R O CZN IKI POLSK IEG O TOWARZYSTWA M ATEM ATYCZNEGO Séria I: PRACE MATEMATYCZNE XXII (1980)
Wo jciec h M. Ko z l o w sk i (Krakôw)
Non-linear operators in Banach function spaces
This paper is an attempt to give a presentation of some general results, which concern the non-linear operators acting from a Banach function space into another.
The reader is referred to [3] and [4] for some similar though only partial results on the Nemytskii operators in Lebesgue and Orlicz spaces.
Sometimes our proofs employ the ideas of Krasnosel’skiT [3]. In this paper the distinguished role is played by the methods of Banach function spaces the theory in which was created mainly by W. A. J. Luxemburg and A. C. Zaanen.
The paper consists of five sections. Section 1 is a brief exposition of some basic concepts of the theory of Banach function spaces.
One of the results of Section 2 is Corollary 2.2, which states that there exists sufficiently large subset Z of X such that the characteristic function of Z is a member of a Banach function space. This section is also de
voted to establishing (in the case of finite measure) a continuous imbedding of a Banach function space E into M (X) the space of all measurable functions of X equipped with topology of convergence in measure. Next definition of correct spaces is presented and discussed.
Section 3 deals with various general properties of non-linear operators.
Also some examples are presented and discussed.
Section 4 contains extension theorem and boundedness theorem. Moreover, we define class of locally uniformly correct spaces.
Section 5 contains continuity theorems, then an important case of the Nemytskii operator is discussed.
One can easily see that the results of the paper can be generalized in many directions. For example one can try to replace (where it is possible) norms by F-norms (in the terminology of [1] and [2]). This would lead us to the theory of modular spaces (for the general facts of this theory see e.g. [10] or [9]). As the simplest interesting generalization one could consider the case of s-homogeneous F-norms, that is the condition
of positive homogenity of the norm is replaced by ||Aw|| = |A |s ||u||
(0 < s ^ 1). This corresponds to the case of s-convex modular spaces and in particular to the case of spaces IF with 0 < p ^ 1. For basic facts of s-convex modulars see e.g. [8].
I should like to express my thanks to Prof. J. Musielak for his continuing help during the preparation of the manuscript and encouregement during the work. Special thanks are extended to Dr. A. Kaminska for numerous improvements of the text and for her reasonable skepticism about some of the results.
1. Preliminaries. Let m be a non-negative countably additive measure in the non-empty point set X . We assume that m is <x-finite, i.e. X is union of an at most countably number of sets of finite measure. It will be also assumed that m is atomless, i.e. there is no sets A of positive measure such that for any measurable set В with В a A, either m(B) = m( A) or m(B) = 0 holds. The assumption of cr-finiteness and atomlessness of the measure is of course a restriction of the generality, we observe, however, that most of measures playing an important role in the theory of integral operators have those properties.
The notation l A will stand for the characteristic function of the set A cz X. Functions defined on X and differing only on a set of measure zero will be identified (in other words, we shall always deal with equiva
lence classes rather than with the individual functions). Automatically, subsets of X whose characteristic functions differ only on a set of measure zero will then also be identified.
Without repeating it every time, we assume that any function on X and any subset of X mentioned in the following is measurable.
The collection of all non-negative functions on X will be denoted by M +(X) or simply by M +. By P we will denote the set of all measurable functions, where we will permit also that function in P assumes the values
± o o on a set of positive measure.
Definition 1.1 (Luxemburg, Zaanen, [6]). The mapping g of M + into the [0, oo] is called a function norm if g has the following properties:
. (1) 0 < g(u) < oo for all u e M + ,
g(u) = 0 if and only if и = 0 almost everywhere, g(u-hv) ^ g(u) + g(v),
g (au) — ад (и) for every finite constant a > 0;
(2) If u , v e M + and и ^ v a.e., then g(u) ^ g(v).
Given the function norm g, we extend the domain of definition of g to the whole of M by defining @(u) = g(|u|) for any u e M . It is easily seen that L e — { ueM; g(u) < oo} is a normed linear space. If Le is com
plete, it is called a Banach function space (normed complete Kothe space).
Definition 1.2 (Riesz-Fischer property). The function norm q is said to
00
have the Riesz-Fischer property, whenever £ q{u„) < oo implies that
00 n ~ 1
e( X un) < » .
n ~ 1
The following two results are due to Luxemburg and Zaanen, [6]:
Theorem 1.3. For a function norm q the following statements are equivalent : (i) q has the Riesz-Fischer property,
(ii) q satisfies the triangle inequality for infinite sums, i.e.
00 00
(*) e( E U„) ^ E б Ы for every sequence (un).
n = 1 n = 1
Theorem 1.4. The normed linear space Le is complete if and only if q
has the Riesz-Fisher property.
If Z <= X is of positive measure the space of all u e L e with (supp и) c Z will be denoted by Le(Z). Clearly L e(Z) is also a normed linear space and if Le is complete, then Le(Z) is complete as well.
A set W ciX is called a carrier of L e(X) if X \ W is the largest in the sense of inclusion set such that ul X\w = 0 for every u e L e. Without repeating it every time we assume that the carrier of L e is equal the whole of X.
