GEORGIAN MATHEMATICAL JOURNAL: Vol. 4, No. 5, 1997, 401-412

SINGULAR NONLINEAR (n − 1, 1) CONJUGATE BOUNDARY VALUE PROBLEMS

PAUL W. ELOE AND JOHNNY HENDERSON

Abstract. Solutions are obtained for the boundary value problem, y^{(n)} + f(x, y) = 0, y^{(i)}(0) = y(1) = 0, 0 ≤ i ≤ n − 2, where f(x, y) is singular at y = 0. An application is made of a fixed point theorem for operators that are decreasing with respect to a cone.

§ 1. Introduction

In this paper, we establish the existence of solutions for the (n − 1, 1) conjugate boundary value problem,

y^{(n)} + f(x, y) = 0, 0 < x < 1,  (1)
\begin{aligned}y^{(i)}(0) & = 0, 0 ≤ i ≤ n − 2, \\ y(1) & = 0, \end{aligned}  (2)

where f(x, y) has a singularity at y = 0. Our assumptions throughout are: (A) f(x, y) : (0, 1) × (0, ∞) → (0, ∞) is continuous, (B) f(x, y) is decreasing in y, for each fixed x,

(C) ∫_{0}^{1} f(x, y)dx < ∞, for each fixed y,

(D) lim_{y→0^{+}} f(x, y) = ∞ uniformly on compact subsets of (0, 1), and

(E) lim_{y→∞} f(x, y) = 0 uniformly on compact subsets of (0, 1).

We note that, if y is a solution of (1), (2), then (A) implies y(x) > 0 on (0, 1).

Singular nonlinear two-point boundary value problems appear frequently in applications, and usually, only positive solutions are meaningful. This is especially true for the case n = 2, with Taliaferro [1] treating the general problem, Callegari and Nachman [2] considering existence questions

401 in boundary layer theory, and Luning and Perry [3] obtaining constructive results for generalized Emden–Fowler problems. Results have also been obtained for singular boundary value problems arising in reaction-diffusion theory and in non-Newtonian fluid theory [4]. A number of papers have been devoted to singular boundary value problems in which topological transversality methods were applied; see, for example, [5]–[10].

1991 Mathematics Subject Classification. 34B15.

Key words and phrases. Boundary value problem, singularity, cone.

1072-947X/97/0900-0401$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1997 Plenum Publishing Corporation

The results and methods of this work are outgrowths of papers on secondorder singular boundary value problems by Gatica, Hernandez, and Waltman [11] and Gatica, Oliker, and Waltman [12] which in turn received extensive embellishment by Eloe and Henderson [13] and Henderson and Yin [14], [15]. In attempting to improve some of these generalizations, the recent paper by Wang [16] did contain some flaws; however, that paper was corrected in a subsequent work by Agarwal and Wong [17].

We obtain solutions of (1), (2) by arguments involving positivity properties, an iteration, and a fixed point theorem due to [12] for mappings that are decreasing with respect to a cone in a Banach space. We remark that, for n = 2, positive solutions of (1), (2) are concave. This concavity was exploited in [12], and later in the generalizations [14]–[18], in defining an appropriate subset of a cone on which a positive operator was defined to which the fixed point theorem was applied. The crucial property in defining this subset in [12] made use of an inequality that provides lower bounds on positive concave functions as a function of their maximum. Namely, this inequality may be stated as:

If y ∈ C^{(2)}[0, 1] is such that y(x) ≥ 0, 0 ≤ x ≤ 1, and y′′(x) ≤ 0, 0 ≤ x ≤ 1, then

y(x) ≥ \frac{1}{4} max_{0≤s≤1} |y(s)|, \frac{1}{4} ≤ x ≤ \frac{3}{4}.  (3)

Although (3) can be developed using concavity, it can also be obtained directly with the classical maximum principle. This observation was exploited by Eloe and Henderson [18], and a generalization of (3) was given for positive functions satisfying the boundary conditions (2).

In Section 2, we provide preliminary definitions and some properties of cones in a Banach space. We also state the fixed point theorem from [12] for mappings that are decreasing with respect to a cone. In that section, we state the generalization of (3) as it extends to solutions of (1), (2). An analogous inequality is also stated for a related Green’s function.

