7 Statement of the principle of the stationary point
Corollary 4 Corollary 4 (criterion for a point spectrum). An operator A has a point spectrum on the imaginary axis if and only if the homogeneous
3. Now we consider the question of the fixed points of non-expansive operators
Let H be a Hilbert space. An operator T :H -> H is called non- expansive if II Txi- Tx211=lixi - x211. Here there is an important connec- tion with monotonicity since the operator A = I - T is monotonic.
In fact
(Axi -Ax2, xi -x2) = (xi - x2, xi - x2) - (Txi - Tx2, X1 - X2) - Tx21111xi - x211.- 0.
Theorem 4. A semigroup U(t) (t 0) of non-linear non-expansive operators has a common fixed point if and only if it is bounded, that is,
sup II U(t)xli < co (x E H).
Similarly, a non-expansive operator T has a fixed point if and only if the semigroup of powers T" (m = 1, 2, .. .) is bounded.
General properties 153 Proof. The necessity is obvious; we prove the sufficiency. First we study the single operator case. We fix an xo E H and set
Ko = ITnxo, n--.- 01, Km = {Tnxo,n ....- m}.
Then we choose a positive number d > diam Ko and denote by S d ( x ) the ball in H of radius d and with centre at x. We define
nm = n sd(x), a = u nm.
xErcin m9
It is obvious that 12m is convex, 12m c 12m , i , and the set 12 is bounded.
Since T is a non-expansive operator we obtain
7-'12m c nm+i, Tf2 c t2. (6)
We introduce a family of operators TE defined by Tex = E (Tx – Tx 0)+ Tx 0 (0 < 6 < 1).
Since xo E 120 and the sets 12m are convex, taking (6) into account we obtain T6f2 c 12. For e < 1 the operator T, satisfies the contraction mapping principle; we denote its fixed point by xs. The boundedness of the set {xs } follows from (6) and the boundedness of 12; without loss of generality we may assume that xs —x (weakly) as E —> 1. Setting As = I – Te we have A sxs = O. Taking the limit in (A s y, y – x s ) -- 0 (y e H) we conclude that Ax = 0 and Tx = x. Thus, the operator T
has a closed convex set of fixed points.
Now we turn to the semigroup U(t). The operator T = U(to) satisfies the boundedness condition; let Ntc, be the closed convex set of its fixed points. From the commutativity of the operators U(t)
(t_- --- 0) it follows that N an invariant set, that is, U(t)Nk, c Nk, (t.-.-- 0). Let N be a minimal (relative to inclusion) closed convex invariant set. It can be proved by the preceding argument that the operator T = U(t0) has at least one fixed point in N, and so it follows that N consists of the common fixed points of the operators U(to).
2 Solvability of the Cauchy problem for an evolution equation
Let E be a separable reflexive embedded space; this means that there is a Hilbert space H such that the embedding EclIcE* is dense and continuous, and that the bilinear form (y, x)
(y e E*, x e E) coincides with the scalar product on H whenever x, y E H. We shall use the following notation for norms: 11x11 is an E-norm, WI* is an E*-norm, and lx 1 is an H-norm.
154 The method of monotonic operators
Our aim is a study of the evolution operator d/dt +A, but first we shall be concerned with constructing the differentiation operator d/dt.
1. Let [0, T] be a finite interval, S be the set of all trigonometric polynomials with coefficients in E,p>l, and p-l+q-i.... i We define D to be the closure of the set S with respect to the norm
„ T ‘ 1 I p „ T ‘ 1/q
IIUIID = ( J o IlUIIP d9 + ( jo Ilu'11,7 di)
= Ilull21(0,T;E)+11u'll2q(0,T;E*)•
The following important proposition holds.
Proposition 4. Each element in D is an H-continuous function on [0, T], so that it is meaningful to talk about the initial condition ul t = 0 = u(0)E H. Moreover, for each uo E H there is at least one u(t) e D such that u (0) = UO.
Proof. Suppose that u E S. Since lui = l lull, by the theorem about the mean there is a point to = to(u) for which
T 1/p
100)1 = Tif IWO ?) dt)---5 T-liitiiip.
o
Therefore, from the formula lu(t)1 2 = lu 40)12 + 2it To (u i, u) dợ, and by using the obvious inequality
T ‘ llq 1/p
Jo Rui, oidt '-- (f iluiii*q do ( Jo 'Fur do
--_iluiii)2
we obtain
sup lu(t)1 2 ---5 (T -1 + 2)liuilD2.
0,-5t---5T
It follows from here that the elements in D are H-continuous func- tions. Then, by an uncomplicated construction, for any uo e H we can find an extension u(t)E D, and Proposition 4 is proved.
We put
V = ..e (0, T ; E), V* = ..cfq (0, T ; E*).
We define the duality between V and V* by the usual formula: the value (y, u) of y E V * at u E V is given by (y, u)= for (v, u) dt.
We consider the operator A= d/dt with the domain {u E D:
u(0) =0}, and shall prove that it satisfies conditions (1)-(3) in § 1.
Cauchy problem for an evolution equation 155
The operator A is monotonic since
T
2(Lu, u)= 2 f (u', u) dt o
=1u(T)1 2 —iu(0)1 2
= lu(T)1 2 --- O.
Because it is obvious that A is a closed operator, we only need to give the construction of the subspaces Vn. For V„, we take the collec- tion of all trigonometric polynomials
n
ti (0 = E bk sin (krta) (bk E E).
Ic---1
It is well known that every element z E V is the sum of its Fourier sine series, and if zn, denotes the nth partial sum of this series, then
liznilv ----. ilizliv. For u E V, we have
T T
(Au, z) = f (u 1 , z) dt = f (u' , zn ) dt.
o o
Therefore,
MAW = sup (Au, z)---. sup (Au, zn) iizIlv=i iiznii---5/
= /113„Aull,
that is, condition (3) is fulfilled.