THE FRIEDRICHS EXTENSION OF CERTAIN SINGULAR DIFFERENTIAL OPERATORS, II

We study the Friedrichs extension for a class of 2nth order ordinary differential operators. These selfadjoint operators have compact inverses and the central problem is to describe their domains in terms of boundary conditions.


Introduction
An unbounded symmetric operator L with domain D(L) dense in a Hilbert space H with inner product (u, v) is called semibounded below if (Lu, u) ≥ c(u, u) for some constant c and every u ∈ D(L).By adding a multiple of the identity operator to L, without loss of generality it may be assumed that c > 0. The largest such c is called the lower bound.Such an operator has a particular selfadjoint extension F , called the Friedrichs extension [9].A characteristic of the Friedrichs extension is that it preserves the lower bound: (F u, u) ≥ c(u, u) for every u ∈ D(F ).Even a cursory examination of the development of the Friedrichs extension in [8,Section XII.5] shows that this extension depends critically on the domain D(L) of the operator.In fact, the domain of the Friedrichs extension of L is obtained by intersecting D(L * ), the domain of the adjoint operator L * , with a subset of the Hilbert space H which is obtained by completing D(L), considered as an incomplete metric space, with the "new" inner product (Lu, u).The proof of the existence of the Friedrichs extension leaves open the possibility of the existence of additional selfadjoint extensions which preserve the lower bound.
For selfadjoint ordinary differential operators, domains are usually specified by the imposition of boundary conditions.Ever since [10], it has been of interest to find a boundary condition description of the domain of the Friedrichs extension of specific operators.
Let τ be the formal differential operator defined by where m(x) > 0, p(x) > 0, m, p are infinitely differentiable on (0, 1] and By varying the initial domain, Baxley [2,3] considered several different semibounded symmetric operators defined by the nth iterate τ n in the Hilbert space L 2 (0, 1; m) of functions u satisfying In each case, the Friedrichs extension was a selfadjoint operator with compact inverse.
The main goal was to describe the domains of these Friedrichs extensions in the usual way in terms of boundary conditions.The motivation for this earlier work was an application to the theory of Toeplitz matrices [4].
We return again to this problem, with a similar motivation.Now the application is to the theory of Toeplitz integral operators associated with the Hankel transform (see [6,7] for related earlier results), work which will appear elsewhere.The Friedrichs extension obtained here is most closely related to the one which was designated G n in [3].It is possible that the Friedrichs extension obtained here is the same as that G n , but we have been unable to verify this conjecture.In any case, the initial domain here is different and is more convenient for our application; the development is also simpler and more self-contained.
In recent years, much attention has been given to characterizing the domains of Friedrichs extensions of ordinary differential operators (see [11,13,14,15]), as well as partial differential operators (see [5], which contains further references).
Proof: Statement 1 follows by integrating by parts using Lemma 2.1 and the definition of C n to see that all boundary terms vanish.
For 2, we use the Schwarz inequality and integration by parts to get, for 0 For 3, we similarly write Proof: In the first case, τ j u ∈ C n−2j and n − 2j ≥ 1.In the second case, and n − 2(j − 1) ≥ 2, so these statements follow quickly from Lemma 2.2.
Proof: Consider the i = 0 case.Since u ∈ C n , from Lemma 2.2 (2 ) we have dt. Letting and integrating we have dt dx = M(τ u, u).
and the above argument give EJQTDE Spec.Ed.I, 2009 No. 5

The Friedrichs Extension
Let C ∞ (0, 1) be the collection of all infinitely differentiable functions on (0, 1) and let ) consist of all functions in C ∞ (0, 1) with compact support in (0, 1).The next lemma is obvious.
We view L n as an operator in the weighted Hilbert space L 2 (0, 1; m) defined earlier.
Theorem 3.1 The operator L n is symmetric and nonnegative for all n ≥ 1.
Proof: Symmetry follows directly from Lemma 2.2 (1 ).We note that for n = 2j even, and for n = 2j + 1 odd, and so L n is nonnegative for all k ≥ 1.
Iterating Lemma 2.4, we see that (u, u) ≤ M n (L n u, u) for any u ∈ D(L n ).Thus M −n is a positive lower bound for L n .
For each n, Theorem 3.1 shows that L n has a Friedrichs extension, which we denote by Ln .
for any integer k ≥ 0, we can use Lemma 2.3 and Lemma 2.4 to conclude, for k = 2j ≤ n, whenever 0 < x 1 < x 2 ≤ 1.Since τ j u i and (τ j u i ) ′ vanish for x = 1, it follows that all the sequences {τ j u i }, for 2j + 1 ≤ n, and {p(τ j u i ) ′ }, for 2j ≤ n, are uniformly bounded and equicontinuous on any compact subset of (0, 1].Using Ascoli's theorem and a familiar diagonalization argument, by passing to a subsequence we can assume without loss of generality that all of these sequences converge uniformly on any compact subset of (0, 1].Since u i − u → 0, then u i → u uniformly on any such compact subset of (0, 1].Thus u(1 dt and take limits using the bounded convergence theorem to get Continuing in this way, we find that each of these convergent sequences converge to the appropriate derivative of the limit function u, and it is easy to see that u (j) (1) = 0 for 0 ≤ j ≤ n − 1.
Letting x 1 = x and x 2 = 1 and integrating, we have (u, u) ≤ M n ( Ln u, u). has a convergent subsequence.From Lemma 3.2 (3 ), u i ≤ M n Ln u i , and so we have ( Ln Then from Lemma 3.2 (2 ), and so with x 2 = 1, Thus the functions {u i } are equicontinuous and uniformly bounded on compact subintervals of (0, 1].From Ascoli's theorem and a diagonalization argument we may assume the sequence converges uniformly to a limit function u on each compact subinterval of (0, 1].Since the last inequality also holds for the limit function u, the dominated convergence theorem guarantees The following lemma is immediate since 0 is not in the spectrum of Ln .