SINGULAR NONLINEAR (n − 1, 1) CONJUGATE PROBLEMS 403

In Section 3, we apply the generalization of (3) in defining a subset of a cone on which we define an operator which is decreasing with respect to the cone. A sequence of perturbations of f is constructed, with each term of the sequence lacking the singularity of f. In terms of this sequence, we define a sequence of decreasing operators to which the fixed point theorem yields a sequence of iterates. This sequence of iterates is shown to converge to a positive solution of (1), (2).

§ 2. Some Preliminaries and a Fixed Point Theorem

In this section, we first give definitions and some properties of cones in a Banach space [19]. After that, we state a fixed point theorem due to [12] for operators that are decreasing with respect to a cone. We then state a theorem from [18] generalizing (3) followed by an analogous inequality for a Green’s function.

Let B be a Banach space, and K a closed, nonempty subset of B. K is a cone provided (i) αu + βv ∈ K, for all u, v ∈ K and all α, β ≥ 0, and (ii) u, −u ∈ K imply u = 0. Given a cone K, a partial order, ≤, is induced on B by x ≤ y, for x, y ∈ B iff y − x ∈ K. (For clarity, we may sometimes write x ≤ y(wrtK).) If x, y ∈ B with x ≤ y, let ⟨x, y⟩ denote the closed order interval between x and y given by, ⟨x, y⟩ = {z ∈ B | x ≤ z ≤ y}. A cone K is normal in B provided there exists δ > 0 such that ∥e_{1} + e_{2}∥ ≥ δ, for all e_{1}, e_{2} ∈ K, with ∥e_{1}∥ = ∥e_{2}∥ = 1.

Remark 1. If K is a normal cone in B, then closed order intervals are norm bounded.

The following fixed point theorem can be found in [12].

Theorem 1. Let B be a Banach space, K a normal cone in B, E ⊆ K such that, if x, y ∈ E with x ≤ y, then ⟨x, y⟩ ⊆ E, and let T : E → K be a continuous mapping that is decreasing with respect to K, and which is compact on any closed order interval contained in E. Suppose there exists x_{0} ∈ E such that T^{2}x_{0} = T (T x_{0}) is defined, and furthermore, T x_{0}, T^{2}x_{0} are orders comparable to x_{0}. If, either

(I) T x_{0} ≤ x_{0} and T^{2}x_{0} ≤ x_{0}, or x_{0} ≤ T x_{0} and x_{0} ≤ T^{2}x_{0}, or (II) The complete sequence of iterates {T^{n}x_{0}}_{n=0}^{∞} is defined, and there

exists y_{0} ∈ E such that T y_{0} ∈ E and y_{0} ≤ T^{n}x_{0}, for all n ≥ 0, then T has a fixed point in E.

In extending (3), Eloe and Henderson [13] first established the following.

Theorem 2. Let n ≥ 2 and h ∈ C^{(n)}[a, b] be such that h^{(n)}(x) ≤ 0, a ≤ x ≤ b, and

h^{(i)}(a) ≥ 0, 0 ≤ i ≤ n − 2,  (4)
h(b) ≥ 0.  (5)

Then h(x) ≥ 0, a ≤ x ≤ b. Moreover, if h^{(n)}(x) < 0 on any compact subinterval of [a, b], or if either (4) or (5) is strict inequality, then h(x) > 0, a < x < b.

We now state the extension of (3) which will play a fundamental role in our future arguments.

Theorem 3. Let y ∈ C^{(n)}[0, 1] be such that y^{(n)}(x) < 0, 0 < x < 1, and y^{(i)}(0) = y(1) = 0, 0 ≤ i ≤ n − 2. Then y(x) > 0 on (0, 1), and there exists a unique x_{0} ∈ (0, 1) such that |y|_{∞} = sup_{0≤x≤1} |y(x)| = y(x_{0}). Moreover, y(x)

is increasing on [0, x_{0}], y(x) is concave on [x_{0}, 1], and

y(x) ≥ \frac{|y|_{∞}}{4^{n−1}} , \frac{1}{4} ≤ x ≤ \frac{3}{4}.  (6)

For the final result to be stated in this section, let G(x, s) denote the Green’s function for the boundary value problem,

−y^{(n)} = 0, 0 ≤ x ≤ 1,  (7)

satisfying (2). It is well known [20] that

G(x, s) > 0 on (0, 1) × (0, 1),  (8)

and

\frac{∂^{n−1}}{∂x^{n−1}} G(0, s) > 0 > \frac{∂}{∂x} G(1, s), 0 < s < 1.  (9)

Also, for the remainder of the paper for 0 < s < 1, let τ(s) ∈ [0, 1] be defined by

G(τ(s), s) = sup_{0≤x≤1} G(x, s).  (10)

The following analogue of (6) for G(x, s) was also obtained in [18].

