Appendix
From now on, we will use the following identities that Laguerre polynomials satisfy
$$ L_{k}^{(\alpha-1)}(x)=L_{k}^{(\alpha)}(x)-L_{k-1}^{(\alpha)}(x), $$
(A.1)
$$ \frac{\mathrm{d}}{\mathrm{d}x}L_{k}^{(\alpha)}(x)=x^{-1}[k \ L_{k}^{(\alpha)}(x)-(k+\alpha) \ L_{k-1}^{(\alpha)}(x)], $$
(A.2)
$$ kL_{k}^{(\alpha)}(0)=(k+\alpha) \ L_{k-1}^{(\alpha)}(0), $$
(A.3)
$$ (k+1)L_{k+1}^{(\alpha)}(x)=(-x+2(k+1)+\alpha-1) \ L_{k}^{(\alpha)}(x)-(k+\alpha) \ L_{k-1}^{(\alpha)}(x). $$
(A.4)
Lemma A.1.
Let j be an integer number such that 0≤j≤n−2 and let
\(d_{j,k}^{n}\)
be as in (
3.5
). Then for all
\(k\in \mathbb {N},\)
\(d_{j,k}^{n}-d_{j,k-1}^{n}=-d_{j,k}^{n-1}\).
Proof.
By equalities (A.3) and (A.1) we have that
$$\begin{array}{@{}rcl@{}} d_{j,k}^{n}-d_{j,k-1}^{n}&=&4^{j}\sum\limits_{i=j}^{n-2}\frac{1}{2^{i}}\left( \begin{array}{c}{i}\\{j}\end{array}\right)(-1)^{n-i-2}\left[L_{k+i+1}^{(n-i-2)}(0)-L_{k+i}^{(n-i-2)}(0)\right]\\ &=&-4^{j}\sum\limits_{i=j}^{n-3}\frac{1}{2^{i}}\left( \begin{array}{c}{i}\\{j}\end{array}\right)(-1)^{n-i-3}L_{k+i+1}^{(n-i-3)}(0)\\ &=&-d_{j,k}^{n-1}. \end{array} $$
□
Lemma A.2.
For
\(i\in \mathbb {N}\)
and
\(k\in \mathbb {N}\)
,
\(D^{i}(L_{k}^{(0)})-D^{i}(L_{k-1}^{(0)})=D^{i+1}(L_{k}^{(0)})\).
Proof.
By (3.6) and \(\left (\begin {array}{c}{i}\\[-2pt]{l}\end {array}\right )+\left (\begin {array}{c}{i}\\[-2pt]{l-1}\end {array}\right )=\left (\begin {array}{c}{i+1}\\[-2pt]{l}\end {array}\right )\), we obtain that
$$\begin{array}{@{}rcl@{}} {}D^{i}(L_{k}^{(0)})-D^{i}(L_{k-1}^{(0)})&=&\sum\limits_{l=0}^{\min\{k,i\}} (-1)^{l}\left( \begin{array}{c}{i}\\{l}\end{array}\right)L_{k-l}^{(0)}-\sum\limits_{l=0}^{\min\{k-1,i\}} (-1)^{l}\left( \begin{array}{c}{i}\\{l}\end{array}\right)L_{k-1-l}^{(0)}\\ &=&\sum\limits_{l=0}^{\min\{k,i\}} (-1)^{l}\left( \begin{array}{c}{i}\\{l}\end{array}\right)L_{k-l}^{(0)}+\sum\limits_{l=1}^{\min\{k-1,i\}+1} (-1)^{l}\left( \begin{array}{c}{i}\\{l-1}\end{array}\right)L_{k-l}^{(0)}\\ &=&\sum\limits_{l=0}^{\min\{k,i\}} (-1)^{l}\left( \begin{array}{c}{i}\\{l}\end{array}\right)L_{k-l}^{(0)}+\sum\limits_{l=1}^{\min\{k,i+1\}} (-1)^{l}\left( \begin{array}{c}{i}\\{l-1}\end{array}\right)L_{k-l}^{(0)}. \end{array} $$
