MATEMATIQKI VESNIK 49 (1997), 129–137

UDK 517.984.5

originalni nauqni rad research paper

SOME ESTIMATES OF THE REMAINDER IN THE EXPRESSIONS FOR THE EIGENVALUE ASYMPTOTICS OF SOME SINGULAR

INTEGRAL OPERATORS

Milutin Dostanić

Abstract. Some estimates are given for the remainder in the expressions for eigenvalue asymptotics of operators of the type of Riesz’ potential and logarithmic potential. As a corollary an estimate is obtained of the spectral asymptotics of an operator playing a role in the thin airplane wing theory. The method is elementary and it is based on the properties of the singular values of compact operators.

0. Introduction and notation

The general method of Birman and Solomyak [1] gives the first term of the convolution operator’s singular values asymptotics if the operator kernels satisfy some conditions. In the general case there is not any known method to determine higher terms of the asymptotics, or at least an estimate of the remainder in the known asymptotic formulas. But in some special cases, it is still possible thanks to the special kernel structure (see [5]).

In this paper some estimates of the remainder in asymptotic formulas for convolution operator’s eigenvalues, where the kernels of the operator are |x − y|^{α−1} and ln |x − y|, are obtained using elementary methods.

Let H be a separable Hilbert space over C and let A be a compact operator. The singular values of the operator A (s_{n}(A)) are the eigenvalues of the operator (A^{∗}A)^{1/2} (or (AA^{∗})^{1/2}).

Denote by λ_{n}(A) eigenvalues of A in the order of decreasing absolute values, taking into account their multiplicity.

By ∫_{a}^{b} m(x, y) · dy we denote the integral operator on L^{2}(a, b) with the kernel m(x, y).

The operator A is a Hilbert-Schmidt one if ∑_{n⩾1} s_{n}^{2}(A) < +∞.

AMS Subject Classification: 47B10

Supported by Ministry of Science and technology RS, grant number 04M01 through Matematički fakultet

A Hilbert-Schmidt operator A = ∫_{a}^{b} m(x, y) · dy satisfies the following equality ∑_{n⩾1} s_{n}^{2}(A) = ∫∫ |m(x, y)|^{2} dx dy.

1. Results

Theorem 1. Let α_{0} = \frac{5 − \sqrt{17}}{4}.
a) If 0 < α ⩽ α_{0}, then

λ_{n}(∫_{−1}^{1} |x − y|^{α−1} · dy) = 2Γ(α) (\frac{2}{nπ} )^{α} cos \frac{απ}{2} (1 + O(n^{−δ})) ,

where δ = \frac{1 − 2α}{3}.

b) If α_{0} ⩽ α < 1, then

λ_{n}(∫_{−1}^{1} |x − y|^{α−1} · dy) = 2Γ(α) (\frac{2}{nπ} )^{α} cos \frac{απ}{2} (1 + O(n^{−δ})) ,

where δ = \frac{3α − 2α^{2}}{2 + 5α − 2α^{2}} .

Theorem 2. λ_{n}(∫_{−1}^{1} −\frac{1}{π} ln |x − y| · dy) = \frac{2}{nπ} (1 + O (\frac{\sqrt{ln n}}{n^{1/5}})).

Consider the operator

Hf(x) = \frac{1}{π} ∫_{−1}^{1} \frac{f′(t)}{x − t} dt,

whose domain consists of the functions f satisfying the following conditions: f′ ∈ L^{1}(−1, 1), Hf ∈ L^{2}(−1, 1), f(−1) = f(1) = 0 [3] (the integral is taken in the sense of principal value). This operator plays an important role in the theory of thin airplane wing.

In [4] M. Kac heuristically deduced the formula

λ_{n}(H) = \frac{πn}{2}(1 + o(1)).

In [3] H. M. Hogan and L. A. Sahnovič gave a strong proof of this asymptotic formula. Applying Theorem 2 we give an estimate of the remainder in the asymptotic formula for λ_{n}(H).

Theorem 3. The following asymptotic formula holds:

λ_{n}(H) = \frac{πn}{2} (1 + O(n^{−1/11})) .

