Certain properties of eigenfunctions of linear pencils - Springer Link

deleting certain places, then x is also enabled in N', and moreover the marking of places in N' will be identical to their marking in N. Therefore, the sequence v is enabled in N I and in the generalized net N2 (without unc. places), and in both cases it will lead to the terminal m a r k i n g . H e n c e at,ate, . . . ae,~ ~-- L (N~) CI L~. (N~) and so ~t = h (~} == g (at,at~ . 9 . at,~) ~ g ( L 1 I~ L~).

Finally, we obtain

L c=_ g (LI ~ iS).

Conversely,

let at.ate. . . a~,~ =,_ L I ~ L*2 := L 1 ~ L c (Nz).

It is obvious that if the markings of some place in N O and of the corresponding place in N I or in the generalized net N 2 are the same (if the place is colored we assume that the colors of the tokens are also the same), then they will also coincide after any transition t in the nets has fired, provided that tokens of the same color are moved in N o and N 2. Therefore, for any k ( l ~ k < n ) , if a sequence tlt 2 ... t k is enabled in both N I and N2, then tk_ I will fire in N O for the same color as in N2, and the whole sequence takes N o to the null marking. Thus, h (tj2

=

. . . t,~)

g (aeyt~.

9 9 at,)

~

L~ (No)

and consequently, g ( L I ~ L ~ ) ~ _ L . Thus, we have proved the implication is obvious. This proves Theorem 2.

~(~). The converse

Theorems 1 and 2 imply COROLLARY. R E - (.~. *)(Fin); moreover, in order to derive any r.e. language from finite languages, it is sufficient to apply the operations of iterated shuffle, intersection with regular languages and morphism, each operation at most twice. The author thanks N. K. Vereshchagin for discussing the work and offering useful comments. LITERATURE CITED i. 2. 3. 4.

J . L . Peterson, Petri Net Theory and the Modeling of Systems, Prentice-Hall, Englewood Cliffs (1981). V . E . Kotov, Petri Nets [in Russian], Nauka, Moscow (1984). M. Jantzen, "The power of synchronizing operations on strings," Theor. Comput. Sci., 14, No. 2, 127-154 (1981). B . L . Budinas, "Solvability of the reachability problem for vector addition systems," in: Problems in Cybernetics [in Russian], Moscow (1988), pp. 22-45.

CERTAIN PROPERTIES OF EIGENFUNCTIONS OF LINEAR PENCILS S. G. Pyatkov

Let us consider the spectral problem of the form L~

~B~,,

(i)

where L is an unbounded symmetric operator in a given separable Hilbert space E, B is an operator self-adjoint in E. Let us assume, for simplicity, that ker B = {0} and t[

,

~o

~.~t~, ~ ) ~ 6 1 1 . . J ! ~

~%~D(L).

6>0,

where (., .) is the scalar product in E, by D(A) we understand the domain of definition of the operator A. Let H I, H 0 be the completion of D(L) according to the norms!l~i[~7 ~(L~ ~i~'~ i!ul1~ -if !B l~a[!, respectively. Let us consider that H l is densely imbedded in E and D (IBll/~). One can show that the operator L-IB: H I § H I is defined and bounded and let us assume that it is fully continuous. Under these conditions in [i, 2] the question was considered of the Riesz basis property in the space H 0 of the eigenfunctions of problem (i), where both necessary and sufficient conditions of the basis property were presented, and also some sufficient conditions. The question arises, which was stated by A. A. Shkalikov: Do there exist pairs of operators L, B, satisfying the conditions presented above, such that the eigenfunctions of problem (i) do not form a Riesz base in the space H0? The second Mathematics Institute, Siberian Branch, Academy of Sciences of the USSR. Translated from Matematicheskii Zametki, Vol. 51, No. i, pp. 141-148, January, 1992. Original article submitted June ii, 1990.

