7 Statement of the principle of the stationary point
Theorem 2. If the compactness condition holds, then a correct operator having the P-property is regular
Proof. We prove that the exponential dichotomy on the half-line J holds. To this end we consider the subspace N2 = N2(0) defined by (13). The compactness condition implies that this subspace is finite-dimensional; suppose N1 is a complement of it. Since property (d 1- ) is already ensured by the estimate (15) and Lemma 1, we prove property (d2-).
As a preliminary we prove uniform stability to the right, that is, the inequality
ilu(t)ii= iiiu(s)11 (t0---. s ---. t ---. 0, u(0) E NO. (18) By assuming otherwise we find a sequence of solutions um with sup Ilum II = 1 and Mum(1m)II -> 0 (am .... t--._ 0). It follows from the com- pactness condition that the set {um (t)} is compact for any t--- 0. Hence, if z(t) is a limit point of the sequence u m (t), then z(t)E C - , Lz =0, and z (0) e Ni, that is, z -= 0 and it follows that um (t) ->ioc0. But then the norm ilum (t)II reaches its maximum value (unity) at a point sm such that sm -> —co and am — s m .- —co. We apply the estimate (7) to Urn (t), setting (a, b)= (am , 0), to = sm , and T = min {15mi, lam —5.1}. As a result we obtain a contradiction which proves (18), and together with it the exponential dichotomy on J, and consequently, the admissibility of the pair (C - , C - ). But then, for every f E C - the equation Lu = f has a solution with the estimate liulic- ---- kilif lic - .
For each f e C we put fm(t) = f(t — t m ), and suppose that vm (t) (t E j - ) is a solution of Lv =fm with the above estimate. Then we set um (t)= v m (t + tm).
174 Linear equations in a Banach space Thus, we have
Lu m = f (t---. — t m ), sup lium(t)II ---Ç. kiiif ilo
t-,-.—t rn
The difference zm,n = um — un satisfies the homogeneous equation on the half-line ( — co, bm,n), where bm,a, = min (—tm, —ta ). From (7) we easily obtain sup t.,..i b„,,„ lizm (t)ii -> 0, that is, in any case the sequence {um } is locally fundamental. By taking the limit we obtain a bounded solution of Lu =f This proves Theorem 2.
2. As well as the operator L we consider the dual operator L*
formally given by L* = —d/dt +A*(t). Regarding L* (since it is refer- red to in general) we assume the condition of left-solvability and an estimate of the form (1). As before, we denote the value of y e B* at X €B by (y, x).
An informal determination that the operators L and L* are duals is contained in Green's formula. Namely, it is assumed that if Lu = f (t E J) and L*v = g (t e J) for f E M 2(B) and g E M 2(B * ), then
t2
(y, ti)I t4 = j. [(g, u)— (v, f)] dt.
ti
Theorem 3. If L is strongly correct and L* is correct, then they are both regular.
Proof. We define L and L* as unbounded operators in the spaces
B 1 = .T2(-00, co; B) and B 1 * = Y 2(-00, 00; B*). It follows from inequalities of the form (2) that WWII -> 0 and Ilv(t)ii -000 as t -0 co for
u E D(L) and y e D(L*). Therefore, Green's formula becomes
t (v, Lu) dt = f (L*v, u) dt
I
(u e D(L), y E D (L*)).
Hence it follows that the operator adjoint to L (relative to the natural duality between B 1 and B 1 *) is some extension of L*. To prove the coincidence of these operators it is enough to establish that L* has a non-empty resolvent set. With this in mind we consider the operator L* — AI, where A >0 is sufficiently large. Since the solutions of the homogeneous equation L*v — Av =0 decrease exponentially as t -4
—co, this operator is regular. Consequently, it is sufficient to prove the invertibility in B i * of a regular operator. To this end we take a function f e B i * with compact support and put v = fr G(t, s)f(s) ds,
Theorems on regularity 175 where G is the Green's function of a regular operator. The solution
y decreases exponentially and satisfies the estimate
t
iiv(t)ii - -- /If . exp (—ci(t—s))lif( Oil ds exp (c i (t — s))iif(s)lids}
t
= /{01(t)+ 02(t)}.
Since oaf +clip, = 'If'', after multiplying by tfri and integrating we have
cif lOt12 dt = f oillfil dt
J J
1/2 1/2
--- (f 11/41 2 dt) (f
J lifil 2 dt) ,
J
that is, liv 6, --- //2ciiitlia,, which (in view of closedness) gives invertibility in B 1 *.
Now we assume that L is not regular. Then we can find a function fo e C with compact support such that Lu = fo has no solutions in C. We prove that the range of L (as of an operator in B 1 ) is not dense in B 1 . Assuming otherwise, we find a sequence urn E B 1 such that
Lu n., = fn., -* fo. Since um E C and lifm — f0lim2= Ilfm —follBi, from strong correctness we obtain Mu m — uolic -O, that is, the equation Lu =fo is
solvable in C. Hence, the range of L is not dense in B 1 .
Since L* is adjoint to L, the orthogonal complement of the range of L consists of the zeros of L*. These are the elements of C (B*) that contradict the correctness of L*. This proves that L is regular.
From what has been proved above the regular operator L is invertible in B 1 . By a theorem of Phillips (Yosida [61], p. 273), a
conjugate operator L* is invertible in B 1 .. This, together with correct- ness, gives regularity, and Theorem 3 is proved.
3. It would be useful to establish the duality of the concepts of
correctness and weak regularity. In this direction we have only been
able to obtain the following result.
Lemma 2. If one of L or L* is weakly regular, then the other is correct.
Proof. We prove that L* is correct if L is weakly regular. Let n(t) be an odd continuous scalar function such that n (t) = 1 for t.- .-- 1. We fix an E >0 sufficiently small that the pair (C, C) remains admissible
176 Linear equations in a Banach space
for the family of operators d/dt +A(t + s) - en MI with a common constant k in an equality of the form (4)• 1
Assuming that L* is not correct, we can find sequences fy m l, {gm }c C(B*) for which L * y,,,, = gm, IIYmII = 1 and ligm lic O. We select a tm EJ such that liYm(tm)lid an_ put vm (t)= ym (t + tm ), gm (t)=
g(t + tm ), and L n, = dt + A(t + tm ).
From Ilvm (0)11-1- and an estimate of the form (2) we obtain immedi- ately