Laguerre polynomials and Perron-Frobenius operators

1 Laguerre polynomials

1.1 Definition and generating functions

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

$$L_n^{\alpha }(x)=\sum _{k=0}^n\frac{\mathrm{\Gamma }(n+\alpha +1)}{\mathrm{\Gamma }(k+\alpha +1)}\frac{{(-x)}^k}{k!(n-k)!}.$$

The generating function for the Laguerre polynomials is

Define

where

$$I_{\alpha }(x)=i^{-\alpha }{J}_{\alpha }(ix)=\sum _{m=0}^{\mathrm{\infty }}\frac{1}{m!\mathrm{\Gamma }(m+\alpha +1)}{(x/2)}^{2m+\alpha }$$

and

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

$W$ satisfies

$$W(x,y,z)=\sum _{n=0}^{\mathrm{\infty }}\frac{n!L_n^{\alpha }(x)L_n^{\alpha }(y)}{\mathrm{\Gamma }(n+\alpha +1)}{z}^n.$$

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 $n\ge 1$,

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

$w$ also satisfies the ordinary differential equation

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

which yields, for $n\ge 1$,

$$D{L}_n^{\alpha }-D{L}_{n-1}^{\alpha }+L_{n-1}^{\alpha }=0.$$ (2)

Using (1) and (2) gives

$$xD{L}_n^{\alpha }=n{L}_n^{\alpha }-(n+\alpha )L_{n-1}^{\alpha },n\ge 1.$$ (3)

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

$$x{D}^2{L}_n^{\alpha }(x)+(\alpha +1-x)D{L}_n^{\alpha }(x)+n{L}_n^{\alpha }(x)=0.$$ (4)

1.3 Integral formulas for Laguerre polynomials

For $\nu >-1$, $a>0$, $b>0$, using the series for $J_{\nu }$ one calculates

$${\int }_0^{\mathrm{\infty }}{e}^{-a^2{x}^2}{J}_{\nu }(bx)x^{\nu +1}𝑑x=\frac{b^{\nu }}{{(2a^2)}^{\nu +1}}{e}^{-\frac{b^2}{4a^2}}.$$ (5)

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

$${\int }_0^{\mathrm{\infty }}{e}^{-t}{J}_{n+\alpha }(2\sqrt{xt}){(\sqrt{t})}^{n+\alpha +1}\cdot \frac{1}{2\sqrt{t}}𝑑t=\frac{{(2\sqrt{x})}^{n+\alpha }}{2^{n+\alpha +1}}{e}^{-x},$$

i.e.

$${\int }_0^{\mathrm{\infty }}{e}^{-t}{J}_{n+\alpha }(2\sqrt{xt}){(\sqrt{xt})}^{n+\alpha }𝑑t=e^{-x}{x}^{n+\alpha }.$$ (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 $n\ge 0$,

$$L_n^{\alpha }(x)=\frac{e^x{x}^{-\alpha /2}}{n!}{\int }_0^{\mathrm{\infty }}{t}^{n+\alpha /2}{J}_{\alpha }(2\sqrt{xt})e^{-t}𝑑t.$$ (7)

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

$${(1-z)}^{-\alpha -1}{e}^{-yt/(1-t)}=\sum _{n=0}^{\mathrm{\infty }}{L}_n^{\alpha }(y)z^n.$$

For , 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

Then one gets, for ,

$$2e^{-x/2}{x}^{\alpha /2}\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{L}_n^{\alpha }(x)z^n=\sum _{n=0}^{\mathrm{\infty }}{z}^n{\int }_0^{\mathrm{\infty }}{e}^{-y/2}{y}^{\alpha /2}{J}_{\alpha }(\sqrt{xy})L_n^{\alpha }(y)𝑑y.$$