90

0001-4346/92/5112-0090512.50

9 1992 Plenum Publishing Corporation

question, which also arises in a natural manner, is if such pairs exist, then are the sufficient conditions presented in [I, 2] near to necessary? The main goal of this paper is to give an answer to these two questions. Let us note that spectral problems of the form (i) arise in many areas of mathematical physics. One can mention the paper [3], where a S t u r m Liouville problem of form (i) is considered and a number of examples presented. Auxiliary Assertions. For the convenience of the reader we now present several definitions and theorems from [i], which we need in what follows. Let Pk be the spectral family of the operator B. Let us introduce the operators E + = I - P0, E- = P-0, U = E + - E- (I is the identity operator). U is an isomorphism of H!0 to H 0, U 2 = I. Let us define the space H_ I as the completion of H 0 according to the norm I I / I ! . _ , --:: II L-1R/IlJr, ~== II ~ / I [ . .

where H' is the completion of E according to the norm

II~!~i = sup](%~)[/l! ~: llr,.

By an eigen-

h~lil

function of problem (i) we understand an element ( ~ H I such that for some % ~ 0 in the space H' (i) is satisfied. Let { ~ } be the eigenfunctions of problem (i) corresponding to positive k t, and negative 1 i eigenvalues, enumerated taking into account multiplicity, so that {i%[ ]}, {[~-~ ]} are nondecreasing sequences. THEOREM i. The eigenfunctions of problem (i) normalized by the equation lltf,~ils0~-= 1 form a Riesz basis in H 0 if and only if kH1, H iI~/~....Ho (IH~H_Ih/2 is the space obtained with the help of the method of complex interpolation [4]). THEOREM 2. Let H s = IHI, H011_ s. Let us assume that there exists s o > 0 such that the operator from Hs0 to Hs0 , is defined and bounded. Then [HI. H_lll,2 == Ho. Thus, the condition of Theorem 2 U ~ - _ ~ (H~, H~) is a sufficient condition of the Riesz basis property of the eigenfunctions (i). Let us denote by M, N the domains of values of the projectors E +, E- in H 0. THEOREM 3. Let us assume that in H 0 there exists a projector P on M or N such that for some s ~ O P, P*~-'t (II~_ H~)i~ ~r (H,, Ho) (P* is the operator adjoint in H 0 to P). Then there exists s o > 0 such that U, E +, A'- ~ (H,o, lifo). Theorem 3 was formulated in [I] as Lemma 2.3, where s = i was taken in [I]. s < 1 the proof of this lemma does not change.

But for

Remark I. The examples presented in [i] show that Theorems 2, 3 are applicable for a sufficiently large class of differential operators L if B is an operator of multiplication by a function. Basic Results. Here we show in example that linear bundles, the eigenfunctions of which do not form a Riesz basis in H 0, exist, and that the conditions of Theorems 2, 3 are close to necessary. For simplicity we will restrict ourselves to the class of linear bundles of the simplest form

u" == ~

(~,) u, u (1)

=

u

(--1)

=

O, x ~

(--1, 1), 2

(2)

"~

~1

where X(x) is some function such that IX(X)[ = i. Here 1.....d/dx', D (L) ....|u (--I, I ) ~ W2~--1, i), E = H 0 = L2(-I , i). We have H I = W~(--], i), H' = W~1(--i, I), and H_ I is the completion of E in the norm

Actually, we have that Bu = Uu = X(X)U. The next example shows, in particular, that if the condition U ~ I ( H ~ , H s ) s > 0 is not satisfied, then the unconditional basis property can not hold. Example i.

Let N I = {i, 3, 5, 7 .... }, N= = {2, 4, 6,,..}. I

z (z) = 1

L e t us n o t e t h a t not satisfied.

~

- - J,

x~(--

'

(n+i)

U

for some

Let

~5----q-' ~

=

1, t ) \ ~ ] + = ~ - .

Hs = W~(-1, 1) ( s < 1 / 2 ) [ 4 ] . L e t us show t h a t t h e c o n d i t i o n I f [Hz, H - ~ I ~ / 2 = H0, t h e n we m u s t h a v e t h e i n e q u a l i t y [4]

o f Theorem 1 i s

91

'/~ !! u II.. "~./ c II u I1,~. !1 u ~_,

V u ~ Hi,

(3) For this it is suf-

where c does not depend on u. Let us show that (3) is not satisfied. ficient to construct a sequence un~ ~ H~ such that

il ..,~,, II H/II

II u~

u , , , I!H,

(i

=

2

dd

"

IlumllH_~..

q~(mx) ( m ~ N 1 ) .

