L_{n}^{\alpha}L_{n-2}^{\alpha}=0.\end{split}$

Using this and (9), for $n\geq 2$,

 $n\int_{0}^{\infty}e^{-x}x^{\alpha}L_{n}^{\alpha}(x)^{2}dx=(n+\alpha)\int_{0}^{% \infty}e^{-x}x^{\alpha}L_{n-1}^{\alpha}(x)^{2}dx.$

Iterating this, for $n\geq 2$,

 $\displaystyle\int_{0}^{\infty}e^{-x}x^{\alpha}L_{n}^{\alpha}(x)^{2}dx$ $\displaystyle=\frac{(n+\alpha)(n+\alpha-1)\cdots(\alpha+2)}{n(n-2)\cdots 3% \cdot 2}\int_{0}^{\infty}e^{-x}x^{\alpha}L_{1}^{\alpha}(x)^{2}dx$ $\displaystyle=\frac{\Gamma(n+\alpha+1)}{n!}.$

### 1.5 Asymptotics for Laguerre polynomials

It can be proved that for $\alpha>-1$, with $N=n+\frac{\alpha+1}{2}$,33 3 N. N. Lebedev, Special Functions and Their Applications, p. 87, §4.22. for $x\in\mathbb{R}_{\geq 0}$,

 $L_{n}^{\alpha}(x)\sim\frac{\Gamma(n+\alpha+1)}{n!}e^{x/2}(Nx)^{-\alpha/2}J_{% \alpha}(2\sqrt{Nx}),\qquad n\to\infty.$

### 1.6 Laguerre expansions

Suppose that $f:\mathbb{R}_{>0}\to\mathbb{R}$ is piecewise smooth in every interval $[x_{1},x_{2}]$, $0, and $f\in L^{2}(d\rho_{\alpha})$. Let

 $c_{n}(f)=\frac{n!}{\Gamma(n+\alpha+1)}\int_{0}^{\infty}f(x)L_{n}^{\alpha}(x)% \rho_{\alpha}(x)dx,$

$\rho_{\alpha}(x)=e^{-x}x^{\alpha}$. It can be proved that44 4 N. N. Lebedev, Special Functions and Their Applications, p. 88, §4.23, Theorem 3. if $f$ is continuous at $x$ then

 $\sum_{n=0}^{N}c_{n}(f)L_{n}^{\alpha}(x)\to f(x),\qquad N\to\infty,$

and if $f$ is not continuous at $x$ then

 $\sum_{n=0}^{N}c_{n}(f)L_{n}^{\alpha}(x)\to\frac{f(x+0)}{2}+\frac{f(x-0)}{2},% \qquad N\to\infty,$

which makes sense because $f$ is a priori piecewise continuous.

Let $\nu>-\frac{1}{2}(\alpha+1)$ and $f(x)=x^{\nu}$. Integrating by parts,

 $\displaystyle c_{n}(f)$ $\displaystyle=\frac{n!}{\Gamma(n+\alpha+1)}\int_{0}^{\infty}x^{\nu+\alpha}L_{n% }^{\alpha}(x)e^{-x}$ $\displaystyle=\frac{1}{\Gamma(n+\alpha+1)}\int_{0}^{\infty}x^{\nu}D^{n}(e^{-x}% x^{n+\alpha})dx$ $\displaystyle=(-1)^{n}\frac{\Gamma(\nu+\alpha+1)\Gamma(\nu+1)}{\Gamma(n+\alpha% +1)\Gamma(\nu-n+1)}.$

Thus

 $x^{\nu}=\Gamma(\nu+\alpha+1)\Gamma(\nu+1)\sum_{n=0}^{\infty}\frac{(-1)^{n}L_{n% }^{\alpha}(x)}{\Gamma(n+\alpha+1)\Gamma(\nu-n+1)}.$

For $p$ a positive integer,

 $x^{p}=\Gamma(p+\alpha+1)\cdot p!\sum_{n=0}^{p}\frac{(-1)^{n}L_{n}^{\alpha}(x)}% {\Gamma(n+\alpha+1)\cdot(p-n)!}.$

Define

 $f(x)=(ax)^{-\alpha/2}J_{\alpha}(2\sqrt{ax}),\qquad\alpha>-1,\quad a>0,\quad x>0.$