If min{k, i} = min{k, i + 1}, then
$$\begin{array}{@{}rcl@{}} D^{i}(L_{k}^{(0)})-D^{i}(L_{k-1}^{(0)})&=&\sum\limits_{l=1}^{\min\{k,i+1\}} (-1)^{l}\left[\left( \begin{array}{c}{i}\\{l}\end{array}\right)+\left( \begin{array}{c}{i}\\{l-1}\end{array}\right)\right]L_{k-l}^{(0)}+L_{k}^{(0)}\\ &=&\sum\limits_{l=0}^{\min\{k,i+1\}} (-1)^{l}\left( \begin{array}{c}{i+1}\\{l}\end{array}\right)L_{k-l}^{(0)}\\ &=&D^{i+1}L_{k}^{(0)}. \end{array} $$
If min{k, i} + 1 = i + 1 = min{k, i + 1} (i <k), then
$$\begin{array}{@{}rcl@{}} {}D^{i}(L_{k}^{(0)})-D^{i}(L_{k-1}^{(0)})&=&L_{k}^{(0)}+\sum\limits_{l=1}^{i} (-1)^{l}\left[\left( \begin{array}{c}{i}\\{l}\end{array}\right)+\left( \begin{array}{c}{i}\\{l-1}\end{array}\right)\right]L_{k-l}^{(0)}+(-1)^{i+1}L_{k-(i+1)}^{(0)}\\ &=&L_{k}^{(0)}+\sum\limits_{l=1}^{i} (-1)^{l}\left( \begin{array}{c}{i+1}\\{l}\end{array}\right)L_{k-l}^{(0)}+(-1)^{i+1}L_{k-\min\{k,i+1\}}^{(0)}\\ &=&\sum\limits_{l=0}^{i+1} (-1)^{l}\left( \begin{array}{c}{i+1}\\{l}\end{array}\right)L_{k-l}^{(0)}\\ &=&\sum\limits_{l=0}^{\min\{k,i+1\}} (-1)^{l}\left( \begin{array}{c}{i+1}\\{l}\end{array}\right)L_{k-l}^{(0)}\\ &=&D^{i+1}L_{k}^{(0)}. \end{array} $$
□
Lemma A.3.
For n≥3, let
\(d_{j,k}^{n}\)
be as in (
3.5
). Then
$$\begin{array}{@{}rcl@{}} (n-j-2) \ d_{j,k}^{n}+(k+n-1) \ d_{j,k}^{n-1}=\frac{j+1}{4} \ d_{j+1,k}^{n}. \end{array} $$
Proof.
By (3.5) and (A.1) we have that
$$\begin{array}{@{}rcl@{}} &&{}(n-j-2) \ d_{j,k}^{n}+(k+n-1) \ d_{j,k}^{n-1}\\ &&{\kern1.3pc}=4^{j}\sum\limits_{i=j}^{n-2}\frac{1}{2^{i}}\left( \begin{array}{c}{i}\\{j}\end{array}\right)(-1)^{n-i-2}(n-j-2)L_{k+i+1}^{(n-i-2)}(0)\\ &&{\kern2.2pc}+4^{j}\sum\limits_{i=j}^{n-3}\frac{1}{2^{i}}\left( \begin{array}{c}{i}\\{j}\end{array}\right)(-1)^{n-i-3}(k+n-1)L_{k+i+1}^{(n-i-3)}(0)\\ &&{\kern1.3pc}=4^{j}\sum\limits_{i=j}^{n-2}\frac{1}{2^{i}}\left( \begin{array}{c}{i}\\{j}\end{array}\right)(-1)^{n-i-2}(n-j-2)L_{k+i+1}^{(n-i-2)}(0)\\ &&{\kern2.2pc}-4^{j}\sum\limits_{i=j}^{n-2}\frac{1}{2^{i}}\left( {}\begin{array}{c}{i}\\{j}\end{array}{}\right){}({}-1{})^{n-i-2}(k{}+{}n{}-{}1)[L_{k+i+1}^{(n-i-2)}(0){}-{}L_{k+i}^{(n-i-2)}(0)]. \end{array} $$