In the sequel Le will be denoted by £ or Я and function norms respectively by || • ||E and || • ||H.
2. Further remarks on the Banach function spaces.
Theorem 2.1. Let m(X) < со, let E be a Banach function space. Then there exists a unit in E, i.e. a function ù e E such that û(x) > 0 a.e. in X .
P roof. For every u e E we denote { x e l : u{x) = 0} by N(u). We shall write и ~ v, whenever N (u) = N(v). Clearly the relation ~ is an equiva
lence relation, it follows now that E can be decomposed into equivalence classes ù, the class ù consists of all v e E such that N (u) = N(v). Let
£ be a collection of all classes ù, where u e E . Now È may be equipped with a binary relation -3 defined by formula v -3 ù о N (и) с N (v) for arbitrary u e ù and v gv.
It is evident that relation -3 is a partial order in È. Let S be an arbitrary linearly ordered subset of È. If S consists of a finite number of elements, then there exists a v e S such that N(i>)czN(u) for every ù e S . Hence ù -3 v for every ù e S , i.e. v is a maximal element of S.
Assume now, that S is an infinite set. Denoting p = in fm(N(ù)), where infimum is taken over all ù e S , we have p e [ 0 , oo). There exists a sequence (ûn) <= S such that m ( N (ù„))lp, as n -► oo. Hence N (1^ ) $ N(ii2) $ •••?
because S is linearly ordered.
Denoting N* = f] N (щ) we observe that i = 1
m (N*) — m( П N (щ)) = lim m(N(ii„)) = p.
i = 1
Since m(N(ùn)) i p , to every lie S there exists a iifc, which has the following property m(iV(ii)) ^ m(iV(iifc)). Hence N(ii) з N( ùk) з N*. Thus iV* c= iV(ii) for every lie S.
Now we can choose a sequence (un) such that u„eù„ and u„ ^ 0.
Since m(N(û„))l p, it follows that m(N(ù„)) < m( X) for n ^ 2, i.e. iin # Ô.
Let us put
v2 ^ M2> — £nMn>
where e„ > 0 is such that s„\\un\\E < i l l^ - ill r - Define u* by
|0 for x e N*, и* (x) =
[u„(x) for x e N (Un-ifyN (u„), n ^ 3.
00
Therefore u* = £ i>„ • We have then
n = 1
00 00
\\u* \ \e — Il Z _ i)\JV(uM>II£ ^ Z WVn ' liV(Mll_ ! >\ДГ(ия)II£
n - 2 n - 2
oo oo
£n \\Un ’ 1 )\iV (i4 lï) l l £ ^ ^ £ n I I I I H
n ~ 2 n— 2
GO J
^ Z лл“ 2 " ' Il W2 II £ = i II w 2 II £ < 0 0 •
n = 2 Thus m* is a member of E.
Let ii* denote the class of u*. Observe now that ti -3 it* for every lie S , since N (ù*) = N* c= N(ii). Finally S is bounded above by ii* and by Kuratowski-Zorn lemma we obtain that È has a maximal element ii**.
Suppose that m(N(it**)) > 0. Since the carrier of E is equal to the whole of X , it follows that, if В is a set of positive measure, then not all functions in E are equal to zero on B. Hence there exists a function w in E and a set А а В (m(A) > 0) such that w(x) Ф 0 for x e A By atom- lessness of m we may assume that A Ç B. Using the same argument to B = N (ù**), we get A ç N (it**) and w. Putting uA(x) = |w(x)|, we have мл (л) > 0 for all x e A . Let v = 1a ua + 1x\au**, then v e E and N(v)
$ N (it**). Hence, ii** -3 i) and v ф ù**. This implies ù** is not the maximal element in È. Contradiction implies m(N (it**)) = 0, hence there exists a u**eii** such that u**(x) > 0 for x e l . This completes the proof.
Let us note that the assertion holds also in the case of ir-finite measure.
Indeed, let X = (J < oo and (Yt) be a disjoint collection of sets.
i = 1
From the above argument it follows that for every i e N there exists a мге £ ( ^ ) such that щ(х) > 0 for x e Yt. Let (a,) be a sequence of positive
00 00
numbers such that £ af | | | | < oo. Put и = £ a.ly.u,-, then ||w||£
ao I = 1 i = 1
I аг||пг||£ < oo, i.e. u e E . For x e l we have u(x) = агм((х) for a cer- i = 1
tain i e N , therefore u(x) > 0.
Co r o l l a r y 2.2. Let E be a Banach function space. Then there exists a sequence (An) of measurable sets such that c: A 2 <= ..., lim m(An) = m( X)
and 1 a„£E for every n e N . "
P roof. By the previous results we get a function u e E such that u(x) > 0 for every x e X . Let us define
A„ = {x gI : u(x) ^ 1 / n} . 00
Clearly (An) is non-decreasing and IJ An = X . Thus we have lim m(An)
n = 1 »
1 1
= m(X). Since — • l AfJ ^ и ■ l An ^ u, it follows that — - l ^ e E and ob-
n n
viously l Ane E, which is the desired result.