=

(5)

= (i'_, Let us evaluate

(4)

, oo.

( x ) ~ I, ][~[]E = 1 and let U m =

Let us take the function ~ (x) ~ Co (--I, I), 0 ~ We have supp um C (--I/m, ~/m) and

il

H_,

--

We have d l" =

~

L-lx/.$,ni =

(% (:~" ~) = Let us evaluate max ]v(x) l.

{(~ (~

- - 1 ) 2, i)..2,

We have ~i

~) X (L~)um (~) d~,

G~: ( x ,

i--I

~.~> ~', < ~'.

.0

G~ (z, [) Xu,,~ (ij) d~ = 1~ + 7:. Let us take, for example, the case x ~ ( - - l , 0). Two variations are possible: x ) > --I/m, x ~ -i/m. For simplicity let us consider that m > i0. Let us consider the first version. There exists l ~ m , l~A~ such that x~(--I/l, --I/([ ~ 2)). We have

,~,j

t,)_I/n

"

=

~-~/(~,-~i)

-

"

,. . . . ,.

,

"

I~) ~ ~ =~,., &.

l,~ = (4-6 + I}_~ -i- I~-2

(6)

If ~ - 8 < m, then the first addend and the corresponding part in the second vanish. the fact that [G~(x..~) I ~ i, I @(x) ] ~ 1 we have

I [:~ + L-~ -i_ L-~ + L I < t/(l

6)

t/(/ + 2) "I 5Oral

Due to

(7)

Let us transform the remaining sums In

"x

--~1

=

,!]in 1 (~ 1) o

C:: x, "~ -- -s

'~1

d'~---

(n,~-'])--lt(n

,~t."-l,'O: -~)

,,

"

(n -r 1)

" ,

"

9 ('T

t

It is easy to verify that for n ( I -- 8

[ 6:.~ (x, " r - -

-.'o)

-- G~ a', ~ --

"~

tn

(n-]

(

1

or n ~- I + 2

-i- i) ) T, ~

~4,:" I)'

l

,

we have

c~ (0, [.,,"It

(8) -

t/'(l'Z -~ ~ )).

-

Furthermore,

l u,;, (1: - - Un) - - u,~ (~ - - 1,/(n -7 ~)11 ""~~ / c.,m/n%.

(9)

Using (8), (9), we obtain I I , , j -%1"c ('l/n 3 ~ re~n4).

Then from (6) will follow "M~176 -

.f

g/

,

/

l ~ n ]

) ,

-'-

In the case .r-~--~ m there will only be present i n t h e sum ( 6 ) t h e t h i r d addend, and the evaluation is conducted similarly. The evaluations are simplified i n t h e c a s e J1 a n d we have 13j j~-~ .]Iv/nz 2, where M I as also M d o e s n o t d e p e n d o n m. If x ~ ( 0 . 1 ) , then let us conduct

92

precisely the same arguments. have

I v I < MJ mf" From this evaluation we will

Thus, we have max

if--~ I~,'[-0dx

,," !e,,, Ih,_, -

% . .

(10)

M~/m-. .

.

.

.

Thus, from (5), (i0) we obtain t1/2

"'

