Recall that in all assertions below condition (1.5) is assumed to be fulfilled.
Lemma 4.1 W is a Hilbert space.
Proof The relation establishes a bijection between W and the Hilbert space . □
Lemma 4.2
The value
satisfies the inequality
Proof This follows from the inequalities
□
Remark 4.1 The estimate (4.1) is accurate, since if , , and , , then
Remark 4.2 If , , , then
In fact, denote . Then . Since , .
Lemma 4.3 Suppose (1.5) holds. Then , the operator is bounded, and the norm of T satisfies the estimate
(4.2)
Proof Since , we can estimate
From this and (4.1)
(4.3)
□
Remark 4.3 If (see (1.6)), then . In fact, if , then . It is possible to find a nonincreasing function such that
Now find u such that : . Since , , and is nonincreasing, by Remark 4.2, . But .
Lemma 4.4 The image is dense in .
Proof The proof is left to the reader. □
The following theorem [[5], p.318] can be used to show compactness.
Theorem 4.1 (Gelfand)
A set
E
from a separable Banach space
X
is relatively compact if and only if for any sequence of linear continuous functionals that converge to zero at each point
the convergence (4.4) would be uniform on the E.
Lemma 4.5 Suppose (1.3) holds. If is bounded, then
uniformly on the set .
Proof Since
it is sufficient to show that
uniformly on . We have
The first term tends to zero because of the inequality
and (1.3). The second term is equal to
This tends to zero because of (4.1) and (1.3). □
Lemma 4.6 Suppose (1.4) holds. If is bounded, then
uniformly on the set .
Proof Since
it is sufficient to show that
when uniformly on . We have
Now we refer to (1.4) and (4.1). □
Lemma 4.7 If (1.3) and (1.4) hold, then T is compact.
Proof Let . We use the criterium of compactness of Gelfand (see Theorem 4.1). Let be a sequence of functionals such that for any . We have to show that uniformly for .
Let . Using Lemma 4.5 choose N such that
for all n and for all . The same we can do with for sufficiently small . For this aim we can use Lemma 4.6.
Now we only have to show that uniformly on the set . Since
where , it suffices to show that . We show that uniformly for . Let
Then . The set is compact. Thus by Theorem 4.1 of Gelfand converges to zero uniformly for these s. So, uniformly for , and uniformly for . □
Remark 4.4 It seems that condition (1.4) is necessary for compactness of T.
Lemma 4.8 Equation (2.1) is equivalent to problem (2.2).
Proof Denote . From (2.1) it follows that
(4.5)
(if we choose v such that the corresponding limits exist). If for , then . From this on . Since the segment is arbitrary, the relation is fulfilled on the whole semiaxis . So, . By the first equality in (4.5)
Now, . Choosing v such that , we obtain . □