nonclassical jacobi polynomials and sobolev orthogonality

Results. Math. 61 (2012), 283–313c© 2011 Springer Basel AG1422-6383/12/030283-31published online March 15, 2011DOI 10.1007/s00025-011-0102-4 Results in Mathematics

Nonclassical Jacobi Polynomialsand Sobolev Orthogonality

Andrea Bruder and L. L. Littlejohn

Abstract. In this paper, we consider the second-order Jacobi differentialexpression

�α,β [y](x) =−1

(1 − x)a(1 + x)−1((1 − x)α+1 y′(x))′ (x ∈ (−1, 1));

here, the Jacobi parameters are α > −1 and β = −1. This is a non-classical setting since the classical setting for this expression is generallyconsidered when α, β > −1. In the classical setting, it is well-known that

the Jacobi polynomials {P(α,β)n }∞

n=0 are (orthogonal) eigenfunctions ofa self-adjoint operator Tα,β , generated by the Jacobi differential expres-sion, in the Hilbert space L2((−1, 1); (1 − x)α(1 + x)β). When α > −1and β = −1, the Jacobi polynomial of degree 0 does not belong to theHilbert space L2((−1, 1); (1 − x)α(1 + x)−1). However, in this paper, we

show that the full sequence of Jacobi polynomials {P(α,−1)n }∞

n=0 forms acomplete orthogonal set in a Hilbert–Sobolev space Wα, generated by theinner product

φ (f, g) := f(−1)g(−1) +

1∫

−1

f ′(t)g′(t)(1 − t)α+1dt.

We also construct a self-adjoint operator Tα, generated by �α,−1[·] in Wα,

that has the Jacobi polynomials {P(α,−1)n }∞

n=0 as eigenfunctions.

Mathematics Subject Classification (2010). Primary 33C45, 34B30, 47B25;Secondary 34B20, 47B65.

284 A. Bruder and L. L. Littlejohn Results. Math.

1. Introduction

For α, β > −1, the special functions properties as well as the spectral proper-ties of the classical Jacobi differential expression

lα,β [y](x) :=1

ωα,β(x)

×[((−(1 − x)α+1(1 + x)β+1

)y′(x)

)′+ k(1 − x)α(1 + x)βy(x)

]

= −(1 − x2)y′′(x) + (α− β + (α+ β + 2)x)y′(x) + ky(x)

where k ≥ 0 is fixed, x ∈ (−1, 1) and ωα,β(x) = (1−x)α(1+x)β are well known.In this case, the nth degree Jacobi polynomial y = Pα,β

n (x) is a solution of theequation

lα,β [y](x) = (n(n+ α+ β + 1) + k) y(x) (n ∈ N0);

details of the properties of these polynomials can be found in [7,16]. The right-definite spectral analysis has been studied in [1,9]. Through theGlazman–Krein–Naimark (GKN) theory it has been known that there exists aself-adjoint operator A(α,β) generated from the Jacobi differential expressionin the Hilbert space L2((−1, 1); (1−x)α(1+x)β) having the Jacobi polynomialsas a complete set of eigenfunctions.

For α, β ≥ −1, let

L2α,β(−1, 1) := L2((−1, 1); (1 − x)α(1 + x)β) (1.1)

be the weighted Hilbert space with usual inner product

(f, g)α,β =

1∫

−1

f(x)g(x)(1 − x)α(1 + x)βdx (1.2)

and related norm ‖·‖α,β . We will study the nonclassical Jacobi case whenα > −1 and β = −1. One main difference to the classical case is that theconstant Jacobi polynomial Pα,−1

0 (x) does not belong to the Hilbert spaceL2

α,−1(−1, 1). However, it is true that the Jacobi polynomials {P (α,−1)n }∞

n=1

still form a complete orthogonal set in L2α,−1(−1, 1). Moreover, by the GKN

theory, there exists a self-adjoint differential operator A(α,−1), generated bylα,−1[·], that is positively bounded below in L2

α,−1(−1, 1), that has the set

{P (α,−1)n }∞

n=1as eigenfunctions. Furthermore, in [12], Kwon and Littlejohnshowed that the full sequence of Jacobi polynomials {P (α,−1)

n }∞n=0 are orthog-

onal in a Hilbert–Sobolev function space Wα with inner product

φ (f, g) := f(−1)g(−1) +

1∫

−1

f ′(x)g′(x)(1 − x)α+1dx. (1.3)

Vol. 61 (2012) Nonclassical Jacobi Polynomials and Sobolev Orthogonality 285

In this paper we prove that the entire sequence of Jacobi polynomials{P (α,−1)

n }∞n=0 are, in fact, complete in Wα. More importantly, we construct

a self-adjoint, positively bounded below operator Tα, generated from lα,−1[·],in Wα having the entire set of Jacobi polynomials {P (α,−1)

n }∞n=0 as a complete

set of eigenfunctions. The general left-definite theory, that was recently devel-oped by Littlejohn and Wellman [13] is, surprisingly, of paramount importancein the construction of this self-adjoint operator.

We note that, for m ∈ N, the Jacobi polynomials {P (α,−m)n }∞

n=0 areorthogonal with respect to inner products of the form (1.3) but whose inte-grand involves the mth derivative of the functions. In this respect, we referthe reader to [2–4,11,12] where general results on the Sobolev orthogonalityof the Jacobi or Gegenbauer polynomials when one or both parameters α andβ are negative integers.

2. Preliminaries: Properties of the Jacobi Polynomials

For α, β > −1, the Jacobi polynomials {P (α,β)n (x)}∞

n=0 are defined by

P (α,β)n (x) := kα,β

n

n∑j=0

(1 + α)n(1 + α+ β)n+j

j!(n− j)!(1 + α)j(1 + α+ β)n

(1 − x

2

)j

, (2.1)

where

kα,βn :=

(n!)1/2(1 + α+ β + 2n)1/2(Γ(α+ β + n+ 1))1/2

2(α+β+1)/2(Γ(α+ n+ 1))1/2(Γ(β + n+ 1))1/2;

see [7,15,16] for detailed discussions of these polynomials. For each n ∈ N0,

y = P(α,β)n (x) is a solution of the Jacobi differential equation

lα,β [y](x) = (n(n+ α+ β + 1) + k) y(x).

The Jacobi polynomials can always be normalized, in any number of ways,and we will assume that the polynomials are normalized in various spacesthroughout this paper. They form a complete orthogonal set in the Hilbertspace L2

α,β(−1, 1). In fact, they satisfy the orthogonality relation

1∫

−1

P (α,β)n (x)P (α,β)

r (x)(1 − x)α(1 + x)βdx

=2α+β+1Γ(n+ α+ 1)Γ(n+ β + 1)

(2n+ α+ β + 1)Γ(n+ α+ β + 1)n!δn,r

for α, β > −1. The derivatives of the Jacobi polynomials satisfy the identity

dj

dxjP (α,β)

n (x) = a(α,β)(n, j)P (α+j,β+j)n−j (x) (n, j ∈ N0), (2.2)


where

a(α,β)(n, j) =(n!)1/2 (Γ(α+ β + n+ 1 + j))1/2

((n− j)!)1/2 (Γ(α+ β + n+ 1))1/2(j = 0, 1, . . . , n),

and a(α,β)(n, j) = 0 if j > n. Furthermore, for n, r, j ∈ N0, we have theorthogonality relation

1∫

−1

dj(P

(α,β)n (x)

)

dxj

dj(P

(α,β)r (x)

)

dxjwα+j,β+j(x)dx (2.3)

=n!Γ(α+ β + n+ 1 + j)

(n− j)!Γ(α+ β + n+ 1)δn,r. (2.4)

When α > −1 and β = −1, the natural setting for the Jacobi polynomi-als {P (α,−1)

n } is still the Hilbert space L2α,−1(−1, 1). However, because of the

singularity in the weight function, P (α,−1)0 /∈ L2

α,−1(−1, 1). Moreover, with thedefinition of the Jacobi polynomials as given in, say [7],

P (α,−1)n (x) :=

n∑j=0

(n+ α

j

)(n− 1n− j

)(x− 1

2

)j (x+ 1

2

)n−j

,

notice that the first Jacobi polynomial P (α,−1)1 (x) is degenerate. However, any

multiple of the first degree polynomial y = x+1 will be a solution of the Jacobidifferential equation

lα,−1[y](x) = (n(n+ α) + k)y(x).