Lemma 3.3
The range of Ln is all of L 2 (0, 1; m).
Because of the previous three lemmas, it follows from the theory of compact selfadjoint operators that all eigenvalues of the operator Ln are positive and have finite multiplicity, and the eigenfunctions span L 2 (0, 1; m).
To prepare for the next lemma, for 0 ≤ x ≤ 1, we define Q 0 (x) = 1 and and we define R 0 (x) = Just as in the proof of Lemma 2.1, it follows by induction that each Q j is well-defined and bounded.Proof: It is easy to verify that and p(τ i R j ) ′ = (−1) i pR ′ j−i for i < j and p(τ j R j ) ′ = (−1) j+1 .
Thus, τ n maps all of the functions Q to 0 and it is easy to show that the functions Q j and R j are linearly independent.Since w ∈ N(L n * ), it follows as in [8, pp. 1291-1294], that w ∈ C ∞ (0, 1) and τ n u = 0. Thus the 2n functions for some constants a j and b j .We need to show that b j = 0 for all j = 0, 1, Repeated integration by parts gives Since u(x) ≡ 0 for 3/4 ≤ x ≤ 1 our equation simplifies to Substituting for w, we see that any term with a product of the form (τ i Q j )p(x)(τ k Q m ) ′ tends to 0, so all terms in w involving the Q j 's vanish and we may as well assume that w = Proof: If Λ is an eigenvalue of L1 and u is a corresponding eigenfunction, then Λ > 0, u is in the null space of L1 − ΛI and we have the equation Again, using [8, pp. 1291-1294], u is infinitely differentiable and τ u = Λu.This equation reduces to where Λ = β 2 .It is easy to verify that the general solution of this equation is u = c 1 x 1/2−ν u 1 (βx) + c 2 x 1/2−ν u 2 (βx), where u 1 , u 2 are any two linearly independent solutions of Bessel's equation.We may choose u 1 (x) = J ν−1/2 (x) and u 2 (x) = Y ν−1/2 (x).Let w 2 (x) = x 1/2−ν u 2 (βx).Known behavior (see [12]) of u 2 (x) as x → 0 shows that lim x→0 x 2ν w ′ 2 (x) = d = 0 for some constant d.Choosing v in the domain of L 1 so that v(x) = 1 in a right neighborhood of x = 0, we can integrate by parts to get (L 1 v, w 2 ) = (τ v, w 2 ) = −d + (v, τ w 2 ), so that w 2 is not in the domain of L * 1 .Thus c 2 = 0.A similar calculation shows that x 1/2−ν u 2 (βx) is in the domain of L * 1 .Then u is in the domain of L1 if and only if u(1) = 0, which occurs if and only if u 1 (β) = J ν−1/2 (β) = 0. Thus β must be a zero z k of this Bessel function.
dt dy is also finite and we are done.Lemma 2.2 If u ∈ C n , the following conditions hold: EJQTDE Spec.Ed.I, 2009 No. 5 Proof: If u ∈ D( Ln ), then from [8, p. 1242], there exists u

Theorem 3 . 2
All eigenvalues of Ln are strictly positive and Ln has a compact inverse.Proof: Let u be an eigenfunction of Ln .Then from Lemma 3.2, (u, u) ≤ M n ( Ln u, u) = M n λ(u, u) where λ is the corresponding eigenvalue.Hence all eigenvalues of Ln are strictly positive.Thus L−1 n exists.To show L−1 n is compact, suppose there is a sequence { Ln u i } such that Ln u i ≤ K for n = 1, 2, • • •.We will show that { L−1 n ( Ln u i )} = {u i } EJQTDE Spec.Ed.I, 2009 No. 5

b
j R j .EJQTDE Spec.Ed.I, 2009 No. 5the proof of Theorem 3.3, one sees that u = w + v, where w ∈ C ∞ (0, 1) (as noted in the proof of Lemma 3.4), and v ∈ D( Ln ), which (as shown in the proof of Lemma 3.2) has at least n − 1 continuous derivatives.We now identify the eigenfunctions and eigenvalues of L1 in one special case that arises in the case of Toeplitz integral operators associated with the Hankel transform.Theorem 3.4 Suppose m(x) = p(x) = x 2ν where ν > 0 is a constant.Then the eigenvalues Λ k of L1 have multiplicity one and Λ k = z 2 k , where z k is the k th positive zero of the Bessel function J ν−1/2 .An eigenfunction corresponding to Λ k is x 1/2−ν J ν−1/2 (z k x).