On discreteness of spectrum and positivity of the Green’s function for a second order functional-differential operator on semiaxis

We study conditions of discreteness of spectrum of the operator defined by \mathcal{L}u=-\frac{1}{\rho (x)}{u}^{″}(x)-{\int }_0^{\mathrm{\infty }}u(s)d_{s}r(x,s), x\in [0,\mathrm{\infty }). The operator has two singularities at the ends of the interval (0,\mathrm{\infty }). The second question is positivity of solutions of the equation \mathcal{L}u=f under boundary conditions u(0)=0, u^{\prime }(\mathrm{\infty })=0. The used abstract scheme is close to the well-known MS Birman’s method in the spectral theory of self-adjoint operators. Conditions for discreteness of spectrum and positivity of the Green’s operator are obtained. The result relates to the MS Birman’s result on the necessary and sufficient condition for discreteness of spectrum of a polar-differential operation. The results may be interesting for researchers in qualitative theory of functional-differential equations and spectral theory of self-adjoint operators.

MSC: Primary 34K08; 34K10; secondary 34K12.

1.1 Problems and a well-known result

Our first objective is to study the conditions for discreteness of spectrum^a of the functional-differential operator defined by

with two singularities: at x=0 and at infinity. Note that one particular case of the expression (1.1) is the following operator with one deviation:

The second question is existence of positive solutions of the equation \mathcal{L}u=f (see Definition 1.1).

Note the result of Birman [[1], (Chapter 2, Section 29)] for the spectral problem

where the operation {\mathcal{L}}_0 is called polar-differential operation. This singular spectral problem is usually considered in the space L_2(R_{+},\rho ) of functions that are square-integrable on R_{+} with the positive weight ρ. Birman showed that a necessary and sufficient condition of discreteness of spectrum of the operator {\mathcal{L}}_0 is

The singularity at the point x=0 is not reflected in this condition. If {\int }_0^1\rho (x)dx=\mathrm{\infty }, condition (1.3) is not sufficient for discreteness. We impose the second condition,

The two conditions (1.3) and (1.4) are sufficient for discreteness of the spectrum of (1.2) (Theorem 2.1).^b It seems that (1.4) is also necessary one.

Part of this work is a continuation of the research in the articles [2–4].

1.2 Assumptions, notation

Everywhere below, except for the independent appendix, we use the assumptions and notation introduced in this subsection.

The function ρ is assumed to be measurable and positive almost everywhere on R_{+}=[0,\mathrm{\infty }) and satisfying the important condition

Remark 1.1 Comparing (1.3), (1.4), and (1.5) we see that the key role is played by the properties of the function

Conditions (1.3) and (1.4) can be written as {lim}_{s\to \mathrm{\infty }}\mathrm{\Phi }(s)={lim}_{s\to 0}\mathrm{\Phi }(s)=0, and condition (1.5) is boundedness of the \mathrm{\Phi }(s). The latter is sufficient (Lemma 4.3) for the inclusion W\subset L_2(R_{+},\rho ) and for boundedness of the operator T (see, below in this subsection). It is close to a necessary condition: if, for example, {lim}_{s\to \mathrm{\infty }}\mathrm{\Phi }(s)=\mathrm{\infty }, then T is unbounded (see Remark 4.3).

Everywhere below the function r(x,s) is assumed to satisfy the following conditions: it is nondecreasing in s\in R_{+} for almost all x\in R_{+}, it is measurable in x for any s\in R_{+}, and

is locally integrable on (0,\mathrm{\infty }). We can assume that r(x,0)=0 for almost all x\in R_{+} (as follows from a property of the Stieltjes integral).

Now, let us introduce the function

Finally, let

• L_2(R_{+},\rho ) be the Hilbert space of all square-integrable with positive almost everywhere weight ρ on R_{+}=[0,\mathrm{\infty }) functions f, i.e., and with scalar product

  (f,g)={\int }_{R_{+}}f(x)g(x)\rho (x)dx,

• W be the set of all locally absolutely continuous on R_{+} functions u satisfying

  (1.9)

and the boundary condition u(0)=0; W is a Hilbert space (Lemma 4.1) with scalar product

and

• T:W\to L_2(R_{+},\rho ) be the operator defined by Tu(x)=u(x), x\in R_{+}. Note that W\subset L_2(R_{+},\rho ) under condition (1.5) (Lemma 4.3).