Theorem 4. Let G(x, s) denote the Green’s function for (7), (2). Then, for 0 < s < 1,

G(x, s) ≥ \frac{1}{4^{n−1}} G(τ(s), s), \frac{1}{4} ≤ x ≤ \frac{3}{4}.  (11)

§ 3. Solutions of (1), (2)

In this section, we apply Theorem 1 to a sequence of operators that are decreasing with respect to a cone. The obtained fixed points provided a sequence of iterates which converges to a solution of (1), (2). Positivity of solutions and Theorems 2–4 are fundamental in this construction.

To that end, let the Banach space B = C[0, 1], with norm ∥y∥ = |y|_{∞}, and let

K = {y ∈ B | y(x) ≥ 0 on [0, 1]}.

K is a normal cone in B.

To obtain a solution of (1), (2), we seek a fixed point of the integral operator,

T φ(x) = ∫_{0}^{1} G(x, s)f(s, φ(s))ds,

where G(x, s) is the Green’s function for (7), (2). Due to the singularity of f given by (D), T is not defined on all of the cone K.

Next, define g : [0, 1] → [0, 1] by

g(x) = \begin{cases}(2x)^{n−1}, & 0 ≤ x ≤ \frac{1}{2}, \\ 2(1 − x), & \frac{1}{2} ≤ x ≤ 1, \end{cases}

and for each θ > 0, define g_{θ}(x) = θg(x). Then for the remainder of this work, assume the condition:

(F) For each θ > 0, 0 < ∫_{0}^{1} f(x, g_{θ}(x))dx < ∞.

We remark, for each θ > 0, that g_{θ} ∈ K, g_{θ}(x) > 0 on (0, 1), and g_{θ}^{(i)}(0) = g(1) = 0, 0 ≤ i ≤ n − 2.

Our first result of this section is a consequence of Theorem 3 and its proof in [18].

Theorem 5. Let y ∈ C^{(n)}[0, 1] be such that y^{(n)}(x) < 0 on (0, 1) and y^{(i)}(0) = y(1) = 0, 0 ≤ i ≤ n − 2. Then, there exists a θ > 0 such that g_{θ}(x) ≤ y(x) on [0, 1].

Proof. Let y be as stated above and let x_{0} ∈ (0, 1) be the unique point from Theorem 3 such that y(x_{0}) = |y|_{∞}. Define the piecewise polynomial

p(x) = \begin{cases}\frac{|y|_{∞}}{x_{0}^{n−1}} x^{n−1}, & 0 ≤ x ≤ x_{0}, \\ \frac{|y|_{∞}}{x_{0}−1}(x − 1), & x_{0} ≤ x ≤ 1. \end{cases}

The proof of Theorem 3 in [18] yields that y(x) ≥ p(x) on [0, 1]. If we choose θ = p(\frac{1}{2}), then

p(x) ≥ p(\frac{1}{2})g(x) = g_{θ}(x) on [0, 1],

and so, y(x) ≥ g_{θ}(x) on [0, 1].

In view of Theorem 5, let D ⊆ K be defined by

D = {φ ∈ B | there exists θ(φ) > 0 such that g_{θ}(x) ≤ φ(x) on [0, 1]},

(i.e., D = {φ ∈ B | there exists θ(φ) > 0 such that g_{θ} ≤ φ(wrtK)}). Then, define T : D → K by

T φ(x) = ∫_{0}^{1} G(x, s)f(s, φ(s))ds, 0 ≤ x ≤ 1, φ ∈ D.

Note that, from conditions (A)–(F) and properties of G(x, s) in (8)–(9), if φ ∈ D, then (T φ)^{(n)} < 0 on (0, 1), and T φ satisfies the boundary conditions (2). Application of Theorem 5 yields that T φ ∈ D so that T : D → D. Moreover, if φ is a solution of (1), (2), then by Theorem 5 again, φ ∈ D. As a consequence, φ ∈ D is a solution of (1), (2) if, and only if, T φ = φ.

Our next result establishes a priori bounds on solutions of (1), (2) which belong to D.

Theorem 6. Assume that conditions (A)–(F) are satisfied. Then, there exists an R > 0 such that ∥φ∥ = |φ|_{∞} ≤ R, for all solutions, φ, of (1), (2) that belong to D.