Using

 $(1-z)^{-\alpha-1}e^{-xz/(1-z)}=\sum_{n=0}^{\infty}L_{n}^{\alpha}(x)z^{n},% \qquad|z|<1,$

we obtain, as $e^{-\frac{xz}{1-z}-x}=e^{-x/(1-z)}$,

 $\begin{split}&\displaystyle(1-z)^{-\alpha-1}\int_{0}^{\infty}e^{-x/(1-z)}(x/a)% ^{\alpha/2}J_{\alpha}(2\sqrt{ax})dx\\ \displaystyle=&\displaystyle\int_{0}^{\infty}e^{-x}(x/a)^{\alpha/2}J_{\alpha}(% 2\sqrt{ax})\sum_{n=0}^{\infty}L_{n}^{\alpha}(x)z^{n}dx\\ \displaystyle=&\displaystyle\sum_{n=0}^{\infty}\left(\int_{0}^{\infty}f(x)L_{n% }^{\alpha}(x)\rho_{\alpha}(x)dx\right)z^{n}.\end{split}$

Doing the change of variable $2\sqrt{ax}=by$ with $b>0$ and then applying (6) with $A^{2}=\frac{b^{2}}{4a(1-z)}$ and $\nu=\alpha$,

 $\begin{split}&\displaystyle(1-z)^{-\alpha-1}\int_{0}^{\infty}e^{-x/(1-z)}(x/a)% ^{\alpha/2}J_{\alpha}(2\sqrt{ax})dx\\ \displaystyle=&\displaystyle(1-z)^{-\alpha-1}(2a)^{-\alpha-1}b^{\alpha+2}\int_% {0}^{\infty}e^{-\frac{b^{2}y^{2}}{4a(1-z)}}J_{\alpha}(by)y^{\alpha+1}dy\\ \displaystyle=&\displaystyle(1-z)^{-\alpha-1}(2a)^{-\alpha-1}b^{\alpha+2}\cdot% \frac{b^{\alpha}}{(2A^{2})^{\alpha+1}}e^{-\frac{b^{2}}{4A^{2}}}\\ \displaystyle=&\displaystyle(1-z)^{-\alpha-1}(2a)^{-\alpha-1}b^{\alpha+2}\cdot b% ^{-\alpha-2}(2a(1-z))^{\alpha+1}e^{-a(1-z)}\\ \displaystyle=&\displaystyle e^{-a(1-z)}\\ \displaystyle=&\displaystyle e^{-a}\sum_{n=0}^{\infty}\frac{(az)^{n}}{n!}.\end% {split}$

Therefore

 $\displaystyle e^{-a}\sum_{n=0}^{\infty}\frac{a^{n}}{n!}z^{n}$ $\displaystyle=\sum_{n=0}^{\infty}\left(\int_{0}^{\infty}f(x)L_{n}^{\alpha}(x)% \rho_{\alpha}(x)dx\right)z^{n},$

whence, for $n\geq 0$,

 $c_{n}(f)=\frac{n!}{\Gamma(n+\alpha+1)}\int_{0}^{\infty}f(x)L_{n}^{\alpha}(x)% \rho_{\alpha}(x)dx=\frac{n!}{\Gamma(n+\alpha+1)}e^{-a}\frac{a^{n}}{n!}.$

Therefore, for $\alpha>-1$, $a>0$, $x>0$,

 $(ax)^{-\alpha/2}J_{\alpha}(2\sqrt{ax})=\sum_{n=0}^{\infty}c_{n}(f)L_{n}^{% \alpha}(x)=e^{-a}\sum_{n=0}^{\infty}\frac{a^{n}}{\Gamma(n+\alpha+1)}L_{n}^{% \alpha}(x).$

## 2 Integral operators

We remind ourselves that, for $\alpha=1$,

 $u_{n}(x)=\rho_{1}(x)^{1/2}L_{n}^{1}(x)=e^{-x/2}x^{1/2}L_{n}^{1}(x).$

$\{u_{n}:n\geq 0\}$ is an orthonormal basis for $L^{2}(\mathbb{R}_{\geq 0})$.