R em ar k . The result of Corollary 2.2 canot be converted into the result l x e £ . For example one may put E = \ u: (0, 1)-»R; f < oo>.
1 (0,1) J
Th e o r e m 2.3 (Luxemburg, [ 5 ] ) . Let E be a Banach function space, m(X) < oo and (u„) <= E. I f lim ||u„||£ = 0, then u„-^ 0, i.e. (u„) converges in measure to zero. "
If we give up the assumption of finiteness of the measure Theorem 2.3 ceases to hold, as it is shown in the following
Example 2.4. Let E = \ u: R 1
я ; J N*)l
R * 2 + l
dm < oo >, let un = l [fI>„ + 1];
then \\un\\E = J [tl,n+ 1]
measure to zero.
x 2 + 1 dm-> 0 as n -> oo, while un does not converge in Definition 2.5. A Banach function space E is said to be correct if,
00 whenever (D„) is a non-increasing sequence of measurable sets, m( П Ai)
n= 1
= 0 and и is an arbitrary member of E, then lim ||1D u||E = 0, i.e. the function norm is absolutely continuous. "
Examples. Lebesgue spaces И for p e [ l , o o ) , Orlicz spaces И for tp
satisfying A2 condition and Lorentz spaces А ф are correct, while L°°, U
for (p which does not satisfy A2 condition and Marcinkiewicz spaces M^ are not correct spaces.
Theorem 2.6. Let E be correct and m (X) < oo. Then to every e > 0 there corresponds a Ô > 0 such that m(D) < Ô implies ||1ви||£ < e.
P ro o f. Assume to the contrary that there exists an e > 0 such that for every Sn > 0 we may obtain a measurable set Dn a X and
(i) m{Dn) < Sn while ||1D|i k||e > e.
Put 0„ — m(X)/2" and by (i) take Dn such that \ \ Id„u\\e > £. Let En
co
— U Dk'i then E 1 => E2 => ... Hence we have
°° 1
^ m( X) lim X -x r = m W l i m 2 1_" = 0.
n k=n 2 n
Since E is correct then (ii) implies (hi) ||1£ u\\E -► 0 as « -MX).
On the other hand Dn c En, hence \\lDnu\\E ^ ||1£ m||e . Therefore ||1D u||£
-> 0. This contradicts (i).
3. General remarks on non-linear operators. Let T: E -* P be an arbitrary operator. For w e E define an operator Tw: E -*■ P by
(3.1) Tw(u)(x) = T (w )(x)-T (w + u)(x).
Note that 7^,(0) = 0.
Definition 3.2. Let c and d be any conditions involving operators acting from E into P. We say that the pairs (T, c) and (Tw, d) are equivalent if the following statement holds:
“ Г satisfies c if and only if Tw satisfies d” .
Theorem 3.3. Let T: E - + P and w e E . The following pairs (T ,c t) and (Tw,dj) are equivalent:
(1) сг : T {A) <= H , where A cz E, T (w) e H and H is a Banach function space, dji Tw(A — w) e H, where A —w = (m —w; u e A } ;
(2) c2 : T is bounded as an operator T : E -> H , d2: Tw is bounded as an operator Tw: E ^ H ; (3) c3: T is continuous at w,
d3: Tw is continuous at 0;
(4) c4: T is m-continuous at w (i.e. continuous in the sense of topology of convergence in measure),
d4: Tw is m-continuous at 0;
(5) c5: \ \ T ( w) - T ( w + u)\\ „!Zl(\ \T(w)-T(w + l w u n H) + H \ \ T ( w ) - T ( w + + l z w)||h)’ where H is a Banach function space, W и Z = X , W n Z = 0 and l: [0, oo] -> [0, oo] is a non-decreasing function such that;
(i) l(t)e R for t e R, (ii) /(oo) = oo ;
d5: ||Tw(ii) ||h l(\\Tw( \z u)\\) + l(\\Tw( lw u)\\);
(6) c6: T ( w ) - T ( w + l z u) = 1 z ( T { w ) - T ( u + w)), d6- Tw(lz u) = 1 ZT > ) .
Proof. Ad (1). From (3.1) it follows immediately that if T(A) ci H, then Tw(u — w) = T (w) — T (u)e H for very и e A .
Conversely, by (3.1) we obtain T(w + u)(x) = T(w)(x)—Tw(u)x. Denoting v = w + u we have
(3.4) T(v)x = T { w ) x —Tw(v — w)(x).
If v e A , then v — w e A — w and consequently Tw(v — w) e H, thus T (v)e H , since T( w) e H .
Ad (2). If T is bounded, then for every bounded set В a H , T ( B ) is a bounded subset of H. We get
Il f v (^)Ilh < 1 |7 > ) ||„ + ||7 > + и)||я < M for every u e B , where M > 0 is some constant existing by boundedness of T(B).
If Tw is bounded, then by (3.4) we obtain
II^(w)IIh < l|7’(w)||w+ ||r w(M-w)||H < К for every u e B , where К is a constant existing by boundedness of Tw.
Ad (3) and (4). Let E and H be any subspaces of M, assume that E and H are both equipped with linear topologies. Observe that the following operators are continuous.