Therefore, by renormalizing the sequence of Jacobi polynomials as follows:

P(α,−1)1 (x) =

√(α+ 1)(α+ 2)

2α+2(x+ 1) ,

and, for n ≥ 2,

P (α,−1)n (x) =

√n (2n+ α)2α (n+ α)

n∑j=0

(n+ α

n− j

)(n− 1j

)(x− 1

2

)j (x+ 1

2

)n−j

,

the sequence {P (α,−1)n }∞

n=1 is an orthonormal set in L2α,−1(−1, 1). We remark

that the full set of Jacobi polynomials {P (α,−1)n }∞

n=0 cannot be orthogonal onthe real line with respect to any bilinear form of type∫

R

fgdμ,

where μ is a (possibly signed) measure. This is an application of Favard’sclassic theorem in orthogonal polynomials. However,


Lemma 1. The sequence {P (α,−1)n (x)}∞

n=1 of Jacobi polynomials of degree n ≥ 1forms a complete orthogonal set in the Hilbert space L2

α,−1(−1, 1). Equivalently,the set of all polynomials p ∈ P[−1, 1] of degree ≥ 1 satisfying p(−1) = 0 isdense in the space L2

α,−1(−1, 1).

Proof. We have

1∫

−1

|f(x)|2 (1 − x)α(1 + x)−1dx =

1∫

−1

∣∣(1 + x)−1f(x)∣∣2 (1 − x)α(1 + x)dx,

that is to say,

f ∈ L2α,−1(−1, 1) ⇐⇒ (1 + x)−1f ∈ L2

α,1(−1, 1),

and in this case,

‖f‖L2α,−1

=∥∥(1 + x)−1f

∥∥L2

α,1.

Let f ∈ L2α,−1(−1, 1) and let ε > 0. Then

(1 + x)−1f ∈ L2α,1(−1, 1);

since the Jacobi polynomials {P (α,1)n (x)}∞

n=0 forms a complete orthogonal setin L2

α,1(−1, 1), there exists q ∈ P[−1, 1] such that∥∥(1 + x)−1f − q

∥∥L2

α,1< ε.

Let p(x) := (1 + x)q(x), so deg(p) ≥ 1, p(−1) = 0, and q(x) = (1 + x)−1p(x).Then

ε >∥∥(1 + x)−1f − (1 + x)−1p

∥∥L2

α,1=∥∥(1 + x)−1(f − p)

∥∥L2

α,1= ‖f − p‖L2

α,−1.

�

3. An Operator Inequality

The following result, due to Chisholm, Everitt, and Littlejohn [6], is impor-tant in establishing our main results. The authors in [6] proved a more generalresult for conjugate indices p and q; for the purposes of this paper, we onlystate the result when p = q = 2.

Theorem 1. Suppose I = (a, b) is an open interval of the real line, where−∞ ≤ a < b ≤ ∞. Suppose w is a positive Lebesgue measurable function on(a, b) and ϕ,ψ are functions satisfying the three conditions:

(i) ϕ ∈ L2loc((a, b);w) and ψ ∈ L2

loc((a, b);w);(ii) for some c ∈ (a, b), suppose ϕ ∈ L2((a, c];w) and ψ ∈ L2([c, b);w);


(iii) for all [α, β] ⊂ (a, b),α∫

a

|ϕ(t)|2 w(t)dt > 0 and

b∫

β

|ψ(t)|2 w(t)dt > 0.

Define the linear operators A and B on L2((a, b);w) and L2((a, b);w), respec-tively, by

(Ag)(x) := ϕ(x)

b∫

x

ψ(t)g(t)w(t)dx (x ∈ (a, b) and g ∈ L2((a, b);w))

(Bg)(x) := ψ(x)

x∫

a

ϕ(t)g(t)w(t)dx (x ∈ (a, b) and g ∈ L2((a, b);w));

then

A : L2((a, b);w) → L2loc((a, b) : w)

B : L2((a, b);w) → L2loc((a, b);w).

Define K(·) : (a, b) → (0,∞) by

K(x) :=

⎧⎨⎩

x∫

a

|ϕ(t)|2 w(t)dx

⎫⎬⎭

1/2⎧⎨⎩

b∫

x

|ψ(t)|2 w(t)dt

⎫⎬⎭

1/2

(x ∈ (a, b))

and the number K ∈ (0,∞]

K := sup{K(x) | x ∈ (a, b)}.Then a necessary and sufficient condition that A and B are bounded linearoperators on L2((a, b);w) is that the number K is finite, i.e.

K ∈ (0,∞).

4. Right-Definite Analysis of the Jacobi Expression Whenα > −1 and β = −1

Recall that, for α > −1 and β = −1, the second-order Jacobi differentialexpression is defined to be

α,−1[y](x) := (1 − x2)y′′(x) − (α+ 1)(x+ 1)y′(x) + ky(x)

=1

wα,−1(x)[((1 − x)α+1y′(x))′ + k(1 − x)α(1 + x)−1y], (4.1)

where wα,−1(x) := (1−x)α(1+x)−1, x ∈ (−1, 1) and k is a fixed, non-negativeconstant.

Both points x = ±1 are regular singular endpoints of lα,−1[·] in the senseof Frobenius. In fact, in the case α ≥ 1 and β = −1, both endpoints x = ±1


are limit-point in L2α,−1(−1, 1), defined in (1.1), and thus the right-definite

GKN self-adjoint operator is unique and there are no boundary conditionsprescribed in the domain of A(α,−1). For −1 < α < 1, the endpoint x = −1 isin the limit-point condition, whereas x = +1 is in the limit-circle case. There-fore, when −1 < α < 1, one boundary condition is necessary to define theself-adjoint operator A(α,−1) having the Jacobi polynomials {P (α,−1)

n }∞n=1 as

eigenfunctions.For α > −1, the maximal domain Δα,−1 of α,−1[·] in the Hilbert space

L2α,−1(−1, 1) is defined to be

Δα,−1 :={f : (−1, 1) → C |f, f ′ ∈ ACloc(−1, 1); f, α,−1[f ] ∈ L2

α,−1(−1, 1)}

={f : (−1, 1) → C |f, f ′ ∈ ACloc(−1, 1);

(1 − x)α/2

(1 + x)1/2f ∈ L2(−1, 1);

(1 − x)−α/2(1 + x)1/2((1 − x)α+1f ′(x)

)′ ∈ L2(−1, 1)}.

The maximal operator T (α,−1)max associated with lα,−1[·] is given by

T (α,−1)max (f) := lα,−1[f ]

D(T (α,−1)max ) := Δα,−1.

The minimal operator is defined as T (α,−1)min := (T (α,−1)

max )∗, the Hilbert spaceadjoint of T (α,−1)

max . The operator T (α,−1)min is closed, symmetric and satisfies

(T (α,−1)min )∗ = T (α,−1)

max .

The deficiency index d(T (α,−1)min ) of T (α,−1)

min is

d(T (α,−1)min ) =

{(0, 0) if β = −1, α ≥ 1(1, 1) if β = −1, −1 < α < 1.

This can be seen from the limit-point/limit-circle classification of the singularendpoints x = ±1:

(i) x = ±1 are limit-point if β = −1, α ≥ 1 and(ii) x = −1 is limit-point, x = 1 is limit-circle if β = −1,−1 < α < 1.

Consequently, by von Neumann’s theory of self-adjoint extensions of symmetricoperators ([8], chapter XII), T (α,−1)

min has self-adjoint extensions in L2α,−1(−1, 1)

for α > −1 and β = −1.We define the operator A(α,−1) : D(A(α,−1)) ⊂ L2

α,−1(−1, 1) → L2α,−1

(−1, 1) as follows:{

(A(α,−1)f)(x) := lα,−1[f ](x) (a.e. x ∈ (−1, 1))f ∈ D(A(α,−1)),

(4.2)


where D(A(α,−1)) is defined by

D(A(α,−1)) :=

{{f ∈ Δα,−1 | limx→1−(1 − x)α+1f ′(x) = 0} if − 1 < α < 1

Δα,−1 if α ≥ 1.

