Theorem 1
Let
(5)
If the problem
(6)
(7)
has only the trivial solution for every , then problem (1), (2) has at least one solution.
The proof of Theorem 1 is based on the following result by Krasnosel’skii (see [[12], Theorem 41.3, p.325]). We will formulate it in a form suitable for us.
Theorem 2 Let X be a Banach space, be a symmetrica bounded domain with . Let, moreover, be a compactb continuous operator which has no fixed point on ∂ Ω. If, in addition,
then A has a fixed point in Ω, i.e. there exists such that .
Furthermore, to prove Theorem 1 we will need the following lemma.
Lemma 1 Let, for every , problem (6), (7) has only the trivial solution. Then there exists such that for any and any , the a priori estimate
(8)
holds, where
Proof Suppose on the contrary that for every there exist and such that
(9)
where
(10)
(11)
Put
(12)
(13)
Then
(14)
and from (10) and (11), in view of (3), (4), (12), and (13), we get
(15)
(16)
On the other hand, from (9) and (12) we have
(17)
whence, according to [[13], Corollary IV.8.11] it follows that
(18)
Therefore, (14), (15), and (18) imply that the sequences () are uniformly bounded and equicontinuous. Thus, according to Arzelà-Ascoli theorem, without loss of generality we can assume that there exist and such that
(19)
Furthermore, (15)-(17) yield and show that it is a solution to (6), (7). However, (14) and (19) result in
which contradicts our assumptions. □
Proof of Theorem 1 Let and for , i.e. , define the norm
Then is a Banach space. Let the operators be defined as follows:
(20)
(21)
(22)
and consider the operator equation
It can easily be seen that problem (1), (2), and (23) are equivalent in the following sense: if is a solution to (23), then () and is a solution to (1), (2); and vice versa if is a solution to (1), (2), then is a solution to (23).
Let be such that the conclusion of Lemma 1 is valid. According to (5) we can choose such that
(24)
Let, moreover,
(25)
Now we will show that the operator A has a fixed point in Ω. According to Theorem 2 it is sufficient to show that
Assume on the contrary that there exist and such that
(26)
Then from (26), in view of (20)-(22) we obtain
where , i.e.
(27)
(28)
Now from (27) and (28) it follows that ,
(30)
(31)
Moreover, since , on account of (25) and (29) we have
(32)
Now the equality (32), according to Notation 1, implies
(33)
(34)
Therefore, in view of Lemma 1, with respect to (30)-(34) we obtain
However, the latter inequality contradicts (24). □