£ э м и T(w)e H,
S2: E э и H и + w e E, S2 (0) = w, S3 : Еэи\~> u — w e E, S3{w) = 0.
Using this notation we have
Tw = Sy — T о S2, T = S T wо S3.
Clearly, if T is continuous at w, then Tw is continuous at 0 and vice versa.
Ad (5) and (6). Evident.
R e m a r k 3.5. If one wants to prove a theorem of the type;
(a) I f T: E -> P satisfies c, then T satisfies d,
and if ( T , c ) , ( T , d) are respectively equivalent to (Tw, e),(Tw, f ) for a certain
we E, then it is sufficient to prove the theorem of the type:
(b) I f S : E -> P, S(0) = 0 and S satisfies e, then S satisfies f .
Pr oof . Actually Tw(0) = 0, therefore for Tw the above assertion holds.
The following diagram shows the method of proof of (a).
(T, с) =ё=. (Г, d)
» Л
(Tw, e ) ~ ( T w, f )
De f in it io n 3.6. (i) If T : E -> P satisfies condition c5 for a certain w e E and for every W, Z such that X = W и Z, W n Z = 0 , then T is called an l-operator.
(ii) If T: E P satisfies c6 for every measurable set Z c I , then T is called an invariant operator.
Ex a m p l e s. Urysohn’s non-linear integral operator T(u){x)= f k(x, y, u(y) ) dm
x is an /-operator. Indeed
T( w) ( x) ~T{ w + u){x) = $ k(x, y , w ( y ) ) - k ( x , y, w(y) + u(y))dm
X
= J k(x, y, w( y) ) - k( x, y, w(y) + u l w(y))dm + w
+ j k(x, y, w{y) ) - k( x, y, w(y) + u{y))dm z
= T(w)(x)—T ( w + l w u)(x)+T(w)(x) — - T { w + \ z u)(x).
Hence, Hammerstein and Volterra non-linear operators are also /-operators.
Observe that, if T is invariant, then T is an /-operator. Indeed T(w)—T(u + w) = l z ( r ( w ) — T(u + w))+ i w (T{w)— T(u + w))
= T ( w ) - T ( l z u-\-w)+T(w)—T ( l w u + w).
Note that the Nemytskii non-linear operator N f (u)(x) = f ( x , u(x)) is invariant. Compute
N f ( w) ( x ) - Nf ( w + l z u)(x) = f (x, u(x))~ f (x, w( x ) + l z (x) ■ u(x)) J ( x , w ( x ) ) - f ( x , w { x ) ) , хф Z,
l f ( x , w { x ) ) - f ( x , w ( x ) + u(x)), x e Z ,
0, x $ Z ,
f ( x , w( x ) ) - f ( x , w(x) + u(x)), x e Z , = l z (x)(N f (w)(x)~ N f (u + w)(x)).
4. Extension and boundedness theorems.
Th e o r e m 4 .1 . Let E be a correct space and H be a Banach function space. Let T: E -* P be an arbitrary operator. I f T is an l-operator and T (B) cz H , where В is an open subset of E, then T{E) cz H.
Proof. By Theorem 3.3 and Remark 3.5 it is sufficient to prove the theorem in the case T (0) = 0 and В is an open neighbourhood of 0.
Hence, there exists a r > 0 such that the open ball with centre at zero and radius r is contained in B.
(A) Consider first the restriction of T to £(Z ), where Z c= X is any set of finite measure, i.e. T : £ ( Z ) - + P ( X ) , recall that E{Z) denotes { ueE; (supp u) cz Z}. It is easy to check that E(Z) is correct, because E is correct.
Let и be an element of E. By Theorem 2.6 there is a ô > 0 such that, whenever m(D) < S then ||lDu||£ < r. By atomlessness of the measure we get a sequence (Z£)f=1 of measurable sets satisfying the properties
p
Z = IJ Z£ and m(Z£) < Ô.
i = 1
Thus ||lz.u||£ < r and consequently ||T(1Z u)\\H < oo.
Since T is an /-operator, it follows that
ЦГ(и)11н< t ' ( m i z.«)||H) < да.
i = 1
oo
(B) Next let T: E -> P. We have X = У X i} where X t cz X 2 c ... and
i = l
m(Xi) < oo. Thus, denoting Щ = X \ X t we obtain Wx з W2 zo ... and 00
П Щ = 0 - Since E is correct it follows now that to each u e E there
i = 1
corresponds i ^ 1 such that ||1^ .m||£ < r, i.e. 1 w. e B. Hence, under the assumption of the theorem, \\T(lWiu)\\H < oo. It follows from part (A) that
\\T(\x.u)\\H < oo. Hence
ЦГ(и)11н « /(1 |Т (Ч и )||в) + /(||Г.(1,,«)||н) < да, i.e. T ( u ) e H.
This completes the proof.
The following counterexample will show that the hypothesis of correctness of E cannot be dispensed with.
Ex a m p l e 4.1.a. Let X = (0, 1) and let m denote the Lebesgue measure on (0, 1). Put E = L°°(0, 1) and H = £ ( 0 , 1). Define T by
Г \u(x)\-l, \\u\\Loo < 1, T (w)(x) = ) |м(х)| — 1
x
It is easy to check that T is an /-operator.