Proof. Assume to the contrary that the conclusion is false. This implies there exists a sequence, {φ_{ℓ}}⊂D, of solutions of (1), (2) such that lim_{ℓ→∞} |φ_{ℓ}|= ∞. Without loss of generality, we may assume that, for each ℓ ≥ 1,

|φ_{ℓ}|_{∞} ≤ |φ_{ℓ+1}|_{∞}.  (12)

For each ℓ ≥ 1, let x_{ℓ} ∈ (0, 1) be the unique point from Theorem 3 such that

0 < φ_{ℓ}(x_{ℓ}) = |φ_{ℓ}|_{∞},

and also

φ_{ℓ}(x) ≥ \frac{1}{4^{n−1}} φ_{ℓ}(x_{ℓ}), \frac{1}{4} ≤ x ≤ \frac{3}{4}.

By the monotonicity in (12), φ_{ℓ}(x_{ℓ}) ≥ φ_{1}(x_{1}), for all ℓ, and so

φ_{ℓ}(x) ≥ \frac{1}{4^{n−1}} φ_{1}(x_{1}), \frac{1}{4} ≤ x ≤ \frac{3}{4} and ℓ ≥ 1.  (13)

Let θ = \frac{1}{4^{n−1}} φ_{1}(x_{1}). Then

g_{θ}(x) ≤ \frac{1}{4^{n−1}} φ_{1}(x_{1}) ≤ φ_{ℓ}(x), \frac{1}{4} ≤ x ≤ \frac{3}{4} and ℓ ≥ 1.

Next, if we apply Theorem 2 to φ_{ℓ}(x) − g_{θ}(x) on [0, \frac{1}{4}], for each ℓ ≥ 1, then φ_{ℓ}(x) ≥ g_{θ}(x) on [0, \frac{1}{4}]. Also, Theorem 3 implies that φ_{ℓ}(x) increases on [0, x_{ℓ}] and is concave on [x_{ℓ}, 1], together implying φ_{ℓ}(x) ≥ g_{θ}(x) on [\frac{3}{4}, 1]. We conclude

g_{θ}(x) ≤ φ_{ℓ}(x), 0 ≤ x ≤ 1 and ℓ ≥ 1.

Now, set

0 < M = sup{G(x, s) | (x, s) ∈ [0, 1] × [0, 1]}.

Then, assumptions (B) and (F) yield, for 0 ≤ x ≤ 1 and all ℓ ≥ 1,

φ_{ℓ}(x) = T φ_{ℓ}(x) = ∫_{0}^{1} G(x, s)f(s, φ_{ℓ}(s))ds =

≤ M ∫_{0}^{1} f(s, g_{θ}(s))ds = N,

for some 0 < N < ∞. In particular,

|φ_{ℓ}|_{∞} ≤ N, for all ℓ ≥ 1,

which contradicts lim_{ℓ→∞} |φ_{ℓ}|_{∞} = ∞. The proof is complete.

Remark 2. With R as in Theorem 6, φ ≤ R(wrtK), for all solutions φ ∈ D of (1), (2).

Our next step in obtaining solutions of (1), (2) is to construct a sequence of nonsingular perturbations of f. For each ℓ ≥ 1, define ψ_{ℓ} : [0, 1] → [0, ∞) by

ψ_{ℓ}(x) = ∫_{0}^{1} G(x, s)f(s, ℓ)ds.

By conditions (A)–(E), for ℓ ≥ 1,

0 < ψ_{ℓ+1}(x) ≤ ψ_{ℓ}(x) on (0, 1),

and

lim_{ℓ→∞} ψ_{ℓ}(x) = 0 uniformly on [0, 1].  (14)

Now define a sequence of functions f_{ℓ} : (0, 1) × [0, ∞) → (0, ∞), ℓ ≥ 1, by

f_{ℓ}(x, y) = f(x, max{y, ψ_{ℓ}(x)}).

Then, for each ℓ ≥ 1, f_{ℓ} is continuous and satisfies (B). Furthermore, for ℓ ≥ 1,

\begin{matrix}f_{ℓ}(x, y) ≤ f(x, y) on (0, 1) × (0, ∞), and \\ f_{ℓ}(x, y) ≤ f(x, ψ_{ℓ}(x)) on (0, 1) × (0, ∞). \end{matrix}  (15)

Theorem 7. Assume that conditions (A)– (F) are satisfied. Then the boundary value problem (1), (2) has a solution y ∈ D.