(4.3)

From the GKN theory, A(α,−1) is self-adjoint in L2α,−1(−1, 1). More impor-

tantly, from our point of view, the Jacobi polynomials {P (α,−1)n }∞

n=1 of degree≥ 1 form a complete set of eigenfunctions of A(α,−1); see Lemma 1. Further-more, the spectrum of A(α,−1) is discrete and given by

σ(A(α,−1)) = {n(n+ α) | n ∈ N}.In order to develop left-definite properties of A(α,−1) which, in turn, are

important to construct the self-adjoint operator Tα in the Sobolev space Wα,we turn our attention to developing properties of functions in the spaces Δα,−1

and D(A(α,−1)).Observe that if f, g ∈ Δα,−1, then α,−1[f ]g ∈ L2

α,−1(−1, 1). Furthermore,for f, g ∈ Δα,−1, Green’s formula gives us

x∫

0

(1 − t)α+1f ′(t)g′(t)dt = (1 − x)α+1f ′(x)g(x) − f ′(0)g(0)

−x∫

0

α,−1[f ](t)g(t)(1 − t)α(1 + t)−1dt

+k

x∫

0

f(t)g(t)(1 − t)α(1 + t)−1dt (−1 < x < 1).

(4.4)

Lemma 2. For α > −1 and f ∈ Δα,−1, we have f ′ ∈ L2(−1, 0). In particular,we can define f at x = −1 such that f ∈ AC[−1, 0]. Moreover, it is necessarythat f(−1) = 0.

Proof. Let f ∈ Δα,−1. Note that

f ′(x) = −(1 − x)−α−1

0∫

x

(1 − t)−α/2(1 + t)1/2

(1 − t)−α/2(1 + t)1/2((1 − t)α+1f ′(t))′dt

+(1 − x)−α−1f ′(0) (x ∈ (−1, 0]).

From the definition of Δα,−1, we see that (1−t)−α/2(1+t)1/2((1−t)α+1f ′(t))′ ∈L2(−1, 1); consequently, we apply the inequality in Theorem 1 on the interval(−1, 0] with

ϕ(x) = (1 − x)−α−1, ψ(x) =1

(1 − x)−α/2(1 + x)1/2(x ∈ (−1, 0]).


Since the function 1 − t is bounded on (−1, 0], there exists a constant M suchthat

x∫

−1

ϕ2(t)dt

0∫

x

ψ2(t)dt =

x∫

−1

(1 − t)−2α−2dt

0∫

x

(1 − t)α(1 + t)−1dt

≤ M

x∫

−1

dt

0∫

x

(1 + t)−1dt = −M(x+ 1) ln(1 + x);

since (x + 1) ln(1 + x) is bounded on (−1, 0], the inequality from Theorem 1implies that f ′ ∈ L2(−1, 0) as claimed. From the identity

f(x) = f(0) −0∫

x

f ′(t)dt (−1 < x ≤ 0),

define

f(−1) := f(0) −0∫

−1

f ′(t)dt.

With f defined this way, we see that f ∈ AC[−1, 0].We now show that f(−1) =0. We suppose that f is real-valued and, by way of contradiction and withoutloss of generality, we assume that f(−1) > 0. Then there exists x∗ ∈ (−1, 0]such that f(t) ≥ c for all t ∈ (−1, x∗], where c is some positive number. Since(1 − x)α is bounded below on (−1, 0] by a constant K > 0, we see that

−∞ >

1∫

−1

|f(t)|2 (1 − t)α(1 + t)−1dt

≥x∗∫

−1

|f(t)|2 (1 − t)α(1 + t)−1dt ≥ Kc2x∗∫

−1

(1 + t)−1dt = ∞,

a contradiction. Hence f(−1) = 0. �

Lemma 3. For α > −1 and f ∈ Δα,−1, we have (1 − x)(α+1)/2f ′ ∈ L2(−1, 0).

Proof. Suppose, to the contrary, that

0∫

−1

|f ′(t)|2 (1 − t)α+1dt = ∞.


We assume, without loss of generality, that f is real-valued. From (4.4), we seethat

0∫

x

(1 − t)α+1(f ′(t))2dt = f ′(0)f(0) − (1 − x)α+1f ′(x)f(x)

−0∫

x

α,−1[f ](t)f(t)(1 − t)α(1 + t)−1dt

+k

x∫

0

(f(t))2(1 − t)α(1 + t)−1dt (−1 < x ≤ 0);

(4.5)

consequently, it must be the case that

limx→−1+

(1 − x)α+1f ′(x)f(x) = −∞,

and hence

limx→−1+

f ′(x)f(x) = −∞.

It follows that there exists x∗ ∈ (−1, 0], f ′(x)f(x) ≤ −1 for x ∈ (−1, x∗].Integrating, we see that

(f(x∗))2

2− (f(x))2

2=

x∗∫

x

f ′(t)f(t)dt ≤ −1

x∗∫

x

dt = x− x∗

so that

(f(x))2 ≥ −2x+ (f(x∗))2 + 2x∗ ≥ −2(x− x∗) (x ∈ (−1, x∗]). (4.6)

With

Mα = mint∈(−1,x∗]

(1 − t)α,

after integrating the inequality in (4.6), we see that

x∗∫

x

(f(t))2(1 − t)α(1 + t)−1dt ≥ −2

x∗∫

x

(t− x∗)(1 − t)α(1 + t)−1dt

≥ −2Mα

x∗∫

x

(t− x∗)(1 + t)−1dt

→ ∞ as x → −1+.


However, this contradicts the fact that f ∈ Δα,−1, and, in particular, that

1∫

−1

|f(t)|2 (1 − t)α(1 + t)−1dt < ∞.

�

Lemma 4. Let f, g ∈ Δα,−1. Then

limx→−1+

(1 − x)α+1f ′(x)g(x) = 0.

Proof. Let f, g ∈ Δα,−1; assume that both f and g are real-valued. We notethat, from Lemma 2, that f(−1) = g(−1) = 0. From (4.5) and Lemma 3, wesee that

limx→−1+

(1 − x)α+1f ′(x)g(x)

exists and is finite. Suppose, then, that

limx→−1+

(1 − x)α+1f ′(x)g(x) = c > 0.

Then we can assume there exists x∗ ∈ (−1, 0] such that

f ′(x) > 0, g(x) > 0, and (1 − x)α+1f ′(x) ≥ c

2g(x)(x ∈ (−1, x∗]).

Multiply this inequality by |g′(x)| to get

(1 − x)α+1f ′(x) |g′(x)| ≥ c

2|g′(x)|g(x)

(x ∈ (−1, x∗]).

Now integrate to obtain

∞ >

1∫

−1

(1 − t)α+1f ′(t) |g′(t)| dt (using Lemma 3)

≥x∗∫

x

(1 − t)α+1f ′(t) |g′(t)| dt ≥ c

2

x∗∫

x

|g′(t)|g(t)

dt ≥ c

2

∣∣∣∣∣∣x∗∫

x

g′(t)g(t)

dt

∣∣∣∣∣∣=c

2|ln g(x∗) − ln g(x)| → ∞ as x → −1+ since g(−1) = 0.

This contradiction gives us the required result. �

We summarize the results from Lemmas 2, 3, and 4.

Theorem 2. Assume α > −1 and β = −1. Let f, g ∈ Δα,−1. Then(a) f ′ ∈ L2(−1, 0);(b) f ∈ AC[−1, 0];


(c) The Jacobi differential expression (4.1) is strong limit point at x = −1;that is to say,

limx→−1+

(1 − x)α+1f ′(x)g(x) = 0;

(d) The Jacobi differential expression (4.1) is Dirichlet at x = −1; that is tosay,

0∫

−1

(1 − x)α+1 |f ′(x)|2 dx < ∞.

As a consequence of this theorem, note that for f, g ∈ Δα,−1, and for−1 < x ≤ 0,

0∫

−1

(1 − t)α+1f ′(t)g′(t)dt = f ′(0)g(0) −0∫

x

α,−1[f ](t)g(t)(1 − t)α(1 + t)−1dt.

(4.7)

We now turn our attention to establishing that the Jacobi differentialexpression (4.1) is both strong-limit point and Dirichlet at x = 1 on D(A(α,−1)),defined in (4.3).

Lemma 5. Let α > −1. Then, for all f ∈ D(A(α,−1)),

limx→1−

(1 − x)α+1f ′(x) = 0.

Proof. From the definition of D(A(α,−1)) in ( 4.3), this result is automaticallytrue for −1 < α < 1. So we assume α ≥ 1. Let f ∈ D(A(α,−1)). Since theJacobi expression is limit point at x = 1 in this case, we know that

limx→1−

(1 − x)α+1[f ′(x)g(x) − f(x)g′(x)] = 0 (g ∈ Δα,−1).