For $x,y\in\mathbb{R}_{>0}$ define

 $k(x,y)=k_{x}(y)=k^{x}(y)=\frac{J_{1}(2\sqrt{xy})}{((e^{x}-1)(e^{y}-1))^{1/2}}.$

For $\phi\in L^{2}(\mathbb{R}_{\geq 0})$ and $y\in\mathbb{R}_{>0}$, define

 $\displaystyle K\phi(y)$ $\displaystyle=\int_{\mathbb{R}_{\geq 0}}k_{y}(x)\phi(x)dx.$

We have established, with $\alpha=1$,

 $J_{1}(2\sqrt{xy})=(xy)^{1/2}e^{-x}\sum_{n=0}^{\infty}\frac{x^{n}}{(n+1)!}L_{n}% ^{1}(y).$

Hence

 $\displaystyle\int_{0}^{\infty}k_{y}(x)\phi(x)dx$ $\displaystyle=\int_{0}^{\infty}\phi(x)(e^{x}-1)^{-1/2}(e^{y}-1)^{-1/2}(xy)^{1/% 2}e^{-x}$ $\displaystyle\cdot\sum_{n=0}^{\infty}\frac{x^{n}}{(n+1)!}L_{n}^{\alpha}(y)dx$ $\displaystyle=\sum_{n=0}^{\infty}\frac{(e^{y}-1)^{-1/2}y^{1/2}L_{n}^{1}(y)}{(n% +1)!}\int_{0}^{\infty}\phi(x)(e^{x}-1)^{-1/2}x^{1/2}e^{-x}x^{n}dx$ $\displaystyle=\sum_{n=0}^{\infty}q_{n}(y)\left\langle\phi,p_{n}\right\rangle,$

for

 $p_{n}(x)=\frac{1}{(n+1)!}(e^{x}-1)^{-1/2}e^{-x}x^{n+\frac{1}{2}}=\frac{1}{(n+1% )!}e^{-x/2}(e^{x}-1)^{-1/2}x^{n}u_{n}(x)$

and

 $q_{n}(y)=(e^{y}-1)^{-1/2}y^{1/2}L_{n}^{1}(y)=(1-e^{-y})^{-1/2}u_{n}(y).$

Then

 $K\phi=\sum_{n=0}^{\infty}q_{n}\left\langle\phi,p_{n}\right\rangle.$

The following states the trace of the operator $K:L^{2}(\mathbb{R}_{\geq 0})\to L^{2}(\mathbb{R}_{\geq 0})$.55 5 cf. A. A. Kirillov, Elements of the Theory of Representations, p. 211, §13, Theorem 2.

###### Theorem 1.

$\mathrm{tr}\,K=\int_{0}^{\infty}k(x,x)dx=\int_{0}^{\infty}\frac{J_{1}(2x)}{(e^% {x}-1)}dx=0.7711\ldots$.

## 3 Hardy spaces

For $x\in\mathbb{R}$ let $P_{x}=\{z\in\mathbb{C}:\mathrm{Re}\,z>x\}$. Let $H$ be the collection of holomorphic functions $f:P_{-1/2}\to\mathbb{C}$ such that for any $x>-\frac{1}{2}$, $f|P_{x}$ is bounded and such that

 $\int_{\mathbb{R}}\left|f\left(-\frac{1}{2}+iy\right)\right|^{2}dy<\infty.$

Define $M:L^{2}(\mathbb{R}_{\geq 0})\to H$, for $\phi\in L^{2}(\mathbb{R}_{\geq 0})$, by

 $M\phi(z)=\int_{\mathbb{R}_{\geq 0}}e^{-zs-s/2}\phi(s)ds.$

For $f\in H$ define

 $P_{\lambda}f(z)=\sum_{k\geq 1}\frac{1}{(z+k)^{2}}f\left(\frac{1}{z+k}\right)% \qquad\mathrm{Re}\,z>-\frac{1}{2},$

called a Perron-Frobenius operator. $\lambda$ denotes Lebesgue measure.