Remark 1. Since the function

α ↦ \{ \begin{matrix}\frac{1−2α}{3}, & 0 < α ⩽ α_{0}, \\ \frac{3α−2α^{2}}{2+5α−2α^{2}} , & α_{0} ⩽ α < 1, \end{matrix}

attains its minimum when α = α_{0}, Theorem 1 implies a more rough but unified estimate for 0 < α < 1, i.e.

λ_{n}(∫_{−1}^{1} |x − y|^{α−1} · dy) = 2Γ(α) (\frac{2}{nπ} )^{α} cos \frac{απ}{2} (1 + O(n^{\frac{2α_{0}−1}{3}} )) .

Remark 2. Part b) in Theorem 1 holds in the case 0 < α < 1 but in the interval (0, α_{0}) the better estimate is given by a). That is the reason why these two cases are separated.

2. Proofs

To prove the previous Theorems we need some Lemmas.

Lemma 1. If A and B are compact operators on a Hilbert space H such that

\begin{aligned}s_{n}(A) & = a · n^{−α} + O(n^{−β}), & a > 0, α < β < α + 1, \\ s_{n}(B) & = O(n^{−β_{1}}), & β_{1} ⩾ \frac{β}{α + 1 − β} , \end{aligned}  (1)

then s_{n}(A + B) = a · n^{−α} + O(n^{−β}).

Proof. Observe that from the assumptions of Lemma 1 it follows β_{1} > β.

For n ∈ N let k = k(n) = [n^{1− \frac{β}{β_{1}}} ] − 1, m = m(n) = [n^{\frac{β}{β_{1}}} ] ([x] denotes the greatest integer not exceeding x). Then n = (k + 1)m + j, where j = j(n) is a

nonnegative integer and j(n) ⩽ n^{\frac{β}{β_{1}}} +n^{1− \frac{β}{β_{1}}} . From the properties of singular values of the sum of two operators [2] it follows s_{(k+1)m+j}(A + B) ⩽ s_{km+j}(A) + s_{m+1}(B) and hence

n^{β} (s_{n}(A + B) − \frac{a}{n^{α}} ) ⩽

n^{β} (s_{km+j}(A) − \frac{a}{(km + j)^{α}} )+n^{β} (\frac{a}{(km + j)^{α}} − \frac{a}{((k + 1)m + j)^{α}} )+n^{β}s_{m+1}(B). (2)

The assumptions (1) and the properties of the sequences k(n), m(n) and j(n) give

\begin{aligned}\overline{lim} n^{β} (s_{km+j}(A) − \frac{a}{(km + j)^{α}} ) & < +∞, \\ \overline{lim} n^{β}s_{m+1}(B) & < +∞, \end{aligned}  (3)

and since β_{1} ⩾ \frac{β}{α+1−β}, we obtain \overline{lim} n^{β} (\frac{1}{(km+j)^{α}} − \frac{1}{((k+1)m+j)^{α}} ) < +∞. Com-

bining (2) and (3) we get \overline{lim} n^{β} (s_{n}(A + B) − \frac{a}{n^{α}} ) < +∞. Since s_{(k+1)m+j}(A) = s_{(k+1)m+j}(A + B − B) ⩽ s_{(k+1)m+j}(A + B) + s_{m+1}(B), i.e. s_{km+j}(A + B) ⩾ s_{(k+1)m+j}(A) − s_{m+1}(B).

In the similar way we conclude \underline{lim} n^{β} (s_{n}(A + B) − \frac{a}{n^{α}} ) > −∞, and so s_{n}(A + B) = an^{−α} + O(n^{−β}).

Remark 3. The proof of the following statement is carried out in the same way.

(a) If A and B are compact operators such that

s_{n}(A) = an^{−α} + O(n^{−β}L(n)), s_{n}(B) = O(n^{−β_{1}}L(n))

and if L is a slowly varying function, L(x) → +∞ when x → +∞, α < β < α + 1, β_{1} ⩾ β/(α + 1 − β), then

s_{n}(A + B) = an^{−α} + O(n^{−β}L(n)).