Construct a real-valued g ∈ C2[−1, 1] such that

g(x) ={

0 for x near − 11 for x near + 1;

for example, we could take

g(x) =

⎧⎨⎩

0 if − 1 < x ≤ 0−16x3 + 12x2 if 0 < x < 1/21 if 1/2 ≤ x ≤ 1.

Clearly g ∈ Δα,−1; moreover, with this choice of g, we have

0 = limx→1−

(1 − x)α+1[f ′(x)g(x) − f(x)g′(x)

= limx→1−

(1 − x)α+1f ′(x).

�


The proof of the following lemma is similar to the proof given in Lemma 4;we omit the details. This result shows that the Jacobi expression is Dirichletat x = 1 on D(A(α,−1)).

Lemma 6. Suppose α > −1. Then, for any f ∈ D(A(α,−1)), we have

(1 − x)(α+1)/2f ′ ∈ L2(0, 1).

We are now in position to prove that the Jacobi expression (4.1) is stronglimit point at x = 1 on D(A(α,−1)). The proof is similar to the one of Lemma 4but different enough to warrant a demonstration.

Lemma 7. Suppose α > −1. Then for f, g ∈ D(A(α,−1)), we have

limx→1−

(1 − x)α+1f ′(x)g(x) = 0.

Proof. Let f, g ∈ D(A(α,−1)). From (4.4), the definition of D(A(α,−1)), andLemma 6, we see that

limx→1−

(1 − x)α+1f ′(x)g(x)

exists and is finite. Without loss of generality, assume that both f and g arereal-valued on (−1, 1); furthermore, by way of contradiction, we assume that

limx→1−

(1 − x)α+1f ′(x)g(x) = c > 0.

Again, we assume that there exists x∗ ∈ [0, 1) such that

(1 − x)α+1f ′(x) > 0, g(x) > 0, (1 − x)α+1f ′(x)g(x) ≥ c/2 (x ∈ [x∗, 1)).

Hence it follows that

∣∣∣((1 − x)α+1f ′(x))′∣∣∣ g(x) ≥ c

2

∣∣∣((1 − x)α+1f ′(x))′∣∣∣

(1 − x)α+1f ′(x)(x ∈ [x∗, 1)).

Integrate over [x∗, x] to get

∞ >

1∫

−1

α,−1[f ](t)g(t)(1 − t)α(1 + t)−1dt− k

1∫

−1

(f(t))2(1 − t)α(1 + t)−1dt

=

1∫

−1

∣∣∣((1 − t)α+1f ′(t))′∣∣∣ g(t)dt ≥

x∗∫

x

∣∣∣((1 − t)α+1f ′(t))′∣∣∣ g(t)dt

≥ c

2

x∗∫

x

∣∣∣((1 − t)α+1f ′(t))′∣∣∣

(1 − t)α+1f ′(t)dt ≥ c

2

∣∣∣∣∣∣x∗∫

x

((1 − t)α+1f ′(t)

)′

(1 − t)α+1f ′(t)dt

∣∣∣∣∣∣=c

2

∣∣ln((1 − x∗)α+1f ′(x∗)) − ln((1 − x)α+1f ′(x))∣∣ → ∞ as x → 1−

by Lemma 5. This contradiction gives us the required result. �


Summarizing Lemmas 6 and 7, we have the following theorem.

Theorem 3. For α > −1 and β = −1, the Jacobi differential expression α,−1[·],given in (4.1), is strong limit point and Dirichlet at x = 1 on D(A(α,−1)). Thatis to say,(a) limx→1−(1 − x)α+1f ′(x)g(x) = 0 for all f, g ∈ D(A(α,−1)), and(b)

∫ 1

0(1 − x)α+1 |f ′(x)|2 dx < ∞ for all f ∈ D(A(α,−1)).

Combining Theorems 2 and 3, we see that Dirichlet’s formula, for f, g ∈D(A(α,−1)), over the interval (−1, 1) reads:

(A(α,−1)f, g)α,−1 =

1∫

−1

[(1 − t)α+1f ′(t)g′(t) + kf(t)g(t)(1 − t)α(1 + t)−1]dt.

(4.8)

In particular, we see that

(A(α,−1)f, f)α,−1 ≥ k(f, f)α,−1 (f ∈ D(A(α,−1))); (4.9)

that is to say, A(α,−1) is bounded below by kI in L2α,−1(−1, 1). The conse-

quence of the inequality in (4.9) is that the left-definite theory, developed byLittlejohn and Wellman, can be applied to A(α,−1). We now briefly review thistheory.

5. General Left-Definite Theory

In [13], Littlejohn and Wellman developed a general abstract left-definite the-ory for a self-adjoint operator A that is bounded below in a Hilbert space(H, (·, ·)).

Let V be a vector space over C with inner product (·, ·) such that H :=(V, (·, ·)) is a Hilbert space. Let r > 0 and suppose that Vr is a vector sub-space of V with inner product (·, ·)r; denote this inner product space by Wr :=(Vr, (·, ·)r). Let A : D(A) ⊂ H → H be a self-adjoint operator that is boundedbelow by rI for some r > 0, that is to say

(Ax, x) ≥ r(x, x) (x ∈ D(A)).

Then, for any s > 0, the operator As is self-adjoint and bounded below in Hby rsI.

Definition 1. Let s > 0 and let Vs be a vector subspace of the Hilbert spaceH = (V, (·, ·)) with inner product (·, ·)s. We say that Ws = (Vs, (·, ·)s is an sth

left-definite space associated with the pair (H,A) if

(i) Ws is a Hilbert space(ii) D(As) is a vector subspace of Vs

(iii) D(As) is dense in Ws

(iv) (x, x)s ≥ rs(x, x) for all x ∈ Vs


(v) (x, y)s = (Asx, y) for all x ∈ D(As), y ∈ Vs.

Littlejohn and Wellman in [13, Theorem 3.1] prove the following result.

Theorem 4. Let A : D(A) ⊂ H → H be a self-adjoint operator that is boundedbelow by rI for some r > 0. Let s > 0 and define Ws := (Vs, (·, ·)s) by

Vs = D(As/2)

and

(x, y)s = (As/2x,As/2y) (x, y ∈ Vs).

Then Ws is the unique left-definite space associated with the pair (H,A).

Definition 2. For s > 0, let Ws := (Vs, (·, ·)s) be the sth left-definite spaceassociated with (H,A). If there exists a self-adjoint operator Bs : D(Bs) ⊂Ws → Ws satisfying

Bsf = Af (f ∈ D(Bs) ⊂ D(A)),

we call such an operator an sth left-definite operator associated with the pair(H,A).

In [13, Theorem 3.2], the authors prove the following existence/unique-ness result.

Theorem 5. Let A be a self-adjoint operator in a Hilbert space H that isbounded below by rI for some r > 0. For any s > 0, let Ws := (Vs, (·, ·)s)denote the sth left-definite space associated with (H,A). Then there exists aunique left-definite operator Bs in Ws associated with (H,A). Furthermore,

D(Bs) = Vs+2 ⊂ D(A).

We refer the reader to other results established in [13]. We note if, inaddition to the hypotheses assumed in this section, that A is bounded thenthe left-definite theory is trivial; specifically, V = Vs and A = As for alls > 0. Only in the unbounded case will there be a non-trivial left-definite the-ory associated with (H,A). Moreover, we note that the authors in [13] proveσ(A) = σ(As) for all s > 0; in fact, the point spectrum σp and continuousspectrum σc match up for A and each left-definite operator As.