1.3 About domain of the operators {\mathcal{L}}_0 and ℒ

If λ is a regular value of {\mathcal{L}}_0, then any solution of -\frac{1}{\rho }{u}^{″}-\lambda u=f satisfies -\frac{1}{\rho }{u}^{″}\in L_2(R_{+},\rho ). So, we have to assume that the domain D({\mathcal{L}}_0) of the operator {\mathcal{L}}_0 consists of all solutions of the equation

These solutions have locally on (0,\mathrm{\infty }) absolutely continuous derivative u^{\prime } and have the form

where G_0(x,s)=\rho (s)min\{x,s\}. In fact, since {({\int }_x^{\mathrm{\infty }}\rho zds)}^2\le {\int }_x^{\mathrm{\infty }}\rho z^{2}ds{\int }_x^{\mathrm{\infty }}\rho ds the integral is convergent, and u^{\prime }(x)={\int }_x^{\mathrm{\infty }}\rho (t)z(t)dt+u^{\prime }(\mathrm{\infty }). From this we have

We impose the boundary conditions u(0)=0, u^{\prime }(\mathrm{\infty })=0. Both of these conditions are necessary for the implementation of the variational method in the space of W. Then the domain is determined automatically (see Corollary A.1 to Lemma A.1 and Lemma 4.8).

Definition 1.1 We say that the boundary value problem

is positively solvable, if it is uniquely solvable for any f\in L_2(R_{+},\rho ), and the implication f\ge 0\to u\ge 0 holds.

Note that positive solvability is equivalent to the positivity of the Green’s function G(x,s), which allows one to represent the solution of \mathcal{L}u=f in the form

or u=Gf, where G is the Green’s operator of ℒ.

2.1 Operator {\mathcal{L}}_0

Instead of direct investigation of the equation {\mathcal{L}}_{0}u=f, where f\in L_2(R_{+},\rho ), consider the equation

Equation (2.1) is the result of the variational method, in which the primary object is the form [u,v]. Equation (2.1) has the short form

defined on the spaces W and H=L_2(R_{+},\rho ). Note that W is a Hilbert space (Lemma 4.1), T:W\to L_2(R_{+},\rho ) is bounded (Lemma 4.3), and T(W) is dense in L_2(R_{+},\rho ). So, we can use the abstract scheme in Appendix 1. According to Corollary A.1 (2.1) is equivalent to an equation {\mathcal{L}}_{0}u=f. By virtue of Lemma 4.8 (2.1) is equivalent to

Theorem 2.1 Suppose conditions (1.3) and (1.4) hold. Then the spectral problem

has a discrete spectrum.

Proof The discreteness follows from Theorem A.1 and Lemma 4.7. □

Remark 2.1 (Estimate of the greatest lower bound of the spectrum)

Let {\lambda }_0 be the greatest lower bound of spectrum of the operator {\mathcal{L}}_0. Thus the problem

for is uniquely resolvable for any f\in L_2(R_{+},\rho ), but it is not if \lambda ={\lambda }_0. Then in view of (A.4) from inequality (4.3) it follows that

This is an accurate estimate. If \rho (x)=1/x^2, then {\lambda }_0\ge 1/4, but \lambda =1/4 is a point of the spectrum, as follows from Example 2.1.

Example 2.1 If \rho (x)=1/(x^2), then s{\int }_s^{\mathrm{\infty }}\rho (s)ds\equiv 1, and the operator T is bounded, but the two conditions (1.3) and (1.4) are not fulfilled. The spectrum of the operator {\mathcal{L}}_{0}u=-x^2{u}^{″} in L_2(R_{+},\rho ), \rho =1/(x^2) is the interval [1/4,\mathrm{\infty }).

In fact, the value is regular (Remark 2.1). Let \lambda >1/4.

By means of the change of variable x=e^t the equation

can be transformed to -u^{″}(t)+u^{\prime }(t)-\lambda u=\phi, where \phi =f(e^t). Since

this equation has to be considered in the space L_2(R,e^{-t}).

The homogeneous equation -u^{″}+u^{\prime }-\lambda u=0 has the solution u=e^{(1/2)t}(c_{1}cos\delta t+c_{2}sin\delta t) for a \delta >0. It is not in L_2(R,e^{-t}). Let u=e^{(1/2)t}sin\delta tv(t)=b(t)v(t). Then

If, for example, v=1/\sqrt{t}, t\ge 1, and v=0, (for 0\le t\le 1 the v(t) may be defined arbitrarily), then \phi \in L_2(R,e^{-t}). But the corresponding solution u\notin L_2(R,e^{-t}). Thus, λ is not a regular value of the operator.