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

$$e^{-x/2}{x}^{\alpha /2}{L}_n^{\alpha }(x)=\frac{{(-1)}^n}{2}{\int }_0^{\mathrm{\infty }}{J}_{\alpha }(\sqrt{xy})e^{-y/2}{y}^{\alpha /2}{L}_n^{\alpha }(y)𝑑y.$$ (8)

1.4 Orthogonality of Laguerre polynomials

Let

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

Let

$$u_n={\rho }_{\alpha }^{1/2}{L}_n^{\alpha },n\ge 0.$$

$u_n$ satisfies the differential equation

$${(x{u}_n^{\prime })}^{\prime }+\left(n+\frac{\alpha +1}{2}-\frac{x}{4}-\frac{{\alpha }^2}{4x}\right)u_n=0.$$

Using this we get

$${x(u_n^{\prime }{u}_m-u_m^{\prime }{u}_n)|}_0^{\mathrm{\infty }}+(n-m){\int }_0^{\mathrm{\infty }}{u}_m{u}_n𝑑x=0.$$

Then

$$(n-m){\int }_0^{\mathrm{\infty }}{u}_m{u}_n𝑑x=0.$$ (9)

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

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

$$n{\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{\alpha }{L}_n^{\alpha }{(x)}^2𝑑x=(n+\alpha ){\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{\alpha }{L}_{n-1}^{\alpha }{(x)}^2𝑑x.$$

Iterating this, for $n\ge 2$,

${\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-x}{x}^{\alpha }{L}_n^{\alpha }{(x)}^2𝑑x$ $={\displaystyle \frac{(n+\alpha )(n+\alpha -1)\mathrm{\cdots }(\alpha +2)}{n(n-2)\mathrm{\cdots }3\cdot 2}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-x}{x}^{\alpha }{L}_1^{\alpha }{(x)}^2𝑑x$

$={\displaystyle \frac{\mathrm{\Gamma }(n+\alpha +1)}{n!}}.$

1.5 Asymptotics for Laguerre polynomials

It can be proved that for $\alpha >-1$, with $N=n+\frac{\alpha +1}{2}$, for $x\in {ℝ}_{\ge 0}$,

$$L_n^{\alpha }(x)\sim \frac{\mathrm{\Gamma }(n+\alpha +1)}{n!}{e}^{x/2}{(Nx)}^{-\alpha /2}{J}_{\alpha }(2\sqrt{Nx}),n\to \mathrm{\infty }.$$

1.6 Laguerre expansions

Suppose that $f:{ℝ}_{>0}\to ℝ$ is piecewise smooth in every interval $[x_1,x_2]$, , and $f\in L^2(d{\rho }_{\alpha })$. Let

$$c_n(f)=\frac{n!}{\mathrm{\Gamma }(n+\alpha +1)}{\int }_0^{\mathrm{\infty }}f(x)L_n^{\alpha }(x){\rho }_{\alpha }(x)𝑑x,$$

${\rho }_{\alpha }(x)=e^{-x}{x}^{\alpha }$. It can be proved that if $f$ is continuous at $x$ then

$$\sum _{n=0}^N{c}_n(f)L_n^{\alpha }(x)\to f(x),N\to \mathrm{\infty },$$

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

$$\sum _{n=0}^N{c}_n(f)L_n^{\alpha }(x)\to \frac{f(x+0)}{2}+\frac{f(x-0)}{2},N\to \mathrm{\infty },$$

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,

$c_n(f)$ $={\displaystyle \frac{n!}{\mathrm{\Gamma }(n+\alpha +1)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{x}^{\nu +\alpha }{L}_n^{\alpha }(x)e^{-x}$

$={\displaystyle \frac{1}{\mathrm{\Gamma }(n+\alpha +1)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{x}^{\nu }{D}^n(e^{-x}{x}^{n+\alpha })𝑑x$

$={(-1)}^n{\displaystyle \frac{\mathrm{\Gamma }(\nu +\alpha +1)\mathrm{\Gamma }(\nu +1)}{\mathrm{\Gamma }(n+\alpha +1)\mathrm{\Gamma }(\nu -n+1)}}.$

