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.