Suppose that | | u | | £ ao < 1. Therefore T(u)(x) = |w(x)| — 1 a.e. Hence
||T(m)||li = j ||m(x)| —1|dm ^ j \u(x)\dm + l ^ 2 < oo.
(0.1) (0,1)
On the other hand put r(x) = 2 for every x e ( 0 , 1), then v e E while
\\T(v)\\H = J — d m = со.
(0,1) x
We have proved that T (В) с= H, but there exists a v e E such that T(v)$ H.
Ex a m p l e 4.1.b. We can not dispense with the assumption, that T is an /-operator. Indeed put E = L1 (R) and H = L°°(R). Define
00
T(u)(x) = X ( J \u(x)\dm)n.
n = l R
Then T acts from E into P and T is not /-operator. Observe that, if \\u\\Li < 1, then T(u)eL®, but for every и satisfying ||m||£i ^ 1 7 » does not belong to L00.
De f in it io n 4.2. A Banach function space E is said to be locally uniformly correct, if E is correct and to every q > 0 and r > 0 there corresponds a function к : [0, oo) -> [0, oo) such that the following conditions are satisfied:
1° к is non-decreasing,
2° k(0) = 0, k ( n ) e N for every n e N ,
3° if m(Y) < oo and || 1 у w||£ < min (nr, q), then there exists a disjoint
k(n)
collection of set (T)f=i such that Y = (J and ||l y u||£ < r for every
i = 1 1
1 < i ^ k(n).
Ex a m p l e s. (1) Let us define the set function AM(A) = \\lAu\\E, where A cz X
is m -measurable and u e E . If Au is a measure and Au m, then E is locally uniformly correct.
Pr oof . It is easily seen that E is correct. Indeed, let u e E and (D„) 00
be a sequence of sets such that Dx =э D2 з ... and m( f| ^ n) = 0; then
n = l
oo
A«( П Dm) = 0, because Au is absolutely continuous with respect to m.
n — 1
It is obvious that Au is a finite measure, hence, ||1D m||£ = AU(D„) 00
—* Au ( 0 I^n) &s» n —> oo.
n= 1
Observe that A„ is always atomless, since m is such and Au <€ m. If AU( X) = ||m||£ ^ nr, then there is a partition (Хг)"=1 of X such that Au(Xi)
= Рх.-мЦе < r.
(2) If || u || £ = (A„(A))1/p, where 1 ^ p < oo and Xu 4 m, then E is locally uniformly correct.
P roof. Correctness of E is obvious by (1). Now (Au(X))1/p < nr implies XU(X) 4 np rp 4 krp, where к is an arbitrary integer number greater than np.
к
By (1) we get a partition X = (J such that А„(ЛГ,) ^ rp, hence
i = 1
iii^ wiie = (^um ) 1/p < r.
(3) If ||m||£ = sup ||u||d, where for every d e D || • ||d is a seminorm satisfying
deD
conditions (1), (2), (3) of Definition 4.2, and E is correct, then E is locally uniformly correct.
P roof. Assume to the contrary that we have q > 0, n e N and r > 0 such that to every к ^ n there exists a set Y of finite measure and ||l y||£
< min {q, nr) holds for some u e E . If (Y])f=1 is a partition of Y, then
|| 1 у.m||e > r for some 1 ^ i 4 k. That is, for a d0eD ||l y.M||d > r holds.
On the other hand ||wly||£ < min (q, nr) implies ||l y M||d < min (q, nr).
Hence, || • ||do cannot satisfy (1), (2) and (3). Contradiction.
From (1) it follows that L1 is locally uniformly correct.
From (2) it follows that IF is locally uniformly correct for p > 1.
From (3) it follows that Orlicz spaces (equipped with Orlicz norm) and Lorentz spaces are locally uniformly correct.
Theorem 4.3. Let E be a locally uniformly correct space and H be a Banach function space. Assume that an l-operator T: E -► P acts from an open subset В of E into the space H and T (В) is bounded. Then by Theorem 4.1, T acts from the whole o f E into H. Moreover, T is bounded, i.e. T(D) is a bounded subset of H , whenever D is bounded.
P roof. Assume that T(0) = 0 and В denotes a ball with centre at 0 and radius r (cf. proof of Theorem 4.1). With no loss of generality we may assume that D is a ball with centre at 0 and radius q, because every bounded set is contained in some ball. Since T(B) is a bounded set, it follows that there exists a p > 0 such that ||u||£ < r implies ||T,(y)||H ^ p.
(A) Let Z с X be a set of finite measure, we shall prove that T restricted to E (Z) (i.e. T regarded as an operator acting from E (Z) into P ( X )) is bounded.
Fix u eD n E(Z), for a certain natural n ^ 0 we have nr 4 ||m||e ^ (n+l)r.
Since E(Z) is locally uniformly correct it follows that there exists a non-decreasing function к such that X is union of (l$= i+1) and ||1У{и||£ < r for every 1 ^ i 4 k(n+ 1).