(b) From Lemma 1 it follows: If A and B are compact operators such that

s_{n}(A) = an^{−α}, a > 0; s_{n}(B) = O(n^{−β_{1}}), β_{1} > α,

then s_{n}(A + B) = an^{−α} + O(n^{− \frac{β_{1}}{1+β_{1}} (1+α)}).

Lemma 2. Let T = ∫_{0}^{1}(x + y)^{α−1} · dy. If α > 0, then
s_{n}(T) = o(n^{−1/2}) (= O(n^{−1/2})).  (4)

If 0 < α < 1, then

s_{n}(T) = O(n^{−α−α( \frac{3}{2} −α)}).  (5)

(Observe that for small (large) values of α a better estimate is given by (4) ((5)), respectively.)

Proof. From ∫_{0}^{1}∫_{0}^{1}(x + y)^{2α−2} dx dy < +∞ (α > 0) it follows that T is a Hilbert-Schmidt operator and we have

s_{n}(T) = o(n^{−1/2}) = O(n^{−1/2}).

Let us now prove the asymptotic relation (5). Let δ > 0 be a fixed number (δ < 1). First, we prove

s_{n}(∫_{0}^{δ}(x + y)^{α−1} · dy) ⩽ C_{1} \frac{δ^{α}}{n^{α}}  (0 < α < 1),  (6)

where the constant C_{1} does not depend on n and δ.

Let J^{α}: L^{2}(0, δ) → L^{2}(0, δ) be the fractional integral operator defined by

J^{α}f(x) = \frac{1}{Γ(x)} ∫_{0}^{x}(x − y)^{α−1}f(y) dy

and let the operator C: L^{2}(0, δ) → L^{2}(0, δ) be defined by

Cf(x) = ∫_{0}^{x}(x + y)^{α−1}f(y) dy.

According to the formula for fractional partial integration [6] we get

Cf(x) = ∫_{0}^{x}(J^{α}f)(y)(−\frac{1}{Γ(1 − α)} \frac{d}{dy} ∫_{y}^{x}(t − y)^{−α}(x + t)^{α−1} dt) dy,

i.e.

C = −\frac{1}{Γ(1 − α)}MJ^{α},  (7)

where M: L^{2}(0, δ) → L^{2}(0, δ) is a linear operator defined by

Mf(x) = ∫_{0}^{x} H(x, y)f(y) dy and H(x, y) = \frac{d}{dy} ∫_{y}^{x}(x + t)^{α−1}(t − y)^{−α} dt.

The function H can be represented in the form (for y > x)

H(x, y) = −(2x)^{α−1}(x − y)^{−α} + (α − 1)y^{−1}Φ (\frac{x}{y}) ,

where Φ(s) = ∫_{1}^{s}(r − 1)^{−α}(r + s)^{α−2} dr, s > 1.

Now, we prove that the integral operator with the kernel H(x, y)θ(x − y) is bounded on L^{2}(0, +∞). (θ(x) = 0 for x < 0, θ(x) = 1 for x ⩾ 0). Since the function H(x, y)θ(x − y) is homogenous of order −1, according to the Hardy-Littlewood-Polya inequality [6] it is enough to prove that k = ∫_{0}^{∞} x^{−1/2}|θ(x − 1)H(x, 1)| dx < +∞, i.e.

k = ∫_{1}^{∞} x^{−1/2}| − (2x)^{α−1}(x − 1)^{−α} + (α − 1)Φ(x)| dx < +∞.

Since 0 < α < 1, from asymptotic behavior of the function Φ when x → 1 and x → +∞ it follows that the previous integral converges. Therefore, we have

∥M∥ ⩽ k.  (8)

Having in mind that s_{n}(J^{α}) ⩽ const · δ^{α}/n^{α} (const. does not depend on n and δ), from (7) and (8) and from the properties of singular values of the sum of two operators it follows s_{2n}(C) ⩽ const · δ^{α}/n^{α}, i.e. s_{n}(C) ⩽ const · δ^{α}/n^{α} (const. does not depend on n and δ). From

s_{2n}(∫_{0}^{δ}(x + y)^{α−1} · dy) = s_{2n}(C + C^{∗}) ⩽ 2s_{n}(C) ⩽ 2 const \frac{δ^{α}}{n^{α}}

we obtain the estimate (6).