6. Left-Definite Theory of the Nonclassical Jacobi DifferentialEquation

Let k > 0. For each n ∈ N, define

V (α,−1)n := {f : (−1, 1) → C|f ∈ AC

(n−1)loc ; f (j) ∈ L2

α+j,j−1(−1, 1), j = 0, . . . , n}


and let (·, ·)(α,−1)n and ‖·‖(α,−1)

n denote, respectively, the Sobolev inner product

(f, g)(α,−1)n :=

n∑j=0

c(α,−1)j (n, k)

1∫

−1

f (j)(t)g(j)(t)(1 − t)α+j(1 + t)j−1dt

(f, g ∈ V(α,−1)n ), and norm ‖f‖(α,−1)

n :=((f, f)(α,−1)

n

)1/2

, where the numbers

c(α,−1)j (n, k) are defined as in [9] by

c(α,−1)0 (n, k) :=

{0 if k = 0kn if k > 0 ,

and, for j ∈ {1, 2, . . . , n},

c(α,−1)j (n, k) :=

{P (α,−1)S

(j)n if k = 0∑n−j

s=0

(ns

)P (α,−1)S

(j)n−sk

s if k > 0,

where the Jacobi–Stirling number P (α,−1)S(j)n of order (n, j) is defined by

P (α,−1)S(j)n :=

j∑r=0

(−1)r+j Γ(α+ r)Γ(α+ 2r + 1)[r(r + α)]n

r!(j − r)!Γ(α+ 2r)Γ(α+ j + r + 1),

(n, j ∈ N; j ≤ n). We extend this definition to include

P (α,−1)S(0)0 := 1, P (α,−1)S(j)

n := 0 if j ∈ N and 0 ≤ n ≤ j − 1,

P (α,−1)S(0)n := 0 for n ∈ N.

In [9], the authors prove that P (α,−1)S(j)n > 0 for n, j ∈ N, j ≤ n. Let

W (α,−1)n :=

(V (α,−1)

n , (·, ·)(α,−1)n

).

We will show that the vector space W (α,−1)n is the nth left-definite space

associated with the pair (L2α,−1(−1, 1), A(α,−1)), where A(α,−1) is the self-

adjoint Jacobi operator defined in (4.2) and (4.3). Using the results from [9]mutatis mutandis, Theorem 6 follows; for a proof see [5].

Theorem 6. Let k > 0. For each n ∈ N,W(α,−1)n is a Hilbert space.

Theorem 7. The Jacobi polynomials {P (α,−1)m }∞

m=1 form a complete orthogo-nal set in each W

(α,−1)n . Equivalently, the set of polynomials P is dense in

W(α,−1)n .

Proof. Fix n ∈ N , and let f ∈ W(α,−1)n , so

f (n) ∈ L2α+n,n−1(−1, 1).


Since {P (α+n,n−1)m }∞

m=0 is a complete orthonormal set in L2α+n,n−1(−1, 1), we

knowr∑

m=0

c(α,−1)m,n P (α+n,n−1)

m → f (n) as r → ∞ in L2α+n,n−1(−1, 1) (6.1)

where c(α,−1)m,n are the Fourier coefficients given by

c(α,−1)m,n =

1∫

−1

f (n)(t)P (α+n,n−1)m (t)(1 − t)α+n(1 + t)n−1dt

for m ∈ N0. For r ≥ n define the polynomials

pr(t) :=r∑

m=max{2,n}

c(α,−1)m−n,n ((m− n)!)1/2 (Γ(α+m))1/2

(m!)1/2 (Γ(α+m+ n)!)1/2P (α,−1)

m (t).

From

dj

dtjP (α,−1)

m (t) =(m!)1/2 (Γ(α+m+ j))1/2

((m− j)!)1/2 (Γ(α+m))1/2P

(α+j,j−1)m−j (t),

we see that, for j = 0, 1, . . . , n,

p(j)r (t) =

r∑m=max{2,n}

c(α,−1)m−n,n ((m− n)!)1/2 (Γ(α+m+ j)!)1/2

(Γ(α+m+ n)!)1/2 ((m− j)!)1/2P

(α+j,j−1)m−j (t).

In particular, by (6.1),

p(n)r (t) =

r∑m=max{2,n}

c(α,−1)m−n,nP

(α+n,n−1)m−n =

r−max{2,n}∑l=0

c(α,−1)l,n P

(α+n,n−1)l

=s∑

m=0

c(α,−1)m,n P (α+n,n−1)

m → f (n)

as r → ∞ in L2α+n,n−1(−1, 1). Furthermore, by Riesz-Fischer, there exists a

subsequence {p(n)rj } of {p(n)

r } such that

p(n)rj

→ f (n) for a.e. t ∈ (−1, 1).

By Dirichlet’s test, the sequence{c(α,−1)m−n,n ((m− n)!)1/2 (Γ(α+m+ j)!)1/2

(Γ(α+m+ n)!)1/2 ((m− j)!)1/2

}∈ 2,

so there exists a gj ∈ L2α+j,j−1(−1, 1) such that

prj→ gj in L2

α+j,j−1(−1, 1). (6.2)


For a.e. a, t ∈ (−1, 1),t∫

a

p(n)rj

(u)du →t∫

a

f (n)(u)du.

Integrate both sides and obtain

p(n−1)rj

(t) → f (n−1)(t) + c1 for a.e. t ∈ (−1, 1) (6.3)

for some constant c1. Passing through the subsequence implies

gn−1(t) = f (n−1)(t) + c1 for a.e. t ∈ (−1, 1).

From (6.3), we see thatt∫

a

p(n−1)rj

(u)du →t∫

a

f (n−1)(u)du+ c1

t∫

a

du,

that is,

p(n−2)rj

(t) → f (n−2)(t) + c1t+ c2 for a.e. t ∈ (−1, 1)

or

gn−2(t) = f (n−2)(t) + c1t+ c2 for a.e. t ∈ (−1, 1).

Continue this process to see that for j ∈ {0, 1, . . . , n− 1} ,gj(t) = f (j)(t) + qn−j+1 for a.e. t ∈ (−1, 1),

where qn−j−1 is a polynomial of degree ≤ n−j−1 and where q′n−j−1 = qn−j−2.

Hence, with (6.2),

p(j)r → f (j) + qn−j−1 in L2

α+j,j−1(−1, 1). (6.4)

For r ≥ n, define πr(t) := pr(t) − qn−1(t). Note that, with (6.4),

π(j)r (t) = p(j)

r (t) − q(j)n−1(t) = p(j)

r (t) − qn−j−1(t) → f (j)(t).

Hence, as r → ∞,

(‖f − πr‖(α,−1)

n,k

)2

=n∑

j=0

c(α,−1)j (n, k)

×−1∫

−1

∣∣∣f (j)(t) − π(j)r (t)

∣∣∣2 (1 − t)α+j(1 + t)j−1dt → 0.

�

Lemma 8. For p, q ∈ P, the vector space of all polynomials p : [−1, 1] → C,and for n ≥ 1,

(p, q)(α,−1)n =

((A(α,−1)

)n

p, q)

α,−1.


Proof. First we note that this may be restated as

(lnα,−1[p], q

)α,−1

=

1∫

−1

lnα,−1[p](x)q(x)wα,−1(x)dx

=n∑

j=0

c(α,−1)j (n, k)p(j)(x)q(j)(x)(1 − x)j+α(1 + x)j−1dx.

(6.5)

Since the Jacobi polynomials form a basis for P, it suffices to prove (6.5) forp = P

(α,−1)m and q = P

(α,−1)r for arbitrary m, r ∈ N0. From

lnα,−1[P(α,−1)m ](x) = (m(m− 1) + k)nP (α,−1)

m (x) (m ∈ N0)

and (P (α,−1)

r , P (α,−1)m

)α,−1

= δr,m (r,m ∈ N0) ,

the left-hand side of (6.5) becomes

(lnα,−1[P

(α,−1)m ], P (α,−1)

r

)α,−1

=

1∫

−1

lnα,−1[P(α,−1)m ](x)P (α,−1)

r (x)wα,−1(x)dx

= (m(m− 1) + k)nδr,m. (6.6)

Upon using (2.3) for α > −1, β = −1 and the recurrence relation for thec(α,−1)j (n, k), that is,

(m(m+ α) + k)n =n∑

j=0

c(α,−1)j (n, k)

m!(m+ α+ j − 1)!(m− j)!(m+ α− 1)!

the right-hand side of (6.5) becomesn∑

j=0

c(α,−1)j (n, k) (6.7)

×1∫

−1

(P (α,−1)

m (x))(j)

(x)(P

(α,−1)r (x)

)(j)

(x)(1 − x)j−1(1 + x)j−1dx

=n∑

j=0

c(α,−1)j (n, k)

m!(m+ α+ j − 1)!(m− j)!(m+ α− 1)!

δr,m (6.8)

= (m(m+ α) + k)nδr,m.