Since the spectrum is a real closed set, [1/4,\mathrm{\infty }) is the spectrum.

2.2 General operator ℒ

The operator (1.1) can be represented as \mathcal{L}={\mathcal{L}}_0-Q, where Q is defined by

The operator Q acts from W to L_2(R_{+},\rho ) and is bounded under certain conditions (see, for example, (5.6)). Along with the general case, let us consider one special case (deviating operator):

where q(x) is assumed to be nonnegative locally integrable function, and h(x) is a measurable function. Note that the notation q(x) in (2.5) corresponds to the definition (1.7), if represent expression (2.5) in the form (2.4).

Theorem 2.2 If conditions (1.3) and (1.4) are fulfilled and Q:W\to L_2(R_{+},\rho ) is bounded, then spectrum of ℒ is discrete. If the function ξ defined by (1.8) is symmetric, i.e. \xi (x,y)=\xi (y,x) for all x,y\in R_{+}, then the spectrum is real and the system of eigenfunctions has the orthogonal basis properties in L_2(R_{+},\rho ).

Proof Conditions (1.3) and (1.4) are sufficient conditions of compactness of the operator T (Lemma 4.7). The symmetry condition of \xi (x,y) allows one to show the identity (Qu,Tv)=(Qv,Tu) (see Section 5). Now we can refer to Theorem A.3. □

Using the estimate (5.6) from Theorem A.3 we have the first main result.

Theorem 2.3 Suppose (1.3) and (1.4) hold and

Then the spectral problem

has a discrete spectrum. If \xi (x,y)=\xi (y,x),^c the spectrum is real, and the system of eigenfunctions has orthogonal basis properties in the spaces W and L_2(R_{+},\rho ).

Remark 2.2 The spectrum is not real in general, because of the non-symmetry of the function \xi (x,y).

The obtained estimate works well in the case of one deviation, if Q is defined by (2.5). From (2.6) we have the following.

Corollary 2.1 Suppose (1.3) and (1.4) hold and

Then the spectral problem

has a discrete spectrum.

Example 2.2 If Qu(x)=q(x)u(x) the operator ℒ has the representation

If (1.3) and (1.4) are satisfied and

then the operator ℒ has discrete spectrum. In this case (2.6) has the form (2.8). It follows also from (5.4). Note that the inequality (2.8) is satisfied, if q(x) is bounded.

Suppose Q:W\to L_2(R_{+},\rho ) is bounded. By the substitution u=T^{\ast }z the equation \mathcal{L}u=f is reduced to the equation

where K=Q{T}^{\ast } is an integral operator with nonnegative kernel,

Thus, if spectral radius of K is less than unit (), then (3.1) is uniquely resolvable and

Since^d z={(I-K)}^{-1}f and u=G_{0}z, the operator G=G_0{(I-K)}^{-1} is positive, and G(x,s)\ge 0. Thus:

Theorem 3.1 Suppose Q is bounded and . Then the boundary value problem (1.13), (1.14) is uniquely resolvable for any f\in L_2(R_{+},\rho ) and the Green’s operator G is positive.

Remark 3.1 Since \rho (K)\le \parallel K\parallel =\parallel T^{\ast }Q\parallel \le \parallel T^{\ast }\parallel \parallel Q\parallel =\parallel T\parallel \parallel Q\parallel the condition is sufficient for positivity of the Green’s operator G.

The second main result is presented in the following statement.

Theorem 3.2 If

then the boundary value problem (1.13), (1.14) is positively solvable (see Definition  1.1).

Proof See Remark 3.1 to Theorem 3.1 and the estimates (4.2) and (5.6). □

Consider the following particular case.

Corollary 3.1 The equation

(q\ge 0) has a positive solution in W for any f\in L_2(R_{+},\rho ), f\ge 0, f\ne 0, if

The following particular case shows that in this estimate the inequality sign < cannot be replaced by ≤.

Example 3.1 From Theorem 3.1 and (5.4) it follows that the equation

(q(x)\ge 0) is uniquely resolvable for f\in L_2(R_{+},\rho ) in W and has positive solution for f\ge 0, if

In particular, if q(x)\le \mathrm{const}=q, and \rho (x)=1/x^2, \parallel T^{\ast }Q\parallel \le 16q^2. Thus, if , the equation -x^2{u}^{″}-q(x)u=f has unique positive solution for any f\in L_2(R_{+},\rho ), f\ge 0, f\ne 0.