By (A.3), (k + n − 1)L
\(_{k+i}^{(n-i-2)}\)(0) = (k + i + 1)L
\(_{k+i+1}^{(n-i-2)}\)(0) and so
$$\begin{array}{@{}rcl@{}} &&{}(n-j-2)L_{k+i+1}^{(n-i-2)}(0)-(k+n-1)[L_{k+i+1}^{(n-i-2)}(0)-L_{k+i}^{(n-i-2)}(0)]\\ &&{\kern-4.3pc}=(n-j-2)L_{k+i+1}^{(n-i-2)}(0)-(k+n-1)L_{k+i+1}^{(n-i-2)}(0)+(k+i+1)L_{k+i+1}^{(n-i-2)}(0)\\ &&{\kern-4.3pc}=(i-j)L_{k+i+1}^{(n-i-2)}(0). \end{array} $$
Thus,
$$\begin{array}{@{}rcl@{}} {}(n-j-2)d_{j,k}^{n}+(k+n-1)d_{j,k}^{n-1}&=&4^{j}\sum\limits_{i=j}^{n-2}\frac{1}{2^{i}}(i-j)\left( \begin{array}{c}{i}\\{j}\end{array}\right)(-1)^{n-i-2}L_{k+i+1}^{(n-i-2)}(0)\\ &=&4^{j}\sum\limits_{i=j+1}^{n-2}\frac{1}{2^{i}}(j+1)\left( \begin{array}{c}{i}\\{j+1}\end{array}\right)(-1)^{n-i-2}L_{k+i+1}^{(n-i-2)}(0)\\ &=&\frac{j+1}{4}4^{j+1}\sum\limits_{i=j+1}^{n-2}\frac{1}{2^{i}}\left( \begin{array}{c}{i}\\{j+1}\end{array}\right)(-1)^{n-i-2}L_{k+i+1}^{(n-i-2)}(0)\\ &=&\frac{j+1}{4} d_{j+1,k}^{n}, \end{array} $$
where in the second equality we have used \((j+1)\left (\begin {array}{c}{i}\\{j+1}\end {array}\right )=(i-j)\left (\begin {array}{c}{i}\\{j}\end {array}\right )\). □
Lemma A.4.
Let
\(j\in \mathbb {N}_{0}\)
. For φ=φ(s), we denote
\(\langle \delta ^{j},\varphi \rangle =(-1)^{j}\frac {d^{j}\varphi }{ds^{j}}|_{s=0}\)
. Then ,
$$ \langle\delta^{(j)},N(-\bar{\mathcal{A}}f)(\cdot,\hat{\lambda})\rangle=\frac{\partial}{\partial\lambda}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle+\frac{j}{4}\langle\delta^{(j-1)},Nf(\cdot,\hat{\lambda})\rangle $$
(A.5)
and
$$ \langle\delta^{(j)},N(\mathcal{A}f)(\cdot,\hat{\lambda})\rangle=\frac{\partial}{\partial\lambda}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle-\frac{j}{4}\langle\delta^{(j-1)},Nf(\cdot,\hat{\lambda})\rangle. $$
(A.6)
Proof.