Let F be a non-decreasing function F: [0, oo) -► [0, oo) such that F(/c(n)) = k{n + 1). Hence
k(n + l )
Г(и)11й « E /(ЦГ(1г,и)||я ) « l(p) k ( n+l )
i = 1
< l(p)-F(k(n)) ^ l(p) ■ F ( k C - ^ J \ ^ l (p) F (к (g)) = X, this implies that T{D) is a bounded subset of H.
(B) Let T : E - > P . Then X = (J Xf, where (X,) is a non-decreasing i = 1
sequence of sets of finite measure.
From || m || £ < q it follows that Ц1* m||£ < q and by (A) we obtain
\\T(lx .u)\\H < K. Since E is correct it follows that there exists an n such that II l*\*n w||£ < r • Thus
\\T(u)\\H ^ l(\\T(lXnu)\\H) + l(\\T(lx,Xnu)\\H) ^ K + p.
Hence, T(D) is bounded. This completes the proof.
We shall show an example of a correct space, which is not locally uniformly correct. Let us consider segment [0, 1] equipped with Lebesgue
0 0
measure. Let [0, 1]. = (J X p, m( X p) = 1/2P and X£ n X } — 0 for i Ф j. Set p= i
00
L1 = {u: [0, 1 ] ^ R ; Z \\lx u\\p < oo}, where the notation || • ||p stands for
P=i p
the usual norm in U ( X p).
00
One can easily check that ||t/||x = Z Ц1* u\\p is a norm, moreover, if
P=i p
is a Banach function space. Indeed, by Theorem 1.4 it is sufficient to prove
00
that || • ||x has the Riesz-Fisher property, assume then Z ||w jx < oo. Hence,
n = 1
00 00
Z Z 111* un\\p < oo, which implies
П= 1 p — 1
0 0 00
(Î) Z ( Z Н Ц л У < oo.
p = 1 n=l
Since I?(X p) is a Banach function space it follows by Theorem 1.3 that
(ii) Z un) i*Jp ^ Z 14n= 1 X B Un W p ‘
Using (i) and (ii) we obtain 00 00
(Hi) I Z u»\U = Z ||( Z wn)-i*J|p^ Z Z lii*PwJP < °°-
n = 1 p = 1 n = 1 p = 1 n=1
Inequality (iii) is the desired result.
Now we shall prove that Zf is a correct space. Let (D„) be a non-
00
increasing sequence of measurable subsets of [0 , 1] such that m( f) A ) — 0 .
n= 1
00
Since \\^Dnnxpu\\p ^ 1|1*ри||р for every n ,p and since £ ||lx u||p < oo it
p= i p
follows that
Hm l|lDnnA:pw|li = lim £ w|lP = Z üm ||1о„пх N ip = ь.
» p » p=i p p=i » p
because L?(XP) is correct for every p and then
l i m | | I n n x D u \\p = 0 .
On the other hand Zf is not locally uniformly correct. Indeed, put up — 4 ■ l Xp, g = 3, r — § and n — 2. Then let У be a set of finite measure such that \\lYUp\\z = 2 < 3 = min (g, nr).
Since ||ly Up\\z = 2 if and only if j 22pdm = 2P it follows that YnXp
(4.4.a) m( Y n X p) = m(Xp).
Let Y = Yx u ... u Yk and ||l y. Mplly < f . Then
(4.4.b) m ( y n Z p) < ( | y ,
because ||l y.wp||y = ( J 22pdm)llp = 4 ■ (m(Y; n Xp))1/p. Using (4,4.b) we get
т ( У п Х р) = Z m(Y, n X p) < I c Q y - X = k ( i r . m(Xp).
i = 1
By (4.4.a) we obtain then
m(Xp) < /c(|)p • m (Xp).
Hence, к > (f)p for every p ^ 1. Therefore к (2) should be greater than (|)p for every p ^ 1, i.e. к (2) cannot be finite.
The following counterexample will show that if we give up the requirement of E to be locally uniformly correct, then Theorem 4.3 ceases to hold even if E is correct.
Let E be Zf [0, 1] and H be L1 [0, 1]. Define T by
r r ( w 4 I N I2 ^ U
(x); 1 < ||w||j < oo and max j |n|pdm = J \u\r dm.
P=1 xP xr
0 0 GO
Observe that, if ||w||2- = £ ( J |u|pdm)1/p < oo, then £ J |u|pdm < oo,
p= i xp p= 1 xp
I —
(|и(х)Г- 1;
7 — Prace Vlatematyczne 22.1
therefore max J \u\pdm = J \u\r dm for a certain r > 0 and consequently T
P=1 xp xr
is well defined.
It is easy to check that T is an /-operator.
For every u e l f we have T ( u ) e l } . Indeed, either T(u)(x) = 0 for every x or ||T (m)||li = J \u\rdm < oo. Moreover, T(B) is a bounded subset of L1,
* r
where В denotes the unit ball in L1. On the other hand put again up(x)
= 4- l ^ x ) . Then ||и ,b = ||Ир||р = 2, but
Il T (wp)||Li = J \u\pdm = J 2P ■2Pdm = 2P -> oo as p -*■ oo.
xp xp
Thus (up) is bounded while (T(up)) is not bounded. Finally T is not a bounded operator.