But if q(x)=\mathrm{const}=1/4, the equation -x^2{u}^{″}-(1/4)u=f may not have a solution for some f (this was considered in Example 2.1).

In concluding this section consider one useful assertion. Let

and

Note that Q(u,v)=(Qu,Tv) (see the equality (5.2)).

Let {\lambda }_0 be the greatest lower bound of spectrum of the operator ℒ. Then by (A.10)

From Theorem A.5 follows the following.

Theorem 3.3 Suppose conditions (1.3) and (1.4) hold. Then {\lambda }_0 is the smallest eigenvalue of the problem \mathcal{L}u=\lambda u, u(0)=u^{\prime }(+\mathrm{\infty })=0, and the following statements are equivalent:

1. 1.

   〈u,u〉 is positive definite,

2. 2.

   {\lambda }_0>0,

3. 3.

   .

Remark 3.2 We do not suppose that {\lambda }_0>-\mathrm{\infty } (but if it is so).

Recall that in all assertions below condition (1.5) is assumed to be fulfilled.

Lemma 4.1 W is a Hilbert space.

Proof The relation u(x)={\int }_0^{x}z(s)ds establishes a bijection between W and the Hilbert space L_2(R_{+}). □

Lemma 4.2 The value A_u={\int }_{R_{+}}\frac{|u(s)u^{\prime }(s)|}{s}ds satisfies the inequality

Proof This follows from the inequalities

 □

Remark 4.1 The estimate (4.1) is accurate, since if u=s^{1/2-\epsilon }, s\ge 1, and u=s^{1/2+\epsilon }, s\ge 1, then

Remark 4.2 If u(s)\ge 0, u^{\prime }(s)\ge 0, u^{″}(s)\le 0, then

In fact, denote B_u={\int }_{R_{+}}\frac{u^2}{s^2}ds. Then B_u=2A_u. Since u(s)={\int }_0^s{u}^{\prime }(t)dt\ge s{u}^{\prime }(s), B_u\ge {\int }_{R_{+}}{u}^{\prime }{(s)}^{2}ds=[u,u].

Lemma 4.3 Suppose (1.5) holds. Then T(W)\subset L_2(R_{+},\rho ), the operator T:W\to L_2(R_{+},\rho ) is bounded, and the norm of T satisfies the estimate

Proof Since (Tu,Tu)={\int }_{R_{+}}{u}^2\rho dx=2{\int }_{R_{+}}\rho (x)dx{\int }_0^{x}u(s)u^{\prime }(s)ds, we can estimate

From this and (4.1)

 □

Remark 4.3 If {lim}_{s\to \mathrm{\infty }}\mathrm{\Phi }(s)=\mathrm{\infty } (see (1.6)), then T(W)\not\subset L_2(R_{+},\rho ). In fact, if u(s)u^{\prime }(s)\ge 0, then (Tu,Tu)=2{\int }_{R_{+}}\frac{u(s)u^{\prime }(s)}{s}\mathrm{\Phi }(s)ds. It is possible to find a nonincreasing function \psi (s) such that

Now find u such that u(s)u^{\prime }(s)=\psi (s): u{(s)}^2=2{\int }_0^s\psi (t)dt. Since u\ge 0, u^{\prime }(s)\ge 0, and u^{\prime } is nonincreasing, by Remark 4.2, . But (Tu,Tu)=\mathrm{\infty }.

Lemma 4.4 The image T(W) is dense in L_2(R_{+},\rho ).

Proof The proof is left to the reader. □

The following theorem [[5], p.318] can be used to show compactness.

Theorem 4.1 (Gelfand)

A set E from a separable Banach space X is relatively compact if and only if for any sequence of linear continuous functionals that converge to zero at each point

the convergence (4.4) would be uniform on the E.

Lemma 4.5 Suppose (1.3) holds. If f_n\in L_2(R_{+},\rho ) is bounded, then

uniformly on the set \{(u,n):\parallel u\parallel \le 1,n\in \{1,2,\dots \}\}.

Proof Since

it is sufficient to show that

uniformly on \parallel u\parallel \le 1. We have

The first term tends to zero because of the inequality

and (1.3). The second term is equal to

This tends to zero because of (4.1) and (1.3). □

Lemma 4.6 Suppose (1.4) holds. If f_n\in L_2(R_{+},\rho ) is bounded, then

uniformly on the set \{(u,n):\parallel u\parallel \le 1,n\in \{1,2,\dots \}\}.