By \(N(\mathcal {A}f)=(-it+\frac {s}{4})Nf\), \(N(-\bar {\mathcal {A}}f)=(-it-\frac {s}{4})Nf\) and definition of distribution δ
(j) we have that
$$\begin{array}{@{}rcl@{}} {}\langle\delta^{(j)},N(-\bar{\mathcal{A}}f)(\cdot,\hat{\lambda})\rangle&=& \langle\delta^{(j)},(-it-\frac{s}{4})Nf(\cdot,\hat{\lambda})\rangle\\ &=&(-1)^{j}\frac{\partial^{j}}{\partial s^{j}}\left( (-\frac{s}{4}-it)Nf(s,\hat{\lambda})\right)\bigg|_{s=0}\\ &=&(-1)^{j}{\int}_{-\infty}^{\infty} \mathrm{e}^{-i\lambda t} \frac{\partial^{j}}{\partial s^{j}}\left( (-\frac{s}{4}-it)Nf(s,t)\right)\bigg|_{s=0} \ \mathrm{d}t\\ &=&(-1)^{j}{\int}_{-\infty}^{\infty} \mathrm{e}^{-i\lambda t} \left[(-it) \frac{\partial^{j}Nf}{\partial s^{j}}(0,t)-j \ \frac{1}{4}\frac{\partial^{j-1}Nf}{\partial s^{j-1}}(0,t)\right]\!\mathrm{d}t\\ &=&\frac{\partial}{\partial\lambda}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle+\frac{j}{4}(-1)^{j-1}{\int}_{-\infty}^{\infty} \mathrm{e}^{-i\lambda t}\frac{\partial^{j-1}Nf}{\partial s^{j-1}}(0,t)\mathrm{d}t\\ &=&\frac{\partial}{\partial\lambda}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle+\frac{j}{4}\langle\delta^{(j-1)},Nf(\cdot,\hat{\lambda})\rangle. \end{array} $$
The proof for (A.6) is completely similar. □
PROPOSITION A.5
Let G be as in (3.2) and \(k\in \mathbb {N}_{0}\). Then
$${}\frac{\partial G}{\partial\lambda}(\lambda,k)= \left\{\begin{array}{ll} -(k+n-1) \mathcal{F}_{n-1}(f)(\lambda,k+q_{n-1})+\mathcal{F}_{n}(-\overline{\mathcal{A}}f)(\lambda,k+q_{n}),&\text{if} \ \lambda>0,\\ \\ (k+n-1) \mathcal{F}_{n-1}(f)(\lambda,k+q_{n-1})+\mathcal{F}_{n}(\mathcal{A}f)(\lambda,k+q_{n}),&\text{if} \ \lambda<0. \end{array}\right. $$
Proof.
By (A.2) and (A.1) we have that
$$\begin{array}{@{}rcl@{}} &&{}\frac{\partial}{\partial\lambda}\left( (-1)^{n-1}|\lambda|^{n-1}{\int}_{\mathbb{R}}{\int}_{s>0}\mathrm{e}^{-it\lambda}\mathrm{e}^{-\frac{s|\lambda|}{4}}L_{k}^{(n-1)}\left( \frac{s|\lambda|}{2}\right)Nf(s,t) \ \mathrm{d}s\mathrm{d}t\right)\\ &&{}=-(n-1)sg(\lambda)(-|\lambda|)^{n-2}{\int}_{\mathbb{R}}{\int}_{s>0}\mathrm{e}^{-it\lambda}\mathrm{e}^{-\frac{s|\lambda|}{4}}L_{k}^{(n-1)}\left( \frac{s|\lambda|}{2}\right)Nf(s,t) \ \mathrm{d}s\mathrm{d}t\\ &&{}+(-1)^{n-1}|\lambda|^{n-1}{\int}_{\mathbb{R}}{\int}_{s>0}(-sg(\lambda)\frac{s}{4}-it)\mathrm{e}^{-it\lambda}\mathrm{e}^{-\frac{s|\lambda|}{4}}L_{k}^{(n-1)}\left( \frac{s|\lambda|}{2}\right)Nf(s,t) \ \mathrm{d}s\mathrm{d}t\\ &&{}-sg(\lambda)(-|\lambda|)^{n-2}{\int}_{\mathbb{R}}{\int}_{s>0}\mathrm{e}^{-it\lambda}\mathrm{e}^{-\frac{s|\lambda|}{4}}\left[k L_{k}^{(n-1)}-(k+n-1)L_{k-1}^{(n-1)}\right]\left( \frac{s|\lambda|}{2}\right)Nf(s,t) \ \mathrm{d}s\mathrm{d}t\\ &&{}=\hspace{.8pt}-sg(\lambda)(k+n-1)(-1)^{n-2}|\lambda|^{n-2}{\int}_{s>0}\mathrm{e}^{-\frac{s|\lambda|}{4}}L_{k}^{(n-2)}\left( \frac{s|\lambda|}{2}\right)Nf(s,\hat{\lambda}) \ \mathrm{d}s \\ &&{}+(-1)^{n-1}|\lambda|^{n-1}{\int}_{s>0}\mathrm{e}^{-\frac{s|\lambda|}{4}}L_{k}^{(n-1)}\left( \frac{s|\lambda|}{2}\right)(-sg(\lambda)\frac{s}{4}-it) Nf(s,\hat{\lambda}) \ \mathrm{d}s, \end{array} $$
So,
$$\begin{array}{@{}rcl@{}} &&{}\frac{\partial G}{\partial\lambda}(\lambda,k)=-sg(\lambda)(k+n-1)(-1)^{n-2}|\lambda|^{n-2}{\int}_{s>0}\mathrm{e}^{-\frac{s|\lambda|}{4}}L_{k}^{(n-2)}\left( \frac{s|\lambda|}{2}\right)Nf(s,\hat{\lambda})\!\! \ \mathrm{d}s \\ &&{}+(-1)^{n-1}|\lambda|^{n-1}{\int}_{s>0}\mathrm{e}^{-\frac{s|\lambda|}{4}}L_{k}^{(n-1)}\left( \frac{s|\lambda|}{2}\right)(-sg(\lambda)\frac{s}{4}-it) Nf(s,\hat{\lambda}) \mathrm{d}s \\ &&{}+sg(\lambda) \sum\limits_{j=0}^{n-3} d_{j,k}^{n}(n-j-2)|\lambda|^{n-j-3}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle\\ &&{}+\sum\limits_{j=0}^{n-2} d_{j,k}^{n} |\lambda|^{n-j-2}\frac{\partial}{\partial\lambda}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle, \end{array} $$
and for λ>0, we get
$$\begin{array}{@{}rcl@{}} \frac{\partial G}{\partial\lambda}(\lambda,k){}&=&{}-{}({}k{}+n{}-{}1){}({}-{}1{})^{n-2}|\lambda|^{n-2}{}{\int}_{s>0}\mathrm{e}^{-s|\lambda|/4}L_{k}^{(n-2)}{}\left( \frac{s|\lambda|}{2}\right)Nf(s,\hat{\lambda}) \mathrm{d}s\\ &&{\kern-.3pc}+(-1)^{n-1}|\lambda|^{n-2}{\int}_{s>0}\mathrm{e}^{-s|\lambda|/4}L_{k}^{(n-2)}\left( \frac{s|\lambda|}{2}\right)N(-\overline{\mathcal{A}}f)(s,\hat{\lambda}) \mathrm{d}s\\ &&{\kern-.3pc}+ \sum\limits_{j=0}^{n-3} d_{j,k}^{n}(n-j-2)|\lambda|^{n-j-3}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle\\ &&{\kern-.3pc}+\sum\limits_{j=0}^{n-2} d_{j,k}^{n} |\lambda|^{n-j-2}\frac{\partial}{\partial\lambda}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle. \end{array} $$
Then, we add and subtract the following singular parts:
$${}(k+n-1){\sum}_{j=0}^{n-3}d_{j,k}^{n-1} |\lambda|^{n-j-3}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle{} \quad\text{and} \quad{} {\sum}_{j=0}^{n-2}d_{j,k}^{n}|\lambda|^{n-j-2}\langle\delta^{(j)},N(-\overline{\mathcal{A}}f)(\cdot,\hat{\lambda})\rangle $$
and by Lemmas A.3 and A.4, we obtain that