Thus

$$x^{\nu }=\mathrm{\Gamma }(\nu +\alpha +1)\mathrm{\Gamma }(\nu +1)\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{L}_n^{\alpha }(x)}{\mathrm{\Gamma }(n+\alpha +1)\mathrm{\Gamma }(\nu -n+1)}.$$

For $p$ a positive integer,

$$x^p=\mathrm{\Gamma }(p+\alpha +1)\cdot p!\sum _{n=0}^p\frac{{(-1)}^n{L}_n^{\alpha }(x)}{\mathrm{\Gamma }(n+\alpha +1)\cdot (p-n)!}.$$

Define

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

Using

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

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 $\nu =\alpha$,

Therefore

$e^{-a}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{a^n}{n!}}{z}^n$ $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\left({\displaystyle {\int }_0^{\mathrm{\infty }}}f(x)L_n^{\alpha }(x){\rho }_{\alpha }(x)𝑑x\right)z^n,$

whence, for $n\ge 0$,

$$c_n(f)=\frac{n!}{\mathrm{\Gamma }(n+\alpha +1)}{\int }_0^{\mathrm{\infty }}f(x)L_n^{\alpha }(x){\rho }_{\alpha }(x)𝑑x=\frac{n!}{\mathrm{\Gamma }(n+\alpha +1)}{e}^{-a}\frac{a^n}{n!}.$$

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

$${(ax)}^{-\alpha /2}{J}_{\alpha }(2\sqrt{ax})=\sum _{n=0}^{\mathrm{\infty }}{c}_n(f)L_n^{\alpha }(x)=e^{-a}\sum _{n=0}^{\mathrm{\infty }}\frac{a^n}{\mathrm{\Gamma }(n+\alpha +1)}{L}_n^{\alpha }(x).$$

2 Integral operators

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

$$u_n(x)={\rho }_1{(x)}^{1/2}{L}_n^1(x)=e^{-x/2}{x}^{1/2}{L}_n^1(x).$$

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

For $x,y\in {ℝ}_{>0}$ 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 $\varphi \in L^2({ℝ}_{\ge 0})$ and $y\in {ℝ}_{>0}$, define

$K\varphi (y)$ $={\displaystyle {\int }_{{ℝ}_{\ge 0}}}{k}_y(x)\varphi (x)𝑑x.$

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

$$J_1(2\sqrt{xy})={(xy)}^{1/2}{e}^{-x}\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{(n+1)!}{L}_n^1(y).$$

Hence

${\displaystyle {\int }_0^{\mathrm{\infty }}}{k}_y(x)\varphi (x)𝑑x$ $={\displaystyle {\int }_0^{\mathrm{\infty }}}\varphi (x){(e^x-1)}^{-1/2}{(e^y-1)}^{-1/2}{(xy)}^{1/2}{e}^{-x}$

$\cdot {\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{x^n}{(n+1)!}}{L}_n^{\alpha }(y)dx$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{{(e^y-1)}^{-1/2}{y}^{1/2}{L}_n^1(y)}{(n+1)!}}{\displaystyle {\int }_0^{\mathrm{\infty }}}\varphi (x){(e^x-1)}^{-1/2}{x}^{1/2}{e}^{-x}{x}^n𝑑x$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{q}_n(y)⟨\varphi ,p_n⟩,$

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

$$K\varphi =\sum _{n=0}^{\mathrm{\infty }}{q}_n⟨\varphi ,p_n⟩.$$

The following states the trace of the operator $K:L^2({ℝ}_{\ge 0})\to L^2({ℝ}_{\ge 0})$.

Theorem 1.

$\mathrm{tr}K={\int }_0^{\mathrm{\infty }}k(x,x)𝑑x={\int }_0^{\mathrm{\infty }}\frac{J_1(2x)}{(e^x-1)}𝑑x=0.7711\mathrm{\dots }$.