Proof Since

it is sufficient to show that

when a\to 0 uniformly on \parallel u\parallel \le 1. We have

Now we refer to (1.4) and (4.1). □

Lemma 4.7 If (1.3) and (1.4) hold, then T is compact.

Proof Let \mathrm{\Omega }=\{Tu:\parallel u\parallel \le 1\}. We use the criterium of compactness of Gelfand (see Theorem 4.1). Let f_n\in L_2(R_{+},\rho ) be a sequence of functionals such that f_n(z)\to 0 for any z\in L_2(R_{+},\rho ). We have to show that f_n(Tu)\to 0 uniformly for \parallel u\parallel \le 1.

Let \epsilon >0. Using Lemma 4.5 choose N such that

for all n and for all \parallel u\parallel \le 1. The same we can do with {\int }_0^a{f}_n(x)u(x)\rho (x)dx for sufficiently small a>0. For this aim we can use Lemma 4.6.

Now we only have to show that {\int }_a^N{f}_n(x)u(x)\rho (x)dx\to 0 uniformly on the set \parallel u\parallel \le 1. Since

where {\phi }_n(s)={\int }_s^N{f}_n(x)\rho (x)dx, it suffices to show that {\int }_a^N{\phi }_n{(s)}^{2}ds\to 0. We show that {\phi }_n(s)\to 0 uniformly for s\in [a,N]. Let

Then {\phi }_n(s)=f_n(z_s). The set \{z_s:s\in [a,N]\} is compact. Thus by Theorem 4.1 of Gelfand f_n(z_s) converges to zero uniformly for these s. So, {\phi }_n(s)\to 0 uniformly for s\in [a,N], and f_n(Tu)\to 0 uniformly for \parallel u\parallel \le 1. □

Remark 4.4 It seems that condition (1.4) is necessary for compactness of T.

Lemma 4.8 Equation (2.1) is equivalent to problem (2.2).

Proof Denote h^{\prime }=-f\rho. From (2.1) it follows that

(if we choose v such that the corresponding limits exist). If v=0 for x\notin [a,b]\subset (0,\mathrm{\infty }), then {\int }_a^b{u}^{\prime }{v}^{\prime }dx={\int }_a^{b}h{v}^{\prime }dx. From this u^{\prime }=h+\mathrm{const} on [a,b]. Since the segment [a,b] is arbitrary, the relation u^{\prime }=h+\mathrm{const} is fulfilled on the whole semiaxis (0,\mathrm{\infty }). So, u^{″}=h^{\prime }=-\rho f. By the first equality in (4.5)

Now, u^{\prime }v{|}_{x=\mathrm{\infty }}=u^{\prime }v{|}_{x=0}=0. Choosing v such that v(\mathrm{\infty })\ge C>0, we obtain u^{\prime }(\mathrm{\infty })=0. □

Here we consider the operator Q defined by (2.4).

5.1 Symmetry of the form (Qu,Tv)

Under the assumptions imposed on the functions r(x,s) and \xi (x,y) in Section 1.2, from Lemma B.1 (in Appendix 2) we obtain the following statement.

Lemma 5.1

In this case X=Y=R_{+}, \mu (e)={\int }_e\rho (x)dx, K(x,dy)=r(x,dy)=d_{y}r(x,y).

Using Lemma 5.1 the form

can be represented in the symmetrical form

Hence, this form is symmetric if the function \xi (x,y) is symmetric: \xi (x,s)=\xi (s,x).

5.2 Unique deviation

Consider first the special case when the operator Q is defined by (2.5), i.e. Qu(x)=q(x)u(h(x)). Using (4.1) we have

From this follows the estimate of \parallel Q\parallel:

In particular, when Qu(x)=q(x)u(x),

5.3 General operator Q

Here we consider the general case of the operator Q defined by (2.4), i.e. Qu(x)={\int }_0^{\mathrm{\infty }}u(t)d_{t}r(x,t). Suppose that the function \xi (x,y)=\rho (x){\int }_0^{x}r(s,y)ds (see (1.8)) is absolutely continuous in y, and \xi (x,y)={\int }_0^{y}p(x,t)dt. Then p(x,t) does not decrease in x.

In this case

where q(x) is defined by (1.7). Using Lemma 5.1, we obtain

The latter step can be done in the same manner as in the relation (5.4).