Comparing (6.6) and (6.8) completes the proof of the lemma. �


Theorem 8. For k > 0, let A(α,−1) be the Jacobi self-adjoint operator inL2

α,−1(−1, 1), defined in (4.2) and (4.3), having the sequence of Jacobi polyno-

mials {P (α,−1)m }∞

m=1 as eigenfunctions. For each n ∈ N, let

V (α,−1)n =

{f : (−1, 1) → C | f∈AC(n−1)

loc ; f (j) ∈ L2α+j,j−1(−1, 1), j=0, . . . , n

}

and, for f, g ∈ V(α,−1)n ,

(f, g)(α,−1)n =

n∑j=0

c(α,−1)j (n, k)

1∫

−1

f (j)(t)g(j)(t)(1 − t)α+j(1 + t)j−1dt.

Then W(α,−1)n =

(V

(α,−1)n , (·, ·)(α,−1)

n

)is the nth left-definite space associ-

ated with the pair(L2

α,−1(−1, 1), A(α,−1)). Moreover, the Jacobi polynomials

{P (α,−1)m }∞

m=1 form a complete orthogonal set in each W (α,−1)n , and they satisfy

the orthogonality relation(P (α,−1)

m , P(α,−1)l

)n

= (m(m− 1) + k)nδm,l.

Furthermore, define

B(α,−1)n := D

(B(α,−1)

n

)⊂ W (α,−1)

n → W (α,−1)n

by

B(α,−1)n f := lα,−1 [f ]

(f ∈ D

(B(α,−1)

n

):= V

(α,−1)n+2

).

Then B(α,−1)n is the nth left-definite operator associated with (L2

α,−1(−1, 1),

A(α,−1)). Lastly, the spectrum of B(α,−1)n is given by

σ(B(α,−1)

n

)= {m(m− 1) + k | m ∈ N0} = σ{A(α,−1)},

with the Jacobi polynomials {P (α,−1)m }∞

m=1 forming a complete set of eigen-functions of each B(α,−1)

n .

Proof. Let n ∈ N. We need to show that W (α,−1)n satisfies the five properties

in Definition 1 in Sect. 5. (i) W (α,−1)n is a Hilbert space (see Theorem 6 and

Ref. [5]). (ii) We need to show

D((A(α,−1)

)n)⊂ W (α,−1)

n ⊂ L2α,−1(−1, 1).

Let f ∈ D((A(α,−1)

)n). Since the Jacobi polynomials {P (α,−1)

m }∞m=1 form a

complete orthonormal set in L2α,−1(−1, 1), we see that

pj → f in L2α,−1(−1, 1) as j → ∞ (6.9)


where

pj(t) :=j∑

m=0

c(α,−1)m P (α,−1)

m (t) (t ∈ (−1, 1)) ,

and, for m ∈ N0,

c(α,−1)m :=

(f, P (α,−1)

m

)α,−1

=

1∫

−1

f(t)P (α,−1)m (t) (1 − t)α (1 + t)−1dt.

Since(A(α,−1)

)nf ∈ L2

α,−1(−1, 1), we see that as j → ∞,

j∑m=0

c (α,−1)m P (α,−1)

m →(A(α,−1)

)n

f in L2α,−1(−1, 1)

where

c(α,−1)m :=

((A(α,−1)

)n

f, P (α,−1)m

)−1,−1

=(f,(A(α,−1)

)n

P (α,−1)m

)α,−1

= (m(m+ α) + k)n(f, P (α,−1)

m

)α,−1

= (m(m+ α) + k)nc(α,−1)m ;

consequently, (A(α,−1)

)n

pj →(A(α,−1)

)n

f

in L2α,−1(−1, 1) as j → ∞. Moreover, by Lemma 8,

(‖pj − pr‖(α,−1)

n

)2

=((A(α,−1)

)n

[pj − pr] , pj − pr

)α,−1

→ 0 as j, r → ∞so {pj}∞

j=0 is Cauchy inW (α,−1)n . SinceW (α,−1)

n is a Hilbert space (Theorem 6),there exists

g ∈ W (α,−1)n ⊂ L2

α,−1(−1, 1)

such that

pj → g in W (α,−1)n as j → ∞.

Furthermore, since

(f, f)(α,−1)n =

n∑j=0

c(α,−1)j (n, k)

∥∥∥f (j)∥∥∥2

j+α,j−1

≥ c(α,−1)0 (n, k)

∥∥∥f (j)∥∥∥2

α,−1= kn (f, f)α,−1 ,

we see that

‖pj − g‖α,−1 ≤ k−n/2 ‖pj − g‖(α,−1)n ,


and hence,

pj → g in L2α,−1(−1, 1). (6.10)

Comparing (6.9) and (6.10), f = g ∈ W(α,−1)n .

(iii) We need to show: D((A(α,−1))n) is dense in W (α,−1)n . Since the set of poly-

nomials is contained in D((A(α,−1))n) and is dense in W (α,−1)n (by Theorem 7),

D((A(α,−1))n) is dense in W(α,−1)n . Furthermore, from Theorem 7, the Jacobi

polynomials {P (α,−1)m }∞

m=1 form a complete orthonormal set in W (α,−1)n .

(iv) We need to show that (f, f)(α,−1)n ≥ kn(f, f)α,−1 for f ∈ V

(α,−1)n . This is

clear from the definition of (·, ·)(α,−1)n .

(v) We need to show: (f, g)(α,−1)n = ((A(α,−1))nf, g)α,−1 for f ∈ D((A(α,−1))n)

and g ∈ V(α,−1)n . This is true for any f, g ∈ P by Lemma 8. Let f ∈

D((A(α,−1))n) ⊂ W(α,−1)n , g ∈ W

(α,−1)n . Since the set of polynomials P is

dense in both W(α,−1)n and L2

α,−1(−1, 1), and since convergence in the space

W(α,−1)n implies convergence in L2

α,−1(−1, 1) (by (iv)), there exist sequences{pj}∞

j=0 and {qj}∞j=0 such that

pj → f in W (α,−1)n as j → ∞(

A(α,−1))n

pj →(A(α,−1)

)n

f

in L2α,−1(−1, 1) as j → ∞ and

qj → g

in W (α,−1)n and L2

α,−1(−1, 1) as j → ∞. Hence, by Lemma 8,((A(α,−1)

)n

f, g)

−1,−1= lim

j→∞

((A(α,−1)

)n

pj , qj

)α,−1

= limj→∞

(pj , qj)n

= (f, f)(α,−1)n .

The results listed in the theorem on B(α,−1)n and the spectrum of B(α,−1)

n followimmediately from the general left-definite theory. �

7. The Sobolev Orthogonality of the Jacobi Polynomials

If we renormalize the Jacobi polynomials as follows:

P(α,−1)0 (x) := 1, P

(α,−1)1 (x) :=

(α+ 22α+2

)1/2

(x+ 1),

and, for n ≥ 2,


P (α,−1)n (x) :=

(2n+ α)12

2n+α/2 (n+ α)

n∑j=0

(n+ α

n− j

)(n− 1j

)(x− 1

2

)j (x+ 1

2

)n−j

,

we obtain the following theorem; a proof can be found in [12]. A key result inestablishing this result is in the following important identity

P (α,−1)n (x) =

(n+ α)!(n− 1)!2n!(n+ α− 1)!

(x+ 1)P (α,1)n−1 (x) (n ≥ 2)

whose proof can easily be checked.

Theorem 9. The Jacobi polynomials {P (α,−1)n (x)}∞

n=0 are orthonormal withrespect to the Sobolev inner product

φ (f, g) := f(−1)g(−1) +

1∫

−1

(1 − x)α+1f ′(x)g′(x)dx,

that is,

φ(P (α,−1)

n , P (α,−1)m

)= δnm (n,m ∈ N0).

8. Jacobi Polynomials and a Self-Adjoint Operator in a SobolevSpace

Definition 3. Define

Wα :={f : [−1, 1) → C | f ∈ AC [−1, 1) ; f ′ ∈ L2

α+1,0(−1, 1)}

φ(f, g) := f(−1)g(−1) +

1∫

−1

f ′(x)g′(x)(1 − x)α+1dx (f, g ∈ Wα).

Write ‖f‖2φ = φ(f, f) for f ∈ Wα.

Theorem 10. (Wα, φ(·, ·)) is a Hilbert space.

Proof. Let {fn} ⊂ W1 be a Cauchy sequence. Hence

‖fn − fm‖2φ = |fn(−1) − fm(−1)|2 +

1∫

−1

|f ′n(x) − f ′

m(x)|2 (1 − x)α+1dx

→ 0 as n,m → ∞.