Now, we prove that if the operator D_{δ}: L^{2}(δ, 1) → L^{2}(0, 1) is defined by

D_{δ}f(x) = ∫_{δ}^{1}(x + y)^{α−1}f(y) dy,

then the following estimate

s_{n}(D_{δ}) ⩽ C_{2} \frac{δ^{α−1}}{n^{3/2}}  (9)

holds (C_{2} does not depend on n and δ).

Let L_{0} = { f ∈ L^{2}(δ, 1) : ∫_{δ}^{1} f(t) dt = 0 } and let Q: L^{2}(δ, 1) → L_{0} be the orthoprojector. Consider the operator A_{0} = D_{δ}Q: L^{2}(δ, 1) → L^{2}(0, 1). Since the range of D_{δ} − A_{0} = D_{δ}(I − Q) is one (I is the unit operator), we have [2]

s_{n−1}(A_{0}) ⩽ s_{n}(D_{δ}) ⩽ s_{n+1}(A_{0}).  (10)

As in [2] we prove that A_{0} = D′_{δ}J′, where D′_{δ}, J′: L^{2}(δ, 1) → L^{2}(0, 1) are the operators defined by

D′_{δ}f(x) = (1 − α) ∫_{δ}^{1}(x + y)^{α−2}f(y) dy, J′f(x) = ∫_{δ}^{x} f(t) dt.

Since D′_{δ} is a Hilbert-Schmidt operator we have

ns_{n}^{2}(D′_{δ}) ⩽ (1 − α)^{2} ∫_{0}^{1} dx ∫_{δ}^{1}(x + y)^{2α−4} dy,

and hence s_{n}(D′_{δ}) ⩽ const δ^{α−1}/\sqrt{n} (const. does not depend on n and δ). Using (10) and

s_{2n}(A_{0}) ⩽ s_{n}(D′_{δ}) · s_{n}(J′) ⩽ const \frac{δ^{α−1}}{n^{3/2}}

(const. does not depend on n and δ) we obtain (9).

Let P_{δ}: L^{2}(0, 1) → L^{2}(0, 1) be the orthoprojector defined by

P_{δ}f(x) = χ_{[0,δ]}(x)f(x)

(by χ_{S} we denote the characteristic function of the set S). Then

T = T (I − P_{δ}) + (I − P_{δ})T P_{δ} + P_{δ}T P_{δ}

and we have s_{3n}(T) ⩽ s_{n}(T (I − P_{δ})) + s_{n}((I − P_{δ})T P_{δ}) + s_{n}(P_{δ}T P_{δ}). Combining this with (6) and (9) we obtain

s_{3n}(T) ⩽ 2C_{2} \frac{δ^{α−1}}{n^{3/2}} + C_{1} \frac{δ^{α}}{n^{α}}

(C_{1}, C_{2} do not depend on n and δ). This inequality holds for each δ ∈ (0, 1).

Substituting δ = n^{α− \frac{3}{2}} we get

s_{3n}(T) ⩽ (2C_{2} + C_{1})n^{−α−α( \frac{3}{2} −α)}

and therefore s_{n}(T) = O(n^{−α−α( \frac{3}{2} −α)}).

Lemma 3. s_{n}(∫_{0}^{1} ln(x + y) · dy) = O (\frac{\sqrt{ln n}}{n^{3/2}}).

Proof. Let V : L^{2}(0, 1) → L^{2}(0, 1) denote the operator defined by

V f(x) = ∫_{0}^{1} ln(x + y)f(y) dy.

The operator V can be expressed in the form

V = V (I − P_{δ}) + (I − P_{δ})V P_{δ} + P_{δ}V P_{δ}.  (11)

In the similar way as for the estimates (6) and (9) we get

s_{n}(P_{δ}V P_{δ}) ⩽ C_{3} \frac{δ}{n},  s_{n}(V (I − P_{δ})) ⩽ C_{4} \frac{\sqrt{− ln δ}}{n^{3/2}}  (12)

(C_{3}, C_{4} do not depend on n and δ).