Thus, the condition of Theorem 1 is not satisfied, that is, [Hz, H-~]~/2 x H0. This entails that the eigenfunctions of problem (2) do not form a Riesz basis in H 0 = Lf(-I, i). In this example the measure of the intervals of fixed sign of the function X(X)I n = (i/(n + i), i/n) decreases as n -2 and U ~ ~ (H~, H~) (Vs > 0). As follows from [5] (see the formulation and proof of Lemma 2), actually for any power decrease of the measures of the corresponding intervals, the operator of multiplication by the function X(X) is not continuous as an operator from ws(-l, i) to W~(--i, l)(Vs~> ()). In this case examples are easily constructed similar to that presented above. In case X(x) changes sign on the interval (-I, i) a finite number of times, the operator U possesses the property that U ~ Y (H~,Hs) for s < 1/2 [4, 5], that is, the conditions of Theorem 2 are satisfied. Let the condition [Hz, H_z]i/2 = H 0 be satisfied. The question arises: how wide is the class of functions X(X) satisfying the condition U (H~, H~) for some s > 0, that is, how close to necessary is this condition? We show here that into this class not only enter functions, changing sign on the interval (--i, l) a finite number of times, but indeed an essentially wider class of functions. In accordance with what was remarked above, it is necessary that the measure of the intervals of fixed sign of the function x(X), enumerated according to decrease of lengths, not decrease in a power fashion. Example 2.

Let us consider problem (2), where we take "> =

-~

U

[(2-=, ~

U (--2

~

n ~N~

] '__,- =_ (0, l)\t~. +, x (x) = i

-

' -

" + .' , - - .

,,,

,)-1l i]

1, x~L>, ~,

-

;- ~ '-)--.

L e t u s show t h a t i n t h e g i v e n c a s e U ~ Lf (I/s, Hs) f o r s < 1 / 2 . L e t us n o t e t h a t f o r s < 1 / 2 , }is = W S ( - 1 , 1) [ 4 ] . L e t u s show t h a t f o r s < 1 / 2 t h e o p e r a t o r E + [E+u = X+(X)U(X), X+(x) = 1 f o r x .~-(.)~ and X+(X) = 0 f o r x.--~ _(.)-- ] i s c o n t i n u o u s a s an o p e r a t o r f r o m w s ( - 1 , 1) t o w S ( - 1 , 1 ) . I t , h e n c e , w i l l f o l l o w t h a t U -~ 7 (II~,Hs) f o r s < 1 / 2 , s i n c e E+ + E- = I . Let us take S = { u ~ 6 ~ 0 (--I. I): ~: ~ Co (1).I) :~ Co (--I. 0)} and obtain for :t,--_-Sthe bounds

II E+,,. II q(o, ,>< ~ II ,, ,,.;

In obtaining the bounds

~

Thus, if the measure of the corresponding interval decreases exponentially, then the condition of Theorem 2 is satisfied, t h a t i s , t h e s e two e x a m p l e s a c t u a i l y give an answer to the questions put in the introduction. And we h a v e l i n e a r pencils, f o r w h i c h H = [H 1, H _ 1 1 1 / 2 ~ H 0. I f t h i s i s s o , t h e n how a r e t h e s e two s p a c e s H a n d H 0 r e i a t e d ? Let us return to the general case, that is, to problem (1). The n e x t t h e o r e m g i v e s a n a n s w e r t o t h i s q u e s t i o n . THEOREM 4. Under the condition H ~ H0 n o t And w h a t i s m o r e , i f o n e o f t h e s e i n c l u s i o n s is

one of the satisfied,

inclusions B'~Ho, t h e n H = H 0.

HO'~H holds.

Proof. L e t u s show t h a t f r o m H 0 c H i t f o l l o w s t h a t H = H0. Let us normalize the eigenfunctions of problem (i) by means of the equation (U(p~, ~'F)H~ + § (/) = :-~ (/. -- ~Tfl. L e t cV U ~F)H.. + + We have [i] ( ~rr , + (~7)H~ = -/, 6~y and

H = { ~ _ H _ ~ : 5~(1~~(,~)

§ 14(,p)l ~) = i l ~ l l ~ < ~ } .

From the inclusion H 0 c H follows the inequality

i! ~p ltH -< ciI ~ i!~. Let us take / =

ZIIL c~+Z~11c~,l~= p+/+ p-/ .

Vm ~ Ho.

(15)

From (15) we have

[(U~,/, V)Ho ! = I ~ = ~ c~ (1) 4 ((p) !'< c I1i IP, II ~ II~~ We hence obtain that

It 1'~1 lira