In particular, since1∫

−1

|f ′n(x) − f ′

m(x)|2 (1 − x)α+1dx ≤ ‖fn − fm‖2φ ,

we see that {f ′n} is Cauchy in L2((−1, 1); (1 − x)α+1).


Since L2((−1, 1); (1 − x)α+1) is complete, there exists g ∈ L2α+1,0(−1, 1)

such that

f ′n → g as n → ∞ in L2

α+1,0(−1, 1). (8.1)

Also, since

|fn(−1) − fm(−1)|2 ≤ ‖fn − fm‖2φ

we see that the sequence {fn(−1)} is Cauchy in C and, hence, there existsA ∈ C such that

fn(−1) → A. (8.2)

Furthermore, since fn ∈ AC[−1, 1)(n ∈ N), we see that

1∫

−1

f ′n(t)(1 − t)α+1dt →

1∫

−1

g(t)(1 − t)α+1dt,

Since g ∈ AC[−1, 1), we may define f : [−1, 1) → C by

f(x) = A+

x∫

−1

g(t)dt.

It is clear that f ∈ AC[−1, 1) and f ′(x) = g(x) ∈ L2α+1,0(−1, 1) for a.e.

x ∈ [−1, 1), so f ∈ Wα. Furthermore, f(−1) = A. Now

‖fn − f‖2φ = |fn(−1) − f(−1)|2 +

1∫

−1

|f ′n(t) − f ′(t)|2 (1 − t)α+1dt

= |fn(−1) −A|2 +

1∫

−1

|f ′n(t) − g(t)|2 (1 − t)α+1dt → 0

as n → ∞ by (8.1) and (8.2). Thus, (Wα, φ(·, ·)) is complete. �

Theorem 11. Let Wα and φ(·, ·) be as before, and

Wα,1 := {f ∈ Wα | f(−1) = 0}Wα,2 := {f ∈ Wα | f ′(x) = 0} .

Then Wα,1 and Wα,2 are closed, orthogonal subspaces of Wα and Wα = Wα,1⊕Wα,2.

Proof. Since Wα,2 is one-dimensional, it is a closed subspace of Wα. Theorthogonal complement of Wα,2 is given by

W⊥α,2 := {f ∈ Wα | φ(f, g) = 0 (g ∈ Wα,2)}.


To see that Wα,1 ⊂ W⊥α,2, let f ∈ Wα,1, g ∈ Wα,2 and consider

φ(f, g) = f(−1)g(−1) +

1∫

−1

f ′(x)g′(x)(1 − x)α+1dx = 0.

The first summand vanishes because f ∈ Wα,1, and the integral is 0 becauseg ∈ Wα,2. Now let f ∈ Wα. We need to find f1 ∈ Wα,1 and f2 ∈ Wα,2 such thatf = f1 + f2. Let f1(x) = f(x) − f(−1) and f2(x) = f(−1); clearly, fi ∈ Wα,i

for i = 1, 2. �

The next result shows that, surprisingly, the space Wα,1 is precisely thefirst left-definite vector space V (α,−1)

1 . It is this theorem that allows us to con-struct a self-adjoint operator Tα in Wα having the entire sequence of Jacobipolynomials {P (α,−1)

n }∞n=0 as eigenfunctions. For n ≥ 2,

P (α,−1)n (x) =

(n+ α)!(n− 1)!2n!(n+ α− 1)!

(x+ 1)P (α,1)n−1 (x).

Theorem 12. Wα,1 = V(α,−1)1 .

Proof. Notice that functions in Wα,1 are defined on [−1, 1) while functions inV

(α,−1)1 are defined on (−1, 1) so the connection between these two spaces is

not immediately obvious.

V(α,−1)1 ⊆ Wα,1: Let f ∈ V

(α,−1)1 . Since V (α,−1)

1 ⊂ Δα,−1 we see, by Lemma 2,that f ∈ AC[−1, 1) and f(−1) = 0 so f ∈ Wα,1.

Wα,1 ⊆ V(α,−1)1 : Let f ∈ Wα,1. From the definition of both spaces, it clearly

suffices to show that f ∈ L2α,−1(−1, 1). For −1 < x < 0,

(1 − x)α/2(1 + x)−1/2

x∫

−1

f ′(t)dt = (1 − x)α/2(1 + x)−1/2f(x)

since f(−1) = 0. We use Theorem 1 on (−1, 0) with ψ(x) = (1 − x)α/2(1 +x)−1/2 and ϕ(x) = 1. Clearly, ψ and ϕ satisfy the conditions of Theorem 1.Moreover

x∫

−1

dt

0∫

x

(1 − t)α(1 + t)−1dt ≤ c

x∫

−1

dt

0∫

x

dt

1 + t= −c(x+ 1) ln(1 + x),

and, since this is a bounded function on (−1, 0), we see that f ∈ L2α,−1(−1, 0).

A similar argument shows that f ∈ L2α,−1(0, 1) and this completes the proof.

�

Theorem 13. The inner products φ(·, ·) and (·, ·)(α,−1)1 are equivalent on Wα,1 =

V(α,−1)1 .


Proof. First of all, (Wα,1, φ(·, ·)) is a Hilbert space, and, by definition,(V (α,−1)

1 , (·, ·)(α,−1)1 ) is a Hilbert space. Let f ∈ Wα,1 = V

(α,−1)1 . Then

‖f‖2φ =

1∫

−1

|f ′|2 (1 − x)α+1dx

≤1∫

−1

[|f ′|2 (1 − x)α+1 + k(1 − x)α(1 + x)−1 |f |2

]dx = (‖f‖1)

2.

By the Open Mapping Theorem, these inner products are equivalent. �

We now construct a self-adjoint operator Tα,1 in the space Wα,1, gener-ated by the Jacobi expression lα,−1, having the sequence of Jacobi polynomials{P (α,−1)

n }∞n=1 as eigenfunctions. Recall that by Theorem 12, V (α,−1)

1 = Wα,1.We also know that the operator

B(α,−1)1 : D(B(α,−1)

1 ) := V(α,−1)3 ⊂ V

(α,−1)1 → V

(α,−1)1

namely, the first left-definite operator associated with (L2α,−1(−1, 1), A(α,−1)),

is self-adjoint and given specifically by

B(α,−1)1 [f ](x) = lα,−1[f ](x) (a.e. x ∈ (−1, 1))

f ∈ D(B(α,−1)1 ) = V

(α,−1)3 ,

where

f ∈ D(B

(α,−1)1

)= V

(α,−1)3 = {f : (−1, 1) → C | f, f ′, f ′′ ∈ ACloc(−1, 1);

(1 − x)(α+3)/2(1 + x)f ′′′, (1 − x)(α+2)/2(1 + x)1/2f ′′, (1 − x)(α+1)/2f ′,

(1 − x)α/2(1 + x)−1/2f ∈ L2(−1, 1)}.More specifically, B(α,−1)

1 is self-adjoint with respect to the first left-definiteinner product (·, ·)(α,−1)

1 which is equivalent to the inner product φ(·, ·). Weshall prove that the operator Tα,1 : D(Tα,1) ⊂ Wα,1 → Wα,1 given by

Tα,1f = B(α,−1)1 f = lα,−1[f ]

f ∈ D(Tα,1) := V(α,−1)3

is self-adjoint in (Wα,1, φ(·, ·)).Lemma 9. Tα,1 in (Wα,1, φ(·, ·)) is densely defined.

Proof. The Jacobi polynomials {P (α,−1)n }∞

n=1 are eigenfunctions of Tα,1 andthey are complete in V (α,−1)

3 . �

Theorem 14. Tα,1 is symmetric in (Wα,1, φ(·, ·)).


Proof. From the previous lemma, it suffices to show that Tα,1 is Hermitian. Letf, g ∈ D(Tα,1) = V

(α,−1)3 . Since V (α,−1)

3 ⊂ V(α,−1)1 and Tα,1f, Tα,1g ∈ V

(α,−1)1 ,

we know that

f(−1) = g(−1) = 0 = Tα,1f(−1) = Tα,1g(−1).

Integration by parts shows that φ(Tα,1f, g) = φ(f, Tα,1g). �

Theorem 15. The operator Tα,1 has the following properties:

(i) Tα,1 is self-adjoint in (Wα,1, φ(·, ·)).(ii) σ(Tα,1) = {n(n+ α) + k | n ≥ 2}.(iii) {P (α,−1)

n }n≥1 is a complete orthonormal set of eigenfunctions of Tα,1 inthe space (Wα,1, φ(·, ·)).