From (11) and (12) and the properties of singular values of the sum of two operators we obtain

s_{3n}(V) ⩽ 2C_{4} \frac{\sqrt{− ln δ}}{n^{3/2}} + C_{3} \frac{δ}{n}.

This inequality holds for each δ ∈ (0, 1). If we put δ = \sqrt{(ln n)/n}, then we have s_{3n}(V) ⩽ C_{5}(\sqrt{ln n})/n^{3/2} (where the constant C_{5} does not depend on n) and therefore

s_{n}(V) = O (\frac{\sqrt{ln n}}{n^{3/2}}) .

Proof of Theorem 1. Let k(x) = |x|^{α−1} (0 < α < 1) and let
A(x, y) = ∑_{n=−∞}^{∞}(k(x − y + 4n) − k(x + y + 4n + 2)).  (13)

By direct calculation it can be proved that

∫_{−1}^{1} A(x, y)φ_{n}(y) dy = k̂ (\frac{nπ}{2} ) φ_{n}(x); φ_{n}(x) = sin \frac{nπ(1 + x)}{2}, n = 1, 2,…,

where k̂(ξ) = ∫_{−∞}^{∞} e^{itξ}|t|^{α−1} dt = 2Γ(α) cos \frac{απ}{2} |ξ|^{−α} and \{φ_{n}\}_{n=1}^{∞} is an orthonormal base of L^{2}(−1, 1).

Let A_{1}, A_{2}, A_{3}, A_{4} be integral operators acting on L^{2}(−1, 1) with kernels

A_{x,y}, k(x − y − 4) + ∑_{\begin{matrix}n≠0 \\ n≠−1 \end{matrix}}(k(x − y + 4n) − k(x + y + 4n + 2)),

−k(x + y + 2) and − k(x + y − 2),

respectively. Then we have

∫_{−1}^{1} |x − y|^{α−1} · dy = A_{1} − A_{2} − A_{3} − A_{4}.  (14)

From (13) it follows

s_{n}(A_{1}) = 2Γ(α) (\frac{2}{nπ} )^{α} cos \frac{απ}{2} .  (15)

Let 0 < α < 1/2. According to Lemma 2 (relation (4)) we have s_{n}(A_{3}) = O(n^{−1/2}), s_{n}(A_{4}) = O(n^{−1/2}). Since the kernel of the operator A_{2} is a smooth function on [−1, 1]^{2}, using the properties of singular values of the sum of two operaors [2] we obtain s_{n}(A_{2} + A_{3} + A_{4}) = O(n^{−1/2}). From this equality, (14), (15) and from Remark 3(b) it follows

s_{n}(∫_{−1}^{1} |x − y|^{α−1} · dy) = 2Γ(α) (\frac{2}{nπ} )^{α} cos \frac{απ}{2} (1 + O(n^{\frac{2α−1}{3}} )) .

Since the operator ∫_{−1}^{1} |x − y|^{α−1} · dy (0 < α < 1) is positive, for 0 < α < 1/2 we obtain

λ_{n}(∫_{−1}^{1} |x − y|^{α−1} · dy) = 2Γ(α) (\frac{2}{nπ} )^{α} cos \frac{απ}{2} (1 + O(n^{\frac{2α−1}{3}} )) .  (16)

Suppose now 0 < α < 1. According to Lemma 2 (relation (5)) we have

s_{n}(A_{3} + A_{4}) = O(n^{−α−α( \frac{3}{2} −α)}) and therefore, using the same line of arguments as before, we have

s_{n}(A_{2} + A_{3} + A_{4}) = O (n^{−α−α( \frac{3}{2} −α)}) .

From (14), (15) and Remark 3(b) it follows

λ_{n}(∫_{−1}^{1} |x − y|^{α−1} · dy) = 2Γ(α) (\frac{2}{nπ} )^{α} cos \frac{απ}{2} (1 + O(n^{−δ})) ,  (17)

where δ = \frac{3α − 2α^{2}}{2 + 5α − 2α^{2}} .