From this

Since

it may be presented in the form

This estimate only works well if for all x deviation is concentrated around the point h(x). For example, if r(x,s)=q(x)r_1(s-h(x)), r_1(+\mathrm{\infty })=1, r_1(-\mathrm{\infty })=0 then

This estimate coincides with (5.3), if r_1(t)=\{\begin{array}{cc}0\hfill & \text{if }t\le 0,\hfill \\ 1\hfill & \text{if }t>0.\hfill \end{array}

In this case

We do not use general spectral theory (see, for example, [6, 7]). But the scheme below is close to the scheme in [[7], Chapter 10] except for using a different notation. We find also convenient explicit use of the embedding T from W to H (see below). This scheme was used also in [2–4].

A.1 Positive form

Let W and H be Hilbert spaces with inner product [u,v] and (f,g), respectively. Let T:W\to H be a linear bounded operator. The equation

has the unique solution u=T^{\ast }f for any f\in H, where T^{\ast } is adjoint operator. Let D_{\mathcal{L}}=T^{\ast }(H).

Assume that

1. 1.

   the image T(W) of the operator T is dense in H,

2. 2.

   dimkerT=0.

Lemma A.1 If the image T(W) of the operator T is dense in H, then T^{\ast } is an injection.

Proof Suppose T^{\ast }f=0 for a f\in H. Then for any g\in T(W)

Since T(W) is dense in H, the f=0. □

Corollary A.1 (Euler equation)

The operator T^{\ast } has an inverse ℒ defined on the set D_{\mathcal{L}}. Equation (A.1) is equivalent to

The spectral problem for the operator ℒ we will write in the form

Let {\lambda }_0 be the greatest lower bound of the spectrum of ℒ. It is well known (see, for example, [[7], Chapter 6]) that

Since (\mathcal{L}u,Tu)=[T^{\ast }\mathcal{L}u,u]=[u,u],

Theorem A.1 The spectrum of ℒ is discrete if and only if T is compact.

Proof Since (A.3) is equivalent to u=\lambda T^{\ast }Tu, discreteness of spectrum of ℒ is equivalent to compactness of T^{\ast }T. But the operators T^{\ast }T and T^{\ast } are compact at the same time [[7], Chapter 10]. □

Theorem A.2 Suppose T is compact. Then (A.3) has a nonzero solution u_n only in the case of \lambda ={\lambda }_n, n=0,1,2,\dots , i.e.

The system u_n forms an orthogonal basis in W. The sequence {\lambda }_n forms a nondecreasing sequence of positive numbers,

and lim{\lambda }_n=\mathrm{\infty }.

Remark A.1 The minimal eigenvalue {\lambda }_0 satisfies the equality (A.4).

A.2 General case

Let

be a symmetric bilinear form, u,v\in W. Assume that Q is bounded in both arguments. Moreover, suppose that this form has the representation Q(u,v)=(Qu,Tv), where Q:W\to H is bounded.^e Then the equation [u,v]-(Qu,Tv)=(f,Tv), \mathrm{\forall }v\in W, is equivalent to

and in the set D_{{\mathcal{L}}_0}=T^{\ast }(H) to

where {\mathcal{L}}_0={(T^{\ast })}^{-1}.

Theorem A.3 Suppose T is compact. Then the equation

has a nonzero solution u_n only in the case of \lambda ={\lambda }_n, n=0,1,2,\dots , i.e.

The system u_n can be chosen to form an orthogonal basis in the space W.

If the form 〈u,v〉=[u,v]-(Qu,Tv) is lower semi-bounded, i.e.

then spectrum of ℒ is semi-bounded, and

is the greatest lower bound of the spectrum [[7], Chapter 6]. Thus we have the following.

Theorem A.4 If (A.9) holds and T is compact then the eigenvalues {\lambda }_n have a minimum and can be put in increasing order

A.3 Positive definiteness and spectral radius of Q{T}^{\ast }

Let \rho =\rho (Q{T}^{\ast }) be the spectral radius of the operator Q{T}^{\ast }. Note that the two operators Q{T}^{\ast } and T^{\ast }Q have the same spectral radius. So

Lemma A.2 The quadratic form 〈u,u〉 is positive definite if and only if .

Proof From (A.11) it follows that \rho [u,u]-(Qu,Tu)\ge 0 and 〈u,u〉=[u,u]-(Qu,Tu)\ge (1-\rho )[u,u]. If then 〈u,u〉 is positive definite. Conversely from the inequality [u,u]-(Qu,Tu)\ge \epsilon [u,u] (\epsilon >0) it follows that \rho \le 1-\epsilon. □