Let

 $h(s)=\left(\frac{1-e^{-s}}{s}\right)^{1/2}$

for $s\in\mathbb{R}_{>0}$, with $h(0)=1$. Because $h\in L^{\infty}(\mu)$, it makes sense to define $S:L^{2}(\mathbb{R}_{\geq 0})\to L^{2}(\mathbb{R}_{\geq 0})$ by

 $S\phi(s)=h\phi,\qquad\phi\in L^{2}(\mathbb{R}_{\geq 0}).$

Define $A:H\to L^{2}(\mathbb{R}_{\geq 0})$ by

 $A=S\circ M^{-1}.$

We prove that $P_{\lambda}$ and $K$ are conjugate.66 6 Marius Iosifescu and Cor Kraaikamp, Metrical Theory of Continued Fractions, p. 9, Proposition 1.1.1.

###### Theorem 2.

$P_{\lambda}=A^{-1}KA$.

###### Proof.

Let $\phi\in L^{2}(\mathbb{R}_{\geq 0})$ and set $f=M\phi$. Then

 $A^{-1}KAf=A^{-1}KS\phi.$

We calculate

 $\displaystyle(S^{-1}KS\phi)(x)$ $\displaystyle=h(x)^{-1}\int_{\mathbb{R}_{\geq 0}}k_{x}(y)\cdot h(y)\cdot\phi(y% )dy$ $\displaystyle=\left(\frac{x}{1-e^{-x}}\right)^{1/2}\int_{0}^{\infty}\frac{J_{1% }(2\sqrt{xy})}{((e^{x}-1)(e^{y}-1))^{1/2}}\cdot\left(\frac{1-e^{-y}}{y}\right)% ^{1/2}\cdot\phi(y)dy$ $\displaystyle=\int_{0}^{\infty}\left(\frac{x}{y}\right)^{1/2}\frac{e^{x/2}}{(e% ^{x}-1)^{1/2}}\frac{(e^{y}-1)^{1/2}}{e^{y/2}}\frac{J_{1}(2\sqrt{xy})}{((e^{x}-% 1)(e^{y}-1))^{1/2}}\phi(y)dy$ $\displaystyle=\int_{0}^{\infty}\left(\frac{x}{y}\right)^{1/2}\frac{e^{(x-y)/2}% }{e^{x}-1}J_{1}(2\sqrt{xy})\phi(y)dy.$

Then

 $\begin{split}&\displaystyle(MS^{-1}KS\phi)(z)\\ \displaystyle=&\displaystyle\int_{\mathbb{R}_{\geq 0}}e^{-zx-x/2}(S^{-1}KS\phi% )(x)dx\\ \displaystyle=&\displaystyle\int_{0}^{\infty}e^{-zx-x/2}\left(\left(\frac{x}{y% }\right)^{1/2}\frac{e^{(x-y)/2}}{e^{x}-1}J_{1}(2\sqrt{xy})\phi(y)dy\right)dx\\ \displaystyle=&\end{split}$

It is a fact that for $\mathrm{Re}\,z>-1$ and for $t\geq 0$,

 $\sum_{k\geq 0}(z+k)^{-2}\exp\left(-\frac{t}{z+k}\right)=\int_{0}^{\infty}(st^{% -1})^{1/2}e^{-zs}\frac{J_{1}(2\sqrt{st})}{e^{s}-1}ds.$

Using this,

 $\displaystyle(MS^{-1}KS\phi)(z)$ $\displaystyle=\int_{0}^{\infty}e^{-y/2}\left(\int_{0}^{\infty}(xy^{-1})^{1/2}e% ^{-zx}\frac{J_{1}(2\sqrt{xy})}{e^{x}-1}dx\right)\phi(y)dy$ $\displaystyle=\int_{0}^{\infty}e^{-y/2}\sum_{k\geq 1}(z+k)^{-2}\exp\left(-% \frac{y}{z+k}\right)\cdot\phi(y)dy$ $\displaystyle=\sum_{k\geq 1}(z+k)^{-2}\left(\int_{0}^{\infty}\exp\left(-\frac{% y}{z+k}-\frac{y}{2}\right)\phi(y)dy\right)$ $\displaystyle=\sum_{k\geq 1}(z+k)^{-2}\cdot M\phi\left(\frac{1}{z+k}\right).$

Thus, as $f=M\phi$,

 $(MS^{-1}KSM^{-1}f)(z)=\sum_{k\geq 1}(z+k)^{-2}f\left(\frac{1}{z+k}\right)=P_{% \lambda}f(z),$

that is,

 $A^{-1}KAf(z)=P_{\lambda}f(z).$