3 Hardy spaces

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

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

$$M\varphi (z)={\int }_{{ℝ}_{\ge 0}}{e}^{-zs-s/2}\varphi (s)𝑑s.$$

For $f\in H$ define

$$P_{\lambda }f(z)=\sum _{k\ge 1}\frac{1}{{(z+k)}^2}f\left(\frac{1}{z+k}\right)\mathit{\hspace{1em}\hspace{1em}}\mathrm{Re}z>-\frac{1}{2},$$

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

Let

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

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

$$S\varphi (s)=h\varphi ,\varphi \in L^2({ℝ}_{\ge 0}).$$

Define $A:H\to L^2({ℝ}_{\ge 0})$ by

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

We prove that $P_{\lambda }$ and $K$ are conjugate.

Theorem 2.

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

Proof.

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

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

We calculate

$(S^{-1}KS\varphi )(x)$ $=h{(x)}^{-1}{\displaystyle {\int }_{{ℝ}_{\ge 0}}}{k}_x(y)\cdot h(y)\cdot \varphi (y)𝑑y$

$={\left({\displaystyle \frac{x}{1-e^{-x}}}\right)}^{1/2}{\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{J_1(2\sqrt{xy})}{{((e^x-1)(e^y-1))}^{1/2}}}\cdot {\left({\displaystyle \frac{1-e^{-y}}{y}}\right)}^{1/2}\cdot \varphi (y)𝑑y$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}{\left({\displaystyle \frac{x}{y}}\right)}^{1/2}{\displaystyle \frac{e^{x/2}}{{(e^x-1)}^{1/2}}}{\displaystyle \frac{{(e^y-1)}^{1/2}}{e^{y/2}}}{\displaystyle \frac{J_1(2\sqrt{xy})}{{((e^x-1)(e^y-1))}^{1/2}}}\varphi (y)𝑑y$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}{\left({\displaystyle \frac{x}{y}}\right)}^{1/2}{\displaystyle \frac{e^{(x-y)/2}}{e^x-1}}{J}_1(2\sqrt{xy})\varphi (y)𝑑y.$

Then

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

$$\sum _{k\ge 0}{(z+k)}^{-2}\exp \left(-\frac{t}{z+k}\right)={\int }_0^{\mathrm{\infty }}{(s{t}^{-1})}^{1/2}{e}^{-zs}\frac{J_1(2\sqrt{st})}{e^s-1}𝑑s.$$

Using this,

$(M{S}^{-1}KS\varphi )(z)$ $={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-y/2}\left({\displaystyle {\int }_0^{\mathrm{\infty }}}{(x{y}^{-1})}^{1/2}{e}^{-zx}{\displaystyle \frac{J_1(2\sqrt{xy})}{e^x-1}}𝑑x\right)\varphi (y)𝑑y$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-y/2}{\displaystyle \sum _{k\ge 1}}{(z+k)}^{-2}\exp \left(-{\displaystyle \frac{y}{z+k}}\right)\cdot \varphi (y)dy$

$={\displaystyle \sum _{k\ge 1}}{(z+k)}^{-2}\left({\displaystyle {\int }_0^{\mathrm{\infty }}}\exp \left(-{\displaystyle \frac{y}{z+k}}-{\displaystyle \frac{y}{2}}\right)\varphi (y)𝑑y\right)$

$={\displaystyle \sum _{k\ge 1}}{(z+k)}^{-2}\cdot M\varphi \left({\displaystyle \frac{1}{z+k}}\right).$

Thus, as $f=M\varphi$,

$$(M{S}^{-1}KS{M}^{-1}f)(z)=\sum _{k\ge 1}{(z+k)}^{-2}f\left(\frac{1}{z+k}\right)=P_{\lambda }f(z),$$

that is,

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

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