(iv) Tα,1 is bounded below by kI in (Wα,1, φ(·, ·)).

Proof. For (iii): We know that {P (α,−1)n }n≥0 is a complete orthonormal set

in (Wα, φ(·, ·)) and we know that Wα = Wα,1 ⊕Wα,2. Also, we have Wα,2 =span{P (α,−1)

0 } and so Wα,1 = W⊥α,2 = span{P (α,−1)

n }n≥1. We next prove that

Tα,1 is closed in (Wα,1, φ(·, ·)). Take a sequence {fn} ⊆ D(Tα,1) = V(α,−1)3

such that

fn → f in (Wα,1, φ(·, ·))T1fn → g in (Wα,1, φ(·, ·)) .

We show that f ∈ D(Tα,1) and Tα,1f = g. We know that B(α,−1)1 is self-adjoint

and hence closed in (Wα,1, (·, ·)(α,−1)1 ), and we know, since φ(·, ·) and (·, ·)(α,−1)

1

are equivalent, there exist constants c1 and c2 such that

c1 ‖f‖φ ≤ ‖f‖1 ≤ c2 ‖f‖φ (f ∈ Wα,1 = V1).

Hence,

‖fn − f‖1 ≤ c2 ‖fn − f‖φ → 0;

that is,

fn → f in(Wα,1, (·, ·)(α,−1)

1

)

and

‖Tα,1fn − g‖1 ≤ c2 ‖Tα,1fn − g‖φ → 0;

hence

Tα,1fn → g in(Wα,1, (·, ·)(α,−1)

1

)


and since Tα,1 is closed in (Wα,1, (·, ·)(α,−1)1 ), we see that f ∈ D(Tα,1) and

Tα,1f = g. Also, we know that, for n ≥ 2,(Tα,1P

(α,−1)n

)(x) = lα,−1[P (α,−1)

n ](x)

= (n(n+ α) + k)P (α,−1)n (x).

This implies

{n(n+ α) + k | n ≥ 2} ⊆ σ(Tα,1).

Since {P (α,−1)n }n≥1 is complete and λn := n(n+ α) + k → ∞, we know that

σ(Tα,1) = {n(n+ α) + k | n ≥ 2}which proves (ii) and (iii). To summarize: Tα,1 is a closed, symmetric oper-ator with a complete set of eigenfunctions. From [14], Tα,1 is self-adjoint.This proves (i). To prove (iv), let f ∈ D(Tα,1). Then, since Tα,1 : V (α,−1)

3 ⊂V

(α,−1)1 → V

(α,−1)1 ,

φ (Tα,1f, f) = (Tα,1f) (−1)f(−1) +

1∫

−1

(Tα,1f)′ (x)f′(x)(1 − x)α+1dx

=

1∫

−1

(Tα,1f)′ (x)f′(x)(1 − x)α+1dx

=

1∫

−1

⎡⎢⎣∣∣∣((1 − x)α+1f ′(x)

)′∣∣∣2(1 − x)α(1 + x)−1

+ k |f ′(x)|2 (1 − x)α+1

⎤⎥⎦dx

≥ k

1∫

−1

|f ′(x)|2 (1 − x)α+1dx = kφ (f, f) .

�

Next, we define the operator Tα,2 : D(Tα,2) ⊂ Wα,2 → Wα,2 by

(Tα,2f)(x) = lα,−1[f ](x)D(Tα,2) := Wα,2.

It is straightforward to check that Tα,2 is symmetric in Wα,2 and since thedomain of Tα,2 is the entire space, it follows that Tα,2 is self-adjoint.

We now construct the self-adjoint operator Tα in (Wα, φ(·, ·)) that isgenerated by the Jacobi differential expression lα,−1[.], having the entire setof Jacobi polynomials {P (α,−1)

n }n≥0 as eigenfunctions and having spectrumσ(Tα) = {n(n+ α) + k | n ∈ N0}. For f ∈ Wα, write

f = f1 + f2


where fi ∈ Wα,i, (i = 1, 2). Define Tα : D(Tα) ⊂ Wα → Wα by

Tαf = Tα,1f1 + Tα,2f2 = lα,−1[f1] + lα,−1[f2] = lα,−1[f ],D(Tα) = D(Tα,1) ⊕ D(Tα,2).

A proof that operators of this form are self-adjoint can be found in [10, Theo-rem 11.1]. Furthermore, since we know explicitly the domains of Tα,1 and Tα,2,we can specifically determine the domain D(Tα) of Tα.

Theorem 16. Tα is self-adjoint in (Wα, φ(·, ·)) and

D(Tα) = {f : [−1, 1) → C | f ∈ AC[−1, 1); f ′, f ′′ ∈ ACloc(−1, 1);

(1 − x)(α+3)/2(1 + x)f ′′′, (1 − x)(α+2)/2(1 + x)1/2f ′′,

(1 − x)(α+1)/2f ′ ∈ L2(−1, 1)}= {f : [−1, 1) → C | f ∈ AC[−1, 1); f ′, f ′′ ∈ ACloc(−1, 1);

(1 − x)(α+3)/2(1 + x)f ′′′ ∈ L2(−1, 1)}.Furthermore, σ(Tα) = {n(n+α) + k | n ∈ N0} and Tα is bounded below by kIin (Wα, φ(·, ·)).For the following theorem let us recall the definitions of the first and thirdleft-definite spaces:

V(α,−1)1 = {f : (−1, 1) → C | f ∈ ACloc(−1, 1);

(1 − x)α/2(1 + x)−1/2f, (1 − x)(α+1)/2f ′ ∈ L2(−1, 1)}= {f : [−1, 1) → C | f ∈ AC [−1, 1) ;

(1 − x)(α+1)/2f ′ ∈ L2(−1, 1); f(−1) = 0}= Wα,1

V(α,−1)3 = D(Tα,1) = {f : (−1, 1) → C | f, f ′, f ′′ ∈ ACloc(−1, 1);

(1 − x)(α+3)/2(1 + x)f ′′′, (1 − x)(α+2)/2(1 + x)1/2f ′′,

(1 − x)(α+1)/2f ′, (1 − x)α/2(1 + x)−1/2f ∈ L2(−1, 1)}= {f ∈ V1 | f ′, f ′′ ∈ ACloc(−1, 1); (1 − x)(α+3)/2(1 + x)f ′′′

(1 − x)(α+2)/2(1 + x)1/2f ′′ ∈ L2(−1, 1)}= {f : [−1, 1) → C | f ∈ AC[−1, 1); f ′, f ′′ ∈ ACloc(−1, 1);

f(−1) = 0; (1 − x)(α+3)/2(1 + x)f ′′′, (1 − x)(α+2)/2(1 + x)1/2f ′′,

(1 − x)(α+1)/2f ′ ∈ L2(−1, 1)}.Theorem 17. Let

D := {f : [−1, 1) → C | f ∈ AC[−1, 1); f ′, f ′′ ∈ ACloc(−1, 1);

(1 − x)(α+3)/2(1 + x)f ′′′, (1 − x)(α+2)/2(1 + x)1/2f ′′,

(1 − x)(α+1)/2f ′ ∈ L2(−1, 1)}.Then D(Tα) = D.


Proof. We first show D(Tα) ⊆ D. Let f ∈ D(Tα) = D(Tα,1) ⊕ D(Tα,2). Write

f = f1 + f2

where f1 ∈ D(Tα,1) = V(α,−1)3 ⊆ D, f2 ∈ D(Tα,2) ⊆ D. Then f ∈ D. To show

that D ⊆ D(Tα), let f ∈ D. Write

f(x) = f1(x) + f2(x)

where

f1(x) := f(x) + f(−1)f2(x) := −f(−1).

Then f1 ∈ D and f1(−1) = 0; that is, f1 ∈ V(α,−1)3 = D(Tα,1). Also, f ′′

2 (x) = 0so f2 ∈ D(Tα,2). Together, f ∈ D(Tα). �

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Andrea BruderDepartment of Mathematics and Computer ScienceTutt Science CenterColorado College14 E. Cache la Poudre StreetColorado Springs, CO 80903USAe-mail: [email protected]

L. L. LittlejohnDepartment of MathematicsBaylor UniversityOne Bear Place #97328Waco, TX 76798-7328USA

Received: November 15, 2010.

Revised: January 18, 2011.

Accepted: January 25, 2011.

nonclassical jacobi polynomials and sobolev orthogonality

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