$$\begin{array}{@{}rcl@{}} \frac{\partial F}{\partial\lambda}(\lambda,k){}&=&{}-(k+n-1) \mathcal{F}_{n-1}(f)(\lambda,k+q_{n-1})+\mathcal{F}_{n}(-\overline{\mathcal{A}}f)(\lambda,k+q_{n})\\ &&{}+{} \sum\limits_{j=0}^{n-3}{} \left[{}(n{}-{}j{}-{}2)d_{j,k}^{n}{}+{}(k{}+{}n{}-{}1)d_{j,k}^{n-1}{}\right]|\lambda|^{n-j-3}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle\\ &&{}+{}\sum\limits_{j=0}^{n-2} d_{j,k}^{n} |\lambda|^{n-j-2}{}\left[{}\frac{\partial}{\partial\lambda}\langle\delta^{(j)},{}Nf(\cdot,\hat{\lambda})\rangle{}-{}\langle\delta^{(j)},{}N(-\overline{\mathcal{A}}f)(\cdot,\hat{\lambda})\rangle\right]\\ {}&=&{}-(k+n-1) \mathcal{F}_{n-1}(f)(\lambda,k+q_{n-1})+\mathcal{F}_{n}(-\overline{\mathcal{A}}f)(\lambda,k+q_{n}). \end{array} $$
On other hand, for λ<0, we have that
$$\begin{array}{@{}rcl@{}} \frac{\partial F}{\partial\lambda}(\lambda,k){}&=&{}(k{}+{}n{}-{}1)({}-{}1)^{n-2}|\lambda|^{n-2}{}{\int}_{s>0}\mathrm{e}^{-\frac{s|\lambda|}{4}}L_{k}^{(n-2)}\left( {}\frac{s|\lambda|}{2}{}\right)Nf(s,\hat{\lambda})\mathrm{d}s \\ &&{}+(-1)^{n-1}|\lambda|^{n-1}{\int}_{s>0}\mathrm{e}^{-\frac{s|\lambda|}{4}}L_{k}^{(n-1)}\left( \frac{s|\lambda|}{2}\right)N\left( \mathcal{A}f\right)(s,\hat{\lambda}) {}\mathrm{d}s \\ &&{}-\sum\limits_{j=0}^{n-3} d_{j,k}^{n}(n-j-2)|\lambda|^{n-j-3}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle\\ &&{}+\sum\limits_{j=0}^{n-2} d_{j,k}^{n} |\lambda|^{n-j-2}\frac{\partial}{\partial\lambda}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle. \end{array} $$
Then, we add and subtract the singular parts
$${}(k+n-1)\sum\limits_{j=0}^{n-3}d_{j,k}^{n-1} |\lambda|^{n-j-3}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle \ \quad\text{and} \quad \sum\limits_{j=0}^{n-2}d_{j,k}^{n}|\lambda|^{n-j-2}\langle\delta^{(j)},N(\mathcal{A}f)(\cdot,\hat{\lambda})\rangle $$
and by Lemmas A.3 and A.4, we get
$$\begin{array}{@{}rcl@{}} \frac{\partial F}{\partial\lambda}(\lambda,k)&=&{}(k+n-1) \mathcal{F}_{n-1}(f)(\lambda,k+q_{n-1})+\mathcal{F}_{n}(\mathcal{A}f)(\lambda,k+q_{n})\\ &&{\kern-.5pc}-{}\sum\limits_{j=0}^{n-3}[(n{}-{}j{}-{}2)d_{j,k}^{n}{}+{}(k{}+{}n{}-{}1)d_{j,k}^{n-1}]|\lambda|^{n-j-3}\langle\delta^{(j)},{}N{}f(\cdot,{}\hat{\lambda})\rangle\\ &&{\kern-.5pc}+{}\sum\limits_{j=0}^{n-2} d_{j,k}^{n} |\lambda|^{n-j-2}\left[\frac{\partial}{\partial\lambda}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle{}-{}\langle\delta^{(j)},N(\mathcal{A}f)(\cdot,\hat{\lambda})\rangle\right]\\ &=&{}(k+n-1) \mathcal{F}_{n-1}(f)(\lambda,k+q_{n-1})+\mathcal{F}_{n}(\mathcal{A}f)(\lambda,k+q_{n}). \end{array} $$
□
Lemma A.6.
Let n≥2 and
\(k\in \mathbb {N}_{0}\). Then
-
(a)
\((k+n-1)L_{k}^{(n-2)}(x)-(k+n)L_{k+1}^{(n-2)}(x)=x \ L_{k}^{(n-1)}(x)-(n-1)L_{k+1}^{(n-2)}(x)\),
-
(b)
\((k+n-1)d_{j,k}^{n-1}-(k+n)d_{j,k+1}^{n-1}=-(n-1)d_{j,k+1}^{n-1}+\frac {j+1}{2}d_{j+1,k}^{n}\).
Proof.
