Proof of Theorem 2.1

First Part

The first part of the proof goes through several lemmas.

Lemma 3.1.

is well-defined and under hypothesis (i) of Theorem 2.1, for all the mapping is of class and one has, for all

For the proof see [19].

We set for all

Lemma 3.2.

Assume that hypothesis (iii) of Theorem 2.1 holds. Then for one has Moreover, if in addition hypotheses (i) and (iv) of Theorem 2.1 hold, then for all the mapping is of class on the ball in and for all one has

Proof.

Let Let be the constant of (iii) with So for all So one has

Assume now that hypotheses (i) and (iv) of Theorem 2.1 hold. Let us show that is of class on Take Let be given. Let be such that Then, for all under (iv), which implies that

Let us now show that is Fréchet-differentiable on Take Let be given. Let be such that Then, for all under (iii). But this implies that Thus is Fréchet-differentiable at and

To show the continuity of at let be the constant of hypothesis (iii) corresponding to Let be given and let be such that So is of class

Lemma 3.3.

Under hypothesis (ii) of Theorem 2.1, for all for all for all there exists such that

Proof.

Let and Hypothesis (ii) of Theorem 2.1 implies for all the existence of such that

Therefore we obtain

Set so and satisfies the required relations.

Lemma 3.4.

Under hypotheses (i) and (iv) of Theorem 2.1,

Proof.

Since the problem is a problem of bounded solutions of first-order linear difference equations.

Let Assume that Then for all there exists a unique such that for all

where

Consider the operator such that for all

where

Recall that the norm of is defined by and that the norm of a linear operator between normed spaces is defined by

So So Since and is invertible so it is surjective.

Set Then under (iv) one has So is surjective that is

Recall that where consists of all singular functionals, see Aliprantis and Border [20]. In fact it consists (up to scalar multiples) of all extensions of the "limit functional" to

If then there exists such that for all ( being the space of convergent sequences having a limit in )

Lemma 3.5 ().

If then where for every So there exists such that for all

Second Part

Our optimal control problem can be written as the following abstract static optimisation problem in a Banach space:

that satisfies all conditions of Theorem 4.3, Ioffe-Tihomirov [17]. So we can apply this theorem and obtain the existence of not all zero, such that:

(AE) denotes the adjoint equation of this problem and (PMP) the Pontryagin maximum principle. They can be written, respectively:

Set where and

(AE) becomes:

So we get

Let be arbitrarily chosen in and let be in Consider the sequence defined as follows:

So one has if hence

Thus, it holds that

Now

Therefore, for all and for all one has

which implies

that is,

(PMP) becomes:

So for all

Consider, for all the sequences defined as follows:

Since the inequality holds for every we obtain

using as is of finite support.

Lemma 3.6.

().

Proof.

Recall we obtained the existence of not all zero, such that:

If then since

Hence We can set it equal to one.

From Lemma 3.6 and the previous results, conclusions (a) and (b) are satisfied.

Conclusion (c) is a straightforward consequence of the belonging of to

Lemma 3.7.

().

Proof.

Indeed we obtained for all Using one has for all Thus

Proof of Theorem 2.2.

Define on such that Under hypothesis (i) of Theorem 2.2, for all the mappings and are of class on The proof can be found in [19].

We consider the proof of Lemma 3.4 and we set Then the proof goes like that of Theorem 2.1.