Proof. We begin by defining a sequence of operators T_{ℓ} : K → K, ℓ ≥ 1, by

T_{ℓ}φ(x) = ∫_{0}^{1} G(x, s)f_{ℓ}(s, φ(s))ds.

Note that, for ℓ ≥ 1 and φ ∈ K, (T_{ℓ}φ)^{(n)}(x) < 0 on (0, 1), T_{ℓ}φ satisfies the boundary conditions (2), and T_{ℓ}φ(x) > 0 on (0, 1); in particular, T_{ℓ}φ ∈ D. Since each f_{ℓ} satisfies (B), it follows that, if φ_{1}, φ_{2} ∈ K with φ_{1} ≤ φ_{2}(wrtK), then for ℓ ≥ 1, T_{ℓ}φ_{2} ≤ T_{ℓ}φ_{1}(wrtK); that is, each T_{ℓ} is decreasiing with respect to K. It is also clear that 0 ≤ T_{ℓ}(0) and 0 ≤ T_{ℓ}^{2}(0)(wrtK), for each ℓ.

Hence when we apply Theorem 1, for each ℓ, there exists a φ_{ℓ} ∈ K such that T_{ℓ}φ_{ℓ} = φ_{ℓ}. The above note implies, for ℓ ≥ 1, that φ_{ℓ}^{(n)}(x) < 0 on (0, 1), φ_{ℓ} satisfies (2), and φ_{ℓ}(x) > 0 on (0, 1). In addition, inequality (15), coupled with the positivity of G(x, s), yields T_{ℓ}φ ≤ T ψ_{ℓ}(wrtK), for each φ ∈ K and ℓ ≥ 1. Thus,

φ_{ℓ} = T_{ℓ}φ_{ℓ} ≤ T ψ_{ℓ}(wrtK), ℓ ≥ 1.  (16)

By essentially the same argument as in Theorem 6, in conjunction with inequality (16), it can be shown that there exists an R > 0 such that, for each ℓ ≥ 1,

φ_{ℓ} ≤ R(wrtK).  (17)

Our next claim is that there exists a κ > 0 such that κ ≤ |φ_{ℓ}|_{∞}, for all ℓ. We assume this claim to be false. Then, by passing to a subsequence and relabeling, we assume with no loss of generality that lim_{ℓ→∞} |φ_{ℓ}|_{∞} = 0. This implies

lim_{ℓ→∞} φ_{ℓ}(x) = 0 uniformly on [0, 1].  (18)

Next set

0 < m = inf \{G(x, s) | (x, s) ∈ [\frac{1}{4}, \frac{3}{4}] × [\frac{1}{4}, \frac{3}{4}]\}.

By condition (D), there exists a δ > 0 such that, for \frac{1}{4} ≤ x ≤ \frac{3}{4} and 0 < y < δ,

f(x, y) > \frac{2}{m}.

The limit (18) implies there exists an ℓ_{0} ≥ 1 such that, for ℓ ≥ ℓ_{0},

0 < φ_{ℓ}(x) < \frac{δ}{2} for 0 ≤ x ≤ 1.

Also, from (14), there exists an ℓ_{1} ≥ ℓ_{0} such that, for ℓ ≥ ℓ_{1},

0 < ψ_{ℓ}(x) < \frac{δ}{2} for \frac{1}{4} ≤ x ≤ \frac{3}{4}.

Thus, for ℓ ≥ ℓ_{1} and \frac{1}{4} ≤ x ≤ \frac{3}{4},

φ_{ℓ}(x) = ∫_{0}^{1} G(x, s)f_{ℓ}(s, φ_{ℓ}(s))ds ≥ ∫_{\frac{1}{4}}^{\frac{3}{4}} G(x, s)f_{ℓ}(s, φ_{ℓ}(s))ds ≥

≥ m ∫_{\frac{1}{4}}^{\frac{3}{4}} f(s, max\{φ_{ℓ}(s), ψ_{ℓ}(s)\})ds ≥ m ∫_{\frac{1}{4}}^{\frac{3}{4}} f(s, \frac{δ}{2})ds ≥ 1.

But this contradicts the uniform limit (18). Hence, our claim is verified. That is, there exists a κ > 0 such that

κ ≤ |φ_{ℓ}|_{∞} ≤ R for all ℓ.

Applying Theorem 3,

φ_{ℓ}(x) ≥ \frac{1}{4^{n−1}} |φ_{ℓ}|_{∞} ≥ \frac{κ}{4^{n−1}} , \frac{1}{4} ≤ x ≤ \frac{3}{4}, ℓ ≥ 1.