From Theorem 4.3 we get
Corollary 4.5. I f an l-operator T: E- +P acts from an open subset В of a locally uniformly correct space E into a Banach function space H and T is continuous, then T is bounded.
5. Continuity theorems.
Lemma 5.1. Assume m(X) < oo, E is a Banach function space and H is a correct space such that l x e H . Let В с: E be an open set. I f T: B ^ H is invariant in the sense of Definition 3.6 and T is m-continuous at u e B , then T is continuous at u.
P ro o f. Without loss of generality we may assume T(0) = О, В is an open ball with center at 0 and radius r and it is sufficient to prove continuity of T at 0. Proof of this fact is analogous to that of Theorem 4.1 and is based on Theorem 3.3 and Remark 3.5
Assume to the contrary that there exists a sequence (z„) с В such that lim ||z„||£ = 0 while lim ||T(zz)||H ф 0.. Let us choose a subsequence (w„)
n rt
satisfying the following conditions:
(i) ||T(w„)||H > a > 0 for every n e N, OO
(ii) Z llw» b < r.
n= 1
(I) Observe that if ( Yn),(vn) с H satisfy (a) m(Y„) -> 0 as n -> oo,
(b) >i<x, where a > 0 , n e N , (c) 111 oo -vn\\H < | a for every n e N ,
U Yi J=n+ 1
00
W„ = Yn\ (J Yi (note that (JTp are disjoint) and v is defined by formula
i = n + 1
v — Z then v ф H.
i = 1 *
Indeed, lim m(W„) ^ lim m(Y„) = 0. Assume that v gH. H is correct, then
П П
by Theorem 2.6 we obtain lim \\\w„vWh — 0- On the other hand
vA h
= 111 00 »JI H — lib .» . - 1 00 v„+i 00
U Yi и Yi U !
i = n +1 i== n + l i = n+ 1
lib . Vn\\H ~~ 111 00 oo Уп11я
u 4 и Yt\Y„
i= n + l i= n + l
l b . Уп11н ““ 111 00 vn Il H'-111 00 vn\Ih
и Yi и Yi\Yn
i! = n+l i= n +1 1 4 C J5 1 111 00 vn\\H~- P 00 Vn\\H
и Yi и Yi
i= n +1 i == n + 1 Contradiction.
(II) Now we shall inductively construct sequences (У„) and (u„) such that (Yn), (T(un)) satisfy (a), (b) and (c).
Put = w1} = X and bt = m(X). Suppose that uk, Yk and bk have been chosen. Since H is correct it follows that there exists a bk+1 such that 2bk + 1 < bk and
(5.2) \\lDT(uk)||я < for D satisfying m(D) < 2bk + l .
By Theorem 2.3 and m-continuity of T it follows from the assumption lim ||w j|£ = 0 that T(wn) - ^ 0. Therefore there exists a k0 e N such that for
n
n ^ k0
m(X\G'„) < bk+1, where G'n = < x e X ; |T(w„)(x)| < - ----} > ■
С о II lx -Ця J
Take any n ^ k0 and put uk + l = w„ and Gk+1 = G'n. Hence
H^Gk+l ^ ( Wk+l)Hn ^ ^
a l|lGk + 1lltf
< ea -
X \ \ H
Defining Ук+1 = X\G k+l we get
IUvk+1 T (wk+1)||H = ||1XT (nk+1) — lGk+1 T (Mjc+i)IIh
^ P x T (мк+1)||и — IUGk+1 T (ик-ц)||я > a — 6a = 6a > t a - Observe that т ч(Ук+1) < bk + l ^ m(Yk)/2. Indeed, assume to the contrary that т(У к) < 2bk +15 we derive from (5.2) that ||l y T (nk)||H < |a , this contradicts the inductive assumption ||l y Т(ик)||н > f a . Hence, lim m(Yk) = 0.
k к
Now we have to verify (c). We have
m( U Yi) < Z m(Yi) ^ Z b> ^ bk+i • Z 1 = 2 ^ + i
i = к + 1 i = f e + l i = к + 1 i ~ 0
Thus 111
i = f c + l i = к + 1
T(wk)||H < ?<*• This completes the inductive step.
i = fc + 1и n
Finally we define и by и = Y W fcWk- Since
*= l
N Ie — I! Z WkMfc||£ ^ Z IIWkMfcllE ^ Z II ^ k II E < r >
к = 1 fc = 1
it follows from the assumption that T ( u ) e H. On the other hand T is invariant, then
k = 1 k = 1
i n U)iiH » 1 1 i ^ r ( « ) | | H = 1 i rawku)\iH = || y. T ( i Wkuk)\\H
k = 1 oo
= II Z Wk ^(мк)||я-
k = 1
The above implies Y but from (II) it follows that 00 k= 1
Y l\vk T(uk) has properties (a), (b) and (c). By (I) then this function does
k = 1
not belong to H. Contradiction completes the proof.
Th e o r e m 5.3. Assume that E is a Banach function space and H is a correct space. Let В с E be an open set. I f T: В -*■ H is invariant and m-continuous at u, then T is continuous at и.
P roof. We may assume that Г(0) = 0 and it is sufficient to prove continuity of T at 0.