Observe that for 0 < α ⩽ α_{0} (α_{0} ⩽ α < 1) the better estimate is given by (16) ((17)), respectively.

Proof of Theorem 2. If G_{1}(t) = \frac{1}{π} K_{0}(|t|) (K_{0} is Mc-Donald’s function) then

∫_{−∞}^{∞} G_{1}(t)e^{iξt} dt = (1 + ξ^{2})^{−1/2}.

It is well known that the function K_{0} decreases exponentially and

K_{0}(x) = (− ln x)A(x) + B(x),  (18)

where A and B are even entire functions and A(0) = 1.

Let

R(x, y) = ∑_{n=−∞}^{∞}(G_{1}(x − y + 4n) − G_{1}(x + y + 4n + 2)).

By direct calculation we obtain

∫_{−1}^{1} R(x, y)φ_{n}(y) dy = Ĝ_{1} (\frac{nπ}{2} ) φ_{n}(x) = (1 + \frac{n^{2}π^{2}}{4} )^{−1/2} φ_{n}(x)

(φ_{n}(x) = sin \frac{nπ(1+x)}{2}, n ∈ N), which implies that ∫_{−1}^{1} R(x, y) · dy is positive. Using

(18) and (1 + \frac{n^{2}π^{2}}{4} )^{−1/2} = \frac{2}{nπ} (1 + O(\sqrt{ln n}/n^{3/2})) and applying technique from Theorem 1, taking into account Remark 3(a) we obtain

λ_{n}(∫_{−1}^{1} −\frac{1}{π} ln |x−y| · dy) = s_{n}(∫_{−1}^{1} −\frac{1}{π} ln |x−y| · dy) = \frac{2}{nπ} (1 + O (\frac{\sqrt{ln n}}{n^{1/5}})) .

Proof of Theorem 3. It is well known [3] that the operator B = H^{−1} is positive and has the form Bf(x) = ∫_{−1}^{1} b(x, t)f(t) dt, where

b(x, t) = −\frac{1}{π} ln |x − t| + \frac{1}{π} ln |1 − tx + \sqrt{(1 − t^{2})(1 − x^{2})}|.

By the method from Lemma 2 and the paper [3] we get

s_{n}(∫_{−1}^{1} \frac{1}{π} ln |1 − tx + \sqrt{(1 − t^{2})(1 − x^{2})}| · dt) = O(n^{−6/5}).

From Theorem 2 it follows

s_{n}(∫_{−1}^{1} −\frac{1}{π} ln |x − t| · dt) = \frac{2}{nπ} (1 + O(n^{−1/11})) .

Applying Lemma 1 (Remark 3(a)) from the previous relation and (19) it follows

s_{n}(B) = \frac{2}{nπ} (1 + O(n^{−1/11})), i.e.

λ_{n}(H) = \frac{πn}{2} (1 + O(n^{−1/11})) .

REFERENCES

[1] M. Š. Birman and M. Z. Solomjak, Asymptotic behavior of the spectrum of weakly polar integral operators, Math. USSR Izvestija, 4, 5 (1970), 1151–1168 (Translations A.M.S. 1971).

[2] I. C. Gohberg and M. G. Krein, Introduction to the Theory of Nonselfadjoint Operators, Amer. Math. Soc. Transl. 18 (1969), Providence, R.I.

[3] H. M. Hogan and L.A. Sahnovič, Asymptotic behavior of the spectrum of a singular integrodifferential operator, Diff. Equations 20, 8 (1984), 1444–1447.

[4] M. Kac, Sbornik perevodov “Mathem.”, 1, 2 (1957), 95–124 (in Russian)

[5] B. V. Palcev, Asymptotic eigenvalue behavior of convolution operators on a finite interval with rational Fourier transformation of kernel, Math. Dokl. 194, 4 (1970), 774–777.

[6] S. G. Samko, A. A. Kilbas and O. I. Maricev, Fractional Integrals and Derivations and Some Applications, Minsk 1987.

(received 05.05.1997.)

Faculty of Mathematics, Belgrade University, P.O.B. 550, 11000 Beograd, Yugoslavia