One can mimic part of the proof of Theorem 6 to show, if θ = \frac{κ}{4^{n−1}} , then

g_{θ}(x) ≤ φ_{ℓ}(x) on [0, 1] for ℓ ≥ 1.

By (17), we now have

g_{θ} ≤ φ_{ℓ} ≤ R(wrtK) for ℓ ≥ 1;

that is, the sequence {φ_{ℓ}} belongs to the closed order interval ⟨g_{θ}, R⟩ ⊂ D. When restricted to this closed order interval, T is a compact mapping, and so, there is a subsequence of {T φ_{ℓ}} which converges to some φ^{∗} ∈ K. We relabel the subsequence as the original sequence so that lim_{ℓ→∞} ∥T φ_{ℓ}−φ^{∗}∥ = 0.

The final part of the proof is to establish that lim_{ℓ→∞} ∥T φ_{ℓ} − φ_{ℓ}∥ = 0. To this end, let θ = \frac{κ}{4^{n−1}} be as above, and set

0 < M = sup{G(x, s) | (x, s) ∈ [0, 1] × [0, 1]}.

Let ϵ > 0 be given. By the integrabilty condition (F), there exists 0 < δ < 1 such that

2M[ ∫_{0}^{δ} f(s, g_{θ}(s))ds + ∫_{1−δ}^{1} f(s, g_{θ}(s))ds] < ϵ.

410 PAUL W. ELOE AND JOHNNY HENDERSON

Further, by (14), there exists an ℓ_{0} such that, for ℓ ≥ ℓ_{0},

ψ_{ℓ}(x) ≤ g_{θ}(x) on [δ, 1 − δ],

so that

ψ_{ℓ}(x) ≤ g_{θ}(x) ≤ φ_{ℓ}(x) on [δ, 1 − δ].

Observe also that, for δ ≤ s ≤ 1 − δ and ℓ ≥ ℓ_{0},

f_{ℓ}(s, φ_{ℓ}(s)) = f(s, φ_{ℓ}(s)).

Hence, for ℓ ≥ ℓ_{0} and 0 ≤ x ≤ 1,

T φ_{ℓ}(x) − φ_{ℓ}(x) = T φ_{ℓ}(x) − T_{ℓ}φ_{ℓ}(x) =

= ∫_{0}^{δ} G(x, s)[f(s, φ_{ℓ}(s)) − f_{ℓ}(s, φ_{ℓ}(s))]ds +

+ ∫_{1−δ}^{1} G(x, s)[f(s, φ_{ℓ}(s)) − f_{ℓ}(s, φ_{ℓ}(s))]ds.

So, for ℓ ≥ ℓ_{0} and 0 ≤ x ≤ 1,

|T φ_{ℓ}(x) − φ_{ℓ}(x)| ≤ M[ ∫_{0}^{δ}[f(s, φ_{ℓ}(s)) + f(s, max{φ_{ℓ}(s), ψ_{ℓ}(s)})]ds +

+ ∫_{1−δ}^{1} [f(s, φ_{ℓ}(s)) + f(s, max{φ_{ℓ}(s), ψ_{ℓ}(s)})]ds] ≤

≤ 2M[ ∫_{0}^{δ} f(s, φ_{ℓ}(s))ds + ∫_{1−δ}^{1} f(s, φ_{ℓ}(s))ds] ≤

≤ 2M[ ∫_{0}^{δ} f(s, g_{θ}(s))ds + ∫_{1−δ}^{1} f(s, g_{θ}(s))ds] < ϵ.

In particular,

lim_{ℓ→∞} ∥T φ_{ℓ} − φ_{ℓ}∥ = 0.

In turn, we have lim_{ℓ→∞} ∥φ_{ℓ} − φ^{∗}∥ = 0, and thus

φ^{∗} ∈ ⟨g_{θ}, R⟩ ⊂ D,

and

φ^{∗} = lim_{ℓ→∞} T φ_{ℓ} = T ( lim_{ℓ→∞} φ_{ℓ}) = T φ^{∗},

which is sufficient for the conclusion of the theorem.

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(Received 4.10.1995)

Authors’ addresses:

Paul W. Eloe Department of Mathematics University of Dayton Dayton, Ohio 45469-2316 USA

Johnny Henderson Discrete and Statistical Sciences Auburn University Auburn, Alabama 36849-5307 USA