(I) Observe that if (Yn),(vn) satisfy
(a) m(Y„) < oo and Yk n Yj = 0 if к Ф j, (b) ||lyfcvk\\fj > %a > 0 for every k e N ,
00
and v is defined by v = Y 1укук> then v is not a member of H.
к = 1
00
Indeed, assume that v e Я , define W„ = (J Yk, note that Wx з W2 з ...
k = n
00 00 00
and П = П U *k = $> because (У„) is a disjoint collection of sets.
n=1 л= 1 fc= n Since H is correct we have
-joe < ||ly vn\\H = ||ly v\\H ^ ||Iw„v\\h 0 as n -> oo.
Contradiction.
(II) Assume to the contrary that there exists (u„) c= В such that ||T(u„)||H
> a > 0 for every n, while lim ||u j|£ = 0 . П
Now we shall construct (Y„) and (T(ukJ ) satisfying (a) and (b).
Using cr-finiteness of the measure and Corollary 2.2 we obtain a non-
00
decreasing sequence of sets (X n) such that X = (J X n> m (Xn) < 00 and n- 1
1 xn€ H f°r every n e N .
It is easy to check that ||1*.Т(м1)||я > a for some i e N . Indeed, assume to the contrary that for every n we get ||lx„ ^ (mi)IIh ^ a - Then we have
« < I I T f u . l l U = J | l j f M T( w x ) + 1 д г \ д г п T( « i) | Ih « l | l J r „ T ' ( « , ) l l e + l | l W | > r ( « 1 ) | | „
^ a + II 1л-\х„ T(u h .
Since H is correct it follows that ||lx\x„ T'(u1)||H -> 0 as n -> со. Hence a < ||Т(м1)||я ^ a. Contradiction.
к
Assume now that Yk and u„k have been chosen. Clearly m( (J Yt) < со,
i = 1
and 1 oo e H . From Lemma 5.1 it follows now that operator T is i=lU П
к
continuous in the intersection of В and the space E( (J У^). Hence, there
i = 1
exists a un, ^ , such that"к + 1
\\T(u„k + l • 1 k )Ilh < i a •
U Yi
i= 1
Similarly as in the case TXtq) we obtain Gk + l с X such that m(Gk + l)
< o o and ||lGfc+1 • T(M„fc + 1)||H > a and Ц +1е Я . Let us put Yk + l =
G k + i \ U Yt. We have i= 1
llyk+i Т (ипк+1)\\н > Пок+1Т(иПк+1)\\н - T (ипк+1)\\Н- U П\с* + 1
i=l
— l|l* T(u„k+1)\\H ^ Il lGk+г Т(м„л+1)||я - и Yt
i= 1
— Il 1 /t Т(и„к + 1)\\н — || 1 к Т(и„к+1)\\н
u n u n
^ et —l a —lot = \ a . This completes the proof.
Example 5.6. Let / be a real valued function of X x R into R, where X <= Rn is a Lebesgue measurable set. By the Nemytskii operator N f induced by / we shall understand the operator
N f (u)(x) = f ( x , u ( x ) ) , where u e E and x e X . One assumes that / satisfies so-called Carathéodorÿs conditions (1 ) / ( x , •): Л э м И / ( х , « ) е Л is continuous for almost every x e X , (2) f { - , u ): Х э х \ - * f ( x , u ) e R is measurable for all u e R ,
which guarantee the measurability of T(u), whenever и is a measurable function.
The following result is due to Nemytskii, [11].
Theorem b y Nemytskii. I f m{X) < oo and f : X x R - * R satisfies (1), (2), then N f is m-continuous.
The above theorem guarantees the fact that all theorems of this paper can be applied to the investigation of the Nemytskii operator, which plays an important role in the theory of non-linear integral equations.
For instance, the Hammerstein-type integral operator T (u )(x )= J k(x, y ) f ( y , u{y))dm
x
can be represented as the composition A = K o N f , where К is the linear integral operator
K(u)(x) ~ j k(x, y) ■ u(y)dm.
x
For another example let V be the non-linear Volterra-type operator V{u){x)= J f ( y , u ( y ) ) d m, x , x0e R .
IXQ,X]
V can be represented as the composition V = R o N f , where R(u)(x) = j u(y)dm.
IXQ,X]
Counterexample 5.5. If we give up the requirement of correctness of H, then Theorem 5.3 ceases to hold.
Let X = [0, 1] and let m denote the Lebesgue measure. Let us put E = 1} [0, 1] and H = L00 [0, 1]. Define / (x, и) = min (1, |u|).
It is easy to check that fy acts from L1 into L00 and i^(0) = 0. For every n e N put un = l[0,i/n)- Thus
IlM je = J \un(x)\dm = \/n -► 0 .
[0,1]
On the other hand
I|N/ (m„)||loo = ess sup j/(x,u„(x))| = ess sup |m„(x)| = 1 for each n.
jce[0,l] xe[0,l]
Combining Theorem 5.3 with Corollary 4.5 we get
Th e o r e m 5.6. Let E be a locally uniformly correct space and H be a correct space. I f m-continuous invariant operator acts from E into H, then T is continuous and bounded.
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