-
(a)
By (A.1), we obtain that
$$\begin{array}{@{}rcl@{}} &&{\kern-2.6pc}(k+n-1)L_{k}^{(n-2)}(x)-(k+n)L_{k+1}^{(n-2)}(x)\\ &&{\kern-2.3pc}~~~=(k+n-1)[L_{k}^{(n-1)}(x)-L_{k-1}^{(n-1)}(x)]-(k+n)[L_{k+1}^{(n-1)}(x)-L_{k}^{(n-1)}(x)]\\ &&{\kern-2.3pc}~~~=-(k+n)L_{k+1}^{(n-1)}(x)+(2(k\!+n)-1)L_{k}^{(n-1)}(x)-(k\!+n-1)L_{k-1}^{(n-1)}(x)\\ &&{\kern-2.3pc}~~~=-(k+1)L_{k+1}^{(n-1)}(x)-(n-1)L_{k+1}^{(n-1)}(x)+(2(k+n)-1)L_{k}^{(n-1)}(x)\\ &&{\kern-1.3pc}~~-(k+n-1)L_{k-1}^{(n-1)}(x). \end{array} $$
Then, by (A.4) and (A.1), we get
$$\begin{array}{@{}rcl@{}} &&{\kern-3.1pc}(k+n-1)L_{k}^{(n-2)}(x)-(k+n)L_{k+1}^{(n-2)}(x)\\ &&{\kern-2.3pc}=-(-x+2k+2+n-1-1)L_{k}^{(n-1)}(x)+(k+1+n-1-1)L_{k-1}^{(n-1)}(x)\\ &&{\kern-2.3pc}~~~~-(n-1)L_{k+1}^{(n-1)}(x)+(2(k+n)-1)L_{k}^{(n-1)}(x)-(k+n-1)L_{k-1}^{(n-1)}(x)\\ &&{\kern-2.3pc}=-(n-1)L_{k+1}^{(n-1)}(x)+(n-1)L_{k}^{(n-1)}(x)+x \ L_{k}^{(n-1)}(x)\\ &&{\kern-2.3pc}=-(n-1)L_{k+1}^{(n-2)}(x)+x \ L_{k}^{(n-1)}(x). \end{array} $$
-
(b)
We take \(\bar {k}=k+i+1\) and \(\bar {n}=n-i-1\). By (a), we have that
$$\begin{array}{@{}rcl@{}} {\kern-1.3pc}(\bar{k}+\bar{n}-1)L_{\bar{k}}^{(\bar{n}-2)}(0)-(\bar{k}+\bar{n})L_{\bar{k}+1}^{(\bar{n}-2)}(0)=-(\bar{n}-1)L_{\bar{k}+1}^{(\bar{n}-2)}(0). \end{array} $$
Therefore, if we replace \(\bar {k},\bar {n}\) then
$$\begin{array}{@{}rcl@{}} &&{\kern-2.3pc}(k+n-1)L_{k+i+1}^{(n-i-3)}(0)-(k+n)L_{k+i+2}^{(n-i-3)}(0)=-(n-i-2)L_{k+i+2}^{(n-i-3)}(0)\\ &&{}~~~=-(n-1)L_{k+i+2}^{(n-i-3)}(0)+(i+1)L_{k+i+2}^{(n-i-3)}(0). \end{array} $$
So, by (3.5), the last equality and \(\left (\begin {array}{c}{i}\\[-3pt]{j}\end {array}\right )(i+1)=\left (\begin {array}{c}{i+1}\\[-3pt]{j+1}\end {array}\right )(j+1),\) we obtain that
$$\begin{array}{@{}rcl@{}} &&{\kern-3.1pc}(k+n-1) \ d_{j,k}^{n-1}-(k+n) \ d_{j,k+1}^{n-1}\\ &&{\kern-2.3pc}=4^{j}\sum\limits_{i=j}^{n-3}\frac{1}{2^{i}}\left( \begin{array}{c}{i}\\[-3pt]{j}\end{array}\right)(-1)^{n-i-3}[(k\,+\,n\,-\,1)L_{k+i+1}^{(n-i-3)}(0)\,-\,(k+n)L_{k+i+2}^{(n-i-3)}(0)]\\ &&{\kern-2.3pc}=4^{j}\sum\limits_{i=j}^{n-3}\frac{1}{2^{i}}\left( \begin{array}{c}{i}\\[-3pt]{j}\end{array}\right)(-1)^{n-i-3}[-(n-1)L_{k+i+2}^{(n-i-3)}(0)+(i+1)L_{k+i+2}^{(n-i-3)}(0)]\\ &&{\kern-2.3pc}=-(n-1) \ d_{j,k+1}^{n-1}+4^{j}\sum\limits_{i=j}^{n-3}\frac{1}{2^{i}}\left( \begin{array}{c}{i+1}\\[-3pt]{j+1}\end{array}\right)(j+1)(-1)^{n-i-3}L_{k+i+2}^{(n-i-3)}(0)\\ &&{\kern-2.3pc}=-(n-1) \ d_{j,k+1}^{n-1}+\frac{j+1}{2} \ d_{j+1,k}^{n}. \end{array} $$
□
Lemma A.7.
Let n≥2 and
\(f\in \mathcal {S}(H_{n})\)
. Then
$$\begin{array}{@{}rcl@{}} &&{}\mathcal{F}_{n}(-\bar{\mathcal{A}}f)(k,\lambda)-\mathcal{F}_{n}(\mathcal{A}f)(k,\lambda)\\ &&{\kern.1pc}=-(-1)^{n-1}|\lambda|^{n-1}{\int}_{s>0}\mathrm{e}^{-s|\lambda|/4}L_{k}^{(n-1)}\left( \frac{s|\lambda|}{2}\right) \ \frac{s}{2} \ Nf(s,\hat{\lambda}) \mathrm{d}s \\ && \ \ \ \ \ \ \ +\sum\limits_{j=0}^{n-3}\frac{j+1}{2} \ d_{j+1,k}^{n} \ |\lambda|^{n-j-3}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle. \end{array} $$
Proof.
$$\begin{array}{@{}rcl@{}} &&{\kern-1.5pc}\mathcal{F}_{n}(-\bar{\mathcal{A}}f)(s,\lambda)-\mathcal{F}_{n}(\mathcal{A}f)(s,\lambda)\\ &&{}=(-|\lambda|)^{n-1}{\int}_{s>0}\mathrm{e}^{-\frac{s|\lambda|}{4}}L_{k}^{(n-1)}\left( \frac{s|\lambda|}{2}\right) [N(-\bar{\mathcal{A}}f)(s,\hat{\lambda})-N(\mathcal{A}f)(s,\hat{\lambda})] \mathrm{d}s\\ &&{}~~~~+\sum\limits_{j=0}^{n-2}d_{j,k}^{n}|\lambda|^{n-j-2}\left[\langle\delta^{(j)},N(-\bar{\mathcal{A}}f)(\cdot,\hat{\lambda})\rangle-\langle\delta^{(j)},N(\mathcal{A}f)(\cdot,\hat{\lambda})\rangle\right]\\ &&{}=(-1)^{n-1}|\lambda|^{n-1}{\int}_{s>0}\mathrm{e}^{-\frac{s|\lambda|}{4}}L_{k}^{(n-1)}\left( \frac{s|\lambda|}{2}\right) \left( -\frac{s}{2}\right)Nf(s,\hat{\lambda}) \ \mathrm{d}s\\ &&{}~~~~+\sum\limits_{j=1}^{n-2}\frac{j}{2} \ d_{j,k}^{n}|\lambda|^{n-j-2}\langle\delta^{(j-1)},Nf(\cdot,\hat{\lambda})\rangle\\ &&{}=-(-1)^{n-1}|\lambda|^{n-1}{\int}_{s>0}\mathrm{e}^{-s|\lambda|/4}L_{k}^{(n-1)}\left( \frac{s|\lambda|}{2}\right) \left( \frac{s}{2}\right)Nf(s,\hat{\lambda}) \ \mathrm{d}s\\ &&{}~~~~+\sum\limits_{j=0}^{n-3}\frac{j+1}{2} \ d_{j+1,k}^{n}|\lambda|^{n-j-3}\langle\delta^{(j)},Nf(\cdot,\hat{\lambda})\rangle, \end{array} $$
where in the second equality we have